^^/^(/. Sheldon & Co77ipany's Text-Hooks, _ _ The Scieiwe of Government in Connection tvith American Institutions, By Joseph Alden, D.D., LL.D., Pres. of State Normal School, Albany. 1 vol 12mo. Adapted to the wants of High Schools and Colleges. Allien' 8 Citizen's 3Ianual : a Text-Book on Government, in 'Connection with American Institutions, adapted to tlie wants of Common Schools. It is in the form of questions and answers. By Joseph Alden, D.D., LL.D. 1 vol. IGmo. Hereafter no American can be said to be educated^ who does not thoroughly ui ------ - ^ — jjaid, th ;hose de S " "" *"~'* -I PKIVATELIBEAEY i ^ OF i '.k Bought-, I VT^ Co3t-,?r a No. h^ ^x«o«jixv,vi. ttiiv.!. axxuiigcu. 0.0 \j\j jiauiiitait; iiiu exinessiori of ideas, and assist in literary composition. By PetePv Mark ROGET. Revised ^1 e dited , with a^Ljst of Foreigii \Vor^s defined in EnglTsh, and other additions, by Barn as Sears, D.D., late President of Brown University. A new American, from the last London edition, with important Additions, Corrections, and Improvements. 12mo, cloth. Faircliilds' Moral rUilosophy ; or, The Science of jQMia^tioiu By J. H. Fairchilds, President of Oberlin C ollege. 1 vol. ISmo. The aim of this volume is to set forth, more fully than has hitherto been doneTTEFHoctrine that virtue, in its elementary form, consists m benevo- lence, and that all forms of virtuous action are modifications of this principle. After presenting this view of obligation, the author takes up the questions of Practical Ethics, Government and Personal Rights f.nd Duties, and treats them in their relation to Benevolence, aiming at a solution of the problems ot rij,'ht and wrong upon this simple principle. University of California • Berkeley The Theodore P. Hill Collection : of Early American Mathematics Books Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/completealgebraeOOolnerich OLNEY'S MATHEMATICAL SERIES. THE COMPLETE ALGEBEA EMBRACING SIMPLE AND QUADRATIC EQUATIONS, PROPORTION AND THE PROGRESSIONS, WITH AN ELEMENTARY AND PRACTICAL VIEW OF LOGARITHMS; A BRIEF TREATMENT OF NUMERICAL HIGHER EQUATIONS. AND A CHAPTER ON THE BUSINESS RULES OF ARITHMETIC TREATED ALGEBRAICALLY Designed to be sufficiently elementary for beginners, and sufficiently thorough and comprehensive to meet the wants of our better high schools, ACADEMIES, AND ORDINARY COLLEGES. BY EDWARD OLNEY, Pro/tssor o/ Mathematics in the University 0/ Michigan. NEW YORK : SHELDON & COMPANY, 8 MURRAY STREET. OLNEY'S MATHEMATICAL SERIES EMBRACES THE FOLLOWIifG BOOKS. "Primary Arithmetic, Ji7e9ne?its of A.rithm^etic, Practical Arit?imetic, Teacher's Mand-^ook of Arithmetical ^Exercises, Science of Arithmetic. Send for fiill Circnlar of Olney's Arithmetics. fntroduction to Algebra, Complete Algebra, JECey to Complete Algebra, University Algebra, £'ey to U'72ire7^sity Algebra, 2'est Examples in Algebra, Elements of Geometry, (Separate.) JSleme7its of Trigo7iometry, (Separate.) introduction to Geometry, ^art I, (Bound separate.) ^lemeiits of Geometry and Trigonometry, (Bound in one Vol.) Geometry a^id Trigonometry , (University Edition.) General Geometry a7id Calculus, Copyright ^^ 1870, 1875, 1878, by Sheldon &f> Co. Blectrotyped by Smith & McDouqal, 82 Beekman St., N. Y. PREFACE. Theee main purposes have controlled the author in the preparation of this work : — First, to provide, in a single volume, of convenient size, an elementaiy treatise upon Algebra, sufficiently simple for the use of beginners in the science— sufficiently full on the subjects em- braced to render their rediscussion in a subsequent volume unnecessary, and to afford a range of topics comprehensive enough to meet the wants of our Common and PubUc High tichools, and for a good pre- paration to enter any of our Colleges or Universities. Second, to train the pupil to methods of reasoning, rather than in mere methods of operating. Third, to present the whole in such a form as to make the book a convenient one to use in the class room. As the mathematical studies are at present pursued in our schools, pupils are expected to have a fair knowledge of Mental, and some knowledge of what is variously styled Practical or "Written Arithmetic, before entering upon the study of Algebra. Without expressing any opinion upon the propriety of this course, the author has accepted it as a fact likely to exist for some time to come, and has attempted to adapt his Algebra to it. This book is, therefore, not a mere child's book, but is designed as a First Book in Algebra for pupils who have the knowledge of Arithmetic usually considered requisite to a com- mencement of this study. The volume contains a more than ordinarily thorough and full treat- ment of the processes of Literal Arithmetic, viz., the Fundamental Kules, Fractions, Involution, Evolution, and the Calculus of Eadicals. It is in these processes that pupils usually find the most difficulty; and in them comparatively few become proficient, especially in the Calculus of Eadicals. Yet no one can become an algebraist, and, if not an algebraist, not a mathematician, without being perfectly mas- ter of these processes. Hence they are treated so much at length and with so much care and thoroughness. It is idle to deceive the pupil with the notion that he is mastering the subject of algebra by obtain- ing a superficial acquaintance with the nature and uses of some of the more simple forms of the equation. The pupil who cannot attain a good degree of familiarity \^dth the Hteral arithmetic, cannot appre- ciate mathematical reasoning, or profit much by this study. VI ■ PREFACE. The treatment of Simple and Quadratic Equations will be found as full as is requisite for the foundation of a good mathematical educa- tion. It is not the author's purpose to rediscuss the themes here con- sidered in his second volume. The matter presented on these subjects does not differ much from what is usually contained in our so-called higher algebras ; and, in fact, is more thoroughly treated than in most of them. Katio, Proportion, and the Progressions have received a proper share of attention. The transformation of a Proportion will be found to be discussed in a somewhat different manner from the usual one, and, it is hoped, in such a way that the pupil who enters into the spirit of the treatment, and becomes familiar with the meth- ods, will feel that this elegant and important mathematical instrument is his own. It is thought that the chapter on the Business Rules of Arithmetic will be particularly acceptable to a large class of students. Some pr£)blems in Interest, Common Discount, and Alligation, which are ordinarily considered intricate, will be found quite within the reach of ordinary minds. Other applications of the equation to the solution of practical questions, are not less extended than in our common treatises. The chapter on Logarithms will give the student a clear conception of the nature of this kind of number, and enable him to understand and use the common tables. No attempt is made at the Discussion of Equations, or the Devel- opment of Series. To discuss an equation well, is a mathematical accomplishment. It is not a thing which the tyro can master. He may follow another through such a discussion ; but he will do little more. The Development of Functions is reserved for the second volume, where the Binomial Formula, and the Logarithmic Series will be seen to be but deductions from a more general, and more easily produced formula. The Binomial Formula is given (though not pro- duced) in this treatise, and much care is bestowed in showing how to use it in developing the common forms to which it is appUcable. It is thought that these topics, with the discussion of them given in the text, will meet the author's first purpose. The attempt to accomplish the second end, viz., to train the pupil to methods of reasoning, rather than in mere methods of operating, has given character to the presentation of every topic. Propositions are clearly stated at the outset, and demonstrations are given in form, and with the rigor of a geometrical argument. That there is some defect in our methods of instruction, in this regard, must be pftinfully evident to every one who has been called to examine large numbers of our youth in this study. The author has examined for admission to PREFACE. Vll college, from 25 to 150 diflferent students from all parts of our country, each year, for the last 16 years, and he has almost invariably found Httle or no knowledge of the processes as arguments, even when a good degree of skill in the use of the processes had been attained. Perhaps a majority of those examined could multiply the square root of 2 by the cube root of 3, but scarce one in 50 could develop the process in a logical form, or, in most cases, give any rational account of it. Now, it need not be said that, in a course of education, this is a fundamental defect ; it is failure just where success is vital. The processes of a mathematical science are of comj)aratively little worth to a great majority of those who study them ; the development of tho reasoning powers to which such studies are addressed, is of the high- est importance to all. By teachers who cannot appreciate these truths, this book will very probably be misunderstood ; but to such as do feel the force of them, the author appeals with the fullest confidence, not indeed that his book will meet the exigency, but that it will be wel- comed as an effort in the right direction, and as a help in remedying this radical defect. To such as use the book the author commends the following SUGGESTIONS TO TEACHEKS. Great pains have been taken to adapt the whole to the use o£ the class room. Definitions and all Propositions are made to stand out clearly, and have been written -with great care. These should be mem- orized by the student, as well as thoroughly comprehended. It is also recommended that the Bules be committed to memory, verbatim ; not, indeed, to be recited as a mere parrot-like performance, but as a means of acquiring the language of the science, and attaining faciHty in clothing its thoughts in a becoming garb. But let the Demonstra- tion of every Proposition and Fade he inseparably connected icith the state- ment of the truth or process. The author has endeavored to make these demonstrations to their propositions just what the demonstrations in Geometry are to theirs ; and the method of use should be the same. The Model Solutions are samples of what the pupil should give at the blackboard in the class room, repeating the reasoning in the * ' expla- nation " of every example, till it is perfectly familiar. These solutions are not mere statements of how the thing is done, but are designed to be a logical presentation of the process as an argument. It is just at 'this point that the great, and well-nigh universal failure referred to above occurs. By a close adherence to the plan here presented, it is hoped something may be done to remedy the e\'il. The Synopses should be made the basis of reviews. No student VUl PBEFACE. should be allowed to think he has mastered a subject, till he can go to the blackboard, put down the Synopsis, and discuss the whole theme therefrom, stating every principle, in good language, and dem- onstrating, in good style, every proposition. The Test Questions are not designed to be full or exhaustive ; they are usually only a few detached questions, such as, if well answered, would satisfy an examiner that a student understood the subject under consideration. The Examples are more than ordinarily numerous, and have been selected from the common sources. In the Key there will be found a very large number of additional examples for the teacher's use in class room drill. It will often be found serviceable to assign these in the class room in preference to those in the text, for reasons which will be apparent to every teacher. As to originahty, though nothing may have been developed which is absolutely new, there are many features to which attention might be called. Among these are the exposition of the doctrine of the signs -f- and — , the theory of signs in multipUcation, the treatment of subtrac- tion, some things in factoring, and the whole spirit of the treatment of the Calculus of Radicals, the exposition of the nature of a Literal Frac- tion, the treatment of Proportion, etc. The very idea of the character and province of the science, as given in the definition of Algebra, may not be without interest. But the Author is in no fear of being called a plagiarist. That he has not borrowed more freely, may be his chief misfortune. Valuable suggestions have been received from many practical teach- ers ; and the Author feels under special obligation to Mr. Thos. Hunter, President of the Female Normal and High School, New York City, who has read the entire work in manuscript, and whose words of criticism, not less than his generous appreciation and approbation, have done much to foster the hope that the work may not be without merit. With these remarks, the Author commits his effort to the judgment of his fellow teachers, not indeed without the keenest sense of its im- perfections, but with the humble hope that it may be found suggestive, and to some extent serviceable in promoting a much needed reform in our methods of teaching this fundamental branch of mathematical science. EDWARD OLNEY. TJNrVEBSITT OF MICHIGAN, Ann Arbor, January, 1870. N. B. — For scope and purpose of tlio Appendix, soe foot-note, p. 3i)l. CONTENTS. INTROD UCTION, SECTION I. A BRIEF SURVEY OF THE OBJECTS OF PURE MATHEMAT- ICS AND OF THE SEVERAL BRANCHES. PAGE PxJEE Mathematics. — Definition {1); branches enumerated {2>3) 1 Quantity. — Definition (4) 1 Number. — Definition (5); Discontinuous and continuous (6, 7,8) 2,3 Definition OF THE Several Branches. — Arithmetic \0)\ Al- gebra {10) ; Calculus {11) ; Geometry {12) 3, 4 Synopsis 4 SECTION 11. L0GIC0-MATHE3kL\TICAL TERMS. Proposition. — Definition (13) 5 Varieties of Propositions. — Enumerated (J4); Axiom {15)', Theorem {16); Lemma {18); Corollarj' {10); Postulate {20); Problem {21) 5-7 Definition OF —Demonstration {17); Rule {22); Solution {23); Scholium {24=) 6, 7 Synopsis 7 ♦-♦-♦ • PART I.-LITERAL ARITHMETIC. OHAPTEE I. FUNDA3IENTAL BULBS. SECTION I. NOTATION. System of Notation. — Definition {2S) 8 Symbols OF Quantity.— Arabic {26); Literal {27); advan- tages of latter {28) •. 8, 9 Examples 10, 11 Laws.— Of Decimal Notation {29); of Literal Notation {30) 12-11 X CONTENTS. PAQK Symbol 00> and its meaning {31) 14 Symbols of Opebation.— (3;2) ; Sign -f- (55); Sign — (54); Sign—' (55); Sign X, etc. (36)\ Sign-f, etc. (57); Signy- {44) 14-15 Definitions.— Power (55) ; Eoot (59) ; Exponent (40), A positive integer {41), A positive fraction [42)^ A negative number {43) ; Examples 15-18 Symbols of Kelation : — Sign : {45^; Sign •• {40)', Sign = {47) ; Signs > < {49), 18-19 Syivibols of Aggeegation. — , (),[], { }, |, (50, 51) 19 Symbols of Continuation {52) 19 Symbols of Deduction {53) 20 Positive and Negative. — How applied {54) ', Double use of signs -j- and — {55) ', Essential sign of a quantity {56) ; Illustrations ; Abstract quantities no sign {57) ', Less than Zero {58) ; Increase of a negative quantity {59) 20, 21 Names of Diffekent Forms of Expeession. — Polynomial (00) ; Monomial, Binomial, Trinomial, etc. {Gl) ; CoeflQ- cient {62) ; Similar Terms (05) 23, 24 Exeecises in Notation 24, 25 ExEEcisES in Reading and Evaluating Expebssions 25, 26 Synopsis and Test Questions 26, 27 SECTION II. ADDITION. Definitions. — Addition {64) ; Sum or Amount {65) 28 Prop. 1. — To add similar terms {66); Examples ; Cor. 1, Sign of the sum {67) ; Sch. Addition sometimes seems like Sub- traction ; Cor. 2, Algebraic sum of a positive and negative quantity {68) ; Cor. 3, Addition does not always imply an increase {69) 28-31 Pkop. 3.— To ndd dir.similar tciraa (TO); Sch. Tlie sign — before a term ; Cor. Adding a negative quantity (71) ', Examples 31-r33 Peob. 1. — To add polynomials [72, ; Sck. 1, Practical method of adding ; Sch. 2, Literal and decimal addition compared ; • Examples 33-37 Prop. 3. — Terms but partially similar (73) ; Examples 37. 38 Prop. 4. —Compound turms {74) ; Examples ^ oS. ."^i) Synopsis and Te?.t Qi'estjoks iO CONTENTS. XI PAGE SECTION III. SUBTKACTION. Definitions. — Subtraction, Minuend, Subtrahend, Difference, or Kemainder (75, 76) 40, 41 Prob.— To perform Subtraction {77) 42 Practical Suggestions, Sch. 1, 2 ; Kemoving a ( ) preceded by the sign — {78) ; Introducing a ( ) {70} ; Several paren- theses inclosing each other {80) 45-48 Examples 42-49 Synopsis and Test Questions 49 SECTION IV. MULTIPLICATION. Definitions. — {81), Cors. 1, 2, and Sch. Character of the factors and product {82, 83, 84) 50 Peop. 1, — Order of factors immaterial {83) 51 Prop. 2. — Sign of product {80) ; Cors. 1, 2, 3, Sign of pro- duct of several factors {87, 88, 89) 51, 52 Prop. 3. — Product of quantities affected with exponents (90) ; Examples 52-54 Prob. 1. — To multiply monomials {91) ; Examples 54-56 Prob. 2. — To multiply poljTiomials [92); Examples 56-59 Three Important Theorems. — Square of the sum {94); Square of the difference {93); Product of sum and differ- ence {96 1 ; Examples 59-61 Synopsis and Test Questions 62 SECTION V. DIVISION. Definitions, — {97) ; Relation to multiplication (.9^, 99) ; 6'or.9. 1-5, Deductions from definition {100-104); Can- cellation {105) 63, 64 Lemma 1. — Sign of quotient {106) 64 Lemma 2. a*" -^ a" {107) ; Examples ; Cor. 1. Exponent {108) ; Cor. 2, How negative exponents arise {109) ; Cor. 3, To transfer a factor from dividend to divisor {110) 64-66 XU CONTENTS. PAGS Feob. 1. — Division of Monomials {111) ; Examples 67, 68 Pbob. 2. — To divide a polynomial by a monomial {112); Ex- amples ; Arrangement of terms {113) 68-70 Peob. 3. — To divide a polynomial by a polynomial {lid); Examples 70-75 Synopsis and Test Questions 76 OHAPTEE n. FACTOItIJ!^G. SECTION I. FUNDAMENTAL PKOPOSITIONS. Definitions.— Factor {115} ; Common Divisor {116) ; Common Multiple (117) ; Composite {118^ ; Prime {119, 120) 77 Peop. 1. — To resolve a monomial {121) ; Examples 78 Peop. 2. — To remove a monomial factor from a polynomial {122) ; Examples 79 Peop. 3.— To factor a trinomial square {123); Examples 79-81 Peop. 4. — To factor the difference of two squares {124) ; Examples 81, 82 Peop. 5. — Given one factor to find the other {125) ; Examples 82 Peop. 6. — By what the sum, and the difference of like powers are divisible {126) ; Examples 83-85 Frop. 6 applied to the case of fractional exponents {128) ; Examples 85, 86 Peop. 7. — To resolve a trinomial {120) ; Examples 86, 81 Peop. 8. — To resolve a polynomial by separating it into parts {130); Examples 87, 88 SECTION II GKEATEST OR HIGHEST COMMON DISTESOR. Definition il31) ^8 Lemma 1.'— The H. C. D. the product of all common factors a32) ; Examples 89. 90 CONTENTS. xiii PAQK Lemma 2. — A divisor of the polynomial of the form Ajc"-}- ^x"-i4-ac"-2 ^ + F {134) 90 Lemma 3. — ^A divisor of a number divides any multiple {135) 91 Lemma 4. — A C, D. divides the sum, and the difference {136) 91 General Ruue for H. C. D. demonstrated and applied {137 y 138) 92-98 SECTION III. LOWEST OK LEAST COMMON MULTIPLE. Pbob. — To find the L. C. M. by resolving numbers into factors {140): Examples 98, 99 Finding the L. C. M. facilitated by the method of H. C. D. Sch 99, 100 Synopsis and Test Questions 101 « • » CHAPTEE III. F B A C T I O N S. SECTION L Definitions and Fundamental Principles. — Of the terms Fraction, Numerator, Denominator, etc, {14:1).. . . 103 The true conception of a literal fraction {142) 103 Value op a Fraction {143) 103 Cor's 1, 3. — Changes in terms of a fraction {144, 145,). . 103 Uses and Definitions of various terms (146-156) 10S-I05 Signs of a Fraction : — Three things to consider {157) 105 Essential character ; Examples {158). 105, lOG SECTION IL REDUCTIONS. Varieties of {159) 107 Prob. 1.— To lovrest terms ; Examples 107, 109 Sch. 1.— The use of the H. C. D. • 108 XIV C0JSTENT8. PAGE ScH. 2. — The converse process ; Examples 109 Pbob. 2. — To improj)er or mixed form (160); Examples 109, 110 Cob. — Use of negative exponents (161) 110 Pbob. 3. — From integral or mixed to fractional form (162); Examples Ill, 112 Pbob. 4. — To common denominator {163)', Examples. , . 112, 113 Cob.— To L. C. D. {164); Examples 114, 115 Pbob. 5. — Complex to simple {165); Examples 115-117 SECTION III. ADDITION. Pbob.— To add fractions {166); Examples 117-119 Cob. — To add mixed numbers {167) ; Examples 119-121 SECTION IK SUBTKACTION. Pbob. — To subtract fractions {168); Examples 121-123 Cob. — To subtract mixed numbers {169); Examples 123, 124 SECTION V. MULTIPLICATION. Pbob. 1. — A fraction by an integer {170); Examples 124, 126 Pbob. 2. —To multiply by a fraction {171); Examples 126-129 ScH. — When no common factors 127 Cos.— To multiply mixed numbers {172) 129 SECTION VL DIVISION. Pbob. 1.— To divide by an integer {173); Examples 130, 131 Pbob. 2.— To divide by a fraction {174:)\ Examples 131, 136 Eeason for inverting the divisor 132 ScH. 1. — To reverse the operation of multiplication 132 ScH. 2. — Using the form of a complex fraction 135 REcrPBOCAii, what {175) 136 Synopsis or Fbactions and Test Questions 137 CONTENTS. XV PAGE OHAPTEE IV. , POWERS AND ROOTS. SECTION L INVOLUTION. Note. — The Fundamental Principle 138 Genekai. Definitions. — Power (j? 7^); Root (177)', Power and Eoot correlatives (178); Names of different Powers { and Roots {179) ; Exponent or Index ; How to read an expo- nent {180) ; Transferring a factor from numerator to denomi- nator of a Fraction (181); Radical {18*^)', Radical Sign {183}; Imaginary Quantity (J^^j; 'Real {183); Similar Radicals {186); Rationalize {187); to affect with an expo- nent {188); Involution {189); Evolution (190) ; Calculus of Radicals {191) 138-141 Involution : Pkob. 1. — To raise to any Power {192) ; Examples 141-143 CoK.— Signs of Powers {193) 143 Peob. 2.— To affect with any Exponent {194=) 143 Examples 144-148 Pbob. 3.— The Binomial Formula {195) 148 Applications 148-154 Form for expansion of (1 -j- ^)" 1^0 Cob. 1. — When the series terminates (196) ... 151 Cob. 2.— Number of terms {197) 151 Cob. 3.— Equal coefficients {198) 152 Cob. 4. — Sum of exponents in any term {199) 152 Cob. 5.— Statement of the Rule {200) 152 Cob. 6. — Signs of the terms in the expansion of (a — &)"» {201) 153 SECTION IL EVOLUTION. Peob. 1. — To extract root of perfect power {202); Examples 154 ScH.— Signs of root {203) 155 Cob. 1. — Roots of Monomials {204:); Examples 155 Cob. 2.— Root of Product {205) 156 Cob. 3.— Root of a Quotient {206) ; Examples 156 XVI CONTENTS. PAOK Prob. 2.— To extract the square root of a Polynomial, Eule, Dem. and Examples (207) 157-160 Pbob. 3. — To extract the square root of a Decimal Number, Kule, Dem. (209) ; Examples 161-164 CoE. — Eoots of Fractions {210); Examples 165 Pbob. 4. — To extract the cube root of a Polynomial {211); Eule, Dem., Examples 165-169 Pkob. 5. — To extract the cube root of a Decimal Number {212); Eule, Dem., Examples 169-173 Peob. 6. — To extract roots whose indices are composed of the factors 2 and 3 {213); Examples 174 Peob. 7. — To extract the mth root of a Decimal Number {214) 1 74 Peob. 8. — To extract the mth root of a Polynomial {215). . . 175 Examples 175-177 SECTION III. CALCULUS OF EADICALS. Reductions : Peob. 1. — To remove a factor {217) ; Examples 178-180 CoR.— To simpHfy a fraction {218) ; Examples 180, 181 Peob. 2. — When the index is a Composite Number {210) Examples 181, 182 Prob. 3.— To any required index {220) ; Examples 182, 183 OoR. — To put a coefficient under a radical sign {221) ; Examples 183-185 Peob. 4.— To a Common Index {222) ; Examples 185-187 Peob. 5. — To rationalize a monomial denominator {223) ; Examples 187-189 Peob. 6. — To rationalize a Eadical Binomial denominator of the 2nd degree {224); Examples 189, 190 Prob. 7. — To rationalize any Binomial Eadical {225) ; Examples 190, 191 Peob. 8.— To rationalize Va -f- VJT + "/c {226) 192 SECTION IV. COMBINATIONS OF EADICALS. Addition and Subteaction : Peob. 1.— To add or subtract {227) ; Examples 192-194 MULTIPWCATION . Prop, I.— Product of lil^e roots {228) 195 CONTENTS. XVU PAGE Prop. 2.— Similar Eadicals (229) 195 Pbob. 2.— To multiply radicals {230); Examples 195-198 DrvisioN : Pbop.— Quotient of like roots (231) 198 Peob. 3.— To divide radicals (232); Examples. 199-201 Involution : Pbob. 4.— To raise to any power (233); Examples 201, 20i Cor. — Index of Boot and Power alike {23d) 202 Evolution : Pbob. 5. — To extract any root of a Monomial Kadical (;255); Examples 202-205 Pbob. 6. — To extract the square root of a + n vT, or m^a ± n^b {237) ; Examples 205, 206 Imaginary Quantities : Definition {238) ; not unreal {239) ; a curious property of {24:0) 206 Prop.— Eeduced to form m^— 1, {241) 207 Examples in Multiplication and Division 207, 208 Stnopsis and Test Questions on the chapter 209, 210 PART IL-ALGEBRA. OHAPTEE I. SIMPLE EQUATIONS. SECTION L EQUATIONS WITH ONE UNKNOWN QUANTITY. Definitions :~Equation {1) ; Algebra {2) ; Members (5) ; Numerical Equation {4) ; Literal Equation {5) ; Degree of an Equation {6) ; Simple Equation (7) ; Quadratic Equation {8) ; Cubic Equation {9)\ Higher Equations {10) 211-213 Transformations. — What {11^ 12)', Axioms {13); Prdb. To clear of fractions {14); Examples 213-216 Illustration by balance 214 Prob.— Transposition {15, 16); Examples 216, 217 Solution of Simple Equations : — What {17) ;When an equa- XVIU CONTENTS. PAGE tion is satisfied (18) ; Verification {19); Proh. 1. To solve a Simple Equation (20) ; Examples 218-230 ScH. 1. — Kinds of changes which can be made (21) 220 CoK. 1. — Changing signs of both members (22) 223 ScH. 3. — Not always expedient to make the transformations in the same order (23) 226 ScH. 4. — Equations which become simple by reduction (J^4) 229 Simple Equations Containing Kadicals : — (25) ; Proh. 2. To free an equation of Eadicals (26); Examples 230-236 SuMMAKY or Pkactical Suggestions {27 f 28) 236, 237 Applications to the solution of examples {29) ; Statement, Solution, What {30)', Knowledge required in making statement {31) ; Directions to guide in making statement {32)', Examples 237-250 Translations or Equations into Pkactical Peoblems (55) ; Examples 250-252 SECTION IL SIMPLE, SIMULTANEOUS, INDEPENDENT EQUATIONS WITH TWO UNKNOWN QUANTITIES. Definitions. — Independent Equations {37) ', Simultaneous Equations {38) ; Elimination {39) ', Methods of Elimina- tion {40) 253, 254 Elimination. — Peob. 1. — By Comparison {4:1) ; Examples. . 254-256 Pkob. 2.— By Substitution {42) ; Examples 257-259 Pkob. 3.— By Addition or Subtraction {43) ; Examples . . . 260-263 Examples for General Practice 263-265 Applications 265-271 SECTION III SIMPLE, SIMULTANEOUS, INDEPENDENT EQUATIONS WITH MOKE THAN TWO UNKNOWN QUANTITIES. Pkob.— To solve {46) ; Examples 272-277 Applications 277-279 Synopsis and Test Questions on Chapter I., Part. II 280 CONTENTS. XIX FAOE OHAPTEE n. RATIO, JPHOrOMTIOK, AND PROGHESSION. SECTION I. KATIO. Definitions.— Ratio (47) ', Sign (48) ; Cor. Effect of multi- plying or dividing the terms (40) ; Direct, and Reciprocal Ratio (50, 31 ) ; Greater, and Less Inequality {52} ; Com- , pound Ratio {33) ; Duplicate, Subduplicate, etc. {54) 281-283 Examples 283-285 SECTION II PROPORTION. Definitions. — Proportion (55); Extremes, and Means {56); Mean Proportional (57) ; Third Proportional {58) ; In- version {59) ; Alternation {60) ; Composition {61) ; Division {62) ; Continued Proportion {64) 285-287 Peinciples of Tbansformation : Peop. 1. — Product of extremes equal product of means (65); Cor. 1. Square of mean proportional {66); Cor. 2. Value of any term {67) 287 Peop. 2. — To convert an equation into a proportion {68); Examples 288 Peop. 3. — "What transformations may be made* without de- stroying the proportion {69) 288 EXAJMPLES OF TeANSFOEMATION I Multiples of Terms 289 Change in Order of Terms 289-291 By Composition and Division 291, 292 Miscellaneous Exercises 292, 293 SECTION III rnOGRESSIONS. Definitions. — Progression, Arithmetical, Geometrical, As- cending, Descending, Common Difference, Ratio {70) ; Signs and Illustrations {71); Five things considered {72}.. 294, 295 XX CONTENTS. PAG» Akithmetical Progkession. — Prop. 1. To find the last term (75) ; Prop. 2. To find the sum {74) ; Cor. 1. These two formulas sufficient (75) ; Examples ; Cor. 1. To insert means {76) 295-299 Geometeical Peogression. — Prop. 1. To find the last term {77); Prop. 2. To find the sum {78); Cor. 1. These form- ulas sufficient {70) ; Cor. 2. A second formula for sum f {80) ; Cor. 3. To insert means {81) ; Cor. 4. Sum of an infinite series {82) ; Exami^les 299-303 Synopsis of Eatio, Proportion, and Progressions, and Test Questions 304 Applications of Ratio, and Proportion 305-313 OHAPTEE EI. BUSINESS RULES [OF ARITHMETIC.^ SECTION L PERCENTAGE. The Problem Stated {83) ; Prob. 1. To express the relation between base, rate per cent., and percentage {84) ; Prob. 2. To express the relation, rate per cent., Amount or Difference, and. Base {85) ; Sch. These two formulas sufficient ; Examples 314-316 SECTION II SIMPLE INTEREST AND COMMON DISCOUNT. pROB. 1. — To express the relation between principal, rate per cent., and time {86)- Prob. 2. To express the rela- tion between amount, principal, rate per cent. , and time {87) ; Sch. These two formulas sufficient ; Examples ; Cor. To find the time required for principal to double, triple, etc. {89) 317-324 Prob. — To find one of the equal periodic payments which will discharge principal and interest {00) ; Examples 324-326 CONTENTS. XXI PAGK SECTION III. PARTNERSHIP. Simple Paetnekship {91); Examples ; ^ch. Rule deduced. . 327-329 Compound Pabtnekship {92) ; Examples ; Sch. Rule deduced 329, 330 SECTION IV. ALLIGATION. Examples and Solutions 330 -334 CHAPTER lY. QUADRATIC EQUATIONS. SECTION I. PURE QUADRATICS. Definitions. —Quadratic {94:) ; Pure {96) ; Affected {97) ; Root {98) 335 Resolution op a Puee Quadeatic Equation {99) ; Cor. 1. A Pure Quadratic has two roots {100) ; Cor. 2. Imaginary- roots {101) ; Examples 335-340 Applications 340-344 SECTION II AFFECTED QUADRATICS. Definition {102) 344 Resolution. — Common method {103); Examples 345-355 Sch. 1.^ — Definition, completing the square 348 Cob. 1.— Two Roots, character of {104) 350-351 Cob. 2. — To write the root of x'^ -\- px = q without complet- ing the square {105) 353 Cob. 3.— Special methods {106); Examples 355, 356 XXU CONTENTS. PAGE SECTION III. EQUATIONS OF OTHER DEGEEES WHICH MAY BE SOLVED AS QUADRATICS. Pbop. 1. — Any pure Equation {107) \ Examples 357, 358 Pbop. 2. — Any equation containing one unknown quantity with only two different exponents, one of which is twice the other {108) ; Examples 358-360 Special Solutions ; Examples {109-111) 360-366 SECTION IV. SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE BETWEEN TWO UNKNOWN QUANTITIES. Pbop. 1. — One equation of the second degree and one of the first {112} ; Examples 367, 368 Pbop. 2. — Two equations of the second degree usually involve one of the fourth after eHminating {113) 368 Pbop. 3. — Homogeneous Equations {111:)', Examples 368-370 Pbop. 4. — "When the unknown quantities are similarly in- volved (115) ; Examples 370-372 Special Expedients {116) ; Examples 372-374 Applications 374-379 Synopsis of Quadratics and Test Questions 380 OHAPTEE V. LOGARITHMS. Definition, Illustrations and Examples {117) 381, 382 A system of Logarithms, what {118) ; Two in use {119) ... 382 The Common Use or Logarithms {120); Prop. 1. Sum of log- arithms equals logarithm of product {121) ; Prop. 2. Dif- ference of logarithms equals logarithm of quotient {122); Prop. 3. Logarithm of a power {123) ; Prop. 4. Logarithm of a root {124) 382 CONTENTS. XXUi FAGB Table of Logaeithms, what {125) ; Prob. How to make {126) 383-385 Prob. To find the logarithm of a number from a table (127) ; Characteristic, Mantissa {128, 129) ; Sample of a Table ; Examples 385-389 Prob. — To find a number corresponding to a given loga- rithm (150) ; Examples 389 Examples in the use of logarithms {ISl) 389, 390 APPENDIX. SECTION /. DIFFERENTIA TION. Definitions 391-394 Rules and Examples 394-401 Indeterminate Coefficients 401 SECTION II. The Binomial Formula demonstrated 402-404 SECTION III. The Logarithmic Series produced 404 409 Production of a Table of Logarithms 409-411 SECTION IV, Higher Equations 411-423 XXiv CONTENTS. SECTION V. PAGE Interpretation or Discussion of Equations 423-436 SUCTION VI. Permutations and Combinations 436-439 INTRODUCTION* FOR THE USE OF VERY YOUNO PUPILS WITH BUT A LIMITED KNOWLEDGE OF ARITHMETIC. SECTION / How Letters are used to represent Numbers. Exercise 1. Three times 5, and 4 times 5, and 2 times 5, make liow many times 5 ? 3 times 7, and 4 times 7, and 2 times 7, are how many times 7 ? 3 times any nnm- ier, and 4 times the same mimher, and 2 times the same mimber, are how many times that number ? Ex. 2. Two times 8, plus 5 times 8, plus 3 times 8, plus 1 time 8, are how many times 8 ? 2 times 17, plus 5 times 17, plus 3 times 17, plus 1 time 17, are how many times 17? 2 times a?iy numher, plus 5 times tlie same numher, plus 3 times the same member, plus 1 time the same numher, are how many times that number ? Ex. 3. Five 23's, plus 4 23's, plus 11 23's, are how many 23's? ^^is., 20 23's. * Pupils who have a fair knowlecli^e of arithmetic, and are somewhat mature in years, will not need to read this introduction. (See the second paragraph of the preface.) Little or no attempt is made in this introduction, at demonstration. It is desifjned to give the pupil simply some insight into the character of the literal notation, to acquaint him with the mariner in which the fundamental operations are perfonned un simple literal expressions, and to inti'oducc the equation as a method of solving problems. Of course the teacher can introduce more or less explanation of principles as he may deem best for the particular class. But with those for whom this part ia prepared, such explanations must be necessarily largely oral. 2 XXVI INTRODUCTION. Im In Algebra tve often use letters to represent, or stand for, mmihers. The following exercises will show how : Ex. 4. Suppose a stands for some number, as in the first exercise for the 5 ; 3 times a, and 4 times a, and 2 times a, make how many times a ? Again, suppose a stands for some number, as 7 in the first exercise ; 3 times a, and 4 times a, and 2 times a, are how many times a ? Again, suppose a stands for any numher, only that it shall mean the same niiniber each time ; 3 times a, and 4 times a, and 2 times a, make how many times a ? Ex. 5. If m stands for (represents) some number; how many times ni, are 2 times m, plus 5 times m, plus 3 times m, plus 1 time m ? Does it make any difference what number m stands for, so that it means the same number all the time ? Compare this with Ex. 2. Ex. 6. Suppose b represents some number (meaning the same number all the time in this exercise), 5 b's, plus 4 Z»'s, plus 11 b^s, are how many b's ? Compare with Ex. 3. 2. Thus we 7nay use any letter to represent any number, only so that it alivays means the sa^ne number in the same exercise. 3. Wien a letter is used to represent a number, the figure which tells hoio many times the mwiber represented by the letter is taken, is just loritten before the letter, the luord " times " being left out. Thus 3« means 3 times a, ib means 4 tiiiies b, ^m means 7 times m, 105a; means 105 times the number represented by x, whatever that number may be. X 4. The number placed before a letter to tell hoio many times the letter is taken, is called a Coefficient. If no figure stand before a letter, the letter is taken once, or its coefiicient is said to be 1. Thus, m means one time m. HOW LETTERS REPRESENT NUMBERS. XXVll Ex. 7. How many times the number represented by h, are 4J, 3J, 6^, and h ? That is, 4 times some number, plus 3 times the smne numher, plus 6 times the same numher, plus one time the same numher, are how many times that number ? Ex. 8. ba, plus a, plus 6tt, plus 8a, are how many times a ? ^?Z5., 20«. Query. — If the a in ^a meant one number, the a alone another number, the a's in 6« and la still other numbers, could you answer this exercise in the same way ? You could not answer it at all. The a must mean the same number all the time, in the same ex- ample. Ex. 9. 10a + 5« + la + 2a, * are how many times a ? Query.— Is it necessary that a should mean the same thing in this exercise that it did in Ex. 8 ? Ex. 10. Za + 2a + ba + 8a, are how many times a? Ans., 18a. How much is this if a is G ? Ans,, 108. How much if a is 11 ? Ans., 198. Ex. 11. Eleven times 8, minus 5 times 8, are how many times 8 ? Ex. 12. 11a — ba are how many times a ? Eleven times any number, minus 5 times the same number, are how many times that number? Ex. 13. 12a; — Ix = how many times x ? How much is this if X represents 3 ? If a; represents 2-J- ? A7is. to the last, 11. Ex. 14. bh + 4:b + 10b — 125 = how many times b ? Ex. 15. How much is dm + ^n — ^m + 6m — bm — 2m? * The pupil 19 presumed to be acquainted with the use of the signs. They are •xplaiued m the body of the full treatise, simply as a part of the science. XXVm INTRODUCTION. We do thi3 thus : Sm + 8m are 11m, ll?7i — 4m are 7m, 7m + 6m are 1dm, 13m — 5m are Sm, 8m — 2m are 6m. Hence the answer is 6m. S. In such examples as the last you can add together all the quantities with the + sign into one sum, making in this case 17m, and all those with the — sign into another, making in this case Ihriy and then subtract. Thus, 17^^ — ll7)i is Qnif the same result as before. This is, generally, the better way to do such examples. Ex. 16. How much is lOx — 4:X — ^x + dx — Sx + llo;? How much is it if x is 3 ? Ex. 17. How much is 10:r — 15:?: ? Sug's. — Of course we cannot take 15a? out of 10a;. But we can take 10.C of the 15x from the first 10a;, and there will then remain 5x of the 15a;, which cannot be taken out of the 10a;. We indicate this by writing it — 5a;. This means that 5a; was to be subtracted, but that we liad nothing to take it from. Ex. 18. How much is Sx — 5x — 2x? Ans.f — 4a;. Query. — What does this answer mean ? Ex. 19. How much is 12a + 3a — 6a - 20a ? Ex. 20. How much is 2 times 3 times a certain number, as 5 ? A71S., 6 times the number. Ex. 21. How much is 5 times 7m ? that is, 5 times 7 times a number which we will represent by m ? Ans., 35 times w?, or 35m. Ex.22. How much is 6 times Sa? 7 times 3a? 10 times 7b? 9 times 8?/ ? 9 times 8 times a number are how many times that number? Ex. 23. How much is 3 times 7m, and 4 times Sm, minus 2 times lOwi, if m represents 6 ? HOW LETTERS REPRESENT NUMBERS. xxix Ex. 24. What is 10a divided by 2 ; that is, what is J of 10a ? 27x divided by 9 is how much ? A71S. to the last, Sx. Ex. 25. How many times a number is 10 times that number, divided by 2 ; that is, J of 10 times a number ? Ex. 26, How much is ^ of 48.r ? 25a; divided by 5 ? y\ of 11a; ? 11a; divided by 11 ? 7x divided by 7 ? Ex. 27. Divide 10a; by 5, then add Sx, then multiply by 2, then subtract 4a;, then divide by 3. What is the result ? [Note.— The teacher should extend such exercises until his pupils can perform them mentally, as rapidly as one would naturally pro- nounce them.] SECTION IL More about representing Numbers by Letters. X 6. Wlien two letters refresenting numlers are written side lij side, as in a word, their product is indicated. Thus, ah means the product of the two numbers repre- sented by a and h, 3abc means 3 time^ the product of the numbers represented by a, I, and c. [Note. — Instead of saying, as above, the number represented by a, we usually simply say " the number «," or, " a," without using the word number at all. Thus we say 3 times the product of a, 6, and c] 7. If we want to rejjresent the jJroduct of a numher rep- resented hy a letter, as a, hy itself a certain numher of times, instead ofivriting aa, or aaa, etc., as we might, we write a^ a^ etc. Thus J* means the same as dhlh. a^ is read " a square ; " a\ "a cube ; " i\ " i fourth power ; " x\ " x fifth power," etc. XXX INTRODUCTION. ^'The little figure iiilaced at the right and a little above the letter is one form of what is called an Exponent ;] hut theiminl must not get the idea that all exjjonents meaiijust what has now been explained. This is the case only ivhen the ex])onent is a whole number loitliout any sign or loith 2 the + sign. Thus, a~^ does not mean aaa. Nor does a^ mean any such thing, although the - 3, and the f are expo- nents. What these do mean will be explained in due time. [Note. — It is of the utmost importance that the pupil be guarded, from the outset, against the notion that an exponent necessarily in- dicates a power. This false notion once in the head, plagues the pupil ever after.] Ex. 1. How much is \a^b, \i a =2, and 5 = 5? How much Za^Wx, if a = 3, Z> = 2, a; = 8 ? How much \W&y\ if « = 2, c = 1, ^ = 3 ? Ex. 2. How much is a^¥ + 2«?y — %, if a = 4, 5 = 3, ^ = 2 ? How much 'Za^bif — lay"" + o5, the letters having the same values as before. How much hby — 'Zab^ + 4a^3/'' -2a? Ex. 3. How many times ab"" is 4«&'* + 2a5' — Zab^ ? How many times a^y is lOa^y + 4«'?/ — Oa^y — a'^y ? Ex. 4. How man*y times ani^y^ is 4 times Sant'y^ ? How much* is 6 times 2anfy^ ? 4 times 7a'^b'^c'' ? 10 times ISimfx' ? Ex. 5. How many times ax is | of 20ax ? | of S5ax ? 102ax divided by 3 ? How much is ^ of 72a'x' ? 125a;y divided by 25 ? ISmf divided by 9 ? 8. We have learned in arithmetic that representing numbers by the figures 1, 2, 3, 4, etc., is called Arabic or Decimal Notation. In like manner, representing numbers ♦ This mcaus the same as the preccdiujj qucstiou, HOW LETTERS KEPKESENT NUMBERS. xxxi by the small letters of the alphabet, as a, h, c, dy x, y, etc., is called Literal Notation. The pupil will see that this Literal Notation is altogether a different thing from the Roman Notation, in which the seven capital letters, I, V, X, L, C, D, M are used. Because the Literal Notation is so much used in Alge- bra, it is often called the Algebraic Notation. But this notation is just as much used in some other branches of Mathematics, as in Algebra. 0. An expression like Ha^x, without any other joined with it by the signs + or — , is called a Term, or a monomial. If there are tivo such terms joined together by either of the signs + or — , the two taken together are called a binomial, as 6JV + 2«?/^, or 10a; — Say. If three terms are joined in this way it is called a Trino- mial, as Sa^y — 2ah + 21a\ Any expression consisting of more than one term is, in general, called a ^olyno- 7nial. Ex. 6. Point out the monomials, binomials, trinomials, and polynomials in the following : '^ax — 3^^ bxy — 6cd -f a — 2y, Sa'^nfxy, & — <:r, a + m, a + b + c — d, 22oaWc'^d^, abed, a — b, ab, c — x^y + ax, x' + ?/% 10^' + Sxy. 10. Terms Avhich have the same letters, affected with the same exponents, are called Similar, Thus 12a^y, Qa^y, and — Sa'^y are similar; but 12ay, ^a^y, Sex, are not similar. Ex. 7. Point out the similar terms among the folloAving: Sax% 2ax, - 5a'x\ ax, 17ax\ 16cy\ - 12cY, Sa\v% — bcy^, Qcy'', l^ax", c^y". 11. Terms having the + sign are called Positive^ and those having the — sign, Negative, XXXll INTRODUCTION. SECTION III, How Numbers are Added in tlie Literal Notation. 12 » RULE. — Write the expressions so that simi- lar TERMS SHALL FALL IN THE SAME COLUMNS. COMBINE EACH GROUP OF SIMILAR TERMS INTO ONE TERM, AND WRITE THE RESULT UNDERNEATH WITH ITS OWN SIGN". The POLYNOMIAL THUS FOUND IS THE SUM SOUGHT. Ex. 1. Add hax — Icy, 3ax + ^cy, cy — 2ax, — 4:ax — dcy, — ax + hey, and 2ax + 2cy. Operation. * Having written the numbers so that similar terms ^ax — 2c7/ fall in the same column, we may begin to add with dax + 4cy any column we choose. Adding the right-hand — 2ax + cy\ column we find it makes + Icy, and write this — ^ax — 3cy sum underneath the column added. In like man- — ax + 5cy ner the other column makes Zax (or + ^ax), which 2ax + 2cy we write without any sign, as + will then be un- — derstood. The sum is Zax + ley. dax + Icy Ex. 2. Ex. 3. bed - 2a + 4cxy 2cd + 3rt - bxy 8a - 2xy Qcd + 14:xy - Za — Ixy + 11a + xy ^cd - Iha ' 10a77i - Mif + 2a^x — Qam + 4^dy^ — lOiO^x Aam — Sdy"" — 2a''x lam - ISdy' + Qa'x — 9am + 2dy^ — ha'x am + ISf/i/' — a'x bed 4- 2a + bxy lam — lOa'x * It is not proposed to attempt demonstrations in this introduction, but merely to acquaint the student with a few of the more elementary facts and processes of the science. Judicious teachers will give more or less oral explanation, accordiptf to the character of the class. t When no sign is expressed before a term, + is understood. SUBTRACTION IN THE LITERAL NOTATION. XXXIU Ex. 4. Add 5rc - 3a + 5 + 7 and - 4a - 3a; + 2^> - 9. Sam, 2x -7a -\- 3b —2. Ex. 5. Add 2a + 35 - 4c - 9 and 5a - 3b + 2c - 10. Ex. 6. Add 3a + 2b — 5, a + 5b — c, and 6a — 2c + 3. Ex. 7. Add 6xy - 12a;', - 4:x' + 3xij, 4a;' - 2xy, and -• 3xy + 4a;'. Sum, 4:xy — Sx'. Ex. 8. Add 3a' + Uc - e' + 10, - 5a' + 6bc + 2^' - 15, and - 4a' - 9bc - lOe' + 21. SECTION IV, How Numbers are Subtracted in the Literal Notation, 13. BULK— Cb.a2^gb the signs of the terms in THE SUBTRAHEND FROM + TO — , OR FROM — TO +, OR CONCEIVE THEM TO BE CHANGED, AND ADD THE RE- SULT TO THE MINUEND. Ex. 1. From 5bij - 6a' + 3a;' subtract 2by + 3a' + x\ Operation. 5by — Qa^ + 3a;' Minuend. — 2by — 3(Z* — X* Subtrdliend with signs changed. The Bemainder sought, which is the sum of the — ^a^ + 2x^ \ minuend and the subtrahend with its signs changed.* Ex. 2. From 3aa; + 51"' y^ — 2m' take 3aa; — b'^y^ — 3m'. Bern., eby + m\ . * If the teacher thinks it best, he can give a familiar explanation of the prin- ciple involved ; yiich as is foiincl on pages 41, 42, 43, of Part I. If any explana- tion is given, it will be well to have it the same in substance as found in the body «)f the work, for obvioua reasons. 2* XXXIV INTRODUCTION. Ex. 3. From 10a — 3b -h 2c — x' subtract b — 5c + x\ Rem., 10a — 4& + 7c - 2x\ Ex. 4. From I2xy — 3c' + ah take Qxy + c' — 2aZ>. Ex. 5. From bc'y — oal) take m,x — 'Hc'y. Rem., '^&y — Sab — mx. Ex. 6. From x^ — lla:?/^ + 3« take — 6xyz -{• 7 — 2a — 5xyz. Rem., x^ -\- 5a — 7. 14. When there is a term in the subtrahend which has no similar term in the minuend, we see that this term ap- pears in the remainder with its sign changed. Ex. 7. From 6«^' —2by' + 4:X - 3cy take - 2ac' + 3b^y — 3x — 3cy + m. Rem., Sac" — 2by^ — 3b^y + 7x — m, Ex. 8. From Sm'x^ -3a — U take a' - b\ Rem., %m\c' - 3a - ^h - «' + b\ Ex. 9* From a" + 2ah + b^ take a^ - 2ab + b\ Ex. 10. From «' + b^ take «' - b\ Ex. 11. From 3cm -y take 2b - 3c. Ex. 12. From x^ — 2xy + 1 take ixy. SECTION V. How Numbers are Multiplied in the Literal Notation. 15. R ULE. — To MULTIPLY TWO MONOMIALS TOGETHER, MULTIPLY THE NUMERICAL COEFFICIENTS AS IN THE DECIMAL NOTATION, AND TO THIS PRODUCT AFFIX THE LETTERS OF THE FACTORS, AFFECTING EACH LETTER WITH AN EXPONENT EQUAL TO THE SUM OF THE EXPONENTS OF THAT LETTER IN BOTH THE FACTORS. If THE SIGNS MULTIPLICATION IN THE LITERAL NOTATION. XXXV OF BOTH THE FACTORS ARE ALIKE, THE PRODUCT MUST HAVE THE + SIGN ; BUT IF THE SIGNS OF THE FACTORS ARE UNLIKE, THE SIGN OF THE PRODUCT MUST BE — . Ex. 1. Multiply 6ax' by 3a'x\ Operation. — If we wish we may write the factors as ia arithmetic, though this is not necessary. Tlie product of Sax"^ the numerical coefficients 3 and 5 is 15. To tliis affixing da^x^ the letters a and x, and as a has an exponent 1* in the mul- tiplicand, and 3 in the multiplier, giving it an exponent 3 loa^x^ in the product, and x an exponent 5 for a like reason, we have 15a^x^, as the product of 5ax'^ x dtC^x^. Ex. 2. Multiply 10 ^nn' by 3^^^';^'. 3xy by 4:x'y\ 7cx by Sex. 2a by a^. Ex. 3. Multiply - 5a' by 6b. Product, - SOa'h - Ex. 4. Multiply - Sa'x by — 2a'x'i/. Product, (ja^x^y. Ex. 5. Find the products of the following : — lla'x by 2axy ; X^c^mx^ by — dvfx ; 9a by 4J ; — am by — xy ; ^c^dj^ hj — ad] — bxy by — x'^y'^. 16. To imdtij^ly tivo factors together ichon one or loth are 2)olynomials. R TILE. — Multiply each term of the multiplicand BY EACH TERM OF THE MULTIPLIER, AND ADD THE PROD- UCTS. Ex. 1. Multiply 2a^^• - Uy + A.m by 2rt'^'m. Operation. — It is immaterial 2«2^ — Zhy + 4?7^ where we write the multiplier, "la^V^m but we may as well write it as in arithmetic. So also it makes no ^w'lfimx — Qd^U^my + Qa^h difference whether we begin at tlie When uo exponent is expressed, 1 is always understood. 37.0„i2 XXXVl INTRODUCTION. right hand or the left to multiply. It is customary to arrange the letters in a term alphabetically-^ thus we write ^a^bhnx, instead of ^¥a^X7n, or au}'- such form. There would be really no difference in the value of tlie terms, however, in whatever order the letters came. i Ex. 2. Multiply Qax — 5a'x + Sax' by 2«V ; 2my - dcy' by 5m^c' ; 4«5 — Scd + x hy ay, a + h — c by a; ; lOa^Di^x — 4:any + Smx' by (dO^y^. Ex. 3. Multiply Wx" - 2aif + bxy by a" - 2xy. Operation.— da^x^ — 2ay^ + 5xy a^ — 2xy Product by a\ da'x^ - 2aY + 5«^iry Product by — 2xy, — Qa^x^y + 4:axy* — lOx'^y'^ Sum of partial products, da*x'^ — 2a^y^ + 5a'^xy — Qa'^x^y + Aaxy* — lOx'^y* There being no similar terms in these partial products, we can add them only by connecting them Avith their proper signs. Ex. 4. Multiply x"" — 2xy + ?/' by 2x — dy. Operation.— x"^ — 2xy + y"^ 2x-Sy Product by 2x, 2x^ - Ax'y + 2xy^ Pj-oduct by - 3y, - Zx'^y + Qxy^ ■ Entire product, 2x'' - Wy + ar^^ - 3y^ Ex. 5. Multiply a+ Jihja — n. Prod, a^ Ex. G. Multiply «* + c^' + c^" + « + 1 by a' — 1. Ex. 7. Multiply «' - ^al + V by a' + ^ah + h\ Ex. 8. Multiply m + n by m + n, Ex. 9. Multiply m — n by m — n, Ex. 10. Multiply 4a;' - 9^' \)y x + 2y, DIVISION IN THE LITERAL NOTATION. XXXVll Ex. 11. Multiply together x — S, x — 5, a; — 4, and x -7. - Ex. 12. Multiply x^ -{• y"" + z^ — xy — xz — yzhjx -{- y + z. Prod, x^ + y^ + z^ - ^xyz. SECTION VL How Numbers are Divided in the Literal Notation, 17. To divide when dividend and divisor consist of the same quantity affected with ex])onents. RULE. — The quotient is the common quantity AFFECTED WITH AN EXPONENT EQUAL TO THE EXPONENT OF THE DIVIDEND MINUS THE EXPONENT OF THE DIVI- SOR. If the signs of the divisor and dividend are ALIKE, THE SIGN OF THE QUOTIENT IS + ; IF UNLIKE, — . Ex. 1. Divide a' by a;. Quot., a\ The student will observe that the product of the divisor and quo- tient must always equal the dividend. In this case, a^ x o? =. a^ Ex. 2. Divide - x' by x\ Quot., - x\ Ex. 3. Divide — m^ by — m^. Quot., m. Ex. 4. Divide ¥ by - 1)\ Quot., - h\ Ex. 5. Give the quotients in the following cases • if -^ y* ; - x'' -i- x'; a' -^ -a'; - c' -^ - c\ 18. To divide one monomial hy anot^ier, RULE. — Divide the numerical coefficient of the dividend by the numerical coefficient of the divi- sor, AND TO THE QUOTIENT ANNEX THE LITERAL FAC- TORS, AFFECTING EACH WITH AN EXPONENT EQUAL < TO XXXVlll INTRODUCTION. ITS EXPOiq-EN^T 1^ THE DIVIDEND MINUS THAT IN THE DIVISOR, AND SUPPRESSING ALL FACTORS WHOSE EX- PONENTS ARE 0. The sign of the quotient will be + WHEN THE DIVIDEND AND DIVISOR HAVE LIKE SIGNS, AND — WHEN THEY HAVE UNLIKE SIGNS. Ex. 1. Divide Ub'x'y by Sbxij. QuoL, hVx, Ex. 2. Divide %\a\i^mf by - Wmnij. Qtiot., — dm^y. Ex. 3. Divide - lOoa'y' by - 21«^ Quot, baif. Ex. 4. Divide —lS7nn''x by Qmnx. Quot.^ — Zn, Ex's 5 to 9. Give the quotients in the following : 12«y -^ 3rt/; - 64ay -^ 16.?y ; ^Ix'xf -^ - mx'f-. 10. To divide one monomial hy another tuhen the coeffi- cient in the dividend is not divisible hy that in the divisor, or the exponent of any letter is greater in the divisor than in the dividend, or there is a letter in the divisor not found in the dividend. RULE. — Write the divisor under the dividend IN THE FORM OF A COMMON FRACTION, AND THEN RE- DUCE THIS FRACTION TO ITS LOWEST TERMS BY CAN- CELLING ALL FACTORS COMMON TO BOTH NUMERATOR AND DENOMINATOR.* The sign of the quotient is determined as in the last case. Ex. 10. Divide l^a'x' by 4«V. Operation, l^a^x^ -r- A.a^x'^ = -7-,— r- Now iu 18 and 4 there is a common factor 2 which can be cancelled ; the a^ in the numerator, ♦ The pupil is presumed to be familiar with this operation iu arithmetic. DIVISION IN THE LITERAL NOTATION. ^XXIX which is two factors of a, will cancel two factors of a from a' in the denominator, and leave a factor a in the denominator, and in like manner x"^ in the denominator cancels x'^ from the x^ in the numerator, leaving x therein. Hence, , „ „ = — , and — is the re- quired quotient Ex. 11. Divide 12wy by lOm'y. 12jnY 6^2 Operation. \2m'^y^ -r- IQm^y = \Onfy 5m 3y Ex. 13. Divide iSaVy' by - 32a'xy. Quot., - \ . Ex. 13. Divide - ^Wx'y by - 40^>V. Qiiot, |^. Ex. 14. Divide Qax by Ihy. Quot, ^, Ex. 15. Divide — Wmy by \^nx. Quot., — tt.— . Ex's 15 to 20. Find the quotients in the following cases : 24a*a;'^ -h Ibmx^ ; — "Hax^ by — 14«??i* ; ' — bab by bxy ; IQa^y"" by — Vla'y'^ ; 7aV by — Sx^ ; — imn^ by Smn. 20. To divide a polynomial by a monomial, RULE. — Divide each term of the polyn-omial DIVIDEND BY THE- MONOMIAL DIVISOR ; AND WRITE THE RESULTS IN CONNECTION WITH THEIR OWN SIGNS. Ex. 21. Divide W'x'y'' - Ua'xy + Iba'xy by SaVy'. Operation.— 3a^.cV) 6a^.i;V' - 12a'a;\iy° + 15a*x^y' 2xY - 4.oxy* + 5a^x^y Ex. 22. Divide 12a*^' - IQaY + 20«y - 2Say by - 4:ay. Ex. 23. Divide loa'bc - 20acy^ + 6cd^ by - babe, Quot,-^a + -^~-^. INTRODUCTION. Ex. 24. Divide IGwiV - 2^m'x'' + Wm'x' by 2mV. Ex. 25. Divide 35«'%' + 2Sa'bY - 4:9ai'y* by- laby. 21. To perform division lolien both dividend and divisor are ^polynomials, BULK — 1st. Arrange both dividend and divisor WITH REFERENCE TO SOME LETTER FOUND IN BOTH, i.e., place that term first at the left hand which has the highest exponent of this letter, the term containing the next highest exponent of this letter next, etc. 2nd. Having arranged dividend and divisor thus, DIVIDE the first TERM OF THE DIVIDEND BY THE FIRST TERM OF THE DIVISOR FOR THE FIRST TERM OF THE QUO- TIENT ; THEN SUBTRACT FROM THE DIVIDEND THE PRODUCT OF THE DIVISOR INTO THIS TERM OF THE QUOTIENT, AND BRING DOWN AS MANY TERMS TO THE REMAINDER AS MAY BE NECESSARY TO FORM A NEW DIVIDEND. DiVIDE AS BEFORE, AND CONTINUE THE PROCESS TILL THE WORK IS COMPETE. Ex. 1. Divide Ga'^x' - 4:ax' - 4:a'x + x' -\- a' hy — 2ax + «* + x\ Operation.— These polynomials, arranged according to the letter a, and placed in the ordinary manner f(H- division, become a^ —2ax + x'^)a* - A^a^x + Qa'^x'' — 4ax^ + x\ar' — 2ax + x^ a* - 2a^x + a'x^ — 2a^x + 5a\i^^ — Aax^ — 2a?x + 4«-.f^ — ^ax^ a'\v^ — 2«.c^ + X* n-X' — 2ax^ + x* The pupil will have no difficulty in following the work, if he com- pares it carefully with the rule. The polynomials might equally well have been arranged with reference to x. Thus x'^ — 2ax + a^) X* — Aax^ + 6aV — Aa^x + a\ This arrangement would give x"^ — 2ax + a^ for the quotient, which is essentially the same aa before. A LITTLE ABOUT FACTORING. xli Ex. 2. Divide x\f -V x' -\- if by xij -^ x^ ^ if. Quot., x^ — xy + f. Ex. 3. Divide a^ + %ah + V by a + Z*. Ex. 4. Divide a^ - 2ah -\' h' hy a - h. Ex. 5. Divide 4:ax + 4t" + a' by 2x + «. Ex. 6. Divide a'b' + h' + a^b' + a' by «' - «J + ^'. Ex. 7. Divide lOac + 3c' + Sa' + 4Z»' + Sab + 8Z>c by 25 + rt + 3c. Ex. 8. Divide a;' — y" by .t — y. Operation. x^ — x*y x*y — If" x^y'^ — y^ Ex. 9. Divide cc' — y' by a; + y. Ex. 10. Divide 4^* - b' by 2ft - 4. Elx. 11. Divide 20ft.^'' + 4.a' - ^x' - 25a'x' by 2a' + 2x' — 6ax. Ex. 12. Show that (a' - Z/') -^ {a' + «& + Z»=) = « - J. SECTION VII. A little about Factoring. 22> Tivo or more numbers tvMch, being multiplied to- gether , 2^'f'oduce a given number, are called its factors. Thus 3 and 4 are the factors of 12, because 3 X 4 — 12. xlii INTRODUCTION. So a -\- h and a — b are the factors of a' — b^, because (a + b) X (a — d) = a'' — b\ Try it, and see. Ex. 1. What is the product of a and b ? AVhat are the factors of ab ? Ex. 2. What is the product of 3, x, and ?/ ? What are the factors of dxy ? Ex. 3. What does «' mean ? What are the factors of a^ ? What of a'b' ? of 5a'b'' ? SuG. — Such an expression as 5a^b'^ may be resolved in a great variety of ways : thus 5, a, a, a, h, and b are its factors ; also, 5, a^, and b^ ; also, 5, «^ «, and ¥ ; also, 5a, a^ b, and b, etc. Ex. 4. Wliat is the product of a and x -^ y? AVliat are the factors of ax + ay? Ex. 5. What is the product of Sa and a — b? What are the factors of 3a' - dab ? of 2a - 2ab ? THE SQUARE OF THE SUM OF TWO NUMBERS. Ex. 1. What is the product of a -h b and a + b? What are the factors of a' + 2ab + b"^? Ex. 2. What is the product of x + y and x + y? What are the factors of x'^ + 2xy + y'? Ex. 3. What is the product of 1 + a: and 1 + x? What are the factors of 1 + 2:c + a;* ? 23. We see from the last examples that tJie square of the Slim of two mimbers equals the square of one of the7n,+ twice the product of the two, + the square of the second. Thus (a + b) X (a -\- b) is the square of the sum of the two numbers a and Z», and is equal to a"^ + 2ab + b'. This principle, and those in 24 and 25, are of great im- portance in factoring. A LITTLE ABOUT FACTORING. xliii THE SQUARE OF THE DIFFERENCE OF TWO NUMBERS. Ex. 1. What is the product oi x — y aud x — y'i What are the factors of x^ — Ixy + ?/* ? Ex. 2. AVhat is the product of m — n and m — n'i What are the factors of i)f — 2mn + if ? Ex. 3. What is the product of 1 — a; and 1 — a; ? What are the factors of 1 — 2x + x^ ? Ex. 4. What is the product of 2 - re and 2 — a; ? What are the factors of 4 — 4:X •{■ x^'^ 24z. From these examples, we see that the square of the difference of two numbers is equal to the square of one of them,— tio ice the2)roduct of the two,+ the square of the other. Thus (x - y) X {x — y) = x* — '^xy + ^'. THE PRODUCT OF THE SUM AND DIFFERENCE OF TWO NUMBERS. Ex. 1. What is the product oi x -\- y and a; — y ? What are the factors of x^ — y^'- Ex. 2. What is the product of a + I) and a - b? What are the factors of <^ — V'^. Ex. 3. What is the product of 1 + a: and 1 — a; ? What are the factors of 1 — ic^ ? Ex. 4. What is the product of 2 + ic and 2 - a; ? What are the factors of 4 — x^ ? 26. We see, from these examples, that the product of the sum and difference of two numbers is equal to the diff'er- ence of their squares. Thus (x -\- y) X (x — y) = x^ — y\ Ex. 1. What are the factors of 2a - 2&.? Ex. 2. What are the factors of 3fl' - ^a^'x ? Ex. 3. What are the factors of & + led + (/' ? Xliv INTRODUCTION. Ex. 4. What are the factors of a^ — 2am + vv" ? Ex, 5. What are the factors of a^ — c^ ? Ex. 6. What are the factors of of — 2x + 1 ? Ex. 7. What are the factors of 9 - a;' ? Ex. 8. What are the factors of «' + 2a + 1 ? SECTION VIII, How Operations in Fractions are performed in the Literal Notation. 2B. For the various operations in fractions in the literal notation, the ordinary rules of arithmetic for the corresponding cases apply, only, that the fundamental operations of addition, subtraction, multiplication, and division are performed by the preceding rules. TO REDUCE FRACTIONS TO THEIR LOWEST TERMS. 27* What is the rule for this operatmi in arithmetic 9 Ex. 1. Reduce — -„^ to its lowest terms. loam X Result, -X — . omx 105^'v' Ex. 2. Eeduce ^ ^ '{ to its lowest terms. looy Ex. 3. Reduce -^--^ — . , ..o to its lowest terms. Wbx + 4a'^' SuG, — Divide numerator and denominator by 4a^5. Ex. 4. Reduce ^-r-^ -^ to its lowest terms. Ix^y Ex.5. Reduce ^^^r:r^^-^ to its lowest terms. OPERATIONS IN FRACTIONS IN THE LITERAL NOTATION, xlv ^2 X^ Ex. 6. Reduce -5 — -^ 5 to its lowest terms. a + 2ax + X SuG. — Try a + x and see if it will not divide both terms of the fraction. Ex. 7. Eeduce -^ ,-^ to its lowest terms. a — b Result a — h Ex. 8. Eeduce — ^ c i to its lowest terms. m — Z}}i)i 4- n SuG. — Will any monomial divide tlie terms of tlie fraction ? IMPROPER FRACTIONS REDUCED TO WHOLE OR MIXED . NUMBERS. — /^ 28. What is the rule for this 02)eratio}i in arithmetic? Ex. 1. Eeduce ^ to an integral form. X — a ^ Result, X — a. SuG.— Divide tlie numerator by the denominator. T-i ft T» 1 10«' — ^^ax — Sx^ , . , , „ Ex. 2. Reduce 7 ^^ to an integral form. 11a — dx -^ ^ ^ , 12c' + Sac^x" - 3acx - 2ftV , Ex. 3. Reduce r-^ to an m- 4c — ax tegral form. Ex, 4. Reduce — to an integral form. oa + X T. . -r^ -, 12c' + Sac'^x^ — 3aox - 2a^x' ^ Ex. 5. Reduce ^ — --: — s to an m- 00 4- 2ax fegral form. Ex. 6. Reduce ^-^ to an integral or mixed form. Xlvi INTRODUCTION. Operation. ^'~ — -. The term ay can be divided by y, giving a + - y a. But we can only indicate llie division of b by y, by writing? it in the form -, and as the sign of both b and 2/ is + , the quotient - is + , y y and is to be added to a. 'jlOx^ XOx 4- 4 Ex. 6. Reduce — ^ to an intes^ral or mixed ox ^ 4 form. Result, 40:" — 2 + — . ox 2a^x — x^ Ex. 7. Reduce — ^ to an intesfral or mixed form. x^ Result, 2ax . a Query. — Why has — the — sign ? Q^ 4- ^' ^* Ex. 8. Reduce '- to an intesjral or mixed form. X* Result, «' — ax ■}- x^ — a + X Ex. 9. Reduce — to an integral or mixed form. X ° -, ah — y^ Also, . MIXED NUMBERS REDUCED TO IMPROPER FRACTION'S. 29' What is the rule for this operation in arithmetic 9 Ex. 1. Reduce a to the form of a fraction. a Operation.— The integral part is a. This multiplied by a makes a^ From this b is to be subtracted, because of the — sign. Thus ^ a'^ -b we have for the result -. a OPERATIONS IN FRACTIONS IN THE LITERAL NOTATION, xlvii Ex. 2. Eeduce 2a: + — to the form of a fraction. Also, a — X . X ci^ 4- '^cix + x^ Ex. 3. Eeduce a + x ~ to the form of a a — X fraction. Sug's. — The integral part a -^^ x, multiplied by the denominator a — X, gives a^ — x^. Now notice that the numerator a' + 2aa; + a-'^ is to be subtracted from this, as the sign before the fraction is — , If we subtract a"^ + 2ax + x^ from a^ — x^, we have — 'lax — 2^1 Hence the result is — - — — . a — x We may also write this thus : a"" + 2ax + x^ a^ - x^ - (a^ + 2ax + x^) ^^ a + X = ^ ^. Now the a — X a — X quantity in the parenthesis is to be subtracted from the a^ — x^. Hence, changing the signs of the terms in the parenthesis, and q} 2;^ Q^ 2ax ic'^ dropping the marks, we have . Adding a — X <>ax ^x"^ similar terms, this becomes — "^^ . Ex. 4. Eeduce 1 5-— — ^i to the form of a frac- a + tion. EesuUf -j — to the i a + X Ex. 5. Eeduce a — x '- to the form of a fraction. Result, . 4^2 __ g Ex. 6. Eeduce Zx — — ^ to the form of a fraction. ox Result, —- -. bx xlviii INTRODUCTION. TO REDUCE FRACTIONS TO EQUIVALENT FRACTIONS HA7' ING A COMMON DENOMINATOR. SO. Give the rules of arithmetic for tins process. "VYe shall use only the method of multiplying both terms of each fraction by the denominators of all the other fractions. Ex. 1. Reduce -, -, - to equivalent fractions having a , . , 7^ ,, ciyz bxz cxy common denommator. Results, -^^, — , — -. xyz xyz xyz Ex. 2. Reduce 7-> — ? t to equivalent fractions having a common denominator. Also, — ; — , , j^. X i- y X — y 3 Mesults of the la^f ^^^ ~ ^ "^^ 2^^ ^^-^ ^^' - ^V" Uesults of the last, ^^^^r^^' Z^T^^^ ^^^'W^ TO ADD FRACTIONS. SI. Repeat the rules of arithmetic for this purpose, Ex. 1. Add -, -, and -. Sum, -^r^. Zoo ou Ex. 2. Add — -, and — - — . Siwi, — - — . o 2 o Ex. 3. Add — - — , and ^r — . Sum, Ex. 4. Add 7, and 7. Swn, a — V a -h b Ex. 5. Add Y^„ and f^^- ^um, ^ _ ^, ' 10 • 2a' + 2b' a' -b' ' 2 + 2:c* OPEKATIONS IN FEACTIONS IN THE LITERAL NOTATION, xlix TO SUBTRACT FRACTIONS. 32* Wliat is the rule given in aritJimetic for this pur- pose f Ex. 1. From — take -., Rem., — . 3 4 '12 Ex. 2. From I take |. Rem., — ^. Ex. 3. From — - — take — - — . Rem., — — — . 6 9 15 Ex. 4. From take . Rem., ^ ^. x-y x+y ^ - y "[ — x^ 1 + x^ — 4a:' Ex. 5. From -— — ^ take =. Rem., V- 1 + X 1 — X • 1 — £C* MULTIPLICATION OF FRACTIONS. 33. How is a fraction multi2)lied by a ivhole nitmber? Ex. 1. Multiply ll by ha\ Prod., ^'. Ex. 2. Multiply If^^ by a H- I>. Prod., |^. Ex. 3. Multiply 1-^ by 1 - a;. Prod., \^. -L — X 1 -]- X Ex. 4. Multiply ^ by 3a. Prod., ^. Ei. 5. Multiply 3a; by ^. Pro(?., ^. 1 I^TBODUCTION. Ex.6.Multiplyg^|5^|by« + J. „ , a' — ^ah-^V. Prod., -J . Ex. 7. Multiply 10a' by j^. Prod., — . Ex. 8. Multiply I by 6 ; by 8 ; by 10. /V Ex. 9. Multiply ^~ by 3 ; — ^ by 5. O 34:. Give the rule for muUij^lying one fraction ly another. Ex. 1. Multiply J by ^. P''^-'W Ex.2. Multiply^ by-. ^'•'"'•'57/ Ex. 3. Multiply i^ by 1±^. ProA., -^^^., Ex. 4. Multiply ?^ byl-5?. Prod, 1^. Ex. 5. Multiply -y-^by ;^— ^. Prod., 3^ • DIVISION OF FKACTIONS. 5*5^. ^o?(; 15 « fraction divided ly a whole number f Ex. 1. Divide — by ^x". Quot, --^. ox i-VX Ex. 2. Divide -^-3 by 2a^b, Quot., j^. HOW PBOBLEMS ARE SOLVED IN ALGEBRA. ]i Ex. 3. Divide ^^—-^ hj a-h Quot, ^^^. Ex. 4. Divide -^-^ \^ix-y. Quot, ^^—--,, 36. Hoiu is any quantify divided by a fraction ? Ans. By rnuUiplyiny it by the divisor inverted. Ex. 1. Divide 6 by—. Quot., — . Ex. 2. Divide ^ by ? Quot., f?. X lo Ex. 3. Divide — 5— by . Quot., .,. . Ex. 4. Divide ^ by . c + d -^ X + y ax -\- ay — X — y Quot.. cm + dm — en — dn Ex. 5. Divide ^ — — by , . ab *' be „ ^ ex — cy Quot., ^ a Ex. 6. Divide ^ by -^ 7^. C^fO^., — 5 . a + b "^ a — b ox SECTION IX. How Problems are Solved in Algebra. Ex. 1. John is 3 times as old as James, and the sum of their ages is 32 ; how old is each ? lii INTRODUCTION. Solution. — This example is a very simple one, and can easily be solved mentally. Thus, we see that since John's age is 3 times James's, both their ages together make 4 times James's age, and this is 33 years. Now 4 times James's age = 32 years. Hence, James's age is i of 32, or 8 years ; and John's age being 3 times James's, is 3 X 8, or 24 years. To solve this by Algebra, we proceed as follows : Let x represent James's age ; then, since John is 3 times as old, Zx will represent his age; and the sum of their ages will be 3^' + x. Now the statement is that the sum of their ages is 32; hence ^x + x =z 32. Or, what is the same thing, Ax = 32. If Ax = 32, « = i of 32, or a; = 8. But X stood for James's age, hence, James's age is §; and John's being 3 times as much, is 3 x 8, or 24. 37' The ex2)ression ^x + x = ^2 is what is called an Equatioii ; and it is hythe use of equations that loe solve pivblems in Algebra. Ex. 2. A merchant said that he had 72 yards of a cer- tain kind of cloth, in three rolls. In the first roll, there were a certain number of yards ; in the second, 3 times as many as in the first ; and in the third, twice as many as in the first. How many yards were there in the first roll ? Sug's.— The equafion is a; + 3a; + 2^ = 72. Now, X + dx + 2x is Qx, hence 6x = 72. And if Qx = 72, x, or Ix, is i of 72, x = 12. Queries. — What does the x stand for ? Answer. The number of yards in the first roll. In this problem which is the most, a; + 3.» + 2a;, or 72 ? To start with, do you know how much a; is ? Then is it a known, or an unknown quantity, at the outset ? 58.— The number which we desire to find as the answer of a problem is represented in the beginning of the solution by one of the latter letters of the alphabet, usually x, if there is need of but one letter, and is called the JJnUnown Quantity. Ex. 3. A boy on being asked how old he was, replied, " if you add to my age 3 times my age, and 5 times my age, HOW PKOBLEMS ARE SOLVED IN ALGEBRA. liii and subtract twice my age, the result will be 49 years. How old was he ? Sug's.— The equation is x + Sx + 5x — 2x = 49. Hence, since x + dx + 5x — 2x is 7x, 7x = 49. If 7« = 49, « = I of 49, or x = 7. Ex. 4. There are three times as many girls as boys in a party of 60 children. How many boys are there ? How many girls ? Ex. 5. In a barrel of sugar weighing 200 lbs, there aro three varieties, A, B, and C, mixed. There is 7 times as much of B as of A, and twice as much of as of A. How much of A is there? How much of each of the other kinds ? A71S., of A, 20 lbs. ; of B, 140 lbs. ; of C, 40 lbs. Ex. 6. There were 4 kinds of liquor put into a cask, 2 times as much of the second as of the first, 2 times as much of the third as of the second, and 2 times as much of the fourth as of the third. The cask sprang a leak, and three times as much leaked out as was put in of the first kind, when it was found that there were 36 gallons remaining. How much was there put in of each kind ? Sug's, — The equation is x + 2x + 4^ + 8x — dx = 36. Ex. 7. In a pasture there are a certain number of cows and 23 sheep, in all 34 animals. How many cows were there ? Solution. As it is tlie number of cows we seek, let x represent the number of cows. Then x + 23 is the number of animals in the pasture, and the equation is a; + 23 = 34. Now the X + 23 means just the same thing as the 34, that is 05 + 23 3= 34. So, if w^e subtract 23 from each, there will be just as much left of one as of the other. Subtracting 23 from x + 23, there remains x, and subtracting 23 from 34, there remains 11. Hence X = 11. Now as x represented the number of cows, we know that there were 11 cows. liv INTRODUCTION. 39' The part of an equation on the left of the sign = is called the First Meuiher, and that on the right, the Second Member. Note. — The pupil must not think that because these examples are so simple that he can " do them in his head" without any al- gebra, and may be with less work, that therefore algebra is a very clumsy method of solving examples. He will find by and by, that though the equation does not really help any in the solution of such simple questions, it will solve a great many very difficult ones about which he might puzzle his brains a great while to no purpose, if algebra did not come to his aid. Stick to it, then, and learn how to use this new instrument, the Equation, and you will by and by find it wonderfully useful. It is a grand patent for solving problems. Ex. 8. In a certain pasture there are three times as many horses as cows, and 20 sheep. In all there are 100 ani- mals. How many cows are there ? How many horses ? Sug's.— The equation is a; + 3a; + 20 = 100. Subtracting 20 from each member, a; + 3a; = 80. Uniting the terms of the first member, 4a; = 80. Dividing each member by 4, x= 20. Hence there were 20 cows ; and, as there were tliree times as many horses as cows, there were CO horses. Ex. 9. In a basket of 60 apples there are 4 times as many red apples as yellow, and 10 green apples. How many yellow apples are there? How many red ? Ex. 10. John and James together have 75 cents. James has 25 cents less than John. How many cents has John ? Sug's. — Let x represent the number of cents which John has. Then, as James has 25 cents less, a; — 25 will represent what he has. But both together have 75 cents. Hence the equation is a; + a; — 25 = 75. Now, if we drop the — 25 from the first member, we make this member 25 greater than it now is, *. «., x + x is 25 greater than HOW PROBLEMS ARE SOLVED IN ALGEBRA. Iv X + x—25. Therefore, if we add 25 to the Becond member, makmg it 100, the members will still be equal * This gives x -\- x = 100, or, 2i;=100. Hence x = 50, the number of cents which John has. Ex. 11. A merchant has 90 yards of cloth in two pieces. The longer piece lacks ten yards of containing 3 times as mnch as the shorter. How much in each piece ? SuG. — The equation is a; + ox — 10 = 90. Ex. 12. Divide the number 50 into two parts so that one part shall lack 10 of being 5 times the other. Sug's. — The parts are represented by x^ and 5x — 10. They are 10 and 40. Ex. 13. Divide the number 50 into 3 parts, such that the second shall be 5 more, and the third 15 less than the first. Sug's.— The equation is a; + aj + 5 + a; — 15 = 50. The parts are 20, 25, and 5. Ex. 14. There are 52 animals in a field. Twice the num- ber of cows + 11 is the number of sheep, and 3 times the number of cows — 13 is the number of horses. How many of each kind ? Aus., 9 cows, 29 sheep, and 14 horses. Ex. 15. A man said of his age, " If to my age there added be One half,t one third, and three times three. Six score and ten the sum will be. What is my age ? Pray show it me." Sug's.— The equation isa: + ^ + 5 +9 = 130. * The word " Transposition" is pui-posely omitted from this introduction. Nor is the idea designed to be presented. It will be better for the pupil to " think out" the process, as above. + Meaning " one half my aj/e," " one third my age.'^ Ivi INTRODUCTION. Subtracting 9 from each member- ..|. 1 = 121. Now, we can get rid of the fractions in the first member by mul- tiplying it by 6, the product of both the denominators. Thus, 6 times the first member is Qx + Sx + 2x. Then, if we also multiply the second member by 6, the products will be equal. For if two quantities are equal,* 6 times one of them is equal to 6 times the other. Hence we have 6x + dx + 2x = 726. Uniting terms, lla; = 726. Dividing by 11, a: = 66. Ex. 16. Mary gave half her books to Jane, and one third of them to Helen, when she had but ^ left. How many had she at first ? Sug's. — Let X represent the number of books Mary had at first. Then she gave Jane -, and Helen - books. And what she gave the otlier girls, added to what she had left,makes all she had in the first place. Hence the equation is XX. 2 + 3 + ^ = "- Multiplying each member by 6, dx + 2x + 12 = Gx. Subtracting 5a; from each member, 13 — x. That is, Mary had 12 books at first. Ex. 17. A boy lost 25 cents of some money which his uncle gave him, and gave half he had left to his brother. He then earned 50 cents, when he had just as much as his uncle gave him. How much did his uncle give him ? Sug's. — Let x=\ the number of cents his uncle gave him. Then he had a; — 25 cents after losing 25 cents. After giving away half of this, he had the other half, or — - — cents, left. He then earned 2 50 cents, and the amount he had was equal to what his uncle gave him, * In this case the two quantities are the two members. t The Bi^n of equality used in this way means the same as the word " reprc- eent." HOW PROBIJSMS ARE SOLVED IN ALGEBRA. Ivii Hence the equation is — ^— ^ + 50 = x. Multiplying each member by 2, a: — 25 + 100 = 2x. Uniting, — 25 + 100 makes 75, and x + 75 = 2x. Subtracting x from each member, 75 = x. Ex. 18. A boy being asked how many marbles he had, said, " If I had five more than I have, half the number subtracted from 30 would leave twice as many as I now have." How many marbles had he ? Sxtg's. — Letting x represent the number of marbles the boy had, the equation is Now there is a little peculiarity about this equation, which the pupil must be careful to notice whenever it occurs, or he will make a great many mistakes. It is this : When we multiply both mem- bers by 2, to get rid of the fraction, we must write 60 — x — 5 = 4x The mistake would be to write 60 — :r + 5 = 4a:. The explanation is, that the — sign before — - — shows tliat it is to be subtracted from 30, hence the signs of the terms composing it, viz., x and 5, must be changed, according to tbe rule for subtraction. But the pupil may think that we do not change the sign of the x. If he does he mis- 2/4-5 takes. The — sign before the fraction — ^ — does not belong to the X, but to the fraction as a whole. The sign of x in the fraction — is +, since when no sign is expressed + is understood. 2 "What then becomes of the — sign before the fraction, if it is not the same as the sign of a; in the equation 60— ic — 5 = 4^? It has been dropped, since the thing signified by it has been performed, and the — sign before the x is the sign of that term in the original equa- tion, changed. " The boy had 11 marbles. Ex. 19. What is the value of x in the equation Sug's. — Multiplying each member by 3, we have Ox]— 2 + 22; = Gl. Hence » = 6. Iviii INTBQDUCTION. Ex. 20. Find the value of x in the equation 3a: - 5 , , 2a; - 4 X + — ~ — = 12 ^ — . X = ^. Ex. 21. What is the value of x in the equation X — 1 X -\- 4: ^^ X + 3 ^ Ex. 22. Show thiit in ^^-^^ = Ans., X = 40A. 23- X 4 + X 5 4 ~i Ex. 23. Two boys were to divide 32 marbles between them so that ^ of what one had should be 5 less than what the other had. How many was each to have ? Sug's. — Letting x ■= what one had, then 33 — a: = what the other had. X The equation is - + 5 = 32 — ar, 2 Z'l-x or + 5 = a:. 2 Query. — Whj'' will either equation answer the purpose ? Ex. 24. AVhat number is that to which if 7 be added, half the sum will exceed \ of the remainder of the number after 3 has been subtracted, by 8 ? ^. X + 1 x — Z ^ Equation, — z — = o. x— id. Ex. 25. The sum of two numbers is sixteen, and tlie less number divided by three is equal to the greater divided by five. What are the numbers ? SuG.— Let x and 16 — x represent the numbers. Ex. 2G. Divide twenty-two dollars between A and B, so that if one dollar be taken from three-fourths of B's share, and three dollars be added to one-half of A's money, the sums shall be equal. How many dollars will each have ? HOW PROBLEMS ARE SOLVED IN ALGEBRA. lix Ex. 27. The sum of two numbers is thirty-three. If one-sixth of the greater be subtracted from two-thirds of the less number, the remainder will be seven. What are the numbers ? Ex. 28. The sum of A's and B's money is thirty-six dollars. If five-eighths of B's, less two dollars, be taken from three-fourths of A's, the difference will be seven dol- lars. How many dollars has each ? Ex. 29. The difference between two numbers is twenty -five: and if twice the less be taken from three times the greater, the remainder will be eighty. What are the numbers ? Ex. 30. A and Bgain money in trade, but A receives ten dollars less than B. If A's share be subtracted from twice B's, the remainder will be fifty-seven dollars. How much money did each receive ? Ex. 31. One number is four less than another, and if twice the less be subtracted from five times the greater, the remainder Avill be thirty-eight. What are the numbers? Ex. 32. Two farms belong to A and B. A has twenty acres less than B. If twice A's farm be taken from three times B's number of acres, the remainder will be one hun- dred acres. How many acres has each ? Ex. 33. One number is seven less than another, and if three times the less be taken from four times the greater, the remainder will be six times the difierence between the two numbers. What are the numbers ? Ex. 34. Anna is four years younger than Mary. If twice Anna's age be taken from five times Mary's, the remainder will be thirty-five years. What is the age of each ? Ex. 35. One number is ten less than another. If three times the less be taken from five times the greater, the re- mainder will be seven times the difference of the two num- bers. What are the numbers ? INTRODUCTION. SECTION L A Brief Survey of the Object of Pure Mathematics and of the several Branches. [Note. —Omit this section till a General Keview is taken. ] 1, I* are Mathematics is a general term applied to several branches of science, which have for their object the investigatioix of the properties and relations of quantity — comprehending number, and magnitude as the result of extension — -and of form. 2. The Several branches of Pure Mathema- tics are Arithmetic, Algebra, Calculus, and Geometry. 5. Arithmetic, Algebra, and Calculus treat of number, and Geometry treats ^f magnitude as the result of exten- sion. 4. Quantity is the amount or extent of that which may be measured ; it comprehends number and magnitude. The term quantity is also conventionally apphed to sym- bols used to represent quantity. Thus 25, m, xi, etc., are called quantities, although, strictly speaking, they are only representatives of quantities. It is not easy to give a philosophical acconnt of the idea or ideas, represented by the word Q'lanUiy as used in Mathematics ; and doubtless, diflferent persons use the word in somewhat diflferent senses. It is obviously incorrect to say that "Quantity is anything which can 2 INTRODUCTION. be measured." Quantity may be afErmed Of any such concept, nevertheless, it is not the thing itself, but rather the amount or extent of it. Thus, a load of wood, or a piece of ground, can be measured ; but no one would think of the wood or piece of ground as being the quantity. The quantity (of wood or ground) is rather the amount or extent of it. The word is very convenient as a general term for mathematical concepts, when we wish to speak of them without indicating whether it is number or magnitude that is meant. Thus we say, "m repre- sents a certain quantity, " and do not care to be more specific. As applied to number, perhaps the term conveys the idea of the whole, rather than of that whole as made up of parts. It is, therefore, scarcely proper to speak of multiplying by a quantity ; we should say, by a number. On the other hand, when we apply the term quan- tity to magnitude, it is with the idea that magnitude may be meas- ured, and thus expressed in number. The distinction between quantity and number is marked by the questions, " How much?" and " How many ?" S, dumber is quantity conceived as made up of parts, and answers to the question, " How many ?" Thus, a distance is a quantity ; but, if we call that distance 5, we convert the notion into number, by indicating that the distance under consideration is made up of parts. Now, the distance may be just the same, whether we consider it as a whole, or think of it as 5 ; i. e. , as made up of 5 equal parts. Again, m may mean a value, as of a farm. We may or may not conceive it as a number (as of dollars. ) If we think of it simply in the aggregate, as the worth of a farm, m represents quantity ; if we think of it as made up of parts (as of dol- lars) it is a number. G. Number is of two kinds, Discontinuous and Contimiotis. 7. T>lscontinuous JViiniber is number conceived as made up of finite parts ; or it is number which passes from one state of value to another by the successive addi- tions or subtractions of finite units ; i. e., units of appreci- able magnitude. THE OBJECT OF PUEE I^IATHEMATICS. 3 8, Continuous Wumher is number which is con- ceived as composed of infinitesimal parts ; or it is number which passes from one state of vahie to another by passing through all intermediate values, or states. Number, as the pupil has been accustomed to consider it in Arith- metic, and as he will contemplate it in this volume, is Discontinuous Number. Thus 5 grows till it becomes 9, by taking on additions of units of some conceivable value ; as when we consider it as passing thus, 5, 5+1 or 6, 6+1 or 7, 7+1 or 8, 8+1 or 9. If the increment were any fraction, however small, the form of the conception loould he the same. As to Continuous Number, this is not the place for a full considera- tion of the idea ; hence, only a single illustration will be given. Time affords one. We usually conceive time as a disco7itinuous num- ber, as when we think of it as made up of hours, days, weeks, etc. But it is easy to see that such is not the way in which time actually grows. A period of one day does not grow to be a period of on© week by taking on a whole day at a time, or a whole hour, or even a whole second. It grows by imperceptible increments (additions). These inconceivably small parts, by which time is actually made up, we call infinitesimals, and number, when conceived as made up of such infinitesimals, we call Continuous Number. 9, Arithinetic treats of Discontinuous JV^um- her, — of its nature and properties, of the various methods of combining and resolving it, and of its appUcation to practical affairs. The leading topics of Arithmetic are : 1. Notation; i. e., methods of representing number, as by the Ara- bic Characters, 1, 2, 3, 4, etc., or by letters, as a, b, m, n, x, y, etc., 2. Properties of Numbers or deductions from the methods of Nota- tion, 3. Reduction, as from one scale to another, from one denomination to another, from one fractional form to another, or,- in short, from any one form of expression to another equivalent form, 4. The various methods of combining number, as by addition, multiplication, and involution, 4 INTBODUCTION. 5. Resolving number, as by subtraction, division, and evolution, And all these processes as effected by the use of any notation, and upon integral or fractional discontinuous numbers of any kind. Arithmetic, therefore, philosophically considered, embraces much that is usually classed as Algebra. Thus all that usually precedes Simple Equations, and all that is embraced in this Part I. , is simply a repetition and extension of the processes of Arithmetic with a new notation — the literal. Again, logarithms are nothing but a new scheme of notation, by means of which certain combinations and res- olutions are more readily effected; and the making of logarithms is but a reduction from one form of expression to an equivalent one in another notation. In the ordinary, or decimal notation, a certain number is represented thus, 256 ; in the logarithmic notation it is 2.40824. On the other hand, much that is ordinarily embraced in Arithmetic is more philosophically, and more economically transferred to Alge- bra, as will appear in the next article. 10* Algebra treats of the Uquaiion, and is chiefly occupied in explaining its nature and the methods of transforming and reducing it, and in exhibiting the manner of using it as an instrument for mathematical investigation. The whole province of the relations of quantity, continuous or dis- continuous number, is covered by Algebra, so far as the equation can be made the instrument of investigation. Much, therefore, of what is found m our Arithmetics can be more expeditiously treated by Alge- bra. Such are the subjects of Ratio, Proportion, the Progressions, Percentage, Alligation, etc. In fact, the equation is the grand instru- ment of mathematical investigation, and demonstrates its efficiency in every department of the science. To hope to get on in mathematics without Algebra, is to expect to walk without feet. 11, Calculus treats of Continuous Number, and is chiefly occupied in ■ deducing the relations of the infini- tesimal elements of such number from given relations between finite values and the converse process, and also in pointing out the nature of such infinitesimals and the methods of using them in mathematical investigation. 12, Geometry treats of magnitude and form as the result of extension and position. n LOGICO-MATHEMATICAL TEBMS. The priof ipal divisions of the science of Geometry are : 1. The ihicient, Special or direct Geometrj^ (the common Geometry of our schools,) including Trigonometry, Conic Sections, and all other geometrical inquiries conducted upon these methods, 2» The Modern, Indirect or General Geometry ^usually called An- alytical, ) and 3. Descriptive Geometry. SYNOPSIS. Subject of Section. Definition of Pure Mathematics : Several Branches of. Subject matter of the several branches. Quantity : Two uses of the term. Number : Illustration of : Kinds of: Illustration of kinds of number. Arithmetic : Topics of. Algebra. Calculus. Geometry. SUCTION 11. Logico-Mathematical Terms. IS. A JPvoposition is a statement of something to be considered or done. Illustration. — Thus, the common statement, " Life is short," is a proposition ; so, also, we make, or state a proposition, when we say, "Let us seek earnestly after truth." — "The product of the divisor and quotient, plus the remainder, equals the dividend," and the re- quirement, " To reduce a fraction to its lowest terms," are examples of Arithmetical propositions. 14:, Propositions are distinguished as Axioms, TJieorems, Lemmas, Corollaries, Postulates, and Problems. lo. A.71 Axiom is a proposition which states a princi- ple that is so simple, elementary and evident as to require no proof. Illustration. — Thus, " A part of a thing is less than the whole of it,'' " Equimultiples of equals arc equal," are examples of axioms. If (5 INTRODUCTION. any one does not admit the truth of axioms, when he understands the terms used, we say that his mind is not sound, and that we cannot reason with him. IG, A. Theorem is a proposition -which states a real or supposed fact, whose truth or falsity we are to deter- mine by reasoning. Illusteation. — " If the same quantity be added to both numerator and denominator of a proper fraction, the value of the fraction will be increased," is a theorem. It is a statement the truth or falsity of which we are to determine by a course of reasoning. 17 • A. Denionstratiofi is the course of reasoning by means of which the truth or falsity of a theorem is made to appear. The term is also applied to a logi- cal statement of the reasons for the processes of a rule. A solution tells how a thing is^ done : a demonstration tells why it is so done. A demonstration is often called 2:>roo/. 18, A. LeTtima is a theorem demonstrated for the purpose of using it in the demonstration of another theorem. Illustration. — Thus, in order to demonstrate the rule for finding the greatest common divisor of two or more numbers, it may be best first to prove that ' ' A divisor of two numbers is a divisor of their sum, and also of their difference." This theorem, when proved for such a purpose, is called a Lemma. The term Lemma is not much used, and is not very important, since most theorems, once proved, become in turn auxiliary to the proof of others, and hence might be called lemmas. 10, A Corollary is a subordinate theorem which is suggested, or the truth of which is made evident, in the course of the demonstration of a more general theorem, or which is a direct inference from a proposition. Illustration. — Thus, by the discussion of the ordinary process of performing subtraction in Arithmetic, the following Corollary might be suggested : " Subtractioa may also be performed by addition, as we can readily observe what number must be added to the subtrahend to produce the minuend." LOGICO-MATHEMATICyij TERMS. 7 20* A. Postulate is a proposition which states that somethiiig can be done, and which is so evidently true as to require no process of reasoning to show that it i^ possible to be done. We may or may not know how to perform the operation. Tt.t. ttstrattox. — Quantities of the same kind can be added together. 21^ A.I*rohlei7l is a proposition to do some specified thing, and is stated with reference to developing the method of doing it. IiiiitrsTRATioN. — A problem is often stated as an incomplete sen- tence, as, " To reduce fractions to a common denominator." 22, A. Mule is a formal statement of the method of solving a general problem, and is designed for practical application in solving special examples of the same class. Of coui'se a rule requires a demonstration. 23, A Solution is the process of performing a prob- lem or an example. It should usually be accompanied by a demonstration of the process. 24, A ScJloliuiil is a remark made at the close of a discussion, and designed to call attention to some particular feature or features of it._ Illusteation. — Thus, after having discussed the subject of multi- pncation and division in Arithmetic, the remark that ' ' Division is the converse of multiplication, " is a scholium. SYNOPSIS. Subject of the section. Proposition. III. Varieties of propositions. Axiom. III. One who will not admit truth of axioms. Theorem. III. Demonstration. Difference tweeu a solution and stratiou. the be- . demon- Lemma. III. Why the term is unimportant. Corollary. III. Postulate. . III. Problem. How stated. III. Rule. Solution. Scholium. III. PART L LITERAL ARITHMETIC CHAPTER I. FUNDAMENTAL RULES. SECTION L Notation. 2^. A System of Notation is a system of symbols by means of which quantities, the relations between them, and the operations to be performed npon them, can be more concisely expressed than by the use of words. SYMBOLS OF QUANTITY. 26, In. Arithmetic, as usually studied, numbers are rep- resented by the characters, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, called Arabic figures, or, simply, figures. * Part first treats of the familiar operations of Addition, Subtraction, Multiplica- tion, Division, Involution and Evolution, and the theory of Fractions. The only dif- ference between the processes hero developed and those with which the pupil is already familiar, grows out of the notation. Hence appears the appositeness of the term Literal Arithmetic. Hence, also, the teacher should be careful that th« pupil see the unity of purpose, aiul the reason for any difforoiico in method of exiy cutiou. NOTATION. 9 j^7. In other departments of mathematics than Arithme- tic, numbers or quantities are more frequently representee! by the common letters of the alphabet, a, b, c, . . . m, n, . . , X, y, z. These letters may, however, be used in Arithmetic ; and the Arabic figures are used in all de^ partments of mathematics. This method of represent- ing quantities by letters is often called the Algebraic method, and the method b}'- the Arabic characters, the Arithmetical. It would be better to call the former the Literal method, and the latter the Decimal. This part [Part I. ] of this volume is nothing but a chapter upon Arithmetic, showing the principles of the Literal Notation, and how the several Arithmetical operations of addition, subtraction, multipli- cation, etc. , are performed by means of this notation. 28, Tlie Literal If otationh^^ some very great ad- vantages over the decimal for purposes of mathemati- cal reasoning. 1st, The symbols are more general in their signification ; and 2d, We are enabled to detect the same quantity anywhere in the process, and even in the result. Thus it happens that the processes become general formuloe, or rules, instead of special solutions. IXiii. — These statements are very important, but it is questionable •whether young beginners can be made to comprehend them fully ; and all will see their meaning better after having studied Algebra some time. But we will try and make them as plain as possible now. To illustrate the first statement, suppose we say a boy has 7 apples ; you know just how many are meant. But when we say a boy has h ap- ples, nobody can tell how many he has. In fact, it is not designed to teU the exact number, but only to say that he has some number. Again, 7 represents the same number of units always ; but a letter may be used to represent any number of units we please ; or, it may be used, as we have just said, without our caring to specify any pre- cise number of units. This may seem to be a very unsatis- factory kind of notation ; but with patience its advantages voll appear. Consider the following examples : 10 FUNDAMENTAL RULES. EXAMPLES. 1. A boy has 8 marbles which he sells for three cents each, and takes his pay in pencils at 6 cents each. How many pen- cils does he receive ? Suppose we answer that he receives 3 times 8 divided by 6, or pencils, without giving the number more explicitly. . . Now take a similar example, using the literal notation : thus, A boy sells a certain number of marbles which we will represent by c, for a number of cents each, which we will call m, and takes his pay in pencils at b cents each. How many pen- cils does he receive ? We will answer as before, and say he receives c times m, divided by h, or — - — pencils. The pupil will notice this difference between the answers ; both, as they now stand, simply tell what operations to perform in order to get the answers ; but, in the former case, we can perform the operations and get the explicit answer, 4, while in the latter case, we can only leave it C X 771 as it is. Such an answer as — j— may seem to the pu- pil to be no answer at all ; and indeed it is not an an- swer in the same sense as he has been accustomed to think of answers ; nevertheless it is often more useful. Notice that the answer 4 is only true for the specific ex- C X 71% ample, while the answer — — is true in every like exam^ pie. No matter what the number of marbles, what the price, or what the price of a pencil, a general answer is C X 772- — T — . We also observe that the quantities 3, 8, and 6 do not appear distinctly in the answer 4 ; but the c, m, and h do, and would, in general, however complicated the problem. NOTATION. 11 2. One boy sold 5 pears at 3 cents each ; another sold 6 apples at 2 cents each ; and a third sold 3 melons at 8 cents each. How much did they all receive ? Ans. 51 cents. 3. One boy sold b pears at c cents each ; another sold m apples at n cents each ; a third sold d melons at g cents each. How much did they all receive ? Ans. bXG-{- mxn -{-dxg cents. Suggestions. — Notice that in the 3d example the several quantities of the problem are distinctly seen in the answer, but not so in the answer to M. 2. Moreover, the answer to Ex. 3 is equally true for any and all values of b, c, m, n, d and g. Consider in like manner the two following : 4. If I buy 5 cords of wood at 4 dollars per cord, and pay for it in cloth at 2 dollars jDer yard, how many yards are required? Aus. 10 yards. 5. If I buy a cords of wood at b dollars per cord, and pay for it in cloth at c dollars per yard, how many yards are required? , axb Ans. yards. 6. A man had a flock of m sheep. He lost 2n of them and raised 10a. After which he sold the flock at $c per head, taking his payment in cloth at $6 per yard. What operations must be performed on these numbers in order to ascertain the number of yards of cloth re- ceived? And how will this number be expressed ? c(m— 2>i+10a) Ans. ■ r * b 7. A man bought 3 horses. He gave for the first twice as much as for the second, and for the third c times as much as for both the others. If x represents the price of the first, how much did he give for the 3d ? Ans. {x-\--)c. 12 FUNDAMENTAL RULES. 20^ In using* the decimal notation certain laws are es- tablished in accordance with which all numbers can be rep- resented by the ten figures. Thus, it is agTeed that when several figures stand together without any other mark, as 435, the right hand figure shall signify units, the second to the left, tens, the third, hundreds, etc. ; also that the sum of the several values shall be taken. This number is, therefore, 4 hundreds + 3 tens + 5 (units). SO, In hke manner, certain laws are observed in repre- senting numbers by letters. FiKST Law. Known Quantities, that is such as are given in a problem, are represented by letters taken from the first part of the alphabet ; while XTnlznoivn Quantities, or quantities whose values are to be found, are represented by letters taken from the latter part of the alphabet. III. — A grocer has two kinds of tea, one of which is worth a cents (any qivtn number being meant by a) j)er pound, and the other h cents. Row many pounds of each must he take to make a chest of e pounds, which will be worth d dollars ? In this problem, a, h, c, and d are the given or known quantities, and hence are represented by letters from the first part of the alphabet. The unknown or required quantities are the number of pounds of each of the two kinds of tea. We therefore represent the number of pounds of the first kind by ic, and of the second kind by y. ScH. — This law is not very" rigidly adhered to, except that letters after and including v, are generally used to represent unknown quan- tities, while the others are used for known quantities. But it is some- times convenient to use a different notation. Thus, in problems in Interest, the principal may be represented by p, whether it is known or unknown, the interest, in like manner, by i, the rate per cent, by r, the time by t, etc. Accented letters, as a', a", a"\ a"", &c., (read "a prime," "a second," "a third," etc.,) and letters with subscripts, aso-i., a^, NOTATION. IS Cj, ^4, etc., (read "a sub 1," ''a sub two," etc.,) are sometime^ used. This form of notation is used when there are sev- eral like quantities in the same problem, but which have different numerical values. Thus, in a problem in vv^hich several walls of different heights, breadths, and lengths, are considered, we may represent the several heights by a\ c." a", etc., or a^, a.., a„ etc. ; 'the thicknesses by h', h",. l'\ etc., or bi, b., h-, etc., and the lengths by l', I", I'", etc., or /i, I2, k, etc. The Greek letters are also often used both for known and unknown quantities. The student will notice a difference between Algebra and Arithmetic, in that, in Algebra, the unknown quanti- ties (what he has called the "Answers" in Arithmetic) are represented in solving a problem, and are used in the solution just like known quantities. This device gives Algebra a great advantage over Arithmetic. Second Law. "When letters are written in connection, without any sign between them, their product is signified. Thus abo signifies that the three numbers represented by a, b, and c are to be multiphed together. ScH. 1. — There is here an interesting difference between this nota- tion and the decimal. There are two points of difference ; 1st, The place of a letter, as at the right or left, has nothing to do with its value ; and 2d, The sign understood between them is that of multi- plication instead of addition, as in the decimal notation. In the deci- mal system, if we write the three figures 5, 4, and 3, as we have v»Titten the letters a, h, and c, thus 5i3 ; 1st, Each figure has a par- ticular value dependent upon the place it occupies, the 5 representing . hundreds, the 4 tens, and the 3 units ; 2d, The amoiint represented i^ 500 -f- 40 -f- 3. Moreover, the several le!:ters are not at all restricted in signification ; « may represent 5, 48, 10.06, or any other number, however small or however large, and integral, fractional, or mixed ; and the same is true of any other letter. In fact, the meaning usually is, that a represents any number, h any other, and c any other, etc 1 J: FUNDAMENTAL EULES. The same letter, however, means the same thing throughout one problem. Such expressions as abc, mnxy, etc., are read by simply naming the letters in order. ScH. 2. — When figures are written with letters, their relation to the letters is the same as that of the letters to each other. Thus 4a6 means the continued product of 4, a, and h. Also 125xy means the continued product of 125, x, and y. 31, A character like a figure 8 placed horizontally, oo , is used to represent what is called Infinity^ or a quantity larger than any assignable quantity. ScH. — As the mathematical notion of infinity is always a trouble- some one to learners, we shall lose no opportunity for making it clear. By an infinite quantity is not meant one larger than any other, or the largest possible quantity. It simply means a quantity larger than any assignable quantity ; i. e., larger than any one which has limits. The mathematical notion concerns rather the manner of conceiving the quantity, than its absolute value. Thus, a series of Is, as 1 1 1, etc., repeated without stopping, represents an infinite quantity, because, from the method of conceiving the quantity, it is necessarily greater than any quantity which we can assign or mention. If we assign a row of 9s reaching around the world, though it is an inconceivably great number, it is not as great as a series of Is extending without limit. Moreover, one infinite may be larger than another ; for a series of 2s extending without limit, as 2 2 2 2, etc., is twice as large as a series of Is conceived in the same way. It is never of any use to try to com- prehend the magnitude of an infinite quantity ; we can not do it ; al- though we can compare infinites just as well as finites. It will be necessary to say much more on this interesting and important, though somewhat puzzling subject, farther on in our course. SYMBOLS OF OPERATION. 32, The Symbols of Operation used in Algebra are the same as in Arithmetic, or in any other branch of mathematics ; but to refresh the memory, we will repeat them. 33, The perpendicular cross, +, is called the plus sign, and read " i)lus." It signifies that the quantities between NOTATION. ■ 15 which it is placed are added together. Thus, a + 2c6 -f- xm is read, " a plus 2cb plus xm." 34, A short horizontal line, — , is called the minus sign, and is read " minus." It signifies that the quantity before which it is placed is to be subtracted. Thus a — 2cb — xm + 12ax is read, " a minus 2cb minus xm plus 12ax" 35, An S-shaped symbol placed horizontally, ^^, is sometimes used to signify the difference between two quan- tities, a ^v^ 6 is read, " the difference between a and 6." This sign differs from the preceding in that it does not indicate which of the two quantities is to be taken as the subtra- hend, while the minus sign requires us to consider tha quantity before which it is placed as the subtrahend. 30, The obHque cress, X, and a simple dot, • , are each signs of multiplication. In the case of literal factors, the sign is usually omitted, according to the second law of notation. Thus, 4 X ^ X ^> ^"^'^j and ^^ac, signify exactly the same thing. 37, The signs of division are, a horizontal hne between two dots, having the dividend at the left and the divisor at the right, as 12ac-i-25; or the dots without the line, as 12ao : 2b ; or the line without the dots, the dividend being writ- 12ac ten above and the di\dsor below it, as ^r ; all of which 2b are read, " 12ac divided by 2b." In performing division, the divisor is sometimes written at the left of the dividend and separated from it by a curved hne ; the quotient is then written at the right and separated from the dividend in the same manner : as, 2a)12ac{Gc, in which 2a is the divisor, 12ac the dividend, and 6c the quotient. Sometimes, espe- cially in Algebra, the divisor is written on the right of the dividend, and separated from it by a vertical line, the quo- IG FUNDAMENTAL RULES, tient in tliis case being written under the divisor. Thus, 12ac I 2a is the last example above, expressed in a diiler- 6c ent form. ScH. — It is very inelegant, though quite common in some j)arts of tlie country, to read such expressions as — — — , " 12ac over 25." Wa should read " 12ac divided by 2^." [Note. — Let great care be taken that the nature of exponents, as ex- plained in the succeeding articles, be clearly comprehended. No lit- tle difficulty arises from an imperfect understanding of this notation. A very common, though very erroneous method of reading such ex- pressions, greatly aggravates the difficulty. ] 38, A JPower of a number is the product which arises from multiplying the number by itself, i. e., taking it a certain number of times as a factor. 30, A. Moot of a number is one of several equal fac- tors into which the number is to be resolved. 40, An Exponent is a small figure, letter or other symbol of number, written at the right and a little above another figure, letter or symbol of number. 41, A JPositive Integral Exponent signifies that the number afi'ected by ic is to be taken as a factor as many times as there are units in the exponent. It is a kind of symbol of multiphcation. III. 2^, read, " 2, third power," signifies that two is to bo taken three times as a factor, i. e., 2 X 2 X 2, or 8. 3*, read, " 3, fourth power," signifies 3 X 3 X 3 X 3, or 81 ; 81 is the fourth power of 3, because it is the product of 3 taken four times as a factor, a^ is aaaaa. x*", read "a, iwth power," or "x, exponent m," is. xajx. . .etc., till m factors of x are taken. 42, A Positive Fractional Bxponent indi- cates a power of a root, or a root of a power. The de- nominator specifies the root, and the numerator the power of the number to which the exponent is attached. DOTATION. 17 III. 8^, read, " 8, exponent f," not " 8, | power," (there is no such tiling as a f power) is the second power of the third root of 8. The third root of 8 being 2, and the second power of 2 being 4, 8=* = 4. "We may also understand the power to be taken first, and then the 3. root, as will be demonstrated hereafter. Thus, 8'* is the third root of the second power of 8. The second power of 8 is 64, the third root of which is 4, which is the same result as was obtained by taking the root first, and then the power. (125) ^ is 5. (125) ^ is 25. (32) ^ is m 8. £c» , read, "x, exponent jn divided by ??," means that x is to be re- solved into n equal factors, and the product of m such factors taken. 43. A Negative Ex2)onent^ i. e., one with tlio — sign before it, — either integral or fractional, signifies the reciprocal of what the expression would be if the exponent were positive, — i. e., had tho + sign, or no sign at all be- fore it. -4 1 1 _3. 1 III. 3 , read " 3, exponent — 4," signifies -^> or ttt" 2 ^^~^^' 1 1 -2 1 1 _:i . 1 or -rr • ar-** is — - etc. Also 8 "^ is — ;' or —r ' m ''is — « • 44, The Radical Sign^ V, is also used to indicate the square root of a quantity. When any other than the square root is to be designated by this, a small figure speci- fying the root is placed in the sign. Thus v^ 5 signifies the 3d, or cube root of 5, and is the same as 5^. v/~34ar^' indicates the 5th root of 34a63 and is the same as (34a63)^. ScH. — Read "v/oac, "the square root of 5ac," not "radical Sac' The latter expression is generic, and appUes as well to F 5ac, or \/dac. Besides it is inelegant. Let the pupil read the following examples and give the signification of each. J^' 18 FUNDAMENTAL RULES. EXAMPLES. 2 1. 25a-26^. Bead, "25, a exponent — 2, 6 exponent f." It means 25 multiplied by — , multiplied by the square of the cube root of b. 2. ar^~^. Kead, "jt, exponent -1." Since — — 1 is n n , X » is the same as a; " , and hence means that n X is to be raised to a power indicated by m — n, and the nth root of this power extracted. 3. Bead and explain ^d^h~''\ 12x"'y ~ ~. SYMBOLS OF RELATION. 45. The Sign of Geometrical Matio is two dots in the form of a colon, : . Thus a : b, is read " a is to b," or, " the ratio of a to 6." It means the same as a-^b. 46. The Sign of Aritlimetical Hatio is two dots placed horizontally, •• • Thus a •• 6 is read, " the Arithmetical ratio of a to 6 and is equivalent to a — b. 47. The Sign of Equality is two parallel hori- zontal hnes, =. Thas, 2cj; =^ xy, is read, " 2gx equals xy" 5ac — 2by == dx- is read, " 5ac minus 2by equals 3x^." I Four dots in the form of a double colon, : :, is the sign of equality between ratios. Thus, a : b : : c : d, read, " a is to 6 as c is to d," means that the ratio of a to b equals the ratio of c to d, and may just as well be written a :b = c : d, CL C or r = -, all of which expressions mean exactly the samo d thing. 48. The Sign of Inequality is a character somewhat like a capital V placed on its side, <^, the open- NOTATION. 19 ing being towards the greater quantity. Thus a'^b is read, "a greater than 6." m<^n is read, " m is less than n." 49. Tlie Sign of Variation is somewhat hke a figure 8 open at one end and placed horizontally. Thus, c c fl CXI J is read, " a varies as -." a d k SYMBOLS OF AGGREGATION. 50. A Vinculum is a horizontal line placed over several terms, and indicates that they are to be taken to- gether. The parenthesis, ( ), the brackets, [ ], and the brace, \ \ , have the same sismification. !!■' Ill, a -^ by^cd — e means that (a -|- &) is to be multiplied by (cd — e). (a -j- &) X (c^ — ^) ^^'^ means the product of (a -j- &) and (cd — e). Brackets and braces are used when one parenthesis would fall within another. Thus, ] 2 + [a + (5 + c)x]?/ i u, signifies that the product of (6 -|- c) multiplied by x, is to be added to a, and this sum multiphed by y ; to this product z is to be added and the sum multiplied by u. 51. A vertical line after a column of quantities, each having its own sign, signifies that the aggregate of the col- umn is to be taken as one quantity. Thus -\-a same as (a — 6 + c)x. — h X is the SYMBOLS OF CONTINUATION. S2* A series of dots, , or of short dashes, , written after a series of expressions, signifies " &c." Thus a : ar : ar^ : ar^ ar" means that the series is to be extended from ar^ to ar» , whatever may be the value of n. 20 FUNDAMENTAL EULES. SYMBOLS OF DEDUCTION. SS, Three dots, two being placed lK)rizontally and the third above and between, . • . , signify therefore, or some ana- logous expression. If the third dot is below the first two, • . • , the symbol is read " since," " because," or by some t ]uivalent expression. POSITIVE AND NEGATIVE QUANTITIES. 54, Positive and Negative are terms primarily applied to concrete quantities which are, by the conditions of a problem, opposed in character. III. — A man's property may be called positive, and his debts nega- tive. Distance up may be called positive, and distance down, nega- tive. Time before a given period may be called positive, and aftei^, negative. Degrees above on the thermometer scale are called posi- tive, and below, negative. 55. The signs + and — are used to indicate the char- acter of quantities as positive or negative, as well as for the purpose of indicating addition and subtraction. This double signification of the signs -(- and — gives rise to some confusion, to avoid which we shall in our elementary illustrations use the small hght lined signs, + - , to mark the distinction of positive and negative, and the common sized characters, -| , to signify addi- tion and subtraction. After the principle has become familiarized, this distinction will be dropped, as it is not used in practice. SO, In problems in which the distinction of positive and negative is made, each quantity in the formidce is to be con- sidered as having a sign of character expressed or under- stood besides the plus or minus sign, which latter indicates that it is to be added or subtracted. The positive sign need not be written to indicate character, as it is customary to consider quantities whose character is not specified as positive. NOTATION. 21 III. 1. — In the expression at; -f m — ex, let the problem out of which it arose be such, that a, m, and x, tend to a positive result, and h and c to an opposite, or a negative result. Giving these quantities theii- signs of character, we have, ( -t-a) X ir^) + ( ■*-»^) — (-c) X ( ■^^^) which may be read, "positive a multiplied by negative &, plus positive m, minus negative c multiplied by positive x. " Suppressing the positive sign this maybe written, a{-b) + m — (-c)x, by also omitting the unneces. sary sign of multiplication. III. 2. — As this subject is one of fundamental importance, let carefu) attention be given to some further illustrations. We are to distinguish between discussions of the relations between mere abstract quantities, and problems in which the quantities have some concrete signification. Thus, if it is desired to ascertain the sum or difference of 468, or m, and 327, or n, as mere numbers, the question is one concerning tha relation of abstract numbers, or quantities. No other idea is attached to the expressions than that they represent a certain number of units. But, if we ask how far a man is from his starting point, who has gone, first, 468, or m miles directly east, and then 327, or n miles directly west ; or if we ask what is the difference in time between 468, or m years B. C, and 327, or n years A. D., the numbers 468, or m, and 327, or n, take on, besides their primary signification as quantities, the additional thought of opposition in direction. They therefore be- come, in this sense, concrete. Again, a company of 5 boys are trying to move a wagon. Three of the boys can puU 75, 85, and 100 pounds each ; and they exert their strength to move the wagon east. The other two boys can pull 90 and 110 pounds each ; and they exert their strength to move the wagon west. It is evident that the 75, 85, and 100 are quantities of an oppo- site character, in their relation to the problem, from 90 and 110. Again, suppose a party rowing a boat up a river. Their united strength would propel the boat 8 miles per hour if there were no cun-ent ; but the force of the current is sufficient to carry the boat 2 miles pei hour. The 8 and 2 are quantities of opposite character in their rela- tion to the problem. Once more, in examining into a man's business, it is found that he has a farm worth m dollars, personal property worth n dollars, and accounts due him worth c dollars. There is a mortgage on his farm of h dollars, and he owes en account a dollars. The m, n, and c are quantities opposite in their nature to h and cX. This opposi- tion in character is indicated by calling those quantities which contribute to one result positive, and those which contribute to the opposite result ne^Or- iive. 22 FUNDAMENTAL RULES. 57. Purely abstract quantities have, properly, no dis- tinction as positive and negative ; but, since in such prob- lems the plus or additive, and the minus or subtractive terms stand in the same relation to each other as positive and negative quantities, it is customary to call them such. III. — In the expression, 5ac — Scd -\- 8xy — 2ad, though the quan- tities, a, c, d, X and y be merely abstract, and have no proper signs of character of their own, the terrns do stand in the same relation to each other and to the result, as do positive and negative quantities. Thus, Sac and 8xy tend, as we may say, to increase the result, while — Scd, and — 2ad tend to diminish it. Therefore the former may be called positive terms, and the latter negative. oSm ScH. — Less than zero. Negative quantities are frequently spoken of as ' ' less than zero. " Though this language is not philosophically correct, it is in such common use, and the thing signified is so sharply defined and easily comprehended, that it may possibly be allowed as a conventionahsm. To illustrate its meaning, suppose in speaking of a man's pecuniary afiairs it is said that he is worth "less than noth- ing ;" it is simply meant that his debts exceed his assets. K this excess were $1000, it might be called negative $1000, or -$1000. So, again, if a man were attempting to row a boat up a stream, but with all his effort the current bore him down, his progress might be said to be less than nothing, or negative. In short, in any case where quan- tities are reckoned both ways from zero, if we call those reckoned one way greater than zero, or positive, we may call those reckoned the other way "less than zero," or negative. 50, The value of a Negative Quantity is conceived to increase as its numerical value decreases. III. — Thus -3 >> -5, as a man who is in debt $3, is better off than one who is in debt $5, other things being equal. If a man is striv- ing to row up stream, and at first is borne down 5 miles an hour, but by practice comes to row so well as only to be borne down 3 miles an hour, he is evidently gaining ; i. e., -3 is an increase upon -5. Finally, consider the thermometer scale. K the mercury stands at 20^ below 0, (marked -20^) at one hour, and at -10° the next hour, the temperature is increasing ; and, if it increase sufficiently will b^ come 0, passing which it will reach -+■ 1°, -*• 2°, etc. In this illustra- MOTATION. / 23 Hon, the quantity passes from negative to positive by passing through 0. This is assumed as a fundamental truth of the doctrine of positive and negative quantities, viz. : That a quantity in passing through CHANGES ITS SIGN. It appears in geometrj^, that a quantity may also change its sign in passing through infinity. Thus the tangent of an arc less than 90^ is positive ; but if the aic continually increases, the tangent becomes in- finity at 90 , passing which it becomes negative. From this illustra- tion it also appears that negative infinity, -co, is, sometimes, at least, to be considered as an increase upon positive infinity, -»- (x. The importance of these considerations appears in Trigonometry and especially in the Calculus. NAMES OF DIFFERENT F0R3IS OF EXPRESSION. 00* A. I*olynoinial is an expression composed of two or more parts connected by the signs plus and minus, each of which parts is called a term. 01* A- 3£oilomial is an expression consisting of one term ; a Sinoniial has two terms; a Trinomial has three terms ; etc. 1 III. 5a-b — cd '^ -{- x — 4 (a + &) is a polynomial of 4 terms. The first three arc monomial terms, and the last is a binomial term. 5ac — ef and x- + is a trinomial. ef and x- + y- are examples of binomials. 2a^x'^y — 125d—^ -\- 12 02, A Coefficient of a term is that factor which is considered as denoting the number of times the remainder of the term is taken. The numerical factor, or the pro- duct of the known factors in a term is most commonly called the coefficient, though any factor, or the product of any number of factors in a term may be considered as coefficient to the other part of the term. III. — In the term 6a, G is the coefficient of a. In ax, a may be called the coefficient of x, or 1 may be called the coefficient of ax. In 6axj/, 6 is the coefficient of axy, 6a oixy, and &ax of y. In 5,a6 — c\ 5 is the coefficient of {ab — c) ; and in {2a^ — cd) xy, i2a' — cd) ia the coefficient of xy. 24 FUNDAMENTAL EULES. . 03, Similar Terms are sucla as consist of the same letters affected with the same exponents. luD. 5ab, 13db, and ah are similar. — IGx^-ip, 5x'y^,&nd — jc*? are also similar to each other. 4a&, Bah-, and — 2ab^ are dissimilar, as are Sax, — 56a;^, 4:cx% and 5xy. EXERCISES IN NOTATION. 1. "Write in mathematical symbols, 5 times the square root of a, added to the cube of the sum of a and b. Result, (a + by + Wa, or {a + by + 5a^. How many terms in this result ? What kind of a term is the first one ? 2. Write the second power of a, plus 3 times the product of c square multiplied by b, diminished by m times the cube root of the binomial, the square root of a minus the cube of b. Result, a- + dc-b — m (a- — ^3) ^, or a- + ^c^b — m^a^^ — 6\ 3. Write three times a into 5, plus the binomial a minus 6, divided by the sum of a square and b cube. 4. Write the fraction, the product of the sum of a and b into the sum of x and \j, divided by the square root of a diminished by the cube root of b. j^esult, (- + ^)(-^y). ^a — a/ mx — y-'^ : im^. NOTATION. 27 Synopsis for Eeview. What? o C o 0/ (Quantity Arabic. — See Ai-ithmetic. (1. More general. Advantages. | 2. Can trace quant's. Examples. { Known quantities j g^j^ ^ I Unknown " ( Literal ^ ^'^ ^^"'^ Acc'ts, subs's, Gr. let's. [Diff. bet. Alg. andArith. ["Letters in connection. 2nd Law \ Sch. 1. Two p'ts of diff. [ Sell. 2. Fig's with lett'rs. Infinity, meaning of. 0/ Operation X, 4-, : , p h)a{c, a\h__ c Definition, Power, Root, ^. \ Integral, Exponent, \ Fractional, !- How read. sarative. iNe: , =,::, X, a. Of Relation | :, •• Of Aggregation \- Of Continuation \ , . Of Deduction | .•.,'.• . What ? ///. How distinguished. Two signs of every quantity. Plus and Minus Terms become Positive and Negative C Meaning. "Less than zero. "J How negatives increase. [ How a quantity changes sign. o ^ r 2 Polynomial. Term, i g J Monomial. Binomial. TrinomiaL o g j Coefficient. ^fg Similar Terms. 28 FUNDAMENTAL RULES, SECTION IL Addition. S4:, Addition is the process of combining several qaantities, so that the result shall express the aggregate value in the fewest terms consistent with the notation. 6S. The Su7n or Amount is the aggregate value of several quantities, expressed in the fewest terms con- sistent with the notation. III. — To add 346, 234, and 15, is to find an expression for their ag- gregate value in the fewest terms consistent with the decimal notation. The sum or amount is 595, because it is such simplest expression for the aggregate. In hke manner the sum of 4ac -)- 5& + ^x, 13ac -|- 26 + 8x, and 126 + 9^, is 17ac -\- 19x -J- 196, because it is the simplest expression for the aggregate value consistent with the literal notation. The aggregate value of 346, 234, and 15, is expressed by simply read- ing them in connection; as, "346 and 234 and 15." But, letting h stand for hundi'eds, t for tens and u for units, this is (3/i, + 4< -j- 6?i) + (2/i -f 3i ,-f h.1) -\- {It -f- 5m) or a polynomial having 2 trinomial and one binomial terms, or 8 terms in all. But 595 is ^h -\- ^t -\- 5u, or simply a trinomial. If the pupil is acquainted with other scales of natation he knows that with radix 100, 595 is expressed by 2 figures. 6G, JProp. !• Similar terms are united by Addition into one. Dem. — Let it be required to add 4ac, 5ac, — 2ac, and — 3ac. Now 4ac is 4 times ac, and 5ac is 5 times the same quantity (ac). But 4 times and 5 times the same quantity make 9 times that quantity. Hence, 4ac added to 5ac make 9ac. To add — 2ac to 9«c we have to consider that the negative quantity, — 2ac, is so opposed in its char- acter to the positive, 9ac, as to tend to destroy it when combined ADDITION. 29 (added) with it. (As if 9ac were property, and — 2ac debts. ) There- fore, — 2ac destro^-s 2 of the 9 times ac, and gives, when added to it, 7ac. In hke manner — Sac added to lac, gives 4ac. Thns the four ' similar terms, 4ac, 5ac, — 2ac, and — Sac, have been combined (added) into one term, 4r/c ,• and it is e\-ident that any other group of similar terms can bo treated in the same manner, q. e. d. EXAMPLES. 1. Add ISm^n, — lOm-n, — 6771271, Bm'^n, and — Am^n. Model Solution. — Adding together 13m^n, and — lOm^n, the — lOm^n destroys 10 of the 13 times m'^n and gives 3m^n. Add- ing 3m^7i and — Qm^n, the 3m^7i destroys 3 of the — Grn"^ ji and gives — dm^n. — Sm^n added to 5m^n destroys 3 of the 5 times m^n and gives 2m^n. 2m'^n added to — Am^n destroj's 2 of the — 4m% and gives — 2m'^n. Hence, the sum of Idm'^n, — lOm^n, — Gm^n, 5m^n, and — 4??i^n is — 27n'^n. 2. Add 18ax^, — 5ax^, — lOax^, 4:ax^ and — Gax'^, ex- plaining as above. Result, ax^^. 3. Add — 5c ^^2^ — 26^ x"', Sc^x^-, 3c^.r2, and — 4:C^x'^, ex- plaining as before. Result, 0. 4. Add 3ajT, ^ax, — ax, 2ax, — lax, and ^ax. 5. Add 2hy\ — (Sbif, — hy\ 8by\ 3by\ and — 2by^. 6. Add 5ax-^, — 2ax^, 3ax-\ — dax'^, and ax^. Sum, — 1ax\ 7. Add 5;r^, — Qx^, — 10^^, + 3x^, and llA 8. Add — Ga2, + 2a\ — Sa^, 4.a\ — Za\ and a\ 9. Add — 2a^x, -f a^^x, — 3av^jr, la^x, and — 4.a^x. ■ Sum, — a^x. 10. Add — 20771^, 4aV^, 3a77i^, and — a'^m. Sum, 4:a'y~m. 11. Add lOa^x^, ~4.a^Vx,— 2a^x^ and 4.a^v'x. 30 FUNDAMENTAL RULES. 12. Add l^m, l^am, — 3a/n, and am. Sum, 2am. 13. Add 21a-, — a\ — 28^2, and — 4a2. Sum, —Ga^ 14. Add 3w^ — ^m\ m-, and — ^m'^. Sam, 2-^-^mK 15. AddTv^^, — 5^x, 12V x, and ^ ^^x. Sum, 11 v"^. 16. Add 96, lb, — f 6, — 86, — |6. Sum, —^b. G7* Cor. 1. — In adding similar termi^, if the terms are all positive, the sum is positive ; if all negative, the sum is nega- tive; if some are positive and some negative, the sum takes the sign of that kind (positive or negative) which is in excess. ScH. — The operation of adding positive and negative quantities may look to the pupil like Subtraction. For example, we say -4-5 and -3 added make -t-2. This looks like Subtraction, and, in one view, it is Subtraction. But why call it Addition ? The reason is, because it is simply putting the quantities together — aggregating them — not finding their difference. Thus, if one boy pulls on his sleigh 5 pounds in one direction, while another boy pulls 3 pounds in the op- posite direction, the combined (added) effect is 2 pounds in the direc- tion in which the first pulls. If we call the direction in which the first pulls positive, and the opposite direction negative, we have -+-5 and -3 to add. This gives, as illustrated, -t-2. Hence we see, that the sum of -t-5 and -3 is -i-2. But the difference between +5 and —3 is 8, as appears in the fol- lowing illustration : Suppose one boy is drawing his sleigh forward, while anotlieris holding back 3 lbs. If it takes just 10 lbs. to move the sleigh itself, the first boy will have to pull 13 lbs. to get it on. But if instead of holding hack 3 lbs., the second boy pushes 5 lbs., the first boy will only have to pull 5 lbs. Thus it appears, that the difference between pushing 5 lbs. (or +5) and holding back 3 lbs. (—3) is 8 lbs. In like manner the sum of $25 of property and $15 of debt, that is the aggregate value when they are combined, is $10. -+-25 and -15 are -t-10. But the difference between having $25 in pocket, and being $15 in debt, is $40. The difference between -+-25 and -15 is 40. 17. A thermometer indicated -+-28° (28° above 0); it then rose 10°, then fell 3°, then rose 2°, and again feU 7°. What was the sum of its movements ; or, how did it stiind at last? ADDITION. 31 Model Solution. — Calling upward movement -t- and downward-, the movements were -t-10, -3, -t-2 and -7, the sum of which is ^-2. Hence it rose 2°. As it originally stood at 28° above 0, and, in the v/hole, rose 2^, it stands at last 30^^ above 0. 18. A party are rowing up a stream, and alternately row and rest. During 3 periods of rowing they advance Smn, 2mn, and 6mn rods. But during the correspond- ing periods of resting, they float down 5mn, mn, and 4m7i rods. What was the result ; did they, on the whole, ascend or descend, and how much ? In other words, what is the sum of -^^mn, -h2mn, -i-6mn, -5mn, -mn, and -imn ? Ans. The}^ ascended mn rods. ( -hmn. ) 19. A man has a farm worth $100c^, on which there is a mortgage of $lDcd ; he has personal property worth $8cd, and accounts due him of $2cd, but owes on ac- count $5cd, and on note ^Icd. What is the sum of his effects? Or what is the sum of -+100cd, -15cd, ■+8cd, -t2cd, -5cd, and -led ? Ans. He is worth $83cd. {-hSScd.) OS, CoE. 2. — The sum of two quantities, the one positive and the other riegative, is the numerical difference, luith the sign of the greater prefixed. 60, CoR. 3. — It appears that addition in mathematics does not always imply increase. Whether a quantity is increased or diminished by adding another to it depends upon the relative nature of the two quantities. If they both tend to the same end, the result is an increase in that direction. If they tend to opposite ends, the result is a diminution of the greater by the less. 70, Pvojy* 2, Dissimilar tei^ms are not united into one by addition, but the operation of adding is expressed by lorit^ 32 FUNDAMENTAL RULES. ing them in succession with the positive terms preceded by the + sign and the negative by the — sig7i. Dem. — Let it be required to add -|- ^cy^, + 3a&, — 2xy, and — mn. 4:cy^ is 4 times cy^, and 3ab is 3 times ab, a different quantity from cy^ ; the sum will, therefore, not be 7 times, nor, so far as we can tell, any number of times, cy^ or ab, or any other quantity, and we can ]onlj express the addition thus : 4:cy^ -\- Sab. In Uke manner, to add rto this sum — 2xy we can only express the addition, as 4c?/ ^ -}" 3a& + ( — 2xy). But since 2xy is negative, it tends to destroy the positive quantities and will take out of them 2xy. Hence the result will be 4ct/^ -{- Sab — 2xy. The effect of — imi will be the same in kind as that of — 2xy, and hence the total sum will be 4c?/^ -|- Sab — 2xy — inn. As a similar course of reasoning can be appUed to any case, the truth of the proposition appears, q. e. d. ScH. — In such an expression as 4ci/^ -|- Sab — 2xy — mn, the — sign before the 7nn does not signify that it is to be taken from the imme- diately preceding quantity ; nor is this the signification of any of the signs. But the quantities having the — sign are considered as opera- ting to destroy any which may have the -\- sign, and vice versa. EXAMPLES. 1. Add together 5ax, — lOcy, 8b, and — n. Model Solution. 5ax and lOcy being dissimilar will not unite into one term, since one is 5 times a certain quantity, and the other is 10 times cy, a different quantity ; therefore I can only express the ad- dition, as 5aic -|- ( — lOcy.) But the lOcy being negative tends to destroy positive quantities, and will take out of such its numerical value. Hence 5ax -{- ( — 10c?/) is 5ax — lOcy. To this adding 86 which is positive and hence will go to increase the result, I have 5ax — lOcy -\- 86. Finally, as n is negative it diminishes the result by its numerical value, and I have for the sum 5ax — lOcy -\- 8b — n. 2. Add together 4am, — 2c'^y, — 8x, and 5bn, explaining as above. ' 3. Add — 2c2?w* + 4c??i, — Gc^m^, and lOc^m, explaining as before. ADDITION. 33 4. Add 7a -^b, — S2x '^, Gmn, and — 5a^, explaining as before, and find the numerical value of the result if a = 3, b = 18, X = 8, m = 2, and n = b. Numerical result, 21. 71» Cor. — Adding a negative quantity is the same as sub- r acting a numerically equal positive quantity ; that is, m + (-nj is m — n, shown as above. 72, JPi^ob, — To add polynomials. RULE. — Wkite the polynomials so that similar TERMS SHALL FALL IN THE SAME COLUMN. COMBINE EACH GROUP OF SIMILAR TERMS INTO ONE TERM, AND WRITE THE RESULT UNDERNEATH WITH ITS OWN SIGN. ThE POLYNO- MIAL THUS FOUND IS THE SUM SOUGHT. Dem, — As the object is to combine the quantities into the fewest terms, it is a matter of convenience to write similar terms in the same column, as such, and only such, can be united into one. {06^ 70,) Now, since in poljTiomials the plus and minus terms stand in the same relation to each other as positive and negative quantities {57), they may be considered as such. The partial sums will then be dis- similar terms and -wall be added by Prop, Art, 70; that is, by connecting them with their own signs, q. e. r>. EXAMPLES. 1. Add 16ac — 2m + xy, 8m — bxy — d — 2ac, — 3xy — 4ac — Q>my and 2mn — 3ac + 8xy. Model Solution. — Writing the first polynomial as it stands, I ar- 16ac — 2m + xy ^^^g^ *^^ «t^e^« «« t^^* — 2ac + 3m — 5xy — d ^^^^^^ t^^^ ^^""^ ^^^^^ ^^^ g^^^ 2j,y in the same column, for _ ^ac + 8^?/ + 2mn convenience in uniting = ; ^^ 1 — ] — -; them. There beinsr no lac, — 5m, + xy, — rt, + 2mn ^ • -i * i o t ' 1 — \ iii —- term similar to + 2mn I lac — 5m + xy —d + 2mn Turing it down, and m hke manner — d. 8xy and — Sxy are 5xy. Bxy and — ^xy are 0. and 34 FUNDAMENTAL liULES. xy, or simply -\- xy, is the sum of the similar terms in xy. Writing this result I pass to the next column. — 6m and -J- 3m are — 3m. — 3m and — 2m are — 5m which being the sum of the similar terms in m, is written down. In hke manner the sum of the terms in ac is lac. The partial sums are, therefore, lac, — 5m, -|- xy, — d, and -\- 2mn. But these being dissimilar terms are added by connecting them with their own signs, (70); whence, the sum of the several polynomials is lac — 5m -{- xy — cZ -f '^mn. In like manner solve and explain the following : 2. Add 6x + 5ay, — 3x -\- 2ay, x — • 6a?/, and 2j7 + ay. Sum, ^x 4" 2ai/. 3. Add 3a?/ — 7, — ay + 8, lay — 9, — 3a?/ — 11, and lOay — 13. • Sum, Hay — 32. 4. Add — 3a6 +7x, 3a6 — 10^, 3a6 — Q>x, —ah + ^x, and 2a6 + 4^. Sum, Aab + 4r. 5. Add — 6a'' + 26, — 35 + 2a^ — 5a^ — Sb, Aa'~ — 2b, and db — 3aK Sum, — Sa^— 26. 6. Add 3a"-b^ — 706^ + ^cixy, — la'^b^ — 2ab' — axy, ab* — laxy + 8a263, — 10a6^ + a-6-^ + 3axy, and — Sa'^b^ + 18a6<. Sum, 0. 7. Add 13a^2 — 14^2 4. ^ac^ — mn'-, 4.ax^ + Wy^ — da^c, 4:y^ — 17a^2 -j. 2ac^ + 2m'^n + 3mn-, and lOy^ — a^c. Sum,15y' + 5ac^ + 2mn^ — 4a3c + 2m^n. 8. Find the sum of 2a3 -|- 4:¥x — c'x% 2c'-x'' + 4a^ — 66% and 26^07 — Ac^x^- -\- 2a\ Sum, 8a^ — 3c"-x^. 9. Find the sum of Sa'^x^ — Sxy, 5ax — 5xy, 9xy — 5ax, 2a"x--\-xy, and 5ax — 3xy. Sum, lOa^a;^ -\- 5ax — xy. 10. Find the sum of 2bx— 12, 3^^— 26j:, 5x^—3^''^ 6^x + 12, x-^+3, and 5x-^ — l\/x. Sum, 14a;2 — 4:^x-\-S. 11. What is the sum of 20a^c^x -f 15a/i — 15 a-'c^x — 23a/i ? Ans., ba'^c^x — 8a/i. ADDITION. 35 12. What is the sum of 16xy — 4/im + llxy + 43/i?n ? Ans., 3dxy + ddhm. 13. What is the sum of Bc^ — 49an + lOc^ + l^a/i ? ^ns., 15c* — 35an. 14 What is the sum of lOx^y^ — llsk + 15^^?/* + 5sk — 4tx'^y^ ? ^ns., 21x'^y'^ — I2sh ScH. 1. — In practice, the partial sums are written at once in connec- tion, with their own signs. Also it is desirable to avoid repeating the quantities added. And, finally, the expert will not take the trouble to arrange the poljniomials, but will simply select and combine the similar terms, wiiting each result in the total sum, at once. Thus, in solving Ex. 7, when the object is simply to find the sum, and not, as above, to explain the process, we proceed as follows : Noting the 13ax^, we cast the eye along till we find the similar term, -|- ^aic^, and say " -|- llax^ ;" again, casting the eye along till we find — Ylax^, we say "0." Therefore nothing is written in the sum for these tenns, as they mutually destroy each other. Again, looking to — 14?/^, and then on to IS^/^, we say "y^," and, passing on to -)- ^y^, say " 5y2," and again pass on to \Qy^ and say "ISt/^." This being the sum of the terms in 2/2, it is then written in the answer. In the same manner the work is carried on to completion : i. e. , ordy naming results, and writing them in the total sum. In this manner, write the answers in the following ex- amples : 15. Add 3^ — 5?/ 4- 4c, 2x — 2y— 3c, and — x + Sy -{- c. Sum, 4:X — 4?/ + 2c. 16. Add 11a + ISx — Id, 4.a — lOx — 2d, and — 9a — ^ + 3d Sum, 6a + 207 — M. 17. Add 3a^ — 4thy + 2mn — 16, 2hy — 5mn -f 11, 3m?j — 2ax + 5, and — hy -\- mn — ax. Sum, mn — Zby. 18. Add 6am^ — 36 + ^cxy — 2ax^, Ab — Sexy + Soa:*, llcxy — xy — 8amx — Sax'^, and 5xy — 6 — b. Sum, 12cxy — 2a772^ -f Axy — 6. 36 FUNDAMENTAL RULES. 19. Add 1x\j — 2.i7-, Zx^ + xy, x/^ + xy, and 4.r* — ^yx, 20. Add 2a^ — 30, 3^-^ — 2ax, ^xr- — 3^^, and Z^lc + 10. /S'l^m, 8^:2 _ 20. 21. Add 8a2j72 — ^ax, lax — ^xy, 9xy — 5ax, and 2a^x^ +xy m. Add 6a.r2 + Wx, — 2ax^ — 6x^, Sax^- — IOj?^ — 7a^- H- 3v/ ^, and a^-2 + llv" x. 23. Add 6xy — 12^^^ — 4^2 _^ 3^^^ 4^2 — 2jti/, and — Sxy + 4^2. 24. Add 4.ax — 130 + 3^^, Bx^ + 3aa; + 9^2, 7^?/ — 4v/^ -f 90, and v^^ + 40 — 6x^ 25. Add 3a-2 + 46c — r^ + 10, — 5a-^ + .66c + 2e2_-15, and — 4a- 2 — 96c — lOc"- + 21. 26. Add 7a — S?/'', 8v^' + 2a, 5if — ^x, and — 9a + 7v^ together. Sum, lA>/x. 27. Add 4m7i + 3a6 — 4c, 3a; — 4a6 + 2mn, and 3m2 — 4p together. >S'a??i, 6m?i — a6 — 4c + 3.r + 3m2 — 4p. 28. Find the sum of 3a2 + 2a6 + 46^ 5a2 — 8a6 + 6^ — a* + 5a6 — 62, 18a2 — 20a6 — 1962, and 14a2 — 3a6 + 2O62. 29. Find the sum of 4:X^ — ha^ — 5a^2 _|. 6a2^, %a^ -f 3^^ + 4ajT2 -]- 2a^x, — Vlx^ + 19a^2 — IBa^x, I'^ax^ — 27a2^ + 18a^ 3a2j7 — 20a'^ + 12.r3, and 31a2.r — 2^?^ — 31a^2 — 7^3. Sam, — 7^^ — a^. 30. Find the sum of 2a6 + 12 — xHj, x^y -\- xy + 10, Zxy^ + 2^2y — xy, 5xy + 11 + x^y, and 17 — 2x^y — x'^y Sum, 2ab + 50 + 5xy — x'^y -\- 4:xv'y. ADDITION. 37 31. Add j-^ -[- ojc — ah,ab — \^x + xy, ax -\- xy — 4a/j, x'^ + \/j; — X, and xy -{■ xy -{- ax. Sum, 2x'^ -\- Sax — 4a6 + ^^y — ■^• 32. Add Ix^y — 2xv'^ + 7, ^xy + 3xy^ + 2, SyV^-— v^^ ,-1 1. A — 6, 9yva; — 4i/^j; — 3, and 1 + Ixy'^ — 2yx^. Sum, ISx^y -f Sxy^ + 1. ScH. 2. — The object and process of addition, as now explained, will be seen to be identical with the same as the pupil has learned them in Arithmetic, except what grows out of the notation, and the consideration of positive and negative quantities. For example, in the decimal notation let it be required to add 218, 10506, 5003, 81, and 106. The units in the several numbers are similar terms, and hence are combined into one : so also of the tens, and of the hundreds. To make this still more evident, let ii stand for units, t for tens, h for hundreds, th for thousands, and t. th for ten thousands. 21'" is then 2/i + 1« + 8m, 10506 is Itth + 5h + 6u, 5003 is 5th + 3u, 81 is 8«, + lu, and 106 is Ih -[- 6u. "Writing these so that similar terms shall fall in the same column, we have the arrangement in the margin. Whence, adding, we get the sum. The process of carrying has no analogy in the literal notation, since the relative values of the terms are not supposed to be known. 2h -{- 4:t -{- oil Again, there is nothing usually It.th -{-oh -\- 6u found in the decimal addition ^*h -\- oil ii]£e positive and negative quan- of -\-.lu titles. With these two excep- tions the processes are essen- l/l 4- Qu It.th + Wi + 9A + 4/! + 4?^ *'^^^^ ^^^ '^°'^- The same may be said of addition of compound 1oJ44 numbers. 73. I^roj), 3, Literal terms, which are similar only with reqject to part of their factors, may be united into one term with a polynomial coefficient. D^^i- — Let it be required to add 5aar, — 2cx, and 2mx. These terms are similar, only with respect to x, and we may say 5a times x and 38 FUNDAMENTAL EULES. — 2c times x make (5a — 2c) times x, or (5a — 2c)x. And then, 5a — 2c times a; and 2m times x make (5a — 2c -|- 2m) times x, or (5a — 2c -f- 2?n)x. q. e. d. EXAMPLES. 1. Add ax'\ — bx^, — 2cx'^, and 4:mx-\ with respect to xK Sum, {a — b — 2c + 4??i)j72. 2. Add 4:xy, Saxy, — lOmxy, and cxy, with respect to xy. Sum, (4 + 3a — 10m + c)j7j/. 3. Add amx + 2S^t^?7i, 3(a + h) -^- H- 56c + f • a-2^~ 3. 3 8. Find the sum of 2xy' + 2x-"'y'^ — 2,/:3, — S^'jr^y^ + 2r'^a?-'"i/^ — a + 6^73, 3^4^^ _ 26u--"* ?/^ + 3a, and — 2^3 ^cy. 74, I^rop, 4, Compound terms which have a common compound, or polynomial factor may be regarded as similar and added with respect to that factor. Dem, 5(a;2 — y"^ ), 2(x2 — y"- ) and — 'd{x'^ — y"- ) make, when added with respect to {x^ — y^ ), 4:{x" —y- ), for they are 5 + 2 — 3, or 4 times the same quantity (x^ — y" ). In a similar manner we may reason on other cases, q. e. d. ADDITION. 39 EXAMPLES. 1. Find the sum of ^x + n — S,r\ 5x^ — 2^ x -{- n, 3v/a7 + n — lx\ and IQx^ + Sv^j- + n. Sum, lOv^j- + n + 5xK % Find the sum of 6a — 6(a — 5) + 7, 3a + 12(a — 6) — 8, and 2(a _ 6) — 3a — 20. Sum, 6a — 21 + 8(a — h). 3. Find the sum of 7(m -f 3) — 16(77i — 3), 8(7?i + 3) + 7(m — 3), and 3(m — 3) — 4(m + 3). >S'wm, ll(m + 3) — 6(m — 3). 4. Find the sum of i-Va _ 36 — ^V^ __ 36 + iVa _ 36 — -J-^V'a — 36. ^'w?7i, t\V« — 36. 5. Add 3 (a — and 5 (a + c) (^ + y?) ., 3r3a + c) Sum, — ^- 6. Find the sum of a(a + 6) + 3v^a — x, — 4a(a + 6) + 7a(a — ^)"^, — ^a^a — ^■+ lla(a + 6,) — 2a(a4- 6) — 2 (a — xy, and 5a(a + 6) + 14v''a — x. 7. Add a^ X — y + hxij -\- c{a + xy, — bxy + (a + c) {a-\- x)"']-{x —y)'^, 2hxy-\-{a — T)^x — y — a{a-{-x)^ Sum, 2a{x — y)^ + "Ihxy + 2c(a + xy. 8. Show that ^x -{-by -{-ax — z + amy + c^x + dz + y = (am + 6+ l)i/ + (c + 1)^^ + {d — l)z + ao:. 40 FUNDAMENTAL KULES. Defs, Props. Synopsis. r Addition. Sum, or Amount. rCor. 1. Sign of sum. I Cor. 2. Sum of Pos. 1. Similar Terms. Dem. ■{ and Neg. I Cor. 3. Addition not [ always increase. Sell. Sign of term. 2. Dissimilar Terms. Dem. \ Cor. Add. of Neg. = Sub. of Pos. Prob. To add Polys. Rule. Dem. Schol. 1. Practical method. Schol. 2. Same as in Dec. Nota. Props. f 3. Terms partially similar. [ 4t. Compound similar terms. Test Questions. — Does addition always imply an increase? When does it not ? When does it ? What is addition ? How are similar terms added ? How are dissimilar added ? Give the Bule for adding polynomials and demonstrate it. SECTION III. Subtraction. 7^» Subtraction is, primarily, the process of tak- ing a less quantity from a greater. In an enlarged sense, it comes to mean taking one quantity from an- other irrespective of their magnitudes. It also compre- hends all processes of finding the difference between- quan- tities. In all cases the result is to be expressed in the few- est terms consistent with the notation used. The quantity to be subtracted is called the Subtrahend, and that fi'om which it is to be taken the Minuend. SUBTRACTION. 41 76, Hie Difference between two quantities is, in its primary signification, the number of units whicti lie be- tween them ; or, it is what mud he added to one in order to produce the other. When it is required to take one quantity from another, the difference is what must be added to the Sub- trahend in order to produce the Minuend. ScH. — The most comprehensive and fundamental notion of differ- ence is this : HaAdng reached any specified point in a scale of numbers, or in estimating magnitude, how (in what direction), and how far must we pass to reach another specified point in the scale of numbers, or in the value of the magnitude. III. — When we ask, ""What is the difference between 3 and 8?" we ordinarily mean, "How far (over how many units) must we pass in reckoning from 3 to 8 ?" That is, * ' How many units added to 3 will make 8 ?" Let us use the following device to illustrate the whole subject. Consider A the zero point on the line BC. Call distances to the right - "^ ^ ■ + B A 1. ^ c — m 1 1 ,— 9,— 8,- 1 -7,- 1 1 1 1 -6,— 5,— 4,— 3,- -l -1, 1 1 +1.+2 ,+3.+4,-t-5.+6 1 1 1 ,+7,+8,-f9,+i 1 of A positive ( -<-), and distances to the left negative (-). Also call reckoning towards the right positive and towards the left negative, from any point on the scale. 1st, The di'fference between 3 and 8, means either, how far, and in what direction must we go to pass from 3 to 8 or to pass from 8 to 3 ? In the first case we pass 5 to the right, and say 3 from 8 is -t-5, un- derstanding that 3 and 8 are both positive. But to pass from 8 to 3, we pass 5 towards the left, and hence say 8 from 3 gives - 5. These two results are expressed thus : 8 — 3 = 5, and 3 — 8 =: - 5. 2nd, The difference between - 3 and - 8 means either, how far, and in ichai direction must we pass to go from - 3 to - 8, or from - 8 to - 3 ? To pass from - 3 to - 8, we pass 5 towards the left, and hence say - 3 from - 8 gives - 5. That is - 8 — (- 3) = - 5. But to pass from - 8 to - 3, we pass 5 to the right, hence - 3 — (- 8) = -t-5. 3d, The difference between - 3 and -^8 means either, how far and in what direction must we pass to go from - 3 to -t-8, or from -t-8 to 42 FUNDAMENTAL RULES. - 3 ? In the first case we get -+-8 — (- 3) = -t-ll, since we pass 11 to the right. In the second case we get - 3 — ( +8) = - 11, since to go from -f- 8 to - 3 we pass 11 to the left. In each and all of these cases, the question is, "What must be added to the quantity conceived as the Subtrahend, in order to pro- duce the Minuend?" considering, in each instance, the number of units between the two given terms as the numerical value of the dif- ference, and its sign as determined by the fact as to whether we reckon to the right (up the scale), or to the left (down the scale), in passing from the Subtrahend to the Minuend. 77. I^VOh, To perform Subtraction. RULE. — Change the Signs of each Term in the Sub- trahend FROM + TO , OR FROM ■ — TO +, OR CONCEIVE THEM TO BE CHANGED, AND ADD THE RESULT TO THE MiNUEND. Dem.— Since the difference sought is what must be added to the subtrahend to produce the minuend, we may consider this difference as made up of two parts, one the subtrahend with its signs changed, and the other the minuend. When the sum of these two parts is added to the subtrahend, it is evident that the first part will destroy the subtrahend, and the other part, or minuend, will be the sum. Thus, to perform the expmple : . From ^ax — G& — 3(Z — 4m Take 2«.t -^ 2b — M + 8m ] jf these Subtrahend with signs changed — 'lax — 26 -j- 5(/ — iim j. three quan- Minuend, bax — 6/> — 3(Z — 4m j tities are Difference, 3o.c — 8f/ -j- 2(2 — 12m added to- gether, the sum will evidently be the minuend. If, therefore, we add the second and third of them (that is the subtrahend, with its signs changed, and the minuend) together, the sum will be what is neces- sary to be added to the subtrahend to produce the minuend, and henco is the difference sought, q. e. d. EXAMPLES. 1. From 4a?> + 3c2 — xy subtract ^xy — 2z + 2a6 — c^. Model Solution. — Writing the subtrahend under the minuend so SUBTRACTION. 43 that similar terms shall fall under each other, for the convenience of combination, I have Aah + 3c2 — xy 2ah -^ c^ + 3xy — 2z 2ab -f 40'"' — ixy -\- 'Iz As it is immaterial where the process of subtraction is commenced, since there is to be no " borrowing," I will commence at the right, as in the decimal notation, for analogy's sake. The first question is, What must be added to — 2^ to produce the corresponding term in the minuend? This is evidently*^ 2z, as there is no corresponding term in the minuend, and I have only to write a term in the differ- ence which Tvill destroy — 2z when added to it. Passing to the next term, I inquire. What must be added to + 3xy to produce — xy ? First, I must add — Sxy (the term with its sign changed) in order to destroy the -{- Sxy, and then I must add — xy in order to make the — xy of the minuend. So in all I must add — Sxy and — xy, or — 4ucy. — ixy is, therefore, the difference sought. Passing to the next term, What must be added to — c2 to make -\- 3c^? I must add -|- c^ (to destroy — c^) and -f- 3c2 (to make up the required term in the minuend), or in all + c^ and -|- ^c^, or -|- Ac'^ ; which is the difference. Finally, the term to be added to 2ab in order to make 4a& is composed of the two parts — 2ab and -f- 4a6, which make 2ab. It thus appears that 2ah -|- 4c^ — 4xy -}- 2z is the difference, since it is what must he added to the subtrahend to produce the minuend. A Shorter Explanation. — The difference sought may be consid- ered as consisting of two parts, 1st, the subtrahend with its signs changed, and 2nd, the minuend itself. The sum of these two parts is the difference, since it is what is necessary to be added to the subtra- hend to produce the minuend. Therefore conceiving the signs of the subtrahend to be changed, and adding, I have 2a& -j- 40^ — 'key -\- 2z, as the difference sought. [Note.— Let the pupil solve and explain the following examples, giving the reason for changing the signs of the subtrahend (to get a quantity which added to it will destroy it) and the reason for adding this result to the minuend. Keep clearly in view the facts that the difference is what is required to be added to the subtrahend to pro» duce the minuend, and that this difference is made up of two parts.] 2. From 4:X — 3¥ + 2ry — 11 take 20- + b^ -\- 2xij — 5. Bern,, 2x — 46"^ - ^ 6. 44 FUNDAMENTAL RULES. 3. From 15ax -j- 2¥y — Ga^x'^ take — 5ax — 4:t2y — Sa^xK Bern., 20ax + 6b^ij — Sa-'xK 4. From Sa-'b--' — 3xy + 15 — 2^xy take 8 + lOxrj — Sah-^ + 2^xy. Bern., lla^6-2 — ISxy + 7 — 4:\^lry. 5. From 4:a^x^ — 3c + Ub^y'^ take Sh'^y'^ + 4^2^^ — 2r/. . Bern., 4.b^y^ — 3c + 2rf. 6. From a^ — 2a6 + b^ take a^ ^ 2ab + 62. i?em., — 4a6. 7. From a^ — 6^ take o^ — 2a6 — b\ Bern., lab. 8. From 24:xy^ — 14m?/ + l^x'^y^ — 14 + 21xz^ tako Vlxy^ — 10m?/ — 4^^^?/2 + 20^2^ — 8. 9. From llpmx^ — 18n^ + 19m^ — 24 take 7pm^2 — 47^3 _|- 107714 — 17. 10. From a^ + 2ab + b^ take a^ — 2ab + 62. 11. From a^ + 3a-^6 + 3a62 ^ 53 take a^ — Sa'^b + 3a52 — 63. 2 1.13. 2. 1.1 2 12. From x'^ + 2,r^?/^ + y'^ take o:^ — 2x^y'^ + 2/"^. 13. From 6v^ + 2/ + 4ar^a;2take 3(ir + ?/)^ — 4:a^xK Diff., 3v'^' + y 4- 8a2^«. Suggestion. — Eegard "/x -f 2/ as one qiiantity, and obserye thai 3(a; -\- y)- IB the same as Sv'aj + y. 14. From 6a; — Bv^^ + 17 (a + &) take 2x + 7v/^ + 4(a + 6.) i>i^., 4a; — lOV'l^j + 13(a + b). 15. From 10 — a + 6 — c+o; take a -\- b — c — x — 90. 16. From Q>7n -\- ^n — §a — 1 take %n — ^/z + ^ — 2. Diff., 4m + n — a + 1. SUBTRACTION. 45 ScH. 1. — When several poljTiomials are to be combined, some by addition and some by subtraction, it ^^ill be found expedient to write them so that similar terms will fall under each other, wTiting the several subtrahends with their signs changed, and then add the quan- tities as they stand. 17. From the sum of ^aif — 26 + ^ax, 2m/^ -\- b — ax, and 36 — a?/« + 3 — y, subtract Say^ + 46 — ax -]- y. Result, — 26 + 3aa7 + 3 —2y. 18. From the sum of 3a- + xy'- — 26?/, 5a- + %r,y^- — 36?/, and Zxy^ + 4a2 + by, subtract 2a^ — xy'^ — 56?/ + 5. BesuU, 10a2 + Sxy^ -^ by — 5. 19. From the sum of a'^¥ — 3]/^ + ^xy, Za"¥ + 3?/^ — 2j7?/, and by- -f 3a'62 — xy -\- Q, subtract 0-6^ -{- xy — y^ — 3. Result, Ga%^ + Gif + j??/ + 9. 20. To 5ax'^ — 7cb + 8»i^ — 2c-", add 3c6 — 4c-" + 2a^2, then subtract lOwi^ — 5xy + 3c6 — 12c ~", add Gm^ — Ilc6 + 3n — 4:ax% subtract — 16aa7^ — 2m'^ + 4c6, add 5m ^ + Gc" ", and subtract 4:xy — 2?i. Result, Idax-' — 22c6 + ll??i^ + 12c-" + xy + 5n. 21. To ley - 1 + 8ax — 56, add 46 — 2ey " f + ?n, then sub- tract Bax — 4??i + 3 and — Sax + 5c y~ ^ — 6, add lOao? _ 2. — 26 + 8?7i — 3, and subtract Sm — lOcy ^ — 2m. Result, lOcy - 1 + 16aa7 — 36 + 12m. ScH. 2. — The proficient solves such examples as the above, and, in fact, all kindred ones, without re-writing the quantities. Thus, to obtain the result in Hr. 20, he looks at 5f?x^, casts liis eye along till he sees 2ax^, says Tax* ; then looks forward to — 4:ax^, and says 3ax^ ; then notices — 16«.r^ in a subtrahend, and hence thinks of U as -\~l&ax^, and says 19ax^. Again taking tip — 7ch, he looks along noticing the similar terms and saj-s, mentally, — 4:cb, — 7cb, — 18c6, — 22c6. Again, he takes up 8?n^ , and running through with the sim- ilar terms, says, — 2m^, 4?7i-, 6ni'^, ll?7i^. And so for all the other terms. 46 FUNDAMENTAL RULES. This is the practical way of doing such examples ; a^id the pupil should exercise himself till he can write out the result at once. He cau gc o\er the preceding examples thus, and should also E-olve, without writing, those which follow. 22. From 13x-' — lax — 952 take 5^^ — ^ax — h\ 23. From 20a.^ — v'^ -f M take ^ax + 5.r^ — d. 24. From 5«6 + 262 _ c + 5c — 6 take 5-^ — lab + he. 25. From ax -'^ — ax -'^ A;- ex — d take ax:^ — ax -^ — ex ■— M. Biff., a{x -3 — ^3) + (c 4- e)x + d. Suggestion. — This would at first become ax -^ — ax^ -\- ex -\- ex -j- d. But this may evidently be written as above. 26. From x'^y — 3\^xy — • 6ay take SxH/ + ^{xy)'-^ — 4ay- 27. From the sum of 4a.r — 150 + 4.r^ 5x-^-i- Sax + 10 v^^, and 90 — 2ax — 12 v^^ ; take the sum of 2ax — 80 + lx% Ix^ — Sax — 70, and 30 — 4^^ — 2xi 4- 4a2^2. Result, Wax +60 — x'^ — ^a'^x'^. 7S. CoE. 1. — WJien a par^enthesis, or any symbol of like signification (SO), occurs in a polynomial, preceded by a — sign, and the parenthesis or equivalent symbol is removed, the signs of all the terms which ivere within must be changed, since the sign — indicates that the quantity within the parenthesis is a subtrahend. 28. Remove the parenthesis fi'om the polynomial 3a^ x -f 252f/2 — {^a^x — 2m22; + Sb-y-) and express the result in its simplest form. Do the work mentally, writing only the result. Besult, 2m'^z — 2a^x — GbH/K 29. Remove the parenthesis from 5a — 45 + 3c — ( — 3a 4- 25 — c). Result, 8a — 65 + 4c. SUBTRACTION. 47 30. Remove the parenthesis from 4a — 5x — (a — 4^;) -\- [x — 8a). Result, — 5a. Queries. — In Kt. 28 is the sign of Sa^x changed ? What is its sign as it stands in the parenthesis ? Is the — sign which appears in the result before 5a^a', the same as the one before the parenthesis in the example ? No. What became of that before the parenthesis ? Ans. The operation which it indicated having been performed, it is dropped. In Ex. 30, why are not the signs of the terms in the last parenthesis changed ? to* Cor. 2. — Amj quantity can he placed within a paren- thesis, p)receded by the — sign, by changing all the signs. The reason of this is evident, since by removing the parenthesis according to the preceding corollary, the expression would return to its original form. 31. Introduce within a parenthesis the 3d, 4th and 5th terms of the following expression : Q>ax — ^cd — 8??j f 5^ — 1y -\- X — 4a. Besult, Gax — 2cd — (8m — 5x -\- 2y) -}- x — 4a. 32. Introduce within a parenthesis the last three terms of ^xy -\-2cb — 8x — 5 + 2b. Result, Axy + 2cb — {8x -\- 5 — 2b). 33. Include in brackets the 3d, 4th, and 5th terms of hax — 2.r2 4- 3a7 — 12ay + 15. Also the 4th and 5th. Also the 2d and 3d. Form of the first and last, 5ax — 2x'^ + (3^7 — 12ay + 15), 5ax — {2x^~ — ^x) —12 ay -\- 15. Query. — In the last is the sign of 2x* changed ? 34. Prove that (3^ — &y) + {4.y _ 4^) + 2 (^ + 2y) =. x-^2y. 35. Prove that ^ (a + 6 — c) + |(6 + c — a) = &. 36. Prove that 6a — 46 — 2 (a + Z>) = 2 (2a — 36). 48 FUNDAMENTAL RULES. 37. Prove that x^ + 6x^y + 3 — (^^ + 4:X^y + 1) = 2{x^j + 1). *38. Prove that ^(a — 56 + ^c) + i(56 — 2a + ^c) = ic — -^^b — ^a. 39. Prove that ^(a^ _ « + i) 4. i(2a2 _ ^ _|_ 2) = -iL(10a2_7a + 10). 40. Prove that ^a + ^b — {^a — ^b) = b, 41. Prove that ^(9 — 15^) —-^(12 — 20^) — 1(45^; —20) = 1 — 4^. 42. Prove that ^{a — ^b + ^c) + ^{a — ^b + ^c) == ^_(90a _ 366 + 25c). SO. CoR. 3, — When several parentheses occur, included the one within the other, begin the reynoval ivith the inside one. 43. Eemove the parentheses and other marks of aggrega- tion from 4a — { — [c — ■ d -{- (4^^ — i) — j^y-j — 3iy|. Result, Aa -{- c — d -{- 4^2 — 1 — xy -\- dy. 44. Show that a — [b — {c — {d ~ e—f))] = a-[b-{c-{d-^e+f)}] = a- [b-{c-d + e-f}] = a — [b~c + d — e +/] = a — 6 + c — d + e — /. 45. Remove the marks of aggregation from the expres- sion 7a — {3a — [4a — (5a — '^(^)'}}, and afterwards reduce the result to its simplest form. Also combine and remove at the same time. Result, 5a. 46. Kemove the marks of aggregation from a -f 26 — {Qa — [36 + (8.r — 2 + 6?/ — x) + 4a] — 36}. Result, 8b — a + Ix — by — 2. ScH. 3. — Terms having common literal factors may be regarded as similar with respect to these, and treated accordingly, the other factors in each term being regarded as coefficients. * This, and the examples wliich follow in this Bection may be omitted and taken in the review after Fractions, if thought best by the teacher. SUBTBACTION. 49 47. From ax -\- by — cz take my — nx -\- 2z. Result, {a + n)x -{- {b — m)y — (c + 2)z. 48. From ax'^ -\- bxy + cy^ take dx^ — hxy + ky~. Result, {a — d)x- + (6 + h)xy -f (c — k)i/-. 49. From v/F+1/ + 3ax — 12 take 4(^7 + y)'^ -\- b — 2ax. Result, 5ax — • 3(^ + y)"-^ — 12 — b. 50. From a-x- ■ — ■ 4,axy -{-4:a-x-y^ take c^ jt- — Sexy + Gx-y\ Result, {a- — c-)x-^ — (4a — Sc)xy + (4a2 — 6)x-^y-. Show that the second term in the last result may be + (8c — 4a). 51. From Vx^^^- — 2(a + ^) ^+ 3 take - 4(^._y.)i_l. Defi Result, {a -\- xy Synopsis for Review. Subtraction. Subtrahend. Minuend. ^■a, \ Scholium. Difference, j m^^stration. Diagram. ^ /> 7 r> t T> n \ Sch. 1. Both add. and sub'i ^}^ General Prob. Kui.E. Dem. | ^^y^. 2. Practical method. 2 ( Cor. 1. — To remove. g Brackets. \ Cor. 2. — To introduce. 5^ I / Cor. 3.— Several. [ Terms partially similar. Sch. 3. ' Test Questions. — "WTiat two answers can you give to the question, " "What is the difference between 10 and 6 ?" "Why do you change the signs of the subtrahend in subtracting ? Why do you add the sub- trahend, -v^ith signs changed, to the miniiend ? When do you change the signs in removing a parenthesis ? Why ? AVhat becomes of the sign before the brackets ? In removing a parenthesis preceded by a — sign, is the sign of the first terci changed as well as the others ? 50 FUNDAMENTAL RULES. State in the briefest manner the theorj^ of subtraction ? Iteply. Sub- traction is finding the difference between quantities, that is, finding what must be added to one qiiantity to produce the other. This difference may always be considered as consisting of two parts, one of which destroys the subtrahend, and the other part is the minuend itself. Hence, to perform subtraction, we change the signs of the subtrahend to get that part of the difference which destroys the sub- trahend, and add this result to the minuend, which is the other part of the difference. SECTION IK Multiplication. 81. JfllltiplicatiOfl is the process of finding the simplest expression consistent with the notation used, for a quantity which shall be as many times a specified quantity, . or such a part of that quantity, as is represented by a specified number. \ The latter number is called the Multi- plier. The quantity to be multiplied is called the Multipli- cand. Taken together, the multiplier and multiplicand are called Factors. The result is the Product. 82. CoR. 1. — The multiplier must always he conceived as an abstract number, since it shows how many times the multii)li- cand is to be taken. Thus, to proj^ose to multiply $12 by $5 is absurd. We can understand that 5 times $12 is $60 ; but what is meant by 5 dollars times? 83. Cor. 2. — The jjroduct is always of the same kind as the multiplicand. S-Jt. ScH. — It is frequently convenient in j^ractice to speak of the multiplier as positive or negative, although, literally understood, this is a contradiction of Cok. 1, which requires the multiplier to be con- ceived as mere number. In a strict analysis, the multiplier in such cases is to be considered, first, -svithout reference to its sign, i. c, us MULTIPLICATION. 51 abstract, and then the sign is to be interpreted as indicating what is to be done with the product, when it is taken in connection with other quantities. SS, JPvojy* 1 • — The product of several factors is the same in whatever order they are taken. Dem. — 1st. a X &, is a taken h times, oxa-\-a-\-a-\-a-\-a to h terms. Now, if we take 1 unit from each term (each a), we shall get h units ; and this process can be repeated a times, giving a times 6, or 5 X a. • ' • aXh = hX a. 2nd. When there are more than two factors, as ahc. We have shown that ab ^=zba. Now call this product m, whence abc = mc. But by part 1st, mc =z cm. . • . ahc = hac = cctb s= cba. In like man- ner we may show that the product of any number of factors is the same in whatever order they are taken, q. e. d. ScH. — If the multiplicand is concrete, the reasoning is still the same. Thus $a X ?> = $« + $a + ^ct + ^«> etc. to b terms. Now take $1 from each of the terms of $a each, and we have $b ; and this process can be repeated a times, giving $6 X «• . * . $a X b = S& X a. Notice that in each case the multiplier is abstract. 80, JPl*op, 2. — IVhen two factors have the same sign their product is positive : ivhen they have different signs their pro- duct is negative. Dem. — 1st. Let the factors he -\- a and + h. Considering a as the multipher, we are to take + 6, a times, which gives -\- ab, a being con- sidered as abstract in the operation, and the product, -\- ab, being of the same kind as the multipHcand ; that is, positive. Now, when the product, + ab, is taken in connection with other quantities, the sign -j- of the multipher, a, shows that it is to be added ; that is, MTitten wi>th its sign unchanged. . • . (-j- b) X (+ <*) = + «&• 2nd. Let the factors be — a and — b. Considering a as the multi- pher, we are to take — b, a times, which gives — ab, a being consid- ered as abstract in the operation, and the product, — ab, being of the same kind as the multiphcand ; that is, negative. Now, when this product, — ab, is taken in connection mth other quantities, the sign — of the multipher shows that it is to be subtracted ; that is, writ- ten with its sign changed. . • . { — 6) X ( — a) = -|- a6. 52 FUNDAMENTAL KULES. 3d. Let the factors be — a and -f- ^- Considering a as the multi- pUer, we are to take -{-b, a times, which gives -{-ab, a being considered as abstract in the operation, and the product, -|- ab, being of the same kind as the multiplicand ; that is, positive. Now, when this product, + ab, is taken in connection with other quantities, the sign — of the multiplier shows that it is to be subtracted ; that is, written with its sign changed. . • . (+ ^) X ( — a) = — ab. 4th. Let the factors be -f- « and — b. Considering a as the multi- pher, we are to take — b, a times, which gives — ab, a being consid- ered as abstract in the operation, and the product, — ab, being of the same kind as the multiplicand ; that is, negative. Now, when this product, — ab, is taken in connection with other quantities, the sign -|- of the multiplier shows that it is to be added ; that is, written with its own sign. . •. ( — Z/) x (-f- ^) = — o^- Q- e. d. 87 • CoE. 1. — The product of any number of positive fac- tors is positive. Thus (+ a) X (+ 6) X (+ c) x (-\-d) = abed, since (+ a) X (+ b) = -\-ab, which, in turn multiplied ^y + c, gives + abc, &c. 88, CoK. 2. — The product of an even number of negative factors is positive ; since ive can multiply them two and two, thus obtaining positive products, which positive products multi- plied together make the complete product positive. Thus ( - a) X ( - 6) X (- c) X C - ^) X ( - e) X ( -/) z= ( -\- ab) X i -\- cd) X { -\- ef ) = -\- abcdef or abcdef 89, Cor. 3. — The product of an odd number of negative factors is negative ; since, by the last corollary, the product of all but one {an even number) of such factors is positive, and then this multiplied by the remaining negative factor gives -f X — , and hence is negative. 00, I^vop, 3, — TJie product of two or more factors con- sisting of the same quantity affected with exponents, is the conn- mi m quantity with an mponent equal to the sum of the expO' MULTIPLICATION. 53 nents of the factors. That is a" x a" = «""+"; or a"" ■a'' a* __ ^m-vn+,^ ^^^ whether the exponents are integral or frac- tional, positive or negative. Dem. — 1st. When the exponents are positive integers. Let it be required to multiply a'-^ by a^. a'-^ = aaa, and a* = aa. . •• a^ X a* = aaa-aa = a^. That is, there are tliree factors each a, in a^, and tico like factors in a^ ; and, as the product consists of all the factors in both multiplier and multiplicand, it will contain^re factors each a, and hence is a^. In general : To multiply a"* by a" mid a', a" = aaaa to rn factors, a" = aaaaa to n factors, and a' = aaaaa to s factors. Hence the product, being composed of all the factors in the quantities to be multiplied together, contains 77i -f- ^^ ~f" s factors of a, which is expressed a*" + " + «, Since it is evident that this reasoning can be extended to any nxmiber of factors, as a"" X a" X a* X cf , etc., etc., the proposition in this case is proved. 2nd. When the exponents are positive fractions. Let it be required to multiply 64^" by 64^. Now 64^ = 4- 4, i. e., 2 of the 3 equal fac- tors which make 64. In like manner 64'' is2 2 • 2 • 2 - 2. Andsince4-4 is2 - 2 - 2 - 2, 64^ X 64^ = 2 • 2 . 2 • 2 X 2 - 2 ■ 2 • 2 - 2 = 2^ or 9 of the 6 .p. equal factors into which 64 is resolvable, and may be expressed 64 ^ . • . 64^ X 64'' = 29 = 64 * =64^ **. In general, let a " be multiphed by a*, a" means m of the n equal factors composing a. Now, if each of these 71 factors be resolved into h factors, a will be resolved into 6n m factors, and to make the quantity a» we shall have to take hm m 6m e instead of m factors. Hence a" = a^". In like manner a* may be en m e bm en shown equal to a'"' ; and a"^ x a^ = «''" x a^'\ This now signifies that a is to be resolved into b)i factors, and bm + en of them taken to m e bm en bm+en »i <• form the product, .'.a"- x a*^ = «*" x «*" = « *» , or a" *^ which proves the proposition for positive fractional exponents, since the m < same reasoning can be extended to any number of factors, as a** x a* X a\ etc. 3d. When tJie exponents are negative. Let it be requii-ed to multiply 54 FUNDAMENTAL EULES. 2-3 by 2 - 2 . 2-3 is -— , and 2 - 2 is-j . • . 2-3x2-2=:; 111 11 -rj X t;! — IT?, or 2 -s So also a-*" Xa-" = — X — . Now, 2 2 2 a'" a" as fractions are multiplied by multiplying numerators together and denominators together, we have — X — = r- by part 1st of the & a'" a" a"' + " "^ -^ demonstration. But this is the same as a ~ <''* + "^ or a - ™ - " . • . -- -^ 1 1 1 a -'"Xa- » = «-"*-". Finally a »Xa ''=-;;^ >< ^7= ~; — ~ a"^ a* a» V a n "«> . Q. E. D. EXAMPLES. 1. Prove as above that 81^ x 81* = 81^ and that 81^^ 9. = 81*. 2. Prove that m" x m* :=m''+*. 3. Prove that 16"* x 16""^ = 16~^. 4 Prove that 25 ~ ^ x 25^ is 1. 5. Prove that a~' x a^ is a. ScH. — The student must be careful to notice the difference be- tween the signification of a fraction used as an exponerit, and its com- mon signification. Thus | used as an exponent signifies that a num- ber is resolved into 3 equal factors, and the product of 2 of them taken ; whereas f used as a common fraction signifies that a quantity is to be separated into 3 equal ^«rte, and the sum of two of them taken. 01, I*TOh, — To multiply monomials. R ULE. — Multiply the numerical coefficients as in the DECIMAL NOTATION, AND TO THIS PKODUCT AFFIX THE LETTERS OF ALL THE FACTORS, AFFECTING EACH WITH AN EXPONENT EQUAL TO THE SUM OF ALL THE EXPONENTS OF THAT LETTEB MULTIPLICATION. 55 IN ALL THE FACTORS. ThE SIGN OF THE PRODUCT WILL BE + EXCEPT WHEN THERE IS AN ODD NUMBER OF NEGATIVE FAC- TORS ; IN WHICH CASE IT WILL BE . Dem.— This rule is bjit an application of the preceding principles. Since the product is composed of all the factors of the given factors, and the order of arrangement of the factors in the product does not affect its value, we can write the product putting the continued pro- duct of the numerical factors first, and then grouping the hteral fac- tors, so that hke letters shall come together. Finally, performing the operations indicated by multiplying the numerical factors as in the decimal notation, and the like literal factors by adding the exponents, the product is completed. EXAMPLES. 1. Multiply together Sa'^-bx, 2cb^y, and 5ac^x. Model Solution. — Since the product must contain all the factors of the given factors, and the order of arrangement is immaterial (85), 3a2 bx X 2c6^y X ^ac^x = S ■ 2 ■ 5 X a^a X hh^ X cc^ X xx X y, which, by performing the operations indicated, becomes dOa'^b^c^x^y. 2. Multiply together dabxy, 2a*bx\ lOcx^, ^y\ and a. Prod., 2^0d'¥cx^y\ 3. Multiply together 5ax% — 2by, — Sa^c, a^, and — 2y^. Prod.j — QOa^hcx''>y\ 4. Multiply together m^", ^rrV^x^, — 3a^^ — 5m', and ^a^x\ Prod., 120a^m' + V + •" + ". 5. What is the product of — 2c"'ci, — lOac", —d^x, — 4a'"ar", aud — c\ Ans. , — 80a"* -^ 'c"* + " + V^^" + \ 6. Multiply Zx^ by 2x^. Prod., Qx^. 7. Multiply 60a^ by 8a^. Prod., 480a«. 8. Multiply ^ah^ by — lah^. Prod., — 21ah^, 56 lUNDAMENTAL RULES. 9. Multiply 5ah^ by la^b-^c. Prod., ^5ah. 10. Multiply 10a' by 3a -2. Prod., 30. 11. Multiply3i/"by 6t/2. 1 + 2h • Prod., 182/~~. 1 1 12. Multiply 13^" by 5x"\ W + n Prod., 65^ '"" . 1 5 13. Multiply — 50a' by — ^a\ 6 Prod., 200a". 14. Multiply 309a'^6'» by 9a™6". 3 Pro6?., 2781a'"6'" + ». 15. Multiply — 5^ - V - ' by — 2^x^y\ Prod., Ibxy. 16. Multiply ^^ by ^ ~ ^.. Prod., x^. 17. Multiply a:™ by a; - "*. Prod., 1. 18. Multiply a^b" by a^b'. Prod., aK 19. Multiply v'o^'by a"^x. Prod., a^'x^. 20. Multiply V^by v^^ Prod., a^. 02, I*VOh, To multiply two factors togetKer when one or both are polynomials. RULE. — Multiply each term of the multiplicand by EACH TERM OF THE MULTIPLIER, AND ADD THE PRODUCTS. Dem. — Thus, if any quantity is to be multiplied by a -{-h — c, if we take it a times (i. e. multiply by o), then h times, and add the results, we have taken it a -j- ^ times. But this is taking it c too many times, as the multiplier required it to be taken a -\-h minus c times. Hence we must muliply by c, and subtract this product from the sum of the other two. Now to subtract this product is simply to add it with its signs changed {77)' But, regarding the — sign of c as we multiply, will change the signs of the product, and we can add the partial pro- ducts as they stand, even without first adding the products by a and h. Q. E D. MULTIPLICATION. 57 EXAMPLES. 1. Multiply 2a — 36 + 4c by 3a + 26 — 5c. Model Sol-dtion. — Writing the multiplier under the multiplicand, as a matter of convenience, I have 2a — 36 + 4c 3a + 2b — 5c 6a2 — dab + 12ac + 4a& — 66* + 86c - lOac + 156c — 20cg 6a^ — dab+ 2ac — 06^= -f 236c — 2Uc=* Novr taking the multiplicand 3a times, I have 6a* — 9a6 -j- 12ac. Taking it 26 times, I have 4a6 — 66* -|- 86c. I have thus taken it too many times, by 5c times. Hence I am to take it 5c times, and then subtract this partial product from the others. Therefore I multiply by 5c, and change the signs as I proceed, and finally add the three partial products. I thus obtain 3a -j- 26 — 5c times the multiplicand. 2. Multiply X -\- y hy X -{- y. Prod., x'^ -f 2xy + y^. 3. Multiply 5x + 4?/ by 3x — 2y. Prod., 15^2 _|_ 207?/ — 8?/2. 4. Multiply x'^ -\- xy — y^ hj x — y. Prod., x^ — 2xy^ + ?/3. 5. Multiply 2ac2 — dby by 2c^ — 3y'K Prod., 4ac5 — Qbc^y — GacHj'^ 4- dbifo 6. Multiply a* _{- 62 _|_ ^s — ab — ac — be by a 4- b + c. 7. Multiply a^ -f- da^x + Sax"" + x^ by a^ — Sa'^x + Sax^ x\ Prod., a'— 3a*x^ + Sa'^x*— x\ 8. Multiply a"" — V" by a"" + l"". I IT III 9. Multiply together {a + 1)), {ci'^ + ah + V"), {a — h\ IV and {a"^ — nh + If^). 58 FUNDAMENTAL RULES. Suggestion. — Perform this first by multiplying together I and II, and men III and IV, and taking the product of these products. 2d, By multiplying the product of I and III into the product of II and IV. 3d, By multiplying the product of I and IV by the product of 11 and III. Besult in each case, a*" — h^. 10. Multiply a3 — ^3 by m^ — w^. Frod., a/m- — tr-x^ — a^n^ + n-x^. 11. Multiply 2fl2 -{- 2a + 5 by a"- — a. Prod., 2a* + Sa^ — 5a. 12. Multiply 2^3 4. 4^2 _^ g^ ^ ig by 3^ — 6. Prod., 6^^ — 96. 13. Multiply a -\-h — c by m — n. Prod., am — an -\- hm — hyi — cm -f- en. 14. Multiply X* -\- x^y^ + y* by ^2 — y\ Prod.y x^ — 2/6. 15. Multiply a;2 — 4^ + 16 by J7 + 5. Prod., x^ -\- x'^ — ^x ^ 80. 16. Multiply a* — a^y + a^y^ — ay^ -f y* by a + y. Prod., a^ + y\ 17. Multiply ^2 _ mx — 100 by ^ + 2. Prod., x^ — 48-2;2 — 200^ — 200. ' 18. Multiply 2x-^ + 3^ — 1 by 2j;2 _ 3j: + 1. Prod., 4^^ — 9a;2 + 6ir — 1. 19. Multiply j:^' — h^ by jt'-^ + 6. Prod., x^ — h^x^-]-hx^ — 6^. 7 _ 2 1 .i 20. Multiply d^ — h ^ hy a^ — h Prod., a^ — a'^h ~ ^— a"^ b + 6^. 21. Multiply 2a~^ — 36^ by 2a "^ + 36^. Prod., 4a ~ ' — 96^. MULTIPLICATION. 59 93333 9 3 3 22. Multiply X* -f x^y ~ * -f x^y '^-^ y'^ hj x^ — y " ^. Frod., x'^ — y~^' 23. Multiply a"' -f 6"' by a" + 6". 1 i ^ 1 Q 2 24. Multiply ^"^ + 2/^ by ^'^ — y^- Prod., xr^ — ij^. 2. 11 3 1 1 25. Multiply m^ + ??i^n^ -f ri^ by ?7i^ — n^. Prod., m — n. 26. Multiply Sa™ - ^ _ 2&" " ^ by 2a — S?;^. Prod., Ga"* — 4a6"- ' — 9a"' " '6^ + 66". 27. Multiply 2a"' -^6 ~ =^ + a " '^6^ by da^V —a'^b- '^. P?'oc?., Ga^'" -p -{- 3a"' " '^6'^ — 2a"' + 2^6" '^' — a^'b -". ^5. Def. — Wlien an indicated operation is performed the expres- sion is said to be expa^ided. 28. Expand (a + 6)(a — 6) ; also {x -f ?/)(^ + y) ; also (^ — a) {x^ -\-ax -{- a-) ; also (7?i -\- n){7n -\- n) — (m — ?i) (w — n). Last Result, 4r)in. THREE IMPORTANT THEOREMS. 04, Theorem. — The square of the sum of two quantities is equal to the square of the first, plus twice the product of the two, plus the square of the second. Dem. — Let a; be any one quantity and y any other. The sum is X -\- y ; and the square is, the square of the first, x^, plus twice the product of the two, 2xy, plus the square of the second, y^. That is (a; + 2/)' =x^ -h 2a-y + y^. For {x + yY = {x -\- yXx + y) which expanded becomes x* -\- 2xy -\- y^. Q. e. d. EXAMPLES. 1. What is the square of 2a2 + 3x^ ? 60 FUNDAMENTAL EULES. Model Solution. — This is the sum of the two quantities 2rt* and 3a;*, and hence by the theorem the square equals the square of the first, 4a'*, plus twice the product of the two, 12a^x^, plus the square of the second, djo\ .-. {2a^ + 3x2)^ = 4a* + 12a''x^ + 9x\ ScH. — The pupil should give mentally the squares of the follow- ing expressions : 2. Square 4a^ + 2x. liesuU, 16a + IQ^c^x + Ix'K _ 2. _ ^ 1 3. Square x ^ -\- xy. Besult, ^ •* + 2x'^y + x'-y\ 4 Square 5a - ' + 3a'^»^ Hesult, 25a-^ -{- SOat^ + da^b*. 5. Square |a6 - ^ _j_ f j? -' 6 - •. Hesult, ^a^b -'^ -\- ^ab~"^x-^ -{- t-^-^b -\ 0^» Theo. — Ulte square of the difference of two quanti- ties is equal to the square of the first, ininus twice the product of the first by the second, plus the square of the second. Dem. — Let a; and y be any two quantities. The difference is jc — y. Now (X — yY = (ic — y){x — y) which expanded gives jc2 — 2xy ■\- y^. Q. E. D. EXAMPLES. 1. Square 2x — By. Resvdi, Ax^ — 12xy + dy^. Model Solution. — Similar to the last 2. Square x~'^y — 2y'. Result, x~^y^ — A:X~'^y^ -J- Ay*. m _1 3. Square 2a^ — 36 -. 2m ml J Result, 4a"' —12a "^6"""+ 96~"". 4. Square m~^ — w~'. Result, m-"" — 2m --^n"' + ?2-^ 5. Square 3a "* — 2b " . 2 1 m 2nt Resu.'t, 9a ^ — 12u mlf~ n -\- 4:b~ "« . MULTIPLICATION. 61 00* Theo. — Tlie product of the sum and difference of two quantities is equal to the difference of their squares. Dem. — Let X and y be any two quantities. Their sum is, x -\- y, and their difference is x — y. Now (x + y) multipUed by (x — y) gives, by actual multiplication, x* — y^, or the difference of the squares of the two quantities, q. e. d. EXAMPLES. 1. Find the product of 2a^ + 36 and 2a2 — 36. Model Solution. — Here I have the sum of the two quantities 2a* and 3h, and their difference. Therefore {2a^ + 3b) X (2a* — 36) is, by the theorem, the square of 2a*, or ia'', minus the square of the second, or 9b-. .-. (2a* X 36) (2a* — 36) =: 4a* — 96*. 2. Find the product of a + 26 and a — 26. Frod., a-^ — 462. 3. Find the product of 2a + 36 and 2a — 36. Prod., 4a^ — 962. 4. Find the product of la + 26 and la — 26. Frod., 49a2 — 462. 5. Find the product of 5a^ + 662 and 5a^ — 66^. Frod:, 25a<^ — 366^ 11 11 6. Find the product of m^ + n^ and m^ — n'^. Frod., m — n. 7. Find the product of 2^x^ + S^y^ and 2^x^ — S^y^, Frod., 2x — 3?/. Frod., da'¥ — ^ab\ 8. Find the product of 30^63 + 2a26^ and 3a-'6a — 2a^bK X 82 FUNDAMENTAL KULES. Synopsis for Eeview. Defs. i [Multiplication. "] Multiplier. Multiplicand Product. Factors. - Corollaries. Fundamental Propositions. 1st. Order of Factors. Dem 2d. Law of Signs. Dem. 3d. Laws of expo nents. Dein ' 1. Multiplier, ab- stract. 2. Product like J multiplicand. "sGH.How the sign of a multi- plier is to be (. understood. Two factors, ) More than V Scho. two. Give + Give Cor's. 1. + Fact's. 2. Even No. — Fact's. 3. Odd No. — Fact's. To perform Multipli- cation. Three important Theorems. Prob. 2. Positive integer. Positive fraction. Negative. I^^^^f^' ^ ( Fraction. What? EuLE. De7n. What ? KuLE. I)e7?i. Square of sum. Dem. Square of difference. Dem. Product of sum and difference. Dem. Test Questions. — State and demonstrate the law of the signs in multiplication. How are expressions consisting of the same quantity- affected by exponents, multiplied ? How many cases arise ? Demon- strate each. What is the difference between the square of the sum, the square of the difference, and the product of the sum and differ- ence of two quantities ? Demonstrate. What is the product of a'-^ and a"* ? Prove it. 2. and a ^ ? Prove it. What of a What of a' ^ and a - 2 ? » and a - "» ? Prove it. What of a' Prove it. DIVISION. 63 SECTION' V. Division. 97* Division is the process of nnding how many times one quantity is contained in another. The JDivi' dend is the quantity to be divided. The Divisor is th3 quantity by which we divide. The Quotient is the result, which shows how many times the divisor is contained in the dividend. The JRemaindev is what is left of the di^ddend after the integral part of the quotient is pro- duced. 98. ScH. 1. — Division is the converse of multiplication. Since a product consists of (contains) as many times the multiplicand as there are units in the multipUer, the multiplier shows how many times the multipHcand, is contained in the product. The product, therefore, corresponds to the dividend, the multiplicand to the divi- sor, and the multipher to the quotient. But, as in multiplication, multiplicand and multipher may change places without affecting the product, either of them may be considered as divisor and the other as quotient, the product being the dividend. 99, ScH. 2. — In accordance with the last schohum, the problem of division may be stated : Given the product of two factors and one of the factors, to find the other ; and the sufficient reason for any quotient is, that multiplied by the divisor it gives the dividend. 100, Cor. 1. — Dividend and divisor may both be multi- plied or both be divided by the same number without affecting the quotient. This truth is axiomatic, but may be illustrated thus : If a given number of apples are divided among any number of boys, each boy will receive just the same number as if twice or thrice .as many were divided among twice or thrice as many boys, or as if ^ or ^ as many were divided among ^ or ^ as many boys. 101, Cor. 2. — If the dividend be multiplied or divided by any number, whih the divisor remains the same, the quotient is multiplied or divided by the same. 64 FUNDAMENTAL BULES. This, too, is an axiom, but may be illustrated as the last. 102, Cor. 3. If the diviHor he multiplied by any number while the dividend remains the same, the quotient is divided by that number ; but if the divisor be divided, the quotient is mul- tiplied. This, like the two preceding, is an axiom, but may be illustrated in a similar manner. 103 • Cor. 4.— The sum of the quotients of tvx) or more quantities divided by a common divisor, is the same as the quotient of the sum of the quantities divided by the same divisor. This is an axiom. To illustrate it, consider that as 2 goes into 8,4 times, and into 6, 3 times, it will go into 8 and 6, or 8 -f- 6, 4 times -f- 3 times, or 7 times, etc. 104, Cor. 5. — The difference of the quotients of two quan- tities divided by a common divisor, is the same as the quotient of the difference divided by the same divisor. This is also an axiom which can be illustrated as the last. 105. Def. — Cancellation is the striking out of a factor common to both dividend and divisor, and does not affect the quotient, as appears from {100). 106, Lemma 1. — When the dividend is positive the quotient has the same sign as the divisor; but when the dividend is neg- ative, the quotient has an opposite sign to the divisor. Dem. — This proposition is a direct consequence of the law of the signs in multiplication, since the dividend corresponds to the pro- duct, and a positive product arises from like signs in the factors, and a negative product, from unlike signs. 107, Lemma 2. — When the dividend and divisor consist of the same quantity affected by exponents, the quotient is the common quantity ivith an exponent equal to the exponent in the DITISION. 65 dividend, minus that in the divisor. That is, a"* -^ a" = a"" ~ ", whether m and n be integral or fractional, positive or negative. Dem. — This is an immediate consequence of the law of exponents in multiplication, since, in the corresponding case the exponent of the product was found to be the su7n of the exponents of the factors. 'Now, as the dividend is the product of the divisor and quotient, it fol- lows that the exponent of the quotient is the exponent of the divi- dend minus that of the divisor, q. e. d. EXAMPLES. ^ 1. Divide a' by a^. Model Solution, a'' -^ a^ = a* ; since a'^ X a^ =a'', and di^^sion is finding a factor which multiplied into the divisor produces the dividend {99). 2. Divide w* by ni^. Quot, m"r^^ 3. Divide n" by n~\ Quot., ?i" or ?i " . 4. 1 Divide (a6)'"' by {aby . Quot., (aby^-n oT{ab) » . 5. Divide a^ by a\ Quot., a~^ or- . a- 6. Divide a " ' by a^ QuoL, a~^ov- . a-' 7. Divide x ^ hj x~\ Quot., x^. 8. Divide x~^hj x ^. Quot., X ^ or — . x'^ 108. CoR. 1. — Any quantity with an exponent is 1, since it may be considered as arising from dividing a quantity by itself. Thus ^o may be considered as x'" -^ ^"^ which is 1, because dividend and divisor are equal. But by the law of exponents x"^ -^ x"^ = x'^ . ' . x'^ = 1. In like manner the significance of any quantity with an exponent may be ex- 6Q FUNDAMENTAL liULES. 52 52 plained. 5" = 1, for -— = 1, and also — - = 5" .* . 5° = 1. 5 0" So, |1 = 1 or 8"; or |^ = 1 or 8«, etc. 100, CoR. 2. — Negative exponents arise from division ivhen there are more factors of any nurnber in the divisor than in the dividend. Thus a^ _i_ a^ == a^-^ {107) = a~'^. A • a^ 1 , 1 , . ^ Ai^am a^ -^ a^ ^= — = — .*. a~= — » wnich accords ° a^ a^ a-i with the definition of negative exponents {4:3). 110, CoR. 3. — A factor may be transferred from dividend to divisor {or from numerator to denominaior of a fraction^ which is the same thing), arid vice versa, by changing the sign 1 of its exponent. Thus -7— = —r- > for - — = -— = — •^ ^ 63 a-b^ 63 63 a* -27-3 • «~' a ^6 , since - — =: 62 63 = a'b-^' Thus also -; — 63 ^xi- ; and 1 63 ■ = b- -3 EXAMPLES. . Show that a' X -%2 62.^2 a^-y^ Model Solution.— Since a-2=:-— and x-^= — , I have cc2 h'X^ a- _ a-^ _ &2 . y3 ^ 52 x'i _ 6-x2 x2 — 1 _ 2 2a '-^x ^?/ 2???/ 2. In like manner show that ^ * It is assiiiued that the pupil has learned in Arithmetic how to multiply and divide fractions, or reduce complex fractions to simple ones. DIVISION. 67 3. Free ,, _., from negative exponents and explain ha 'x-ij 'z ^ the process. Ill, I^vdb, 1, To divide one monomial by another. RULE. — Divide the numerical coEmciENT of the divi- dend BY THAT OF THE DIVISOR AND TO THE QUOTIENT ANNEX THE LITERAL FACTORS, AFFECTING EACH 'VN^TH AN EXPONENT EQUAL TO ITS EXPONENT IN THE DIVIDEND MINUS THAT IN THE DIVISOR, AND SUPPRESSING ALL FACTORS WHOSE EXPONENTS ARE 0. The sign of the quotient will be -J- when dividend and DIVISOR have like SIGNS, AND WHEN THEY HAVE UNLIKE SIGNS. Dem. — The dividend being the product of divisor and quotient, con- tains all the factors of both; hence the quotient consists of all the fac- tors which are found in the dividend and not in the divisor. Or, the correctness of the rule appears from the fact that it is the converse of the corresponding operation in multiphcation, so that quotient and divisor multiphed together produce the dividend. The law for the sign of the quotient is demonstrated in (106). Q. e. d. EXAMPLES. 1. Divide 12a^x'*b by 3a^x^. MODEL SOLUTION. oPEBATioN. 3a-X')12a^x^b 4ax'-6 Explanation. — In the divisor there is a factor 3, hence there must be a factor 4 in the quotient, to produce 12 in the dividend. So, also, since there are 3 factors of a in the dividend, and 2 in the divisor, there must be 1 in the quotient in order that the product of divisor and quotient may be the dividend. In like manner 4 factors of x in the dividend, and 2 in the divisor, require 2 in the quotient. There being 1 factor h in the dividend, and none in the divisor, one factor of 5 must appear in the quotient. Hence 12a^x^h -^ da-x- = 4a.i--&, which quotient containing all the factors of the di^ddend not found in the divisor, will, when multiplied into the divisor, produce the dividend. 68 FUNDAiyiENTAL RULES. 2. Divide 12b^c^y^ by AbC'y^. QuoL, d¥c. 3. Divide 8cdx by dx. QuoL, 8c. 4 Divide lOa'^cx^y by Ba^x^. QuoL, 2acy. 5. Divide lSa%^x by — ^a'bx. QuoL, — Qab. 6. Divide — 20x^y* by — Bx'^yK QuoL, 4:xy-. 7. Divide — ^2a^m^-y^ by la-m^. QuoL, — Qay"". 8. Divide — Sla'bx'^y by da'by. QuoL, — dax\ 9. Divide — l^x'^y- by — 13xy\ QuoL, x. 10. Divide Sa-c^ by — Sa^c^. QuoL, — 1. 11. Divide Sa^m by da"m. QuoL, 1. 12. Divide — ^^^/^ by a; " ^?/ " ^ QuoL, — xY\ 13. Divide x"* by a;". C^wo^., ^"^ - ". 14. Divide 5a " W by 2a6. ^mo^., — . 15. Divide Gah^ by 3 A^. QuoL, 2a " ^6 16. Divide lla~h~^x'^ by 11a " '6 -'j:^. 17. Divide a^Z^'c" ' by ab^c^ QuoL, a'-'b'-'c-'^'+'l 112* I^rob, 2. To divide a jjolynomial by a monomial. B ULE. — DrviDE each teem of the polynomial dividend by THE MONOMIAL DIVISOR, AND WRITE THE RESULTS IN CONNECTION WITH THEIR OWN SIGNS. Dem. — This rule is simply an application of the corollaries {103 f 104:) ; since to divide a poljTiomial is to find the quotient of the sum or diiference of several quantities, which by these corollaries is shown to be the sum or difference of the quotients of the parts, q. e. d. DIVISION. 69 EXAMPLES. 1. Divide l^a^x^ + 24^2^35 — Vla^xc by Zax\ MODEL SOLUTION. opEEATioN. ^ax^)l^a?x'^ + 24«2^3^ — Vla^xc 5a-x~ ^ + y^ — 4a^^ ~ *c* ( Explanation. — I write the divisor on the left of the dividend, and the quotient underneath as a convenient form. Considering the first two terms, the quotient of their sum is the sum of their quotients {103)', hence I divide each separately and add the results, obtaining 5a^x~^ -f- ^^^' Again, the quotient of the difference of these two terms and the third is the difference of the quotients (104:)', hence I subtract from the quotient of the first two the quotient arising from dividing 12a^xc, which is 4a*a;~^c, and have for the entire quotient 5a^x-' + 8a6 — 4a4a;-2c. 2. Divide da^b'- — 18a'¥ + Ga^'b by Sab. QuoL, a-b — 6a^b + 2a. 3. Divide 24:X'iy^ __ 8^:4^5 — 24j??/2 by 8^7. QuoL, dxy- — x^y^ — 3y\ 4. Divide 21a^x^ — la^x^ + 14a^ by lax. QuoL, Sa-x-^ — ax -{- 2. 5. Divide 42as — lla'^ + 28a by la. QuoL, 6a2 _ JJ-a -f 4. 6. Divide dk^' — 2ik^c + 48^-^ by 3^^. QuoL, 3^'o — 8^'c + Wk. 1. Divide 12a^c^ — 48a'c4 — S2a^c^ by Wa^cK QuoL, fa^c"^ — Sa'G — 2a'^c-\ 8. Divide 36m^ — 48m* by 4m^. QuoL, drn^— Uvi^. 3 3 2 1 1 7 2 9. Divide m* — m^n^ by m*. QuoL, wJ — mJ'^n^. 10. Divide a^ — a^b'' + a^ by A QuoL, a^ — b^ + a^. 70 FUNDAMENTAL RULES. 11. Divide lla^ — 33a^ by Hal QaoL, a^ -^ Sa"^. 12. Divide 72m ^^ — 60m^/i^ by 24m'^. QuoL, 3m^ — f/i^. 13. Divide 2a* + 4ta^b + 2a26^ by 2a''. QuoL, a-' + 2ab + 62. 14. Divide Ua'^x^ — dab ■{- ISx^ by 3aa7. QuoL, ^ax'^ — r + 67. 15. Divide «'" + ' — a'» + =^ — a'"^^ — «'" + •' by a'. g^o^., a'"-' — a---' — a'" — a"* + \ 16. Divide 5^"— 10^-" + l^x"y by 527^ 1 —2m QuoL, X " — 2.^-^'" ^■''^ -f 3.r " y. 113* Def. — A polynomial is said to be arranged wth reference to a certain letter when the term containing the highest exponent of that letter is placed first at the left or right, the term containing the next highest exponent next, etc., etc. L:.L.— The polynomial Q>x^y^ -\- ^xy^ -f- ix'-^y + y' + «*, when arranged according to the descending powers of y, becomes y* -j" ^^2/^ -}- 6x^y^ + "Lc^y -{- x*. In this form it also clmnces to be s>-frauged with reference to the ascending powers of x. 114:, I^voh* 3* To 'perform, dividon when both diu^" dend and divisor are polynomials. R ULE. — Having arranged dividend and divisor with ee' TERENCE TO THE SAME LETTER, DmDE THE FIRST TERM OF THB DIVIDEND BY THE FIRST TERM OF THE DIVISOR FOR THE FIRST TERM OF THE QUOTIENT. ThEN SUBTRACT FROM THE DIVIDEND THE PRO- DUCT OF THE DIVISOR INTO THIS TERM OF THE QUOTIENT, AND BRING DOWN AS MANY TERMS TO THE REMAINDER AS MAY BE NECES- SARY TO FORM A NEW DIVIDEND. DiVIDE AS BEFORE, AND CON- TINUE THE PROCESS TILL THE WORK IS COMPLETE. DIVISION. 71 The demonstration of this rule will be more readily comprehended after the solution of an example. Ex. — Divide 6a'^x^ + x* — 4a^3 _|_ at — ia^x by a;^ -f- a'^ —2ax, / ^' MODEL SOLUTION. DIVISOR. DrVTDENa). QUOTIENT, a^ — 2ax -f a;2)a-« — 'ia^x + Ga'^x"^ — 4a^s + ^^(a^ — 2a*" + x^ a* — 2a^x + a-x^ — 2a:-^x + 5a-x'^ — 4:ax^ 2a^x + 4a2^2 — 2 ax3 a-x-^ — 2ax^ + x* a^x^ — 2ax^ + x^ Having arranged the dividend and divisor with reference to the descending powers of a, and placed the divisor on the left of the divi- dend, as a matter of custom, I know that the highest power of a in the dividend is produced by multiplying the highest power of a in the divisor by the highest power in the quotient. Therefore, if I divide a*, the first term of the arranged dividend, by a^ , the first term in the arranged divisor, I get a^ as the highest power of a in the quotient. Now, as I want to find how many times a^ — 2ax -f- *^i^ contained in the dividend, and have found it contained a^ times (and more), I can take this a^ times the divisor out of the dividend, and then proceed to find how many times the divisor is contained in what is left of the dividend. Hence I multiply the di^dsor by a'^ and sub- tract it from the dividend, leaving — 2a^x -f- 5a'x^ — 4ax^ + x*'. The same course of reasoning can be applied to this part. Thus I know that the next highest power of a in the quotient will result from divid- ing the first term of this remainder by the first term of the divisor, etc., etc. When this process has terminated I have taken a^, and — 2ax, and x* times the divisor out of the dividend, and finding nothing remaining, I know that the dividend contains the divisor just a* — 2ax + ^^ times. We will now give tho demonstration of the rule. Dem. — The arrangement of dividend and divisor according to the same letter enables us to find the term in the quotient containing the highest (or lowest if we put the lowest power of the letter first in our arrangement) power of the same letter, and so on for each succeeding term. 72 FUNDAMENTAL RULES. The other steps of the process are founded on the principle, thai the product of the divisor into the several parts of the quotient is equal to the dividend. Now by the operation, the product of the divi- sor into the first term of the quotient is subtracted from the dividend ; then the product of the divisor into the second term of the quotient ; and so on, till the product of the divisor into each term of the quotient, that is, the product of the divisor into the whole quotient, is taken from the dividend. If there is no remainder, it is evident that this product is equal to the dividend. If there is a remainder, the product of the divisor and quotient is equal to the whole of the divi- dend except the remainder. And this remainder is not included in the parts subtracted from the dividend, by operating according to the rule. EXAMPLES. [Note. — The following examples are arranged for division. Let the pupil explain his solutions according to the model preceding the demonstration. ] 1. Divide x^ — 3ax- + Sa'^x — a^ hy x — a. ' Quot, x^ — 2ax -f a2. 2. Divide 2if — Idy^ +26?/ — lQhjy — 8. QuoL, 2y^ — Sy-\-2. 3. Divide x'^ — a^x-^ + 2a^x — a* by x^ — ax + a^. QuoL, x'^ -\- ax — aK 4. Divide a^ + 45^ by a^ — 2a6 + 2h\ OPERATION. { a-\-x-\-x^-\-x^-\-xi a- — ax ax -{- [a — l)x^ ax — a;2 ax- + (« — l)ic^ ax- — x^ ax-^ -\- (a — l)x* ax^ — x^ ax^ — x" ax^ — x^ 39. Divide x{x — l)a^ + (^3 -^ 2x — 2)a^ + (3^^ — x^)a — x^ by fl-x + 2a — x"^. QuoL, {x — l)a + x^ 76 FUNDAMENTAL RULES. ScH. —This process of division is strictly analogous to ' Long Di- vision " in common arithmetic. The arrangement of the terms corre- Bponds to the regular order of succession of the thousands, hundreds, tens, units, &c., while the other processes are precisely the same in both. Synopsis for Eeview. Division. Dividend. Divisor. Quotient. Remainder. Sch. 1. Eelation of div'n and mult'n. Sch. 2. General reason for a quotient. Cor. 1. Multiplying or dividing divi- dend and divisor. Co7\ 2. Multiplying or dividing div'd. Cor. 3. " " " divisor. Cor. 4. Quotient of sum. Cor. 5. " " difference. Def. Cancellation. 1 I f Law of Signs. Dem. § -b I ^ ( Cor. 1. Meaning of exp't 0. i i^ §^ [ Law of Expt's. Dem. •< Cor. 2. Negative expt's. ( Cor. 3. Transfer'ng expt's. ■i) s2 ~ '1. To divide one monomial by another. Rule. Dem. 2. To divide a polynomial by a monomial. Rule. De7n. 3. To divide one polynomial by another. Rule. Dem. Scholium. Test Questions. — How do negative exponents arise, and what do they signify ? What is the value of any quantity with for its expon- ent ? Why ? How do you divide, when dividend and divisor consist of the same quantity affected by exponents ? Why ? Give the Gene- ral Rule (2*ro&, 3) and its demonstration. FUJS'DAMENTAL PROPOSITIONS. 77 CHAPTER n. FACTORING. SECTION I. Fundamental Propositions. lis. The Factors of a number are those numbers which multiphed together produce it. A Factor is, there- fore, a Divisor. A Factor is also frequently called a iniea- sure, a term arising in Geometry. 116. A Common Divisor is a common integi-al factor of two or more numbers. The Greatest Common Divisor of two or more numbers is the greatest common integral factor, or the product of all the common integral factors. Common Measure and Common Divisor are equiv- alent terms. 117. A Com^non MultiiJle of two or more num- bers is an integral number which contains each of them as a factor, or which is divisible by each of them. The Least Common Multiple of two or more numbers is the least in- tegral number which is divisible by each of them. 118. A ComiJOsite JS'uniher is one which is com- posed of integral factors difierent from itself and unity. 119. A I* r hue Number is one which has no in- tegral factor other than itself and unity. 120. Numbers are said to be Prime to each other when they have no common integral factor other than unity. ScH. 1. — The above definitions and distinctions have come into us« from considering Decimal Numbers. They are only applicable to lit- eral numbers in an accommodated sense. Thus, in the general "view which the literal notation requires, all numbers are composite in the sense that they can be factored ; but as to whether the factors are greater or less than unity, integral or fractional, we cannot affirm. 78 FACTORING. ScH. 2. — Skill in resolving numbers into factors, though of great importance, can only be attained by much practice, and habits of close observation. Yet, if the learner masters the following propositions, he will lay the foundation for such an acquisition. 12 H. ^vop, 1, A monomial may he resolved into literal factors by separating its letters into any number of groups, so that the sum of all the exponents of each letter shaWmake the exponent of that letter in the given monomiaL III. 5a%x^ may be resolved into 5a • ah^ • 11X1111 a'^b^x^ X a'^b'^x^ X x^, or into any number of fac|ors, in a like manni^jv Dem. — This is a direct result of the principle that monomials are mul- tipUed by writing the several letters in connection, and affecting esu^ with an exponent equal to the sum of the exponents of that letter in the factors. EXAMPLES. 1. Separate 12a"bx^ into all the possible factors with pos- itive integral exponents. Factors, 3, 2,2,a, a, b, x, x, and x. 3. 2 2. Separate l^a^x^ into two equal factors. _3_ 1 _3_ 1 Factors, 4ai ^x^, and 4ai ^x^. 2. 3. Separate %x^y^ into two factors, one of which is 4txy. 4. Bemove the factor 2{ax)'^ from (Sa'^x. 5. Remove the factor 3a^ from ISoc^. Factors, 4txy, and 2x ^y\ Result, 3a^A Result, 5^c-. 6. Resolve m into two factors. ^ m, ni^ • 771-% 771* • 771*, I 1 2 13 1 Results, 771^ • 771^", v/ 7n - v/ 77) fji^ . 771^, 771* . 77^^, 777^ -771^, etc. FUNDAMENTAL PROPOSITIONS. 79 7. Resolve x into two equal factors. Into 3. Into 5. 12^, Prop, 2, Any factor ichich occurs in every term of a jDolynomial can he removed by dividing each term of the polynomial by it. Dem. — This is the ordinary problem of division by a monomials since divisor and quotient are th© factors of the dividend. EXAMPLES. 1. Factor 3a — 36. Result, ^a — h). 2. Factor ax — bx -{- ex. Result, {a — b -\- c)x. 3. Factor 5 — 5?/. Result, 5(1 — y). 4. Factor Qa'-y^ — l%ay\ Result, 6ay^{ay — 3). 5. Factor 42^^?/ — 14:xy + Ix^y^. Result, 7xy{Gx — 2 -]- x-y). 6. Factor Slan-^x — 6dcmx\ Residt, 9cma;(9m — Ix^). 7. Factor la^x^y — 21ax'^y\ Factoids, lax^y and a^ — ^y\ 8. Factor llab'^x^ — Mab^x'- + OGa^fc^a;^. Factors, Vlab'^x^ and 6x — 76 + ^• a Factor 924£i^b'^c^ — lOlSa^b^c^ + 12S2a^b^c\ Factors, lAa^b^c^ and 66a — lib + 88c. 123 • JPtoJ), 3, If two terms of a trinomial are positive and the third term is twice the jjroduct of the square roots of these two, and positive, the trinomial is the square of the sum of these square roots. If the third term is negative, the tri- nomial is the square of the difference of the two roots. Dem. — This is a direct consequence of the theorems that, "The square of the sum of two quantities is the sum of their squares plus twice their product ;" and ' ' The square of their difference is the sum of their squares minus twice their product. " {94, 05), 80 FACTORING. EXAMPLES. I. "What are the factors of a^ -^ ^ah -[- h-'i Model Solution. — I observe that the two terms a^ and h'^ of thia trinomial are both positive, and that the other term, 2a6, is twice the product of the square roots of a'^, and 6^, and positive. Hence a * + 'lah -|- &2 _ (■^ _|_ ^)2 — (^a -\- h){a -\- h). The factors are a + 6 and a + h. 2. What are the factors of x'^ — 2ax + a^ ? 3. What are the factors of m* -\- n^ -\- 2m-n'^ ? SuG. — Here m* and n^ are both positive, and 2m^n^ is the product t)f their square roots. Aris. {m^ -\- n^)(m^ -\-n^). 4. What are the factors of IGa' — 8a + 1? 5. What are the factors oi m -{- 2^y mn -{- n '} Ans., (v/ 7?i + v/ri)(\/ m + v^n). 2 4 X 2 6. What are the factors of ^^ + ?/^ — ^x'-^y^ 1 1 2 1 ^ Ans., x'-^ — 2/^ ^^<^ ^ — V""' 7. What are the factors of a"h + a^^ + 2a%^ ? 8. What are the factors of x"^ + 2xy — y-^. Query. — Can the last be factored according to this Trob. ? Why ? 9. If 4a'-', 1662, and IGab are the terras of a trinomial, what must be their respective signs so that the trinomial can be factored ? 10. Factor I6a^b^m'' — Sa^b'^m + 1. Factors, {ia-^b^m — • 1) and {4:a%^m — 1). II. Factor da-' + Sab + ^b^ Factors, (3a + ^) and (3a + ^b). 12. Factor 49a^62 — -^f-a6^ + ib^. Factors, {lab — p^) and {lab — %b'^]. V FUNDAMENTAL PROPOSITIONS. 81 13. Factor -^ + 2 + — ' Factor 14. Factor -^x^^ + yV^/^ — \oc^y'- Factors, (fa;^ — iy.) and {^x^ — \y'). 15. Factor— + ^ — 2. ?/-> x^ \y- X- / \?/2 X / 16. Factor a^ ^ 2av/^ + ^. Factors, {a + ^) and (a + v/j;). 17. Factor j; — 2b^ + 6-^. Factors, {^x — h) and (v/a; — h). 18. Factor ^ — 2Vx ■ y + ?/. Factors, {^x — v^y) and (v^^ — ^y)- 19. Factor a'h — "lax^h + x^. Factors, {a^b — x) and {a^ — x). 20. Factor a + 2v/a + 1. Factors, {\^a +1) and (v^a + 1). 124, J^vop, 4z. The difereiice between two quantities is equal to the product of the sum and difference of their square roots. Dem. — This is an immediate consequence of the theorem, that "The product of the sum and difference of two quantities is the difference of their squares. " Thus a- — h- = {a-\-b){a — b). Also, if we have m — n, m is the square of ??i, and n, of n' .'. m — n =^ {m^ -\- n^){m- — n^), etc. EXAMPLES. 1. Factor 16^'^ — 9y\ Factors, {4:X — By) and (Ax + 3y), 82 FACTORING. 2. Factor aHf — h'^xK Factors, {ay — hx) and {ay -\- hx). 3. Factor IQa^x'^ — 2562?/2. Factor Sy {^ax — 56y) and (4a^ + 56i/). 4. Factor x^^ — y*. Factors, {x^ — ^/'O ^^^ (^^ + 2/^)- 2 2 5. Factor x^ — y^; also x^ — if ', also a;^ — y'"^ ; also j; -4 — y-*) also 4a~^ — 96~\ 11 1. 1. Factors of last three, {x'^ — ij^) and {x^ + y^) ; {x -2 — y--^) and {x-^ -{- y -^) \ and {2a-^ — W-^) and{2a-^ -\-3b-^). 6. Factor m — n, _ Factors, {^ m — Vn) and (v^m + ^n). 7. Factor 2a — 462. Factors, 2(v/ a — ^^26) (^ a + ^^26). a Factor 25^" — 3?y^". ni m Factors, {bx'^ — ^'6if){^x~^ -^ ^'^y''). 125. I*rop, o. When one of the factors of a quantity is given, to find the other, divide the given quantity by the given factor, and the quotient loill be the other. Dem. — This is the ordinary problem of division, since the divisor and quotient are the factors of the dividend. Any problem in division af- fords an example. ' EXAMPLES. 1. Resolve 2a^ — SQa^b into two factors one of which is 2aK 13 19 2. Remove the factor — from — . x- ?/3 x^ y^ o -r» ^y e j^ a^ 3c ^ . 9(?2m-2 3. Remove the factor -r- + t: — from a^b -* 6 ' 2m 4 9 12 4 4. Remove the factor 3a — - + 2^ -• from \- \- — ;. a* a'^x x-^ FUNDAMENTAL PROPOSITIONS. 83 126» I*TOp, 6» The difference between any two quanti- ties is a divisor of the duterence between the same poivers of the quantities. The SUM of two quantities is a divisor of the diffeeence of the same even powers, and the sum of the same odd poivers of th-e quantities. Dem, — Let X and y beany two quantities and n any positive integer. First, X — y divides x" — y" . Second, if n is even, x -{- y divides K" — y*. Third, if n is odd, x -^ y divides x" -f- V"- riKST. Taking the first case, we proceed in form with the division, till four of the X — t/)x" — y" (^x" - '^ -\- x" - "y -\- x" - ^y- -\- x" - ^y^ -\- &g. terms of x" — x "" 'y ~ "^~~" the quotient x" ~" '?/ — y" (enough to deter- x" - ^y — x" - "^y^ mine the law) x" ~ -y^ — y" are found. We find x" ~ ^y- — x" - ^y ^ that each remainder ' x" "" y — j/" consists of two terms, the second of x" - ^y^ — x" — ^y* which, — y", is the second term of x" ~ ^y^ — y" the di^'idend constantly brought down unchanged ; and the first contains x with an exponent decreasing by unity in each successive remainder, and y with an exponent increasing at the same rate that the exponent of x decreases. At this rate the exponent of x in the nth remainder becomes 0, and that of y, n. Hence the ?ith remainder is y" — y" or : and the division is exact. SECOND AND THIED. X -|- j/)x" 4-2/" ('x" - 1 — X" - ^v -j- X" - Hj' — X" - *y^ , &c. X" -\- x"- ^y — x"-i?/ + 2/" — X" - hi — X" - "^y^ X" - h)^ ± y^ Taking x -\- y x" - -y"^ - f- x" - 'y^ for a divisor, we — x" — '^y^ + y"^ observe that the exponent — x" - ^y^ — x» — *y* of X in the successive re- x'^ — *y* -^ y" mainders decreases, and that of y in- creases the same as before. But now we observe that the fii'st term of 84 FACTORING. the remainder is — in the odd remainders, as the 1st, 3rd, 5th, etc., and -j- i^i the even ones, as the 2nd, 4th, Gth, etc. Hence if n is even, and the second term of the dividend is — y", the nth remainder is y* — 2/" ^^ ^' ^^^ the division is exact. Again, if n is odd, and the second term of the dividend is -j- y'\ the nth remainder is — 2/" 4~ !/" or 0, and the division is exact, q. e. d. ' 127 » ScH. — The pupil should notice carefully the form of the quo- tient in each of the above cases, and be able to write it without divid- ing. He should also be able to tell at a glance, whether such a divi- sion is exact or not, and if not exact, what the remainder, or fractional term of the quotient is. EXAMPLES. Write the quotients in the following cases : 1. (^5 _ y5) ^(x — y). QuoL, X* -f x^y + x'iy^ -f xy^ -f- y*. 2. (a3 4- 63) -^ (a + h). QuoL, a^ -— a6 -f b^ 3. (a3 _ 53) -^ (a _ b). 4. {a* — m") -^ (a -{- m). 5. {a^ — 771") -i- (a — m), 6. (m7 -f ?i7) -^ (tti + n). 7. (m8 — 6«) ^ (m -f b), also by (m — b). SuG. — Notice that x^ ^ is the 4th (an even) power of x^, and y'' * is the same power of y^. It may be best for the pupil to obtain this quotient thus. Put x^ = a, and y^ = b. Then x' =^ = a"* and y^^ = b'^. But (a^ — b'^) ~ (a-}-6) = a^ — a^b -{- ab'^ — b\ Whence restoring the valaes of a and b, we have (x^^ — y^^) -r- {^^ -h V^) = x^ — x^y'^ -f- ^^y^ — 2/^' since a^ = x^, — a^b= — x^y'^, -\-ab'^ = + '^^y^, and — 9. (tti'o -f ni«) ~ (m2 -f n^). Quot, mP — m^n^ -f m^w^ — m^^ -f- n^. 10. (a;'8 — ?/'8) -:- {x^ — y^), also by {x^ + y^). 11. (a%i« — 2/^) -^ (a?n' — y). FUNDAMENTAL rEOPOSITIONS. 85 Sua— Notice that a^m^ is the third power of am^, as y^ is of y. The quotient mav be obtained by first using a single letter for «m^, as x, \\Titing out [x^ —y')^iz — y)== x^ -^xy ^y-, and finally restor- ing the value of a*. After a Uttle practice the pupil will not need to°go through such a process, but can write the quotient at once. This quotient is a'^m^ -\- am'^y -f- y^- 12. (1 - 7/) -^ (1 - y). . 13. {ye — l)^{y + 1), also by (y— 1). 14 {a'x^ + 5io?/'o) -i- {ax + b^y"^). 15. {16x* — 8Uf) ^ {'2x 4- 3y2), also by {2x — S?/^). [Note.— In the following examples be careful to observe in which cases the division is exact ; but if it is not exact, the quotient should be WTitten with the fractional term. ] 16. Is a^ + b\ exactly divisible hj a -\- b, or a — b? 17. Is m6 — y^, exactly divisible by m + y, or m — ?/ ? 18. Write the following quotients {x"'"'—y"") -^ {x -\- y), also hj X — y. (a;-™ + ^ -^ y"'"^'^) -^ {^c -\- y),m being an in- teger. QuEET. — Is the division exact in both cases in the last example ? Is 2m even or odd ? Is 2m -}- 1 even or odd? FOR REVIEW OR ADVANCED COURSE. [Note. — The following corollary and examples may be omitted till after the pupil has been through the chapter on Powers and Boots, if thought desirable. ] 128, CoR. — The last proposition applies equally to cases I JL involving fractional or negative exponents. Thus, x^ — y^ A. 4. divides x^ — y^, since the latter is the difference between the 4th powers of x'^ and y'^. So in general x~^ — y~~ an ax divides j;~™ — y~~, (^ being any positive integer. This n s becomes evident by putting x~m = v, and y~T =^ w; 86 FACTORING. an an whence x ~ ^ = v", and y~ T = nf. But i/* — lu" is divisi- an as n s ble by v — lu, hence x ~ ^ — y~T i^ divisible hj x~~^ — y ^. EXAMPLES. 19. What is the quotient of (a"' — &"') -^ (a~' — &-')? 20. What is the quotient of (^~^— y""^) -^ (^~^+ 2/"^)? ^ns., ^ ^ — X ^y ^ -\- X ^y ^ — y *. 21. What is the quotient of [^ — if~\ -^ f- 4 2/' 1 ? — X — JL _i 22. Is ;r '^ — ?/ ~" '' divisible by a; ^ — y~"^or^ ^ + 2/~^? 23. Is ^73 — 2/3 divisible by ^x -\- ^y or "^x — \^y ? 24. Is ^3 _[- 2/3 divisible by ^^x -\- ^y or ^x — ^y ? 25. What is the quotient of (ax^ + h\j)^{a^ ^x + V^?/^)? 120, Pvo])» 7. -4 trinomial can be resolved into iico bi- nomial factors, when one of its terms is the product of the tquare root of one of the other two, into the sum of the factors of the remaining term. The two factors are severally the alge- braic sum of this square root, and each of the factors of the third term. III. — Thus, in a;' -f- '^''^ + 10, we notice that 7a; is the product of the square root of x* , and 2 -f- ^ (tiie sum of the factors of 10). The factors of x^ -\-1x -\- 10 are X -f- 2 and x -\-^. Again x^ — 3x — 10, has for its factors x -f- 2 and x — 5, — 3x being the product of the square root of x^ (or x), and the sum of — 5 and 2, (or — 3), which are factors jof ^ 10. Still again, x^ -f 3x — 10 = (x — 2)(x -f- 5), determined in the same manner. Dem. — The truth of this proposition appears from considering the product of X -|- ^ ^y ^ + ^' which is x^ + (a -f- &) * + «&• ^ this product, considered as a trinomial, we notice that the term (a -\~ h) x, is FUNDAMENTAL niOPOSITIONS. 87 the i)roduoi of ^ x* and a + ?>, the sum of the factors of ah. lu hke manner (cc -f- o.){x —b):=x'^ -j- (a — b)x — a&, and (x — a){x — b) =: x^ — {a -{-b)x -}- «^> both of which results correspond to the enun- ciation. Q. E. D. [Note. — In application, this proposition requires the solution of the problem : Given the sum and product of two numbers to find the numbers, the complete solution of which cannot be given at this stage of the pupil's progress. It will be best for him to rely, at present, sim- ply upon inspection, ] EXAMPLES. 1. Show that, according to this proposition, the factors of ^2 _|_ 3^ _|_ 2 are x -{- 1 and ^ + 2. Verify it by actual multiplication. 2. In hke mannei' show the following : a;2_ 7^_^ 12 ^ (j;_3)(^_4). x^ — j;— 12 = (j7 + 3)(j; — 4). ^2 4_^__12 =^ (^_3)(^ + 4). v^ — 5^ + 6 — - a;2 + 207 — 35 =^ 130, I^vop, S. W(y can oft m detect a factor by separat- ing a polynomial into parts. Dem. — The correctness of this process depends upon the principle that whatever divides the parts, divides the whole. EXAMPLES. 1. Factor x^ + 12a? ^ 28. Model Solution. — ^The form of this polynomial, suggests that there may be a binomial factor in it, or in a part of it. Now x'~ — -Lx-j-'L is the square of x — 2, and {x^ — 4ic -(- 4) + (16x — 32) makes x'^ + 12x — 28. But {x^ _ 4.C + 4) 4- (16x — 32) = (x — 2)(x — 2) + (X — 2)16 = (X — 2)(x — 2 -f 16) = (X — 2)(x + 14). Whence X — 2, and x -^ 14: are seen to be the factors of x* + 12x — 28. 88 FACTORING. 2. Resolve 12a-b + 3&?/- — 15aby into its factors. Solution. 3b is seen to be a common factor, and removing it, we have 4a^ -\- y'^ — 5f^y- This latter polynomial may be separated into the parts a* — 2ay -f- y^, and 3a^ — Say. Hence, 4a^ -\- y^ — Say = {a' _ 2ay + 2/2) + (Sa^ — 3ay^ = (a — y)(a - ij) + («— t/)3rt = (^a — y){a — y -{- 3a) = (a — 2/)(4a — y). Hence, the factors of X2a^b + 3by^ — 15aby, are 36, a — y, and 4a — y. 3. Factor Ga-^ + 15a^6 — 4:a^c'' — 10a-bc\ Solution, a- is evidently a factor. Bemoving this, we have 6a^ -f" 15a-6 — 4ac- — 106c'-. The latter expression may be written 3a-(2a -f- S^) — 2c^(2a + 56) = (3a'2 — 2c-2)(2a + 56). Henoe 6a'^-{-15a^h — 4.a^c^ — 10a26c2 = a2(3c»2 — 2c-2)(2a + 56). 4. Factor Sa'^ — 2a — 1. Factors, a — 1, and 3a + 1. 5. Factor 5a^ — Sax + 'Sx'\ Factors, 5a — 3x and a — x. 6. Factor 20ax — 2ban — \(Sbx -f 2Qbn. Factors, 4:X — 5/1 and 5a — 46. 7. Factor Sa^ + 22ab + 1562. Factors, 2a + 36, and 4a + 56. 8. Factor \2x^ — Sxy — ^x^y^ + &y^. Factors, 3ar2 — 2i/, and 4a; — 3?/2. SECTION 11. y Greatest or Highest Common Divisor. 131, Def. — It is scarcely proper to apply the term Greatest Com- mon Divisor to literal quantities, for the values of the letters not being fixed, or specific, great or small cannot be affirmed of them. Thus, whether a^ is gi'eater than a, depends upon whether a is greater or less than 1, to say nothing of its character as positive or negative . So, also, we cannot with propriety call «•' — y^ greater than a — y. If a = |-, And y = \, a'^ — y'^ = ^^, and a — ?/ = i 5 .* . in this case a^ — 2/"^'^^2 _i5a62y -f. 12a^by -f 'Shy' — 15ahy-, and Qa-'b^ — ea'b^'y - '^b^y' -j- 2ab-y- -j- Ga'by — Qa^by^ — ^by" -f 2abyK HIGHEST C. D. 93 OPEBATION. 12a«6* + 3b^y^ — 15ah'y + Vla-by + dby"^ — 15aby^ (A). Gaib^—6a^b^y—2b^y^ -\-2ab^y^ - ^6a^b y—Ga ^by^~ 2by'^-{-2aby-' - (B). (C). (E). (F). (G) (II) 4a2 — 5ya + y^)Sa^ — Sya^^ + y-a — y^ ■i ia-6 H- by- - 3a^b — 3a-' by - oaJjy -f- 4:a''y -j- y^ — — by-^ + aby^ + Sa^; - 5ai/^ - - y — ^a'y'- -y' ■ + «?/' - - (^■h'j -j- 4.y,a-^ — (•% + 3.VV/^ - {5by -f- oy-)a-]- {by- (3by + 3y')a^ + (^v* + r) - + ?/■■*)« — ■ ihy' + y^) --- (ij -- (K) - . 12a^ - . 12a3 - - 15?/a2 + 3y-'a (L) - ■ '^y^i''^ 4-y'^ — V 4 (M) - CN) - - (0) -. ■ - Reject l'Ji/2 a — v)4a=^ — G i-f ?/*(4a — y. 4^2 _ 4 j/tf - 2/« + y' — ya-\- y^ .-. The H. C. D. of (A) and (B) is {b){b + y){a — y) = ab^ + f% liEASONING. 1st. Removing Z> from both (A) and (B), and reserving it as a factor of the H. C. D. (Lem. 1), and rejecting 3 from (A), and 2 from (B), since they are not common factors, and hence cannot enter into the H. C. D., we get (C) and (D). 2nd. Arranging (C) and (D) ^vdth reference to a, the letter with re- spect to which the H. C. D. is sought, both for convenience in divid- ing, and to observe if any other common factor appears, we have (E) and (F). In these we readily discover and remove the common factor, (6 -\-y), reserving it as a factor of the H. C. D., and get (G) and (H)i 3rd. Having now found two of the common factors of (A) and (B), and removed some which were not common, it remains to determine whether there are any more common factors, that is, whether there is a C. D. in (G) and (H). (G) is its own H. D., Lem. 2 ; hence if it divides (H), it is the H. C. D. of (G) and (H). "We, therefore, try it. Dividing, 4a^ goes into 94 FACTORING. 3a^, fa times ; but, to avoid fractions, we multiply (H) by 4, since, as 4 is not a factor of (G), the H. C. D. of (G) and 4 times (H) is the same as of (G) and (H). We thus obtain (I). . • . K (G) divides (I) it is the H. C. D. of (G) and (H). Trying it, we find the remainder (L). Now, any C. D. of (G) and (I) is a divisor of {K)[a multiple of (G)], Lem. 3, and of (L), Lem. 4. And we now have to find the H. C D. of (G) and (L), upon which we reason just as upon (G) and (H) Thus, as (G) is its own only divisor of the 2nd degree, if it divides (L) or (M) [since (M) = 4 (L) and 4 is not a factor in (G)], it is the H. C D. of (G) and (H). Trying, we find a remainder (O). Now, any di- visor of (G) and (M) is a divisor of (N), Lem. 3, and of (0), Lem. 4, The question is therefore reduced to finding the H. C. D. of (G) and (O) Upon these we reason as before. Hejecting Idy^ since it is not a fac tor of (G), and hence, cannot enter into the H. C. D., we have (P) The H. C. D. of (G) and (P) cannot be higher than (P), and as (P) is its own only divisor of its own degree, if it divides (G, ) it is the H C. D. sought. It does. . •. a — y is the R. G. D. of (G) and (H). Finally, h, {h -{- y), and (a — y) are all the common factors of (A) and (B) and hence, 6 (& + y){'^ — V) = «^^ + «^2/ — &*2/ — ^V^ is their H. C. D. ScH. — It often occurs that one or more of the above steps are not re- quired, especially the removing of a compound factor from the given polynomials. EXAMPLE. Find the H. C. D. with respect to x, of a:' — 8x^ -}- ^Ix" — 20^ + 4, and 2x^ — 12^2 _J_ 21^; _ 10. Model Solution. — Calling the 2nd, (A) and the 1st (B), I have the following (A) OPERATION. (B) 2x^ —12^2 + 21x— 10) J7^ — 8x^ + 21^72— 20^ + 4 2 fCJ 2x*~ Wx^ + 42^2 — 40^- -f- 8(07 — 2 2.774 _■ 12j:3 _|_ 2Lr^ — lO.r — 4.x ' -\-'21x'^ — 'SOx + 8 — 4:x^ + 24^3 _ 42^ -f 20 (D) Keject — 3 3^ ^ +12.r — 12 (J'V j.i_^ 4^ + 4" HIGHEST C. D. 95 X' — (^) (A) ■ 4:x + 4)2.r3 — 12^2 + 21a: — 10(2a; — 4 2^3 _ 8.^-^ H- Sx — 4a;^ + rSx — 10 — 4.^2 + IQx — IG (F) - - - Keject _ 3 - - — 3j; + G f ^j (GJ ^ _ 2)a;2 — 4a; + 4(0; - X-' — 207 — 2a; + 4 — 2a; + 4 -2 REASONING. The two given poljoiomials being an-anged with reference to x, and no common, or other factor, appearing, I proceed at once to determine by successive divisions their H. C. D. The H, C. D. cannot be higher than (A), the lower of the two ; and, as it is its own only divisor of the 3rd degree, if it divides (B), it is the H. CD. As 2x^ is contained in x*, ^.r times, to avoid fractions I multiply (B) by 2, and, since 2 is not a factor of (A), the H. C. D. of (A) and (B), is the H. C. D. of (A) and (C). Now if (A) divides (C) it is the H. C. D. Trj-ing it, I find a remainder, (D). But the H. C. D. of (A) and (C) is also a divisor of (D), for (D) is the difference be- tween (C) and (x — 2) times (A), both of which are divisible by the H. C. D. of (A) and (C). The question is now reduced to finding the H. C. D. of (A) and (D). Upon which I reason exactly as before upon (A) and (B). Thus, since — 3 is a factor of (D), and not of (A), it can be rejected ; and the H. C. D. of (A) and (D) is the H. C. D. of (A) and (E). This cannot be higher than (E) ; and as (E) is its own only divisor of the 2nd degree, if it divides (A), it is the H. C. D. Trying it, I find a remainder, (F). Upon this remainder, and (E), I reason as before, upon (D) and (A). Thus, the H. C. D. of (E) and (A) is a divisor of (E) and (F), since (F) is the difference between (A) and a multiple (2x — 4 times) of (E). The question is then reduced to find- ing the H. C. D. of (E) and (F). • Upon these I reason as before, re- jecting — 3, which is not a common factor, and hence, forms no part of the H. C. D. of (E) and (F), and finding by trial that (G) is a di- visor of (E). Therefore, x — 2 is the H. C. D. of (A) and (B). ScH. — The pupil should be careful to notice that at each step in this process, we show that the H. C. D. sought, cannot be higher than the divisor used. Hence the divisor which terminates the vwrk, is the H. C. D. 96 FACTORING. GENERAL DEMONSTRATION OF THE RULE FOR FIND- ING THE H. C. D. Let A and B represent any two polynomials whose H. C. D. is sought. 1st. Arranging A and B with reference to the same letter, for con- venience in dividing, and also to render common factors more readily discernible, if any common factors appear, they can be removed and reserved as factors of the H. C. D., since the H. C. D. consists of all the common factors of A and B . 2nd. Having removed these common factors, call the remaining fac- tors C and D. "We are now to ascertain what common factors there are in C and D, or to find their H. C. D. As this H. C. D. consists of only the common factors, we can reject from each of the polynomials, C and D, any factors which are not common. Having done this, call the re- maining factors E and F. 3rd. SupiDOse polynomial E to be of lower degree with respect to the letter of arrangement than F, (If E and F are of the same degree, it is immaterial which is made the divisor in the subsequent process. ) Now, as E is its own only divisor of its own degree (Lem. 2), if it di- vides F, it is the H. C. D. of the two. If, in attempting to divide F by E to ascertain whether it is a divisor, fractions arise, F can be mul- tiplied by any number not a factor in E (and E has no monomial fac- tor), since the common factors of E and F would not be aftected by the operation. Call such a multiple of F, if necessary, F'. Then the H. C. D. of E and F', is the H. C. D. of E and F. If, now, E divides F', it is the H. C. D. of E and F. Trying it, suppose it goes Q times, with a remainder, E. 4th. Any divisor of E and F' is a divisor of R, since F' — QE = E, and any divisor of a number divides any multiple of that number (Lem. 3), and a divisor of two numbers divides their difference. The H. C. D. divides E, hence it divides QE, and, as it also divides F', it divides the difference between F' and QE, or E. Therefore, the H, C. D. of E and F', is also the H. C. D. of E and E. 5th. We now repeat the reasoning of the 3rd and dth paragraphs concerning E and F, with reference to E and E. Thus, E is by hypo- thesis of lower degree than E ; hence, dividing E by it, rejecting any factor not common to both, or introducing any one into E, which may be necessary to avoid fractions, we ascertain whether E is a divisor of E. Gth. Proceeding thus, till tvv'o numbers are found, one of which di- I HIGHEST C. D. 97 vides the other, the last divisor is the H. C. D. of E and F, since at every step we show that the H. C. D. is a divisor of the two numbers compared, and the last divisor is its own H. D. 7th. Finally, we have thus found all the common factors of A and B, the product of which is their H. C. D. q. e. d. EXAMPLES. 1. Find the H. C. D. of 14aa; — 8a + a.x^ — lax'^ and IQa'^x'^ + Qta-x"^ — ISa^-x^, and give the reasoning as above. The H. G. D. is ax — 4a. 2. Find the H. C. D. of a^ + Sa'b + dab^ + h^ and 5a' + 5b% giving the demonstration. The H. a D. is a -{- b. 3. Find the H. C. D. of 36a' + da^ — 27a^ — 18a' and 21a^b- — da%- — 18a%^, giving the demonstration. The H. a D. is da* — 9a\ 4. Find the H. C. D. of 4:xy^ — 2y^ + 6x-^y and 4:X^y + 8^^ — Axy"^, and give the demonstration. The K a D. is 2x + 2y. 5. Find the H. C. D. of Zx' — IQx^ + 15^ + 8 and x^ — 2x* — 6^3 4. 4^2 _j_ 13^ ^ 6. The E. a D. is x^ -f 3a;2 + 3^7 -f 1. 6. Find the H. C. D. of 2a^x^ — 2a'bx^y + 2ab^x^y^ — 2b'^xy^ and 4:a^b^x-^y'^ — 2ab'^x^y^ — 2b'*xy\ The H. a D. is 2ax-^ — 2bxy. 138. JProh, To find the H. 0. D. of three or more poly- nomials. RULE. — Find the H. C. D. of any two of the given polynomials by one of the foregoing methods, and then find the h. c. d. of this h. c. d. and one of the eemaining polynomials,. and then again compare this last h. c. d. with another of the polynomials, and find their h. c t). Continue this process till all the polynomials have been USED. yb FACTOllING. Dem. — For brevity, call the several polynomials A, B, C, D, etc. Let the H. C. D. of A and B be represented by P, whence P contains all the factors common to A and B. Finding the H. C. D. of P and C, let it be called P'. P', therefore, contains all the common factors of P and C ; and as P contains all that are common to A and B, P' con- tains all that are common to A, B and C. In like manner if P" is the H. C. D. of P' and D, it contains all the common factors of A, B, C, id D, etc., etc. Q. e. d. EXAMPLES. 1. Find the H. C. D. of 2x* + Qx^ + Ax% 3x^ + 9^^ _j_ 9^ + 6, and 3^3 _|. Sx^^ ^ 5x -\- 2. The K a D. is x -\- 2. 2. What is the H. C. D. of lOa^ + lOa^b"- + 20a^b, 2a^ + 263, and 4&« + 12a''b^ + 4:a^b + 12a63 ? Ans., 2(a + b). ^^^^ SECTION III. Lowest or Least Oommon Multiple. 139 » Def. — In speaking of decimal numbers, the term Least Com- mon Multiple is correct, but not in speaking of literal numbers. For example, the numbers (a-f-&)^ and (a"^ — 6^) are both contained in (a-j-6)2 X (a — b), and in any multiple of this product, as m{a-{-b)^ (a — 6). But whether m{a -f- by~(a — 6) is greater or less than {a-\-b)'{a — b) depends upon whether a is greater or less than 6, and also whether m is greater or less than unity. In speaking of literal numbers, we should say Loicest Common Multiple, meaning the multi- ple of lowest degree with respect to some specified letter. 140. JProh. To find the L. G. M. of two or more numbers. RULE. — Take the LrrEKAii number of the highest de- gree, OR the largest decimal number, and multiply it by ALL the factors FOUND IN THE NEXT LOWER WHICH ARE NOT EST IT. Again, isiultiply this product by all the factors FOUND in the next LOWER NUMBER AND NOT IN IT, AND SO CONTINUE TILL ALL THE NUMBERS ARE USED. ThE PRODUCT THUS OBTAINED IS THE L. C. M. LOWEST C. M. 99 Dem." — Let A, B, C, D, etc., represent any numbers arranged in the order of their degrees, or values. Now, as A is its own L. M., the L. C. M. of all the numbers must contain it as a factor. But, in order to contain B, the L. C. M. must contain all the factors of B. Hence, if there are any factors in B which are not found in A, these must be introduced. So, also, if C contains factors not found in A and B, they must be introduced, in order that the product may contain C, etc., etc. Now it is evident that the product so obtained, is the L. C. M. of the several numbers, since it contains all the factors of any one of them, and hence can be divided by any one of them, and if any factor were removed it would cease to be a multiple of some one or more of the numbers, q. e. d. 1. Find the L. C. M. of {x^ — l),{x^- — 1), and {x + 1). MoDEii Solution. — The L. C. M. must contain x^ — 1, and as it is its own L. M., if it contains all the factors of the other two, it is the required L. C. M. The factors of x^ — 1 are (x — l)(x* +, ^ + !)• But this product does not contain the factors of (x^ — 1), which are [x -f- l)(a5 — 1). Hence we must introduce the factor [x + 1), giving {x^ — l)(x -f- 1). as the L. C. M. of x* — 1 and x* — 1. Now as this product contains the third quantity it is the L. C. M. of the three. 2. Find the L. C. M. of {a + by, a^- — h\ {a — by, and a^ + Sa'b + 3b-'a + b\ Suggestions. — The last is (a -\- h)'^ which contains the factors of the 1st, but neither of the factors of the 3rd. Both factors of (a — 6)* must, therefore, be introduced, giving [a -\-b)'^{a — b}''^ as the L. G. M. of the 1st, 3rd and 4th. And as it contains both factors of a^ —h^, viz. ; (a + b)(a ~ 6), it is the L. C. M. of all. 3. Find the L. C. M. of {x^- — 4a2), {x + 2a)^ and {x — 2a)3. L. a M., {x^ — 4a-^)3. 4 Find the L. C. M. of x-^ + 2xy + y^ and x^ — xyK L. a if., {x + y){x^ — xy^). 5. Find the L. C. M. of l^a-ma;^, ^^a^bm'^, and Ib^x"^. L. a M., 10a^b^m-x\ 6. Find the L. C. M. of lGx^y\ 10a"-x, and ^Oa^x^. L. G. 31., 80a^x^y\ ScH. — In applying this rule, if the common factors of the two num- bers are not readily discerned, apply the method.of finding the H. C. D., in order to discover them. 100 ^ACTOEING. 7. Find the L. C. M. of x^ — 2ax'i + 4:a-x — %a\ x^-{- '2ax^ + 4a2^ + 8a^, and x"^ — 4a-. Model Solution. — The L. C. M. of tliese numbers must contain x^ — 2ax^ -\- 4:a^x — 8a'^ ; and as it is its own L. M., if it contains all the factors of x^ -\- 2ax^ -}- ia^x -\- 8a'^, it is the L. C. M. of these two polynomials. But as the common factors of these numbers, if they have any, are not readily discerned, we apply the method of H. C. D. and find that x^ -)- 4a^ is the H. C. D. of the two. Since, then, a;* — 2ax^ -j- 4ia^x — 8a^ contains the factor x' -j- 4m^ of the second number, it is only necessary to introduce the other factor in order to have the L. C. M. of the two. Now (x^ + 2ax2 + 4:a^x -f 8a^) 4- {x^ + 4a«) =x-\-2a. Hence (x^ — 2ax^ + 4m^x — 8a^){x + 2a) or X* — 16a'* is the L. C. M. of the first two numbers, since it contains all the factors of each, and no more. Now, to find whether x'* — 16a* is a multiple of the remaining number, x^ — 4a^ , or, if it is not, what factors must be introduced to make it so, we proceed in the same way as with the first two numbers. But our first step (or 124) shows us that X* — 16a* is a multiple of x* — 4a*. . • . x* — 16a'' is the L. C. M. of the three given numbers* 8. Find the L. C. M. oi Gx"- -~ x — 1 and 2x^ + Sot — 2. L. a M., (2a;2 4- 3^ — 2) (3^ + 1). 9. Find the L. C. M. of x^ — ^x^ + 23a; — 15 and x-^ — 8^7 + 7. L. a M., {x^ — dx^ + 2307 — 15)(«; _ 7) = o:^ — 16^:3 + 86x^ — llGx + 105. 10. Find the L. C. IVL of 2x — 1, 4:X^ — 1, and Ax^ -f 1. 11. Find the L. C. M. of x^ — x, x^ — 1, and x^ + 1. L. a 31., x{x<^ — l)=x-' — x. 12. Find the L. C. M. of x^ — 6x^ + llo; — 6, x^ — 9x^ + 2607 — 24, and x^ — 8x^ + Idx — 12. L. a J/., {x — l){x — 2){x — 3){x — 4.) = X* — 10o;3 + 35o7» — 5007 + 24. I SYNOPSIS. 101 Synopsis for Eeview, f Common. Factor. — Div isor. — Meas ure. a [h. Q* p' j- Distinction, i Common. MuUiple. \^^10.M^^ [Distinction. i Composite. Numbers. < Prime. ( Prime to each other. ^ Scholium. — How these terms are applied. Prop. 1. A monomial. Dem. Prop. 2. A poly, with mon. factors. Prop. 3. «2 +2ab-\-h'\ Dem. Prop. 4. a^ — h^. Dem. Prop. 5. One factor given. Dem. Prop. 6. (a"* ± h-^)^{a ±b). Dem. - Dem. Sch. f^ ft Prop. 7. A trinomial. When ? Prop. 8. Separating into parts. Form of quo- tient. Cor. Frac. and [ neg. exp'ts. How ? Dem. Distinction between G. C. D. and H. C. D. ri. Dem. Lemmas. \ \ ^JJ)^; [ 4. De')n. General Method. — Prob. 1. Rux-e. Dem. Prob. 2. Of more than 2 numbers. RuiiE. Dem. ^ i Distinction between Least and Lowest C. M. Q ■< Prob. Rule. Dem. ^ ( Scholium. By means of H. C. D. Test Questions. — What are the factors of a* — b^ 1 Of a* -}" 2a& + 62 ? Of a* — 2ah + ?>2 ? Of 1 — 2ic + .x^ ? Of Sa" + Ga^x + Sa^x"? Of x* + 1/2+2x2/? Of x^ — x — 12? Of m? Of a^h^yl Of x3 — y^ ? State the general rule for testing the divisibil- ity of the sum or difference of like powers. Prove one of the cases, as (a™ — 5'^' ) + (a — h). Distinction between H. C. D. and G. C. D. Between Lowest and Least C. M. Explain the process of finding each by factoring. Give the General Eule in each case, and its demon- stration. 102 FKACTIONS. OHAPTEE III. FRACTIONS. SUCTION I. Definitions and Fundamental Principles. 141, A. FractioUf in the literal notation, is to be considered as an indicated operation in Division. It is written, as in common arithmetic, with one number above another and a line between them. The number above the line, i. e., the dividend, is called the Numerator; and the number below the Hne, i. e., the divisor, is called the Denom- inator. ^, 2a — 5m.x^ Numerator or Dividend. Thus: p- 3c + 4oX'^ Denominator or Divisor. This expression means nothing more than (2a — 5mx^) -^ (3c + 4^^). Taken together numerator and denominator are called the Terms of the Fraction. 142. ScH. — In the literal notation it becomes impracticable to consider the denominator as indicating the number of equal parts into which unity is divided, and the numerator as indicating the number of those parts represented by the fraction, since the very genius of this notation requires that the letters be not restricted in their signification. Thus in -, it will not do to say, h represents the number of equal parts into which unity is divided, since the notation requires that whatever conception we take of these qiiantities should be sufficiently comprehensive to include all values. Hence b may be a mixed num- ber. Now suppose b = 4|. It is absurd to speak of unity as divided into 4| equal parts. DEFINITIONS. 103 143, The Value of a Fraction is the quotient of the numerator divided by the denominator. 14:4:, Cor. 1. — Since nuraerator is dividend and denom-' inator divisor, it follows from {100 , 101^ 102) that divid- ing or muUijjlying both terms of a fraction by the same quan- tity does not alter its value ; that multiplying or dividing the numerator multijMes or divides the value of the fraction ; and that multiplying or dividing the denominator divides or multi- plies the fraction. 146, CoE. 2. — A fraction is multiplied by its denominator by simply removing it. Thus, to multiply j by 4 it is only necessary to remove the denominator, which gives 3. This is the same as dividing the denominator by 4, for that gives -, or 3, as the product. So also to multiply - by ^ is to drop the denominator, x, which gives a. It is evident that a is ^ times as much without being divided by x as when a divided : that is, a is ^ times -. X 146, The terms Integer or Entire Number, Mixed Num- ber, Proper and Improper, are applied to literal numbers, but not with strict propriety. Thus, whether m + n is an integer, a mixed number, or a fraction, depends upon the values of m and n, which the genius of the literal notation requires to be understood as perfectly general, until some restriction is imposed. For convenience, we adopt the following definitions : 147, A number not having the fractional /orm is said to have the Integral Form ; as m -\- n, 2c^d — 8a~'^x + ^x^y\ 14S, A polynomial having part of its terms in the frac- tional and part in the integral form, is called a 3Iixed JS^uniber; as a — .r + ^^^ """ ^•^ _. am- 104 FEACTIONS. 14:9, A Proper Fraction, in the literal notation, is an expression wholly in the fractional form, and which cannot be expressed in the integral form without negative exponents. By calling such an expression a proper fraction, we do not assert anything with reference to its value as compared with unity. Thus - is a proper fraction, though it may be greater or less than unity. It may also be written ab ~\ 150, An Improper Fraction is an expression in the fractional form, but which can be expressed in the integral or mixed form without the use of negative expo- nents. Thus, --, — = 2a -— r : the former n — 6ab n — Sao of which is called an improper fraction. 151, A Simple Fraction is a single fraction with 2^ ■^^ both terms in the integral form. Thus — - — -— r— ,, and ^ 4y + 2cc? 2cm^2 • 1 i? i.- • — - are simple fractions. x — 'ly 152, A Comxyouncl Fraction is two or more fractions connected by the word of; but the expression is not generally applicable in the literal notation. Thus we may write t^ of -r with propriety, but not - of -, unless a 34 xj.*' on and h are integral, so that the fraction - may be considered 2 as representing equal parts of unity, as - does. If the o word of is considered as simply an equivalent for X, the notation is, of course, always admissible. But it is scarcely a simple equivalent. FUNDAMENTAL PRINCIPLES. 105 153. A Complex Fraction is a fraction having in one or both its terms an expression of the fractional form. c ma^ ^m — - —J- Thus , and are complex fractions. 5 - — a y lo4:, A fraction is in its Lowest Terms when there is no common integral factor in both its terms. lSo» The Loivest Common Denominator is the number of lowest degree, which can form the denomi- nator of several given fractions, giving equivalent fractions of the same values respectively, while the numerators re- tain the integral form. 156, Hedlictionf in mathematics, is changing the form of an expression without changing its value. SIGNS OF A FRACTION. lo7» In considering the signs of a fraction, we have to notice three things, viz.; the sign of the numerator, the sign of the denominator, and the sign before the fraction as a whole. This latter sign does not belong to either the nu- merator or denominator separately, but to the whole ex- pression. Thus, in the expression — — H^ — _, in the nu- ^ 2x-{- 4?/^ merator the sign of 4a is -{-, and of 5cd — . In the denom- inator, the sign of 2x is -\-, and of 4:y- -\- also. The sign of the fraction is, — . These are the signs of ope- ration. {SOf 33 f 34,) 158, The essential character of a fraction, as positive or negative, can only be determined when the essen- tial character of all the numbers entering into it is known. It may then be determined by principles already given. {86, 106,) 106 FEAGTIONS. EXAMPLES. 1. Is the fraction — — essentially positive, or negative, when a, m, x, and y are each negative ? Model Solution. — Since ( — a)^ =: a^, 4a^ is essentially positive. Since ( — m)( — x)= mx, the term 3mx, in itself, is positive, and the nu- merator becomes 4a ^ — (-f Smx), or 4,a^ — Smx {7S). Now, whether 4a^ — Smx gives a -f- or a — result, depends upon the numerical val- ues of a, m, and x. K 4a^ ^ Swicc, 4a^ — Smx is -\- ; but, if 4a'^<^ Smx, 4a^ — 3?nx is — . Again, since ( — x)'^ = — ic•^ the first term of the denominator, 2ic^, is essentially negative. And since ( — yY =2/^' the term 4?/^ is essentially positive and the denominator becomes — 2x"* + (4- 4^/2), or — 2x^ + 4ty^. Whether this is + or — , de- pends upon the relative values of x and y. K we suppose 4a2 >> Smx the numerator becomes -\-, and if 2x^ be greater than 4^/^ the denom- inator becomes — , and we have , which gives a positive result. 2a'^x- -f- 4:1/^ 2. What is the essential sie^n of —--, , ' , when a = ^ 3b"- — Ax — S, X = — 2, 2/ = — 4, and b = — 5? Ans., — . . ^x^y -}- CLX'^ 3. What is the essential sign of ~ , when oCL^X „ X = — 2, fl = — 1, and ?/ = 3 ? Ans., — . 4. What is the essential siefn of ^ , when ^ — 2y' — 3cx X =^ 4:, a = — 2, y= — 1, and c = Q^ Ans., -f . SECTION IL Eeductions. 159. There are five principal reductions required in operating with fractions, viz. ; To Lowest Terms, — From Improjjer Fractions to Integral or Mixed forms, — From In- tegral or Mixed Forms to Improper Fractions, — To forms having a Common Denominator, — and from the Comi)lex to the Simple Form, BEDUCTIONS. 107 I^VOh. !• To reduce a fraction to its loioest terms. R TILE. — Reject all common factors from both terms ; OR DFSmDE BOTH TERMS BY THEIR H. C. D. Dem. — Since the numerator is tlie dividend and the denominator the divisor, rejecting the same factors from each does not alter the value of the fraction. {100.) Having rejected all the common factors, oi*, , what is the same thing, the H. C. D. (which contains all the common factors), the fraction is in its lowest terms. (154:.) EXAMPLES. 1. Reduce — -— to its lowest terms. Model Solution. — Resolving the terms of the fraction into their prune factors, I have ; ; ^ = — <^ — -. 'da* -f- ija-^x -\- '6a^x^ ^JX- a(a..^>f=^)[a -\- x) Now, cancelling the common factors, 3, a, and a -\- x, which is divid- ing dividend and divisor by the same quantity and hence does not alter the value of the fi-action, I have — , or — ■ . Since a{a -f- X) a'^ -j- ax in this there is no factor common to numerator and denominator, it is in its lowest terms. {154:,) 2. Reduce — -, -— ^, — , and — to their Ibcx'^y^ 55i)m^xy= dba^x^ 15a^c^x lowest terms. Results (not given in order), - — , — , Za-x and — . Qox'^ omy~ bcx^y o r» n a — X ax -{- x^ . 2x^ — 16^ — 6 , ^. . 3. Reduce , — — , and — , to their a2 — x-^ ab-^ 4- t^x 3x^ — 24^ — 9 lowest terms. X 2 Results (not in order), — , -, and 6= ' 3' a-{-x . T> , n2 — 2n + 1 3a3 — Sab^ ^ x'^ — ¥x 4. Reduce , ■ and to 71^ — 1 ' 5ab + 56^ ' x-^ + 2bx + h^ their lowest terms. T, 7, ^- — bx n — 1 ,3^2 — Sab Results, — -, — — -, and . X -{- b n -]- 1 56 108 FRACTIONS — 2xy-\ xi — 2/" />»2 ^orxi -\- 11'^ 5. Reduce — to its lowest terms. X 7/ Result, ~. X -^ y 6. Reduce '- — — r — to its lowest terms. 3x^ + t.y+Sr BesulL^f^. X -i- y ^ ^ - 5w2 — lOn + 5 ^ .^ , 7. Reduce to its lowest terms. ^, 7n^' — 7 5(n — 1) ^''^^'^ tT^TTT)- 8. Reduce ^ to its lowest terms. y2 — 2xu 4- X , , Result, ^ ^ - ■ ^^ y — X 72,3 — 2n2 9. Reduce r to its lowest terms. n2 — 4/1 + 4 -r, 7 ^' Result^ -. n — 2 10. Reduce '- to its lowest terms. ^' — ^^ o^ + tf^a:2 4- X* Result. -~ . a" -H •^■- ScH. 1. — Since the H. C, D. is the product of all the common fac- tors {116), the above process is equivalent to dividing both terms of the fraction by their H. C. D. Whenever the common factors of the terms are not readily discernible, the process for finding their H. C. B. {137), niay be resorted to. -. ^ T^ 1 ^^ — «^ «^ — h^ x^ -\- y^ _ fls — 53 11. Reduce , r-, —, and r ^^ their x^ — a* a — b x^ — y^ a* — a-b'^ lowest terms. ^ , , , x^ — xy 4- y^ a^ -{- ab -\- b^ , 1 Results, a2 + a6 + b\ ' , — , and ■ . ' X — y a^ + a^b x*^a'' ,^ ^ , 6^2 _ 7^ _ 20 3a^2 _ lOax + 3a 12. Reduce — — — — , -7—. and 4a;3 — 27^; + 5 ba'x-^ — ba'^x — BOa^* 24^^ — 22^2 + 5 to their lowest terms. 48a;i -4- l^x^ — 15 „ ,^ 2j;2 — 1 3^ — 1 , 3^ + 4 Results, -, ■— — , and 4.07=^ + 3 10a + 5ax 2x'^ + 5x — 1 EEDUCTIONS. 109 ScH. 2. — The opposite process is sometimes serviceable, viz : the introduction of a factor into both terms of a fraction which will give it a more convenient form. 13. What factor will chanofe to ? Ans., X — 1. 14. What factor will chanefe ; to ? ° a — b a2 — ^2 Ans., a -\- h. 15. What factor will change — o^o^ ■ a :nT *o a» — lijx* a3 — - 2a-x + 4aa;2 — Sx^ Ans., a + 2x, a' - - 2a'^b + 4«52 — Sb' 16. What factor will chansfe — ; to [Note. — It requires no special ingenuity to solve such problems, since, if the factor does not readily appear, it can be found by divid- ing a term of one fraction by the corresponding term of the other. ] 160* I*rob, 2. To reduce a fraction from an im- proper to an integral or mixed form. MULE. — Perform the division indicated. (141,) Dem. — The operation is explained in the same manner as the cor- responding case in division. EXAMPLES. 1. Keduce — '■ — r to an intef^ral or mixed form. 2x Model Solution. — This being an indicated operation in Division, I have but to perform the division. Now, since the sum of the quo- tients is equal to the quotient of the sum, I have but to divide each term of lOax -\- c — l!> by 2x, or divide such as I can and indicate the division of the others, and add the results {112). Thus I find that i r= 5a + -^ . 2x 2x 110 FBACTIONS. ^ ^ _ 3a' — 9ac -\- X — Ga 5xy -{- ab -{- .x 2. lieduce , '■ , and Sa X 15^2 _ 4^ + 6 to integral or mixed forms. oa 4(2 — 6 ah X Besults, 3a , 5y + 1 H , and a — 3c — 2-\ — - . oa X "da ^ _ ^ x'' + 12.r + 18 a2 +4o6 + 462 +c j?^ — -^3 o. Keduce ; , , ^~ and X -{- 6 a -\- zh X -\- y to inteoral or mixed forms. a -i- X ^ 2^3 9 Besults, x^ — j;v + V^ — > ^ + 9 or 6 +2x ^, and a + 26 + ^ ^ + 3 a -\-2b , ^ ^ a3 — 1 ^5 — y5 a^ ^ 3^96 _(- :-3Qr?)2 ^ 53 4. Keduce — , --, — ; , and a — 1 X — y a-^ + 2a6 + 62 to intef^ral or mixed forms. x-^y Besult, a -{- h, X* -{- x^y + ^-y^ + ^y^ + 2/^ x^ — x"y + xy^ — y^, and a^ -f a + 1. 1S1» Cor. — By means of negative indices {exponents) any fraction can he expressed in the integral form. EXAMPLES. 5. Express — ; -; — - in the inteeral form. ^ m{a-\- b)~^ ^ Model Solution. ^- = {a 4- h) X - X . But- = m — ^, and -— =(a4-h)^. Hence — 1=: (a 4- h){a + &)'w-i = (a + 5)' m-\ ora-^m-^ -f 3a26w^-^4- 6. Express , — —, -, --, — r-^-— — , and — -^ ■ — -- in Sx^y {x -\- y)^ a~^x~^y^ {m — ?i)-^ tlie integral form. Besults, {x + 2/)~ ', a-x^y, m^x^ — m^nx'^ — m.n^x^ -f- n?x^ and 7 X 8~^a;~Y REDUCTIONS. 11 1 1G2. Prob, 3. — To reduce numbers from the integral or mixed to the fractional form. RULE. — MuiiTiPLY t:he integbal part by the given de- KOailNATOK, AND ANNEXING THE NUMERATOR OF THE ERAC- TIONAIi PART, IF ANY, WRITE THE SUM OVER THE GIVEN DENOMINATOR. Dem. — In the case of a number in the integral form, the process consists of multiplying the given number by the given denominator and indicating the division of the product by the same number, and hence is equivalent to multiplying and dividing by the same quantity, which does not change the value of the number. Tile same is true as far as relates to the integral part of a mixed form, after which the two fractional parts are to be added together. As they have the same di- visors, the dividends can be added upon the principle that the sum of the quotients equals the quotient of the sum {103). EXAMPLES. 1. Keduce 2a — x- + — to a fractional form. a — X Model Solution. — Multiplying 2a — a^ by a — x, I have 2a* — ax* — 2aa; -j- x^, which divided by a — x, of course equals 2a — a* ; or 2a* — ax* —2ax-\-x^ „ „ , 3ax — 4a* 2a — ic* = ! . .-. 2a— x* H = a — X 'a — X 2a* — ax* — 2ax + x^, 3ax — 4a* _ . h . But, as the sum of these two a — X a — X quotients equals the quotient of the sum, I have, after uniting eim- ., ^ x^ — ax* 4- ax — 2a^ liar terms, ■ . 62 2. Eeduce a — h 4- to the form of a fraction. a-\-b ^ ^ a^ Mesult, a + b 1x 3. Beduce 1 + to the form of a fraction. y — ^ y 4. X Result, y-JL±, y — X 4. Reduce a-\-b — to the form of a fraction. « — ^ b b Result, r or b b — a 112 FRACTIONS. 5. Reduce ^ + 2 + —r. — to the form of a fraction. Mesuit, -. X — 2 2x 5 6. Reduce ^x — to the form of a fraction. Result, X A- \ 7. Reduce x -\- 1 -\- —^— to the form of a fraction. Result, la^^hc 4- 9< 8. Reduce 2a — 36 + 4c + 3 Q. X 2a"^bc + 9a62c — 12ahc^ ^«^^ Result, fa. ^ ^ , 2abx — 2b^x , , .. 9. Reduce x ■. 7 — to a fraction. a + b x'i v^ + 7 7 10. Reduce a: + y — Result, . ^ X — y y — X x^ — ^x'iy + ^xy^ — y^ 11. Reduce x'^ + 2xy + 1/2 ^ , ' • X -\-y 12. Reduce -^ {a^x^ + oPx). Result, - a'i — x'^ d' — ^ 13. Reduce 3a — 9 —7—. Result, a + 3 ' ' aH-3 163. JProb, 4, To reduce fractions having different denominators to equivalent fractions liaving a common de* nominator. RULE. — Multiply both terms of each fraction by the DENOMINATORS OF ALL THE OTHER FRACTIONS. Dem. — This gives a common denominator, because each denomina- tor is the product of all the denominators of the several fractions. REDUCTIONS. 113 The value of any one of the fractions is not changed, because both nu- merator and denominator are multiplied by the same number {100). EXAMPLES. X 2 b 3a X 1. Eeduce the fractions -? j, and 7 to equivalent y a + b a — fractions having a common denominator. Model Solution. — Multiplying both terms of the fraction — by a -f & and a — 5, or by a* — h^, I have -5 — — which has the same value as — , since the numerator and denominator have been multiplied by the same number. In like manner mtdtipljdng both terms of 6 , , , T , 2av — ahiJ — 2by 4- b^v .-. - by V and a — b, I have — : -^ —- -, the value of a A- b a^v — b'-y 2-6 ■which is the same as ■ , since, etc. Finally, multiplying both „ 3a — X , , , , ^ , 3a^ y — axy 4- ^aby — bxy terms of ,-? by y and a + ^. I bave r^ r^ '-, a — b^ a^y — b^y which has the same value as — since, etc. These fractions have a — 6 the common denominator a^y — b'^^y as in each case the new denom- inator is the product of all the old ones. 2. Reduce ■^, ■—, J^ and — to forms having a C. D. 2?/ 2c Ix ?i2 *= 3x Queries. — By what are both tei-ms of --— to* be multiplied ? By what both terms of jr— ? By what both terms of — ^? 42c?i-j:2 'JOhn^xy 12cn-y^ , 2^cmxy ' 2Scn'Xy' 28cn^xy* 28cn^xy 28cn^xy' rp rp _1_ 1 1 rj* 3. Reduce -, — - — , and to forms having a C. D. ij O J- ~t~ X ^ , 5a; + 5^2 3 4- 6jt +8a;2 , 15 — Ihx Results, -— -— -, ^ ^ , ., ^ , and. -, _ , -, ., . 15 + 15j7 15 + Ibx 15 + 15j; 4. Reduce r, and r to forms having a C. D. a + 6 a — h ^ ^^ a^' — ab . ab + h^ JtiesuUs, —, and 7-- a* — b- a- — 0' 114 FRACTIONS. ScH. — Practically, this method consists in multiplying all the denom- inators together for a new denominator, and each numerator into all the denominators except its own for a new numerator. But it is much better to repeat the rule as given above, and let that be the form of conception, as it keeps the principle constantly before the mind. 164, Cor. To reduce fractions to equivalent ones having the Loioest Common Denominator , find the L. CM. of all the denominators for the netv denominator. Then multiply both terms of each fraction hy the quotient of that L. CM. divided hy the denominator of that fraction. Dem.— The purpose in getting the L. C. M. is to get the lowest num- ber which can be divided by each of the denominators. That the pro- cess does not change the value of the fractions is evident from {100), the same as under the general rule. EXAMPLES. a a^ a^ 5. Eeduce ? r-> and 7- ^> to equivalent frac 1 — a (1 — «)2 (1 — ay ^ tions haA' ing the L. C. D. MODEL SOLUTION. Opebation.— The L. C. M. of 1 — a, (1 — a)«, and (1 — a)^ ip 3 a X (1 — «)" ^ a — 2a^ + a' a' X (1 — a) ^ ^^ * (1— a)X(i— aj^ 1 — 3a + 3a^ — a^'(l— a)2x(l— a) a^ — a^ _ a* a* and ^ ^nr 1 — '6a -{-'6a^ — a-*' (i — a;-^ 1— ;ia-}-3a — a Explanation. — ^By inspection I observe that (1 — a)^ is the L. C. M. of the denominators, since it is the lowest number which contains it- self, and it also contains each of the other denominators. Now, to make the denominator of , (1 — a)^, I must multiply it by (1 — a)^ 4- (1 — a) ; i. e., by (1 — a)'. But to preserve the value of the fraction, I must multiply the numerator by the same quantity. Thus " . a(l-°l' « -2a^+a-' .,,._,,, 1 — a (1 — a)'* 1 — 3a-\-'6a'^ — a-* 6. Eeduce —> —to forms having the L. 0. D. x~y x + y x^—xif— xHi + ?/' , x^+ xHj + xip -f y' R,sulis, ;^._ y/ and ^^^i -. REDUCTIONS. 115 7. Reduce > — > > to forms having the L. C. D. x+y x^ -\- y^ X + y „ -^ a i{x^-xy+y^) , c{x-'— xy-\-y^) ResuUsj J - — T-r—r-^^ and -^ - ^ / ' , x^+ y^ x^+ y^ x^ -h y^ 8. Eeduce nm, ? ? to forms having the L. C. 1). m + n 7)1 — u Suggestion. — Regard mii as — -. „ ,, m^7i — mn^ (m — nY , (7^ + ny Results f ?^^- f-jand-^^— '-. 9. Eeduce > 1? - — ^—^ to forms having the L. C. D. 1— :c 1 — a;^ 1—x^ The L. C. D. is 1 + x—x^ — x*. ] 0. Reduce r' and 7 ^ to forms having the L. C. D. a^—b'' {a—b}'^ 11. Reduce ? -^ ? ^to forms having the L.C.D. m-\-7i m^+n^ m + n 9 Q 2.7; 3 12. Reduce — ? ^ -? 'and ; to forms having the L.C.D. x 2x—l 4^-2 _i lOo, J^vob. o. To Eeduce Complex Fractions to the form of Simple Fractions. R ULE. MULTIPIiY NUMERATOK AND DENOMINATOR OF THE COM- PLEX FRACTION BY THE PRODUCT OF Alil. THE DENOjMINATORS OF THE PARTIAL FRACTIONS FOUND IN THEM : OR, MULTIPLY BY THE L. C. M. OF. THE DENOanNATORS OF THE PARTIAL FRACTIONS.* Dem. — TMs process removes the partial denominators, since each fraction is multiplied by its own denominator, at least, and this is done by dropping the denominator. It does not alter the value of the fraction, since it is multiplying di%'idend and divisor by the same quan- tity. * The pupil is supposed to have obtained suflacient knowledge of fractions in -ionimon arithmetic to perform these operations. 116 FRACTIONS. EXAMPLES. 1. Beduce — to a simple fractional form. 5a2 MODEL SOLUTION. 2x OPEBATION. = % X 5.' X 36^ ^^^,^ Explanation. — In order to free tlie numerator of its denominator, 2ic 36*, I multiply the numerator -—; by 36'^ ; but, in order tliat this may ob not change the value of the fraction, I also multiply the denominator 4w by the same. In like manner to free the denominator -^ of its de- nominator, I multiply it by 5a* ; but, in order that this may not change the value of the fraction, I also multiply the numerator by the same. 2r |^X5a* X36*^ Indicating these operations I have . To multiply |,X5a^X36^ 2x_ 36^ ba^ gives for the new numerator lOa^x. So, also, I obtain the new denominator by dropping 5a* and multiplying hj by 36*, getting there- 10a* X by 126* 2/. Therefore the simple fraction is ^ which reduced to QC to a simple form. Result, — 5-^ by 36* I drop its denominator and have 2x, which multiplied by low( . 5a*a; jst terms is 77^5-. 66*?/ 2. Beduce m 3. Beduce «+' 4. Beduce cm cm m to a simple form. Result, — , ^ » acn -\- hn to a simple form. Result, ^,^^^ ^^^- ADDITION. 117 5. Reduce — '■ — to a simple form. y 5 3(1 6. Reduce — -r \ — to a simple form. 10 + -i^o; a X b Tj 7. Reduce to a simple form. a X m n b — a a + 8. Reduce — to a simple form. ab — a2 SECTION III, Addition. 166, I^rob, To add Fractions. RULE. — Reduce THEM to a common denominatok, if they HAVE not such A FOEM, AND THEN ADD THE NUMEKATOKS, AND WKITE THE SUM OVEK THE COMMON DENOMINATOE. Dem. — The reduction of tlie several fractions to a common denom- inator, if they have not one, does not alter their values (163), and hence does not alter the sum. Then, when they have a common de- nominator (divisor), the sum of the several quotients is equal to the quotient of the sum of the several dividends divided by the common divisor, or denominator {103). EXAMPLES. «„,.,, ^1 -{- X 1 -\- X^ ,1 + 0^3 1. What is the sum of _ , , and ■ . 1 — X 1 — x^ 1 — x^ 118 FRACTIONS. MODEL SOLUTION. Opeeation.— The L. C. M. of 1 — x, 1 — x'^ and 1 — x^ is (1 — x^) X (1 + x) = 1 + X — x3 _ a;4. (1 4- a;) X (1 + 2x + 2x^ + x^) _ 1 -f 3x 4- 4x2 + 3x3 4. 3.4 (1 — X) X (1 4- 2ic -f 2x^ + x3) ~ 1 -j- X — x^ — X' ~ (1 -{- X-') X (1 + X + x2) _ 1 + X -f 2x3 -f x3 -)- x^ (1 — x2) X (1 + X + x^) ~ Hf X — x^ — X"* • (1 + x3) X(l 4- x) 1 4- X + x" + x^ (1— x3) X (1 -\-x) "^ 1-l-x — x3 — X4' 1 .f X , 1 4- a52 , 1 + x3 _ 1 -I- 3x 4- 4x2 -1-3x3 +x4 X^ 1 — X3 1 4 * — ^^ — ^'^ 1 4-x 3 -j_ 5a; 4. 6x2 _|_ 5a;3 _j_ 3a;4 1 -f- X 4 2x2 4- x3 4 x^ 1 _}_ a; 4- x3 -f- x-« ^ 1 4- X — X* — x^ '' 1 -|- X — x3 — x-»^ 1 4" ic — X-' — X* Explanation. — Explain the reduction to a common denominator as under {163) unless that is already sufficiently familiar. Having reduced the fractions to the L. C. D. I find (read A). Now since the sum of these quotients is equal to the quotient of the sum of the several dividends, or numerators, divided by the common divisor, or denominator, {103) I add the numerators and write the sum ,, _ .,....' 345x46x'45x3+3ic* over the common denominator, which gives — —- — r ^ h " "^ 1 -f- X — X-* — X* for the sum of - — ■ — , - — ■ — 7, and - — ' — — . 1 — xl — x^ 1 — X"* ^.-_a7c — a . c -\- a „ 6^ 4 7c — a 2. Add — , -— — , and -j — . Sum, — . a c 6 a a^h^ ch + h^ -^ a'^ 3. Add -, — r, -, and -r-. Sum, ■ — r . h ah a ¥ a¥ ^•^^^3'i'12'i8'6'"^^9- ^^^''^• ^.^^fl — 3 ,54a ^ 3a — 1 5. Add — - — and — - — . Sum, — - — . 6. Add , , and Sum, 14 a' 1 — a' 1+a '1 — a ADDITION. 1 19 7. Add and . Sum, -— . 1 — a* 1 + a^ . ' 1 — a2 8. Add—— — and . Sum, 1 +x 1 — x ' 1 — X-' 9. Add and Sum, as — a"-b — ab-^ + 63* 10. Add — ^— , -4- and — i— . Sum, ■ — — —^. x-^ — y^ 107 » Cor. — Expressions in the mixed form may either be reduced to the improper form and then added, or the integral parts may be added into one sum, and the fractional into an- other, and these results added. J. 2 3jc 4- 4 1. Add Ix + — — - and Hx -\ . O uX FIRST FORM OF OPEBATION. iC — 2 22x — 2 7x 3a; 4- 4 _ 40x-^ 4- 3x + 4 ' 5x 5x (22r — 2) X 5a; _ 110 x'^ — lOx 3 X 5x ~~ lox (40a;2 -I- 3x 4- 4) X 3 _ 120x^ -\-'dx-J^n 5x X^ ~~ 15x 110x2 _ lOx 120x-^ -f 9x + 12 _ 230xg — x + 12 15x 15x 15x SECONT) FORM OF OPERATION. 7x 4- 8x = 15x (x — 2) X 5x _ 5x2 _ lOa; 3 X 5x ~ 15x (*3x-|-4> X 3 _ 9 x + 12 5x X~3 ~ 15x 120 FRACTIONS. 9^+12 , , 5^2 _ j; 4. 12 + -^- = 1'^ + — ^x — • EXPLA.NATION. — Since the sum of several numbers is the same in whatever order their parts are added, I take the integral parts first. Adding Ix and 8ic I have locc. Reducing the fractions to a common . , X — 2, 5^2 — lOx , 3x + 4, 9ic4-12 denominator, — ~ — becomes — -— , and — ~ — becomes —-1 . 3 15x 5a; 15x 5.1-2 .T 4- 12 Adding these I have — which added to 15x, the sum of 15x 5a;2 x 4- 12 the integral parts, gives for the entire sum 15x -| --z — — — . Idx 2. Add ^Xy 3^ -}- — -, and x -\- -^. Sum, 6x + — — . 5 9 45 « . -, -. ^^- . . 2a^ 3. Add a ^- to 5 + . ^ ^ ^ , 2abx — 3cj;2 bum, a 4- -f 7 . DC 4. Add Ix + - — - — , and dx . 6 ox Sum, Wx + - 15a; 5. Add 6a; —, — 8a;, and 3a; — -. Sum,x- --. «»i-i« • 0, 4- b ^ a — b - . 6. Add 3ma; -, and 7 — 2ma; + 4. a — a + 6 Sum, 4 + mx fl2 7. Add 6x^y^ — 3a; ^-r-^, and 5a; — 2x%^' -f x-i — yi Sum, 407^7/^ -)- 2a; + 07^ ^■^ r pq ' pr ' qr 8. Add ^^— ^, "^ ^, and ^^ -. Sum, 0. SUBTBACTION. 121 05 3a - lax ' 4a 9. Add , , and . Sum, a — X a^x a^ — x'^ a^x -„.,, a^h 6 + c c + a 10. Add — -, -7 —, and (6— c)(c— a)' (c— a)(a— 6) (a — b){b — c) Sum, 0. SuG. — The L. C. D. is {a — h)(b — c){c — a), since this contains all the factors of each denominator, and no more. The terms of the 1st fraction must be multipUed by a — 6, of the 2nd, by 6 — c, and of the 3rd, by c — a. I SUCTION lY. Subtraction. ^""^^^^ 168, JPvoh* To subtract fractions, RULE. — KhDUCE THE FRACTIONS TO A COMMON DENOMINATOR, IP THEY HAVE NOT THAT FORM, AND SUBTRACT THE NUMERATOR OP THE SUBTRAHEND FROM THE NUMERATOR OF THE MINUEND, AND PLACE THE REMAINDER OVER THE COMMON DENOMINATOR. Dem. — The value of the fractions not being altered by reducing them to a common denominator, their difference is not altered. After this reduction, we have the difference of two quotients arising from divid- ing two numbers (the numerators) by the same divisor (the common denominator). But this is the same as the quotient arising from divid- ing the difference between the numbers by the common divisor(i04). EXAMPLES. 1. From subtract '-. X — y X -\- y MODEL SOLUTION. OPEKATION. {X — y){x -\- y) ^ X^ — T/2 (a; 4- ?/) X (x + y) x'- 4- 2xy + y' {x — y}X cx + y) (x — y) X (X — y) x:^ — y^ x2 _ 2xy + it/2 (X 4- y) X (X — y) - {x' - 2xy -f- y^) = x-\-y x — y x:' —y^ = ^xy 4.xy (x' 4 2xy + y') x — y x-\-y 122 FRACTIONS. Explanation. — The L. C. M. of x — y and x -{-y is, tlaeir product, since they have no common factor. Hence x^ — y^ is the L. C. D. To X I XI reduce -■ ' to this denominator I multiply both its terms by a; -j- ^Z' 3.2 _j_ 2a;w -4- w* which gives ^ — 2 • ^^ ^i^® manner multiplying both terms X y 2« 71 0*2 2icw -4— ?y^ '' of — — ^ by X — y, I have — '-^ — . I have now to subtract X -j- 2/ x' —y' ±ZpLplf,om^^±p-pl. Since the difference of the ^ — y ^ — y quotients of two numbers divided by the same number, is the same as the quotient arising from dividing the difference between those num- bers by the common divisor, I take the difference of the numerators (the quantities to be divided) which is ixy, and dividing it by x* — y^ I have — — ^-^, for the remainder of— ^i^ less — — ^. x^ — y^ x — y x-^y __ 1+^^,1 X „ ., 4:X 2. i rom take . Eemainder, 1 — X 1 -\- X 3. From take . Remainder, a — X a -\- X X ^ _i_ 3 4. From — - take . Remainder, X — 6 X 5. From 3^ take '-. Remainder, 5 o 3x Stjg. — Regard 3x as — . 6. From 9v take — - — -. Remainder, — =— . ^8 8 7. From r take — -. Rem.. — ; ■; . a — b a' — 2a6 + 62 ' {a— by 0-1:1 1.1 2 — ^2 _ 3^ 8. From take '1 — x^ 2x a2 — x^ 9 x;^ ■ — 3^' 3x — 3a X + 4: x^ -j-lOx -\- 24' (^4- 2)2 Remainder, x^ + 10^7 + 24 * ^ ^ 3x-{-2 ^ . lax— 10a 9. From ■ — take . ^ ^' 4(3 — x) Remainder, — -. SUBTRACTION. 123 10. Combine the foUowmsf fractions -:: f- ^ 2 — ^ 2 + a; ^ 16x — x^ ^ , 1 Besult, x^ — 4: ' x-\-2'^ 11. Combine — 7—; + -— }- a{a— b){a— c) ^ h{b — c){h — a) ^ — -, —. Result, — -. c{c — a){c — h) abc ^- ^ , . 3a — 4& 2a — h — c 15a — 4c 12. Combine + ^^ a — 46 ^ ,^ 81a — 4^ BesulL 21 ^*^"^^''' 84 13. Combine + ; . a -\- h a-' — 62 a:^ -\- b'i Result, 14. Combine 15. Combine a* — h* 3 7 4 — 2007 1 —2x l-^r 2x 4x^—1 Result, 0. 5 1 24 2{x + 1) 10(a;— 1) ~ 5(2;^ + 3)* 2^ 3 ResuU, ' (^2 _ l)(2;r + 3)' 16. Combine \- ^ . x^—yi {x-^yY {x~yY _ . x-2 — 4^V — y2 ResuU, -— ~. {x^ — y-^Y 169, CoR. — Mixed numbers may be subtracted by annex- ing the subtrahend with its signs changed, to the Tninuend, and then combining the terms as much as may be desired. The reason for the change of signs is the same as in whole numbers. {77.) EXAMPLES. 17. From 3x + ^±^ take x — ^LZlL., ^ ^ 2ar Remainder f 2x H . 124 FRACTIONS. -lo-ni 4a — b . , , „ 3a — 2b 18. From x — subtract 7x . 2 S Bemainder, — , — 6^ — a. b - „ Ti o X 1 3j: + 12a ^ . , 3a — 3^ 19. From Sa take . Bemainder, 5 5 20. From 2x + ^^=1^ take 3x — "^fL+l, ^ ^ ^ . , 16^ + 23 liemainder, . I SUCTION V, k Multiplication, 170, I^voh* 1, To multiply afi^action by an integer. B VLE. — Multiply the numerator or divide the denomi- nator. Dem. — Since numerator is dividend and denominator divisor, and the value of the fraction is the quotient, this rule is a direct conse- quence of {101, 102). \ EXAMPLES. . , m — n - 1. Multiply — by m + n. MODEL SOLIJTIOW. m — n , , ^ m OPEBATION. — X {m -\- n) dxy dxy Explanation. — Since m — n is divided by Sxy if I multiply it by / m -{-n 9,nd then divide (or indicate the division), I shall have m -\yn times as large a quotient as at first. But the value of a fraction is ifne quotient of the numerator divided by the denominator. Hence Bhul- »YI /vy 771 71 '^ / tiplying the numerator of — by m -f n, I have — wh^ch is m — n j m -\-n times — ;r . - ' MULTIPLICATION. 12i 2. Multiply ^^ by 3a. MODEL SOLUTION. 2mx „ 2ma; Explanation. — Since 2mx is to be divided by Sa^b^ if I divide the divisor by 3a, thus making it 3a times as small, it will go into the dividend 3a times as many times as before. Hence — r-^ is 3a times 2mx , Za'h^ 3. Multiply Y hj X -{- y. dC 4 Multiply ^^^ _^^ hjx — y. 5. Multiply by a- + 1. Prod., . Suggestion, a^ — a z= a{a* — 1) = a{a^ — !)(«'* + !)• 6. Multiply — ^-±i^- by (a + yy. Suggestions. — Multiply by one of the factors of (a -\-yY byreject- (a -f- VY immator, gi\T.ng ■ — - — ing it from the denominator, giving ^ ^^ ^^ ^ , and this product by the other factor, giving am — my- 3 7. Multiply by a^ — x^ Prod., 3a + 3^. a — X 8. Multiply -^^^ by 2a + 2y. Prod., ^^'. 9. Multiply 5^—^ by 36. Prod., 3a — 2y. ScH. — A fraction is multipHed by its own denominator by removing 3a— 2,v it In the last example it is evident that 3a — 2?/ is 36 times 3a^ 10. Multiply by a: — y. 3b 126 FRACTIONS. 11. Multiply — ^— by 2ax{x — y) . Prod., Qa^x"^. X y 3c 12. Multiply by x'^ — 1. X — X 13. Multiply ^ ^ by x^ — 2xy + y\ X y 17 1» J*TOb. 2, To multiply by a fraction. RULE. — ^Multiply by the numerator and divide by the DENOMINATOR.* Dem. — Let it be required to multiply m, which is either an integer or a fraction, by — . 1st. Suppose a and h are both integers. Multiplying m by a gives a product 6 times too large, since we were to multiply by only a 6th part of a ; hence we divide the product, am, by h, and have -r-. 2nd. When either a or h, or both, are fractions. Let c be the fac- tor by which numerator and denominator of - must be multiplied to make — a simple fraction {165). Then will — be a simple fraction, i. e., ac and he are each integral ; and the multiplication is effected as in Case 1st, giving -yr- "^^^^ reduced by dividing both terms by c gives — . Hence we see that in any case, to multiply by a fraction, we have only to multiply the multipHcand by the numerator of the multiplier, and divide this product by the denominator. It is also to be observed that this reasoning apphes equally well whether the mul- tiplicand is integral or fractional. EXAMPLES. 1. Multiply -^ by 7. __ _^ . — ^ — ■ I * It is assumed that the pupil knows how to divide a fraction by an integer, from his study of arithmetic. Nevertheless the problem will be introduced hereafter for the purpose of familiarizing the pupil with the literal operations. MULTIPLICATION. 127 MODEL SOLUTION. OPEEATION. -^ X (X — y) a;3 — y3 -^ x^-j-xy+y . (0—6)2 a_^ Again,—— ■ — - — (a — 6) : ^^-\-^y-^y'' ' ic^ + a-t/ + y2- (a — 6)2 X — y a — 6 x3 — y3 a — 6 X- -j- xy -f- y^ Explanation. — In order to multiply — ; 7- hx ' f, Ifirstimil° ^ -" x-^ —y^ "^ a~b tiply by X — y. This is effected by dividing the denominator, x^ — y^, hjx — y, {102), and gives ^ — — - — j. But, since this multipHer was to be divided by a — 6, the product now obtained must be di- vided by the same. Dividing —, — ; ■ — - by a — 6 by dividing the ^ x' -\-xy + y^ ^ •" ^ numerator {101), I have for the complete product ^ — - — j. x -f- ^y -{- y 2. Multiply— by-. Prod., — . « ,.,.., 2a2 11^3^ 22^7-2 3. Multiply .— — by — --. Prod., . 4. Multiply - -y by 3- . Prod., - —. 5. Mulhply - 3^- by - y,-. Prorf.. ^^. ScH, — When there are no common factors in the numerators and denominators of the fractions to be multipUed together, the process consists simplj^ in multiplying numerators together for a new nume- rator and denominators for a new denominator, which process is equivalent to multiphdng hy the numerator of the multipHer and divid- ing by the denominator. As it is immaterial, as far as the result is concerned, which of these operations is performed first, that one should be which is most convenient when there is any choice. 2771 6. Multiply 4^2 — 9^2 by 2j7 — 3?/ 128 FRACTIONS. Suggestion. — In this case divide first, obtaining 2x -\- dy. Multi' ply this by 2m. 7. Multiply- by ^^. 8. Multiply —- by ^ Prod. J 4:mx + Gm^/. Proc?. x^ -\~ ax ■' a2 + ac ProtZ., X* — b* b^'C + bc<^ ' Frod., a'^ -f 6^ / . 7 \ -• 6c 6 + c 9. Multiply ^ by — ; — -. ^.^^., . , ,. . SuG. — In the last example both operations are performed upon the denominator of the multipUcand. ,«-,,., 3;r2 — 5x ^ la „ , 3a,'r — 5a 10. Multiply -^^- by ^^^-^. Prod. j;^^-. 11. Multiply-^ "^yai^;;)- ^'•"'■'i8- 12. Multiply -^^ by ^-^. Prod., ——^ 13. Multiply 1^ by 1^. Prod, il^. 14 Multiply -^ by — . Prod., — — „. 15. MiUtiply - by '-. Prod., — -„. 2/ **' y 16. Multiply «*- - >y -6^ -bi' Prod., ai + hi «*- J -bi' 3c ~^x^ ^c^x 17. Multiply by -^ and write the result 5a^y~'^ la-^y~^ _, , Qicx*y without negative exponents. Frod., -or~7- MULTIPLICATION. 129 18. Multiply together, — , — ( ^ + vY ^^^ ___ Sm _ , 3m Prod., 19. Multiply ,^ ■ by — —- — ^~-. Prod., ■ — . X + '6 172* Cor. — To multiply mixed numbers, Jii^st reduce them to improper fractions. 20. Multiply 1 — ^ by - — 2. ^ , Ixy — 2^2 — Qy2 x(7y—2x) Prod., —^ ^ or -1-^.^— -Z_ 2. 3y- Sy- 21. Multiply together — yipy. ^ ~ ^" > and 1 ^^ ^^^. ?/- 1 Prod., X ^2 d X (V* X* 22. Multiply a - - by - + ^. Prod., ^^. 23. Multiply 1 - ^-^by2+^-. Prod, ^"'^ jr2 _ ^2 24. Multiply a;^ _ j; + 1 by — + - + 1. Prod., ^= + 1 + — . OPEIUlTION. 072 ^ + 1 X x^ J7 — 1+ - X x^ — ,r 4- 1 x^ +1 H , Product. 130 TRACTIONS. SECTION VL Division. 173, J*vob, 1, To divide afrmtion hy an integer. RULE. — Divide the numebator or multiply the de- nominator. Dem. — Since numerator is dividend, and denominator divisor, and the value of the fraction the quotient, this rule is a direct consequence oi {101, 102,) examples. 1. Divide -r r- by a — x. zmx — 1 Solution. — Since the value of this fraction is the quotient of 3(a* — a*) divided by 2mx — 1, if I divide the dividend, 3(a2 — a^), by a — X, I divide the fraction. Hence — • -^ (a — a) = 3a 4- 3.x 2mx — 1* 2. Divide ^"^ ~" t^^ by a + 6. a — 6 Solution. — Since the value of this fraction is the quotient of 3m — 4xT/ divided by a — &, if I multiply the divisor, a — b, hj a •■{- h, I divide the fraction. Hence — '■ ^ -^ (a 4- b) = — 3 :^ a — 6 a'* — 6* 3. Divide ?J^ by 3^. . ^. ., 15a2 — 15j;2 , ^, , ^ 3(a — x) 4 Divide by 5{a + x). QuoL, -. X — y X — y x^ — 1 5. Divide hj x — 1. X-' + 6 -^ DIVISION. 131 6. Divide — m x — y. X -{-y 7. Divide — by ^ + y. X -i-y 8. Divide by x^ + 4. x^ — 4 9. Divide ~ hj x^ — yK X + y SuG. (X* — V*) = (a; 4- y)ix — y). Divide by the factor x — yhy . . a* + x^y 4- x^y^ 4- X2/3 ^_ ^4 dividing the numerator, giving ^ — '—• r^ow X -\- y divide this quotient by the other factor, « -|- y, by multiplying the de- x^ 4- x^V + ^""V^ + xv^ + x^ VioTnTnfl.rn'r onvrnc ' nominator, giving ' ^x+y)^ 10. Divide ^^y\ by 5 mn^ A QuoL, ^ ^ J- -4 ^D?W , ^ ^. . , a2 — 2a6 + &2 11. Divide -r by a" — ft'. 12. Divide ?^g^by42(a-6)=. ' 174. Proh. 2, To divide by a fraction. RULE. — Divide by the numerator and multiply the QUOTIENT BY THE DENOMINATOR. Or, WHAT IS THE SAME THING, INVERT THE TERMS OF THE DIVISOR AND PROCEED AS IN MUL- TIPLICATION. Dem. — The correctness of the first process appears from the fact that division is the reverse of multiphcation, and, hence, as we multiply by the numerator and divide by the denominator in order to multiply by a fraction, to divide by one we must divide by the numerator and) multiply by the denominator. The process of inverting the divisor and then multiplying by it is seen to be the same as the other, since this multipUes the dividend by the denominator of the divisor and divides by the numerator. 132 FRACTIONS. Again, this process may be demonstrated thus : Inverting the divisor shows how many times it is contained in 1. Then if the given divisor is contained so many times in 1, it will be contained in 5, 5 times as many times ; in f , f as many times ; in ax^, ax* times as many times ; or in any dividend as many times the number of times it is contained in 1, as is expressed by that dividend, whether it be integral, fractional or mixed. (The author prefers this demonstration.) ScH. 1. — Since to multiply one fraction by another we may multiply the numerators together for the numerator and the denominators for the denominator, and since division is the reverse, we may perform division by dividing the numerator of the dividend by the numerator of the divisor, and the denominator of the dividend by the denomina- tor of the divisor. This method will coincide with the others when they are worked by performing the operations by division as far as practicable, and this is worked by performing the multipHcations equivalent to the divisions when the latter are not practicable. EXAMPLES. 1. Divide ^^' by -^ a;3 -f 2/^ ^ + MODEL. SOLUTIONS. Operation by the First Method. 2yg _ 2y a.3 _j_ yi — y—^T^Ty 3' a.3_f_y3 Xix + y)_ ^, _^y_^y. Explanation. — I first divide -— -^ — - by y, by dividing the numera- * + y tor {101). But the given divisor is y divided by a; -f- y ; and as divid- ing the divisor multiplies the quotient, {102), I must multiply this 2v quotient, -j-~— j by x 4- !/• Performing this by dividing the denom- X -\- y inator, I have for the true quotient ^^ \ f III. X' —xy-^y-' Operation by the Second Method. 2y^ . y _ 2.v2 x -^ y 2.y ic -\- y* ' X '{-y x^ -f- y^ y ^^ — ^ DIVISION. 133 Explanation. — By inverting the divisor and indicating the multipli- cation of the dividend by it, I indicate that the dividend —t-~ is to x-^ + y^ be multiplied by x -[- V and divided by y, which are the operations required. In this instance I perform the multiphcation hy x -{- y, by dividing the denominator, and the division by y, by dividing the numerator. The operation by the third method is of the same form as the last. Explanation by the Third Method. I am to find how many times — - — is contained in , Now x + y ic^ + y' y • . . x 4- V . . 1 — ; — is contained in 1, '- times, since y goes into 1, — times, and ^-hy y ^^ y —^, — goes X 4- V times -, or times. Hence if — - — c:oes into 1 ^+y "^ y y x+y ^ times, it goes into — -^ -, —-^ times — —^ times, which y x' + y' X' -{-y^ y 2v I find by the rules of multiphcation to be . ic- — ^*/ + y' 2. Divide by . QuoL, -. (a + ^'Y' ' a^ — x^ ^ x{^a + x) 3. Divide ^^-^ by -5"A_ Q^ot, 2fa-by b{a -\- b) *" a-' — 6^ -^02 4 Divide ■ by — - — ■ x-^ — y^ -^ X-' -^ xy + y^ Quot., ^±i:. x — y 5. Divide 5^ by ^. ■ Quot.,^-^. a-' — zab 4- b'^ '' a — b ^ a 134 FRACTIONS. 9. Divide —-7 f by --. 10. Divide x 2x hjx 2x x — 3 '*' X — 3' Suggestion. — Reduce to improper fractions. QuoL, 4(a + X) 3(c — x)' QuoL, 11. Divide x* X'* by X SuG. — This quotient should be written by inspection in the same manner as {x* — y^) ^ {x — y), and is x'"* -f x -j j -. Or it may be performed as follows : — x2 X ■ X a;:i _|_ X + i + L. 1-^ 12. Divide 1 -{- X 1 — X X , 1 by:; 1 + X -. QuoL, 1. l + x ' 1 13. Divide SUGGESTIONS. 1 + a-2 1 — ^-- and 1 — ;^^ l_ic 1 4-x 1 -^ + ?by X — y X -\- y X — X + y r, , ^' + y^ DIVISION. 135 -4 _ 1 _ 2 _ 1 30 tA/ 14. Divide — — by 15. Divide '"^-'^%y^"- ^•^. ^1^0^., 3;r2 — xy — 2?/2. ScH. 2. — It is Bometimes convenient to write the divisor under the dividend in the form of a complex fraction, and then reduce the result to a simple fraction by {105). 16. Divide 1 + - by 1 -. a a- OPEEATION. 1 + -1 X a2 a- ■f- a a 1- In Xaa a2 J a' — 1 a — 1 17. Divide - -\ — r a ab^ by6 + i-l. OPEEATION. rl -a ab^ ab^ ?>3 + 1 "6-f ab^ -^ ab' — ab'^ b+ 1 ^3 + 1 Suggestions. 18. Divide x ~^y — 1/2 j^y j??/ -2 -|- j^ -2^. a;;/ -' 4. x-'j/ rjc_ _^ f] ^ ^.,,^, X* + xy: 19. "Free -r— - — ' from nef2:ative exponents. ab~^ -^ x~^ ° , b^x^y^ — b'^x Result, ^ . axy^ + b^y- 136 FRACTIONS. 20. Free —^^ -r—z: from negative exponents. Itesult, 21. Free — — - — r- from negative exponents. Ans. 22. Free — ^ from neefative exponents. 1 — a:~^y-'-^-\-x~' *' a-x-^y- — a--\-a^xy'^ %1CII 23. What is the reciprocal of — ^' ' , • ? a'^ — b- Solution. — The reciprocal of a quantity being the quotient of 1 di" vided by that quantity, the reciprocal of —— is 1 -^ —, oi a- — b- This is analyzed thus : dividing 1 by 2xy the quotient is ^— . But, dividing the divisor, 2xy, by a^ — ?>-', is equivalent to multiplying the 2.TV 1 quotient : hence the quotient of 1 divided by '-- is - — X («"^ — b-), ^ ■ a- — b^ 2xy 2xy 24. AVhat is the reciprocal of — j-^ — - ? Ans., ^' . (a 5)-5 25. What is the reciprocal of ■ •, ? Ans., a-' — 3a2?) + 3a62 — 63. 175* ScH. — The reciprocal of a quantity being 1 divided by that quantity, the reciprocal of a fraction is the fraction inverted. Thus , , . 1 ^ o •> • 1 .2a . ^x^ „ a—b . a-j-b „ 1 , the reciprocal of 3a';^ is r— ,of — is — , of is — ■ — , of — - is dx' ^x"' 2a a-^b a — b 2x 2x, etc. {174, Explanation of 3rd method). SYNOPSIS. 137 Synopsis of Fractions. O t C I CQ t5 J FracHo7i. i Numerator. / Denominator, { Integral. Foi-ms. < Fractional, ( Mixed, ' Tei*ms. Value of fraction. Cor. 1. mult, or div. num. or denom. Cor. 2. Remov. denom. (Proper, /Simple, I proper, jg-P-'l. Lowest tei-ms. — L. C. D. — Reduction. Of numerator, of denominator, of fraction. Essential sign of fraction. f Prob. 1. To lowest terms. Rule. Bein. — Sch. By H. C. D. Fi'ob. 2. From impr. to int'g. or mx'd forms. Rule. I)e)n. — Cor. Neg. exp't. Pi'ob. 3. From integ. or mixed to frac. forms. Rule. Bern. Prob. 4. To com. denom. Rule. Bein. — Cor. L. CD. Devi. [ Prob. 5. Complex to simple. Rule. Devi. Add. — P?'ob. Rule. Dem. — Cor. Mix'd Nos. Subfn. — Prob. Rule. Devi. — Cor. Mixed Nos. f Prob. 1. Frac. by integer. Rule. Devi. § 1 3Iult. Divis. Frac. by integer, Sch. P7'ob. 2. Any No. by Frac. Rule. De7n. Sch. — Cor. Mix'd Nos. 1. Frac. by integer. Rule. Devi. 2. Any No. by Fraction. Rule. 1. Devis. -12. \- Sch's. •{ 2. {Prob J Prob III Test Questions. — Upon what five principles in Division are most of the operations in fractions based ? Why does the process of reducing to a common denominator not change the value of a fraction ? Give the rules for Multiphcation and Division of Fractions, and the reasons for them. 138 POWEKS AND ROOTS. CHAPTEE IV. PO WEUS AND HOOTS. SUCTION' I. Involution. [Note. — The subjects treated in this chapter are among the most difficult, if not actually the most difficult for the pupil in the whole science. In the examination of hundreds of students from all parts of the country, the author has found that the rule is that they are deficient hi knowledge of Badlcals. An attempt is here made to assist the teacher in remedying this defect, by constantly holding the attention to the one central principle, that The operations in Radicals are all based upon the most elementary principles op factoring, Many of the demonstrations can be given in more concise and elegant forms ; but it is thought better to hold the attention to the single line of thought. If the student learns how to use this key, he can unlock all the mj^steries of the subject. The desire to impress the central principle has led to several repetitions. ] GENERAL DEFINITIONS. 176. A I*OW€ri^ i\'2^'f'oduct arising from multiply- ing a number by itself. The Degree of the power is indicated by the number of factors taken. Thus 2, 4, 8, 16, and 32 are, respectively, the 1st, 2nd, 3d, 4th, and 5th powers of 2. ScH. — It will be seen that a power is a species of composite number in which the component factors are equal. 177* A. Hoot is one of the equal factors into which a number is conceived to be resolved. The Degree of the root is indicated by the number of required factors. INVOLUTION. 139 Thus, 2 is the 1st root of 2, the 2nd root of 4, the 3rd root of 8, the 4th root of 16, the 5th root of 32, etc. 178. ScH. 1.— Poioer and J^oo^ are correlative terms. Thus 32 is the 5th power of 2, and 2 is the 5th root of 32. 179- ScH. 2. — The Second Power is also called the Square; the Third Power, the Cuhe ; and, sometimes, the Fourth Power, the Biquadrate. In like manner the 2nd root is called the square root ; the third root, the cube root ; the fourth root, the biquadi-ate root. These are Geometrical terms which have been transfen-ed to other branches of mathematics. The second power is called the square, because, if a number represents the side of a square, its second power represents its area, or the square itself. Conversely, if a number represents the area of a square, the square root represents the side. Also the third power represents the volume of a cube, the edge of which is the first power or cube root. Biquadrate means twice squared, and hence the fourth power. . 180. An Exx>onent or Index, is a number written a Uttle to the right and above another number, and indicates 1st. If a Positive Integer, a Power of the number ; 2nd. If a Positive Fraction, the numerator indicates a Power, and the denominator a Boot of the number ; 3d. -ZT a Negative Integer or Fraction, it indicates the Reciprocal of what it would signify if positive. III. 4^ is the 3rd power of 4, or 64. a"' is the mth power of a, if m is an integer. 4-^ (read " 4, exponent — 3 ") is — or — • .1 3. . a ~ " is — . S'^ is the cube root of the square, or the square of the cube root of 8, or 4. a" is the ?jith power of the nth root of a, or the nth root of the mth power, if m and n are both integers. a~ ";; is — • a"" ScH. — It is obviously incorrect to read 4^, "the f power of 4." There is no such thing as a 2-fifths power, as will be seen by consider- in ing the definition of a power. Kead 4*^, "4 exponent!;" also a", 140 POWEES AND HOOTS. m, " a exponent -" ; a~ "5^, "a exponent — ^ ." These are abbreviated forms for, "a with an exponent — "^j" etc. In this way any exponent, however compHcated, is read without difficulty. 181, Cor. — A factor can he tmmf erred from numerator to denominator of a fraction, or vice versa, hy changing the sign of its exponent, without altering the value of the fraction. „,1 ^ a"* ^^ a^x-- a^y^ „ a'^x,-- "" x^' _a^ _ or_ V' _ Thus - — - = -r^- ; lor -7-^3- = — 7" — "; ^» ■^ a — 182, A Hadical JS'iiniber is an indicated root of a number. If the root can be extracted exactly, the quantity is called Rational ; if the root can not be ex- tracted exactly, the expression is called Irrational, or Surd. Thus the radical v''25a"2 is rational, but ^V6a is surd. 18 S. A Boot is indicated either by the denominator of a fractional exponent, or by the Hadiccil SigUf V- This sign used alone signifies square root. Any other root is indicated by writing its index in the opening of the V part of the sign. Thus Vam, Vam, are the 3rd and 5th roots of am, and the same as {am)^, (am)'^. 184. An Imaginary Qwaii^fif^y is an indicated even root of a negative quantity, and is so called because no number, in the ordinary sense, can be found, which taken an even number of times as a factor, produces a negative quantity. Thus V— 4 is imaginary, because we can not find any factor, in the ordinary sense, which multiplied by itself once produces — 4. Neither + 2 nor — 2 produces — 4 when squared. For a hke reason V — 3a2, v^ — bx, or v^ — IM^xy"^ are imaginaries. 18S' AH quantities not imaginary are called Real. INVOLUTION. 141 186. Similar Iladicals_a.re like roots of like quantities. Thus 4v^5a, 3^^5a, and (a* — x^)^ba are similar radicals. 187. To nationalise an expression is to free it fi-om radicals. 188. To affect a number tvitJi an Ex- ponent is to perform upon it the operations indicated by that exponent. Thus to affect 8 with the exj^onent f is to extract the cube root of the square of 8, or to square its cube root, and gives 4. 180. Til vol lit ion is the process of raising numbers to required powers. 100, JEvoltttion is the process of extracting roots of numbers. * 101. Calculus of Radicals treats of the processes of reducing, adding, subtracting, or performing any of the common arithmetical operations upon radical quantities. INYOLUTIOX. 102. JProh. 1. To raise a number to any required power . RULE. — Multiply the number by itself as many thies, LESS ONE, AS THERE ARE UXITS IN THE DEGREE OF THE POWER. Dem. — Since the number of factors taken to produce a power, is equal to the degree of the power {170), it follows that to obtain the 2nd power we take two factors, or multiply the number by itself once ; to obtain the 3rd power we take three factors, or multiply the number by itself tuoice ; and in hke manner to obtain the ?ith power we take n factors, or multiply the number by itself n — 1 times. EXAMPLES. 1. What is the 3rd power of la' ? 142 POWERS AND ROOTS. Model Solution. — Since the 3rd power of 2a* is the product aris- ing from taking it 3 times as a factor, I have ^a^ X 2a^ X Sa^ = 8a ^. 2. What is the 4th power of — lOa^ ? (— lOa^j X (— lOa^) X ( — lOtt^) X (— lOa^) = lOOOOa^. Ans. 3 to 6. What is the square of — hm'^n ? Of 6a6 " ^ ? Of 5 b¥ ff2 9 9 Answers, 25m^n% 367—, -r^^, -T^a}'^h~*. 25a-" 25 7 to 10. What is the cube of — 2ab'- ? Of 12m~ ^ ? Of _2 „, of-^? S on • -3 1728 8 m6 Ansivers, — 8a''6«, 1728m ^ = -, — — - x^", 11. What is the square of 2a ■ — 3^? SuG. (2a — 3x) X (2a — 3x) = ia^ — 12ax -\- 9x«. Ans. 12. What is the 3rd power of 2a-='+ 3^? Ans., 8a« + 36a 'a; + 54a-'^2 + 27 xK 13 to 17. Expand the foUowing : (1 + 2^ + 3^2)2^ (1 — a- + ^•-' — x^y, {a + b — cy, (1 + 2a; + 072)3, an.l ( 1 — 3^' + 3^2 _ j;3)2. BesuUs, 1 + 4^ + 10^2 _|_ 12^3 _|_ 9^4^ 1 _ 2a: + 3^» — 4a:3 + 3x^ — 2a;^ + x% 1 -{- Gx + 15x^ + 20a:3 + 15a7-« 4- Gx^ + x% and 1 — 6a; + 15a;2 — 20a;3 + 15x^ — 6x' + a;«. 18. Show that ^ — — + -7-^^-7 ^- 64a'6^ 64a26'< = 62. •19. What is the square of 9a; -f - ? Ans., 81a?2 + 18+ — . INVOLUTION. 14.3 20. Expand as above (4 — x^y. Result, 16 — So;"^ + x. 21. Expand (a"^' + c^)\ Remit, a^ + ia^c"^ + Qac + 4a^c^ + cK 103* Cor. — Since any number of positive factors gives a positive product, all p)owers of positive monomials are positive. Again, since an even number of negative factors gives a posi- tive product, and an odd number gives a negative product, it follows that even powers of negative numbers are positive, and odd powers negative. 194, Froh, 2, To affect a monomial with any exponent. R CLE. — Perform upon the coefficient the operations indicated by the exponent, and multiply the exponents of the letters by the given exponent. Dem.— 1st. Wlcen the exponent by which the monomial is to he affected n is a positive integer. Let it be required to aflfect 4a'»&'"x— « with the exponent p ; or in other words raise it to the pth power, p being a n n positive integer. The pth power of 4a'"6'"a; — * is 4a"'6'x — * X n n 4a"»6''x— * X 4a'"6'"«""* top factors. But as the order of the arrangement of the factors does not aflfect the product (83), it may be considered as, p factors each 4, into p factors each a"', into p fac- n tors each 6% into p factors each x—\ Now p factors each 4 give 4'' by definition, p factors each a"' are expressed a p"^, since a"* is m factors each a, and p factors containing m factors each, make in the whole pm n pn n factors, or aP". Again p factors each b' are expressed & '• , since b'' is n factors each &', and p factors, containing n factors each, are pn factors i ?!? 1 111 each b\ or &^ And since a;-* = -, p factors, or - x - x to p factors make — , as fractions are multiplied by multiplying nume- rators together for a new numerator and denominators for a new d*» 144 POWERS AND ROOTS. nominator, and ic* X ic" X «* to p factors are x p'. But -^ = x — PK Hence collecting the factors we find that {4:a^b^x~^)p = pn Ap ap'^b " x -p^ Q. E. D. 2nd. When the exponent is a positive fraction. Let it be required to -- p — affect 4a'" &'" ic -^ with the exponent -. This means that 4a"' 6 *■ x-*' is q to be resolved into q equal factors and p of them taken. Now, if we n separate each of the factors of 4a"' & '■ x — * -into q equal factors, and then take p of each of these, we shall have done what is signified by the exponent — . q l_ By definition, 4^ represents one of the q equal factors of 4. To obtain one of the q equal factors of a "*, we take one of the q equal factors of a from each of the m factors represented. But one of the q equal factors of a is represented by a'^, and m of these is ai by defi- nition. n n To separate h »' into q equal factors, we notice that b~ is n of the r equal factors of h. Now, if we resolve each of these r factors into q equal factors, b is resolved into rq equal factors ; doing the same with each of the n factors represented, and taking one from each set, we have b resolved into i^q equal factors and n of them taken ; that is ?>'-^is 071C of the q equal factors of & '' . To resolve x -* = — into q equal factors, we consider that a fraction Xs ^ is resolved by resolving its numerator and denominator separately. But one of the q equal factors of 1 is 1 ; and one of the q equal factors of X* is X'' as seen in the resolution of a ™. Hence one of the q equal 1-1 -- factors of X —* or - IS - ^ a; < • X* f XI Collecting these factors we find that one of the q equal factors of 4a"'6'x-*is 4'(a''6'''-ar ''• And finally p of these being obtained ac- p pm pn j)S n cording to Case 1st, gives 4 '' a '' h''' x "^ , as the expression for Aa'-b~x~ - INYOLUTION. 145 affected with the exponent^ ; which rssult agrees with the enun- q ciation of the rale. 3rd. When the exponent is negative and either integral or fractional. n Let it be required to affect 4a'» b'x-' with the exi^onent — t. This, by definition of negative exponents, signifies that we are to take the reciprocal of what the expression would be if t were positive. But n '" 4a'"&''x~* affected with the exponent t (positive) is 4:'a""7> ^x~*^ by the preceding cases, whether t is integral or fractional. The reciprocal of this ig . But since these factors can be transferred to the 4'a'"'f) »■ X -'* numerator by changing the signs of their exponents, we have fn "_ 4 -*a —^"'b~ r x*' , as the result of afi'ecting 4a"' hr x-' with the expo- nent — t, which result agrees with the enunciation of the rule. [Note. — The above demonstration contains the fundamental prin- ciples of the whole subject of the Theory of Exponents, and it is of the highest importance that it be made perfectly famihar. The applica- tion of the rule is so simple in practice as to afford no difficultj^ but in each of the following examples the reasoning should be given in full. The purpose is to fix in the mind these principles of the theory. After this is done, of course, the expert merely performs the operations. The danger is that the how being so simple, the why will be disre- garded.] EXAMPLES. 1. Affect 2a25^c~* with the exponent 5 : that is, raise it to the 5th power. MODEL SOLUTION. 2. 1 ( Operation. {^la'^h^c - *y = ^la^^lf^ Explanation. {la^h^c-'^Y is 2a^ly'c-'>' X 2a«6^c-^ X 'la^ly'c-''- to 5 factors. This gives 5 factors of 2, or 32 ; 5 factors of a^ or 2. 10 factors of a, a^" ; 5 factors of 6 •^ which I obtain by considering 2. -L 2 that h^ is 2 factors of h'\ and hence 5 factors of h^ is 5 times 2 factors of &•* or 6 ^~ ; and since c~* = — r-. 5 factors of it aje — - X — r - - to 146 POWERS AND ROOTS. 5 factors = -7-7 = c- 20, as fractions are multiplied by multiplying 2. numerators and denominators. Hence I have (2a^b'^c~*)^ =i LQ. <2 2. Affect 3a^ with the exponent m ; that is raise it to the mth power, m being an integer. 3. 2. 3. 3. Explanation. (3a^)"' = 3a=^ x3a^X3a^ to m factors. This 3. gives m factors of 3 and m factors of a^. But m factors of 3 are rep- resented by 3™. To obtain m factors of a\ I consider that a is 2 J- " ^ . factors of a^ ; hence m factors of it are 2m factors of a**. That is a resolved into 3 equal factors and 2m of them taken. Therefore (Sa**)"* = 3™a 3 . 3. Affect 16a'5^~V with the exponent f; that is, represent 2 of the 3 equal factors of 16a^^x~^. Explanation. — I Avill first resolve IGa^^x-^ into 3 equal factors (or indicate it when I cannot perform it). To do this I take one of the 3 equal factors of 16, which I represent, as I cannot resolve it, and write L (16)'*. Again, one of the 3 equal factors of a^ * is a^, as a^ ^ is 15 factors of a, and consequently when resolved into 3 equal factors one of the 3 contains 5 factors of a. Thus a X « X « to 15 factors when put into 3 equal groups becomes aaaaa X aaaaa X aaaaa, one of which is a'. To resolve «— ^ I consider it as - =- X-X— X-X — X -• ♦C »C »l/ tC- »c *c »c Separating this into 3 equal groups it becomes —^X—^X-^', hence 11 one of the 3 equal factors of— -r is-, or x~^. Therefore 16a '"iir-^ ^ x*' x^ L being resolved into 3 equal factors, one of them is {lG)'^a^x-^. But I am to take 2 of these, as the exponent f indicates. This gives me (re- peat the reasoning of Ex. 1) (16)-^a^"c— *. 4. Affect 3a"6~^ with the exponent 7 ; that is, take m of the n equal factors of it. INVOLUTION. 147 Explanation. 3 n represents one of the n equal factors of 3. One of the n equal factors of «» is a, as a" means n factors of a. 6-"' =: -=-XrX7-X torn factors. Each of these being separated h'^ b b b into n equal factors 7X-X- to n factors = —. And taking b~ b~^ b^ one of the factors - , from each of the m factors -, I have m J.- ^ On factors of — or — r= 5 « . Therefore one of the n equal factors of ]_ m 3^»5— m is ^~ab " • But I am to take m such factors which gives (re- m m" peat the process of Ex. 2.) 3"a"'b " • [Note. — These specimens of analysis are given to show how each example should be shown to depend for its solution upon the elemen- tary principles. Let it be made sure that the pupil can do this in any given case, after which he should simply apply the rule, until he be- comes expert in doing it. ] 5. What is the square of -aPm ? Of — - — ,or — - abx ~' ? 2a'-^x 2 J^ _i The cube of —, or — -a'^m ^x? 3m^ ^ 2 6. Affect 8ai2^~^ with the exponent -. jt a s -* 4 7. Affect 32a2o^i5 ^rith the exponent — -. Besidty 8. Affect 13x~'^y~^ with the exponent — 3. 148 POWERS AND ROOTS. 9. Perform the following operations and explain each as a process of factoring : {d2a^^b-'')~^ , (lOOfl-^^*)"^, (IW^x'^y-'f^, and (a"'6rj;-^)~T. Results, — — , 10 — , ^— , and. ^^' ^ (11)% aTh-^ 19S, J^rob. 3. To eximnd a binomial affected ivith any exponent. It ULE. — This rule is best stated in a formula. Thus, let a, 6, and m be any numbers whatever, positive or negative, integral or fractional, then will (a + 6)"* represent any BINOMIAL, AFFECTED WITH ANY EXPONENT, AND 1 - /& mini — l)(m — 2) ,,„ + 1-2-3 ^ ' "'^ m{m-l) (m-2) (m-3) + 1-2-3-4 "" ^ m{m-l) (m-2) (m-3) (m-4) + 1-2-3-4-5 '' ^ "^ ®^^- Dem. — This formula is the celebrated Binomial Theorem discovered by Sir Isaac Newton, who also demonstrated it. There are several elementary demonstrations, but they are somewhat prolix and difficult for young students, and the author's experience has satisfied him that little or no good comes to most students from their study. The dem- onstration is, therefore, not inserted here, but will be found in Appendix I. If the pupil learns the formula, and learns to apply it with facility, it is all that is thought best for him to attempt at this stage of his progress. EXAMPLES. 1. Expand {x-\-yY by the Binomial Formula. Model Solution.— To apply the B. F., a; = a of the formula, 5 /^ ^\ y = h, and 5 = w. .'. I have {x + yY = a;* + 53^~^y + — INVOLUTION. 149 6 2 2. 5(5 - 1)(5 - 2) ^ 3 ^ , 5(5-l )(5-2)(5 -3) , . ^ 5<5 — 1)(5 — 2)(5— 3)(5 — 4) . , - , ,.^ , .i , i -j — : x'^—^y", at which term the develop- ment becomes complete since the next coefficient would hare a factor 5 — 5 = which would destroy the term. Perforcaing the operations indicated I have (x + t/)^ = x^ -f 5x*y -\- lOx^y^ -f- 'i-Ox^y* +5.r?/-* -\- y^. (In practice, this result should be written out without writing the preceding, by simply applying the formula mentally. ) 2. Expand (x — ijy by the B. F. SuGOESTioNs. ic = a, — 2/ = ^ and G = m. . •. (x — y)^ = x^ -\~ 6.=( - ,) + 1^..( _ ,). + ^ x^( - ,)» + ^.^ ,^ . 6-5-4-3.2 ^ ^, , 6.5-4-3-2-1 „^ ^'(-y)^+ 2.3.4.5 ^(-y)^+ 2.3-4.5.6 - -'( - V^' = a;6 _ 6x^2/ + 15x V — 20x^2/' + ISx^y^ — 6xi/= + y^. 3. Expand {2m"- — 3n^)^ by the B. F. Suggestions. — Make a = 2m*, 6 = — Sn""' and m = i^ .«. We have (2?7i* — 3n^)* = (27n2)4 -f- 4 (2m2)3( — 3n') -|- ^-^ (2m2)2 ^-^'^'^' + i^(2m*)(-3n^)^ + ^2^-^'^'^''^°^ - ^^'^' = 1 (by performing the operations indicated) 16m* — 96m^n* -f-216m'*?i — 216j;i2/i* + Sin*. 4 Expand {x -{- y)-* hj the B. F. Suggestions, a = x, b =z y, and m = — 4. . • . (x -f- y)~^ = x — ' , . .. _i 1 , -4(— 4-1) , , , , _4(-4-l)(-4 — 2^ x—^—'y^ -j- etc. This series does not terminate since no factor ever becomes ; but the development can be carried to any desired extent. Performing the operations indicated, we have (x -j- y)~~^ = x — * —4x--'y + 10x-«2/2 _ 20x--!y^ + 35x-s?/* — , etc. 2. 5. Expand (m — ?i)^ by the B. F. Suggestions. — Making the proper substitutions in the formula, we 3. 2. S.. ■3 - ' ^(|-l)„f-^ have (m — n)'^ = 7?i -f- fm (— n) -j- ^^^^ — m ( — n)^ + 150 POWERS AND ROOTS. llt^Mri™J-V,.)» + MfMz!Hi^J-(_„). +, etc. A series which ruever terminates, since no coefficient reduces to 0. Per- 2. % forming the operations indicated we find that (m — nY = m* — — 4- — i -7- - 1^ |m ^n — ^m ^n^ — A ^ ^^^ — F4U^ '^'^ — > ^^c. 6. Expand (1 + ^)" by the B. F. Bes., (1 + 07)"= 1 + wa; + -^—^ — ^-x^ + '2~ 8 ^' n(n-l)(n-2)(n-3) _ _ n(n- l) + 2^7-3—7—4 ^ +' ^*^- 2 "^ ScH. — This expansion is in itself a very useful formula, and should be memorized. 7. Expand (3 — y')^ by the B. F. -J. _3 -5 JSeswZ^, (3 — 2/2)2 3^ 32 _ f — _ £_ y. _ £ e _ 5-3"' 128 -y'—^^^' 8. Expand (1 + x^-y by the B. F. Besult, 1 + 5^2 + 10.27^ + 10a;c + 5x» + a;io. 9. Expand (1 — a'^)~^ by the B. F. ■10 pr OPT Besult, 1 + - a2 + _ a. 4- — a«+ — a8+, etc. 10. Expand v^a^ — as^a by the B. F. •i Besult, ^a- — a'^e'^ = a^l — e^ = a{X — 62)"2 = ^ ,, 1 1 1-3 13-5 <^~r'-Y:i''- 27176 '' ~ 2:7:^ ^' -^ ^^-)- 11. Expand - — ^—— by the B. F. Besult, = (c + xy =: 0-^—20 ~*x + 3c -*x^ — {c-^ xy ^ ^ INVOLUTION. 151 4:C-'x^ + &c. = —(1 + +, etc.). 12. Expand 1 by the B. F. {1 + x) Besult, 1 = (1 + ^) =1 -\ (1 + X? 5 25 ll;r3 44a:4 1 , etc. 125 ^625 13. Expand (a; + 2/ + c)^ by the B. F. Suggestions.— Put (a; + y) — z .-. (x 4- 2/ + c)* = (z + c)* = z* -^ 423c -j- 63* c'* 4- "^c^ -f- c* = (restoring the value of z) (X + y)* + 4(x + yYc + 6(x + y)2c^ + 4(x + !/)c' + c^ But (X + y)* = X* -f 4xV + 6x*y* + 4xy^' + y\ (x + v)' = x=^ + 3x«y -j- 3xy* 4- y-', and (x + y)^ = x* + '^*!/ + y^- Whence by substitu- tion we have (x -f- 2/ 4- c)* = x* + 4x^y + Gx^y* -\- 4xy^ + y* + 4cx=» + 12cx*y 4- 12cxy2 -f- 4cy^ + 6c«x« -f 12c'xy 4- Gc^y^ -f 4c^x -}- 4c^y + c*. 14. Expand (2a — 6 + 0^)3 by the B. F. Besuli, 8a3 — 12^26 4- 606^ _ fcs 4. 12^202 — 12a&c2 + 362C2 4- 6ac^ — 36c^ 4- c^. 100. Cor. 1. — T/2e expansion of a hinomial teiminates only when the exponent is a positive integer, since only when m is' a positive integer will a factor of the form m(m — 1) (m — 2)(m — ?>)etc. become 0, as is evident by inspection. 19 7 > Cor. 2. — When m is a positive integer, that is when a binomial is raised to any power, there is one more term in the development than units in the exponent. Since the first coefficient is 1; the 2nd,?7i; the 3rd, — — --, the 4th, m{m — l)(m — 2) . , m{m — l)(m — 2)(m — 3) __ _ ; the 5th, 2 • 3 ^-l ' i&c, we notice that the last factor is m — (the number of the term — 2) ; and the number of the term, therefore, which 152 POWERS AND ROOTS. lias m — TO as a factor is the (m + 2)th term. But this is 0. Hence the (m + l)th term is the last. 108. Cor. 3. — When m is a positive integer the coefficients equally distant from the extremes are equal; since (a + 6)"* = (5 _|_ a)"» ; the former of which gives a"* + moT-^b + '^^i''''—^) ar-'^y^ 4., etc., and the latter 6"* + mb'^-'a -\- — 1^ — I 5'»-V _^j qIq^ "Whence it appears that the first half of the terms and the last half are exactly symmet- rical. 100. Cor. 4. — The sum of the exponents in each term is the same as the exponent of the power. ScH. — The last two corollaries apply to the form (x -\- y)"", and not to Buch forms as (2a"' — 36'^)'", after the latter is fully expanded. 200. Cor. 5. — A convenient rule for writing out the pow- ers of binomials may he thus stated : 1st. Tliere is one more term in the development than there are units in the exponent of the power. 2nd. The first contains only the first letter of the binomial^ and the last term only the second, while all the other terms con- tain both the letters. Srd. The exponent of the first letter of the binomial in the first term of the development is the same as the exi^onent of the required power and diminishes by unity to the right, tvhile the exponent of the second letter begins at unity in the second term of the expansion and increases by unity to the right, becoming, in the last term, the same as the exponent of the jMiver. Uh. The coefficient of the first term of the expansion is unity ; of the second, the exponent of the required power; and that of any other term may he found by multiplying the coeffi- cient of the jjreceding term by the exponent of the first letter in that term, and dividing the product by the exponent of the sec- ond letter + 1. INVOLUTION. 153 Dem.— This rule is a deduction from the formula (a + 6)™ = m{m — l)(m - 2)(m - 3) ^^_^^ _^ ^^ 2-3-4: ^ 1st. Is proved in Cob. 2, Eepeat it. 2nd. Is proved in Cob. 3. Kepeat it. 3d. The law of the exponents is directly observable from the formula. 4th. The coefficients of the first and second terms are seen in the formula to be as stated. The coefficient of the third term may be writ- ten m X — : . which is the coefficient of the second, or preceding 2 term (m), multiplied by the exponent (m — 1) of the first letter in that term, and divided by 2 which is the exponent (1) of the second letter -f- 1. In like manner noticing any other term, as the 5th. Its coefficient m(m — l)(m — 2) ?n — 3 m{m — l)(m — 2) may be written — - X — j • l^ut ^ — is the coefficient of the Ith term, m — 3 is the exponent of the first letter in the 4th term, and 4, the divisor, is the exponent of the sec- ond letter (3) -f 1. 15 to 20. Write out by the above rule the expansions of the following: {m + n)', {x + y)\ {a + cy, {x + my, 2 X % 1 Suggestions upon the last.— Eegard a' and m^ as simple num- 2. JL bers represented by letters without exponents. Thus (a'* -\- vi^)^ = (a^)4 _|. 4 {(^^{m^) 4- 6 (a')2(7n*)2 4-4 (a^)(m'^)»+ (m^)*. Nowper- ^1 ^ fi -1- forming the operations indicated, we have (a"* ^-^'^ )* = «^ + 4a"^r/i^ i. i 3. -f 6aSji + 4aS7i^ + m^. 201, CoR. 6. — If the sign between the terms of the bino- mial is minus, «s (a — b)'", the odd terms of the expansion are -\- and the even ones — . This arises from the fact that the odd terms involve even powers of the second or negative term of the binomial, and the even terms involve the odd powers of 154 POWERS AND ROOTS. the same. Thus the second term involves ( — h) which makes its sign — ; the 4th term has ( — hy, the 6th term ( — hy, &c. But the first term does not involve ( — 6), and the 3rd has ( — hy or h\ the 5th has ( — 6)^ or b*, etc. 21 to 24. Write out the expansions of (m — n)^,{x — y)', (^2 — ^3)3^ (a^— h'^y. Besult of the last, (a^— 6^)^ = a^ — 4a%^4- 6a6^— 4:ah + Z>^. SECTION II. Evolution. 202, I^TOh, 1, To extract any root of a perfect power of that degree. R ULE. — Resolve the number into its prime factors, and SEPARATE these INTO AS MANY EQUAL GROUPS AS THERE ARE UNITS IN THE DEGREE OF THE ROOT REQUIRED ; THE PRODUCT OF ONE OF THESE GROUPS IS THE ROOT SOUGHT. Dem. — Since the mth root {i. e., any root) of a number is one of the m equal factors of that number, if a number is resolved into m equal factors, as the rule directs, one of them is the mth root. EXAMPLES. 1. Extract the cube root of 74088. Model Solution. — Resolving 74088 into its prime factors I find them to be 2 •2.2.3-3 -3. 7- 7- 7. These arranged in 3 equal groups give 2.3.7 X 2.3.7 X 2.3.7. Hence 2.3.7 = 42 is the cube root of 74088, since it is one of its 3 equal factors. 2 to 5. Extract in this manner the following : v/492804, ^592704, V248882, v' 456583. ^^ots, 702, 84 and 12. 2 _i 6. Extract the square root of Ula''x~^y'^z s. Model Solution. — The two equal factors of 81 are 9 • 9 ; of a*, a* 'Ci^ ; of 05-2, x-i -x-i ; of ^/^ y* - y^ ; of z~^ , z~ ^ '^ . z~ . Hence EVOLUTION. 155 81a*xr-Yz = 9a^x~yz~^X Qa^x-^z'^, and consequently, da'ar-^z ^° is its square root. 7 to 11. Extract ^^25ai^, J64a-'A ^^^xy^, ^14Aa<)n', l^a^ 3. Ill SJsGrrm ^^^^^'' ± ^^'^'' ±8a-'x^, ±lx^y^, +12aW, I 12 to 15. Extract Vl25m6a;»2, 3 Ul^^x'^y^, ^— 'd'la'^y -\ 2 JL _ 1 JSoo^s, ^m^x\ 12j;^2/^> — Sa^;?/-^, and 4- 2n "^^/^ QuEEY. — Why the ambiguous sign to the last? 203, ScH. — The sign of an even root of a positive number is am- biguous (that is, + or — ) since an even number of factors gives the same product whether they are positive or negative {87 f 88). The sign of an odd root is the same as that of the number itself, since an odd number of positive factors gives a positive product and an odd number of negative factors gives a negative product {88, 89). 204» Cor. I. — The roots of monomials can be extracted by extracting the required root of the coefficient and dividiyig the exponent of each letter by the index of the root, since to extract the square root is to affect a number with the exponent ^, the cube root -^, the nth root 7, etc. (194,) EXAMPLES, 16 to 21. In this manner write J25a'^b", ^ — 343;zr^?/~*, Nj ^' \Q4.m^y<^ N 243%-'" f Roots, ± 5a\ — Ix'^y-' ± Sm'-n'^x^, 3c^m~^V^, ^-^ ^ ' 4?7iy^ 2a^x^ and — • 'dbiy-' 156 POWERS AND ROOTS. 205, Cor. 2. — The root of the product of aexjeral numbers is the same as the product of the roots. Thus, '^yabcx = v/a - Vft • v/c • 'Vx, since to extract the mth root of abcx we have but to divide the exponent of each letter by m, whicii 1 1 2. 1 gives fl»«5m^m^r« ^^ ^^ . V^ ' Vc - Vx. 206. Cor. 3. — The root of the quotient of two numbers is the same as the quotient of the roots. Thus, ^ 1^ is the same as 'Sjn -~, since to extract the rth root of H}. we have but to ex- tract the rth root of numerator and denominator, which oper- ation is performed by dividing their exponents by r. Hence \m m~ 1 _ vm V\ EXAMPLES. 22. Show that ^8 x 27 = v/8 x ^21. MoDEjj Solution.— We may show this in two ways. 1st, — Experimentally. Thus 1/8 X 27 = ^"216 = 6. Again ?/8~ X ^27"= 2X3 = 6. Hence ^8 X 27 (or the cube root of the product) = \/8 X v^27 (or the product of the cube roots). 2nd, — Analytically. \/8 X 27 signifies that the product of 8 and 27 is to be resolved into 3 equal factors, which is accomphshed by resolving 8 into its prime factors, and 27 into its prime factors, and then separating these factors into three equal groups (202). This will .give the same result as resolv- ing the product of 8 and 27, or 216, into 3 equal factors, since the prime factors of 216 are the same as those of 8 and 27. Ja-"'67 = ^«— Xllbn 23. Show that .. la-"'6« Suggestions. — The 5 equal factors of a-"'6'' are a ^ &^", a ^ P^^'j a Fb^n^ a~r&s^, and a Th^^, since by the rules of multiplication these EVOLUTION. 167 1 m 1 multiplied together make a-^h~. But a" '^b^ is the product of one of the 5 equal factors of a-"*, or Va-"\ and one of the 5 equal factors of - fT 6", or ° I 5« • 24. Extract the square root of a^C^ -\- 2a^bc^ + a-b'^cK Solution. — The factors of this are readily seen to be a^, c^, and 4r+2a6 + 6^ which separated into two equal groups give ac{a+b) and ac{a+b). Hence ac{a + b) or a^c + abc is the required root. 25. Extract the square root of m^ — 2m^x + 'in^^^. Booty ?7i*(l — • a:) or m- — m^x. ScH. — The extraction of roots by resolving numbers into their factors according to this rule, is limited in its apphcation for several reasons. In the case of decimal numbers we can always find the prime factors by trial, and hence if the number is an exact power, can get its root. But in case the number is not an exact power of the degree re- quired, we have no method of approximating to its exact root by this rule, as we have by the common method already learned in arithmetic. In case of Hteral numbers the difficulty of detecting the poljmomial factors of a i^oljTiomial is usually insuperable. Hence we seek general rules which ynR not be subject to these objections. 207* I* rob, 2, To extract the f + cy = [(a + &) + c]3 = (a J^ 6)3 _|_ [3(a _f 6)2 4. 3(a + h)c + c=^]c = a'' _|_ [.3a! -f 3ab-\- 62]?) + [3(a + h)'' + S^t + h)c + c*]c. C. (a + & + c + fZ)3 = [^a 4- 6 + c) + f7]^ = (a + 6 + cj^ + [3(a + 6 -f c)* + 3(a + 6 + c)d + d')d = (1) (2) (3) a' + [3a2 + dab -]-h^]h -{- [3(a + &)« +3(a + &)c + c^]c + [3(a -f 6 -f c)'^ 4- 3{a + h + c)d + d*]t/. Hence it appears ; 1st, That the cube of a polynomial is made up of as many parts as there are terms in the root ; 2nd, that the first part is the cube of the first term of the root ; 3d, That the second part is three times the square of the first term of the root -{- 3 times the first term into the second term -j- the square of the second term, multiplied hy the second ienn of the root ; 4th, That any one of the parts of the power, as the nth, is Three times the square of the n — 1 preceding terms of the root, -{- 3 times the product of these terms into the next, or nth term, -f- the square of this last or nth term, all these terms being multiplied hy the last, or nth term of the root. Finally, it is evident that, if the work does not terminate by this process when the letter of arrangement disappears from the remainder, it can never terminate, since the divisor always contains this letter. ScH. 1. — If the first term of the arranged polynomial is not a per- fect cube the root cannot be extracted. ScH. 2. — If at any time no term of the remainder is exactty divisi- ble by the first term of the trial divisor, the root can not be extracted. EVOLUTION. 16? + + + CO , CO + + 1 8 S CO 1^ + 4- CO 1 H (M CO p + GO + 1 1 s .2 ■« ^ £ CO ±, ft I :« 1 ^ rC> + + 8 1 ^ S> « c CO CD + 'S :^ s s ^ _l_ 1 i^ CM rO cB e 1 + Ci H 4- + H ei c e rO 1? CO G<1 'S" rC 1 C^ ~ \ 1 1 %> « i •^ co^ s l- co' CO s "^^ o 1 ■■§ ■ft 1G8 POWEES AND ROOTS. Explanation — 1st. I arrange this polynomial with reference to a, and thus see at once the first two terms. But the terms 36a'*c and 27a'*6c'^a; are of the same degree with respect to a, and hence to determine which is to have tho precedence, I notice that the first term in the root mil be 3«^ c, and as the second term of the polynomial divided by 3 times the square of this gives the second term of the root, I observe that the terms containing a and c are all to have^. precedence over those containing h andic. Hence I write 36a "^c — 8a'^' next. The remaining terms I arrange, giving a the precedence and noticing that as x will be in the last term of the root, its higher powers will stand last, 2nd. As 27a^c^ is the cube of the first term of the root, that term is 3a^c, which I consequently place in the root, and subtract the term 27a^c'-^ from the polynomial. 3rd. As the second i)art of a cube of a polynomial is 3 times the square of the first term of the root, plus other terms, into the second term of the root, I take 3 times the square of this first term of the root or 27a "'c^, for a trial divisor. Dividing, I find the second term of the root to be — 2a. But the True Divisor, or leading factor in this second part of the power, is 3 times the square of the former part of the root, -f- three times that part into the last tenn found, -}- the square of this term. Hence I add 3 times 3a^c multi- plied by — 2a, and — 2a squared, to complete the divisor. Having completed it, I multiply it by the last term of the root found, — 2a, and thus form the second part of the power of the root, which I sub- tract from the given polynomial. 4th. The explanations of the next and succeeding steps, when there are more, are identical with the last, and can be supplied by the student. 2. Extract the cube root of a^ — 85^ -\- 12a¥ — Ga-b. Boot, a — 2b. 3. Extract the cube root of 5^;^ — 1 — 3^^ _|_ ^-c — 3^7. Jloot, x^ — X — 1. 4. Extract the cube root of Q^Q^xA + 1 — GSj;^ —9^-1' 8j7« — 36.i;^ + 33^2. . Boot, 1x^ — 3:r + 1. 5. Extract the cube root of GOc^^^ + 48ca;5 — Tic^ -f lOScs^ — 90c^^2 4. 8j^6 _ 80c3j;3. Boot, 2x^ + 4cd; — 3c2. I EVOLUTION. 1G9 .• '^ ^ ^ + 1 V 1 c^ J- O ^ t- H (M O (M c5 O CO , ^ 1 + 1 + ' «1- cr^ ^S'JI CO •f + 1 + 1 %^1 4- H> o S 1 00 4- + CO + o CO GO + 2 + + + + Ho* GO 00 i 1 CO CO + + I I I + I 1 + + + + I I ft w lii ft < 02 H 12; ^ P "S tf w ^ > ?^ o ■i: "2 w s « O ft « 5z; 170 POWERS AND EOOTS. 2nd. Take the cube root of the greatest cube in the LEFT HAND PERIOD, AND WRITE IT AS THE HIGHEST ORDER IN THE ROOT. Subtract the cube of this figure from the period USED, AND TO THE REMAINDER ANNEX THE NEXT PERIOD FOR A NEW DIVIDEND. 3rd. Take three times the square of the root already FOUND, REGARDING IT AS TENS, FOR A TRIAL DIVISOR, BY WHICH DIVIDE THE NEW DIVIDEND. ThE QUOTIENT IS THE NEXT FIGURE IN THE ROOT OR A GREATER ONE. To OBTAIN THE TrUE Divisor add to the trial dwisor 3 times the product of the last root figure by the preceding part of the root, regard- ed as tens, and also the square of the last figure in the ROOT. Multiply this true divisor by the last root figure AND SUBTRACT THE PRODUCT FROM THE LAST NEW DIVIDEND, AND BRING DOWN THE NEXT PERIOD. 4tli. Take three times the square of the root already FOUND REGARDED AS TENS, FOR A TRIAL DIVISOR, AND PROCEED AS IN THE 3rd STEP. KePEAT this PROCESS TILL ALL THE PERIODS HAVE BEEN BROUGHT DOWN. If THE NUMBER IS A PERFECT CUBE THE REMAINDER IS ZERO. If IT IS NOT A PERFECT CUBE ANNEX PERIODS OP 3 ZEROS EACH AND CONTINUE THE OPERATION TILL THE REQUIRED DEGREE OF ACCURACY IS ATTAINED, MARKING THE FIGURES THUS OBTAINED AS DECIMAL FRACTIONS. ScH. 1. — In pointing off decimal fractions, or the fractional part of mixed numbers, make full periods of three figures each, annexing O's if necessary. ScH. 2. — If, at any time, the trial divisor is not contained in the dividend to be used according to the 3rd paragraph in the rule, annex a to the root and also two zeros to the trial divisor, bring down the next period, and then divide. ScH. 3. —When the work does not terminate with the last period of significant figures it will not terminate at all, and the number is a surd. This is evident since the right hand figure in any subtrahend arises from cubing the corresponding digit in the root, and the cube o/ no digit produces in unit's place. EVOLUTION. 171 Deu. — 1st. That this method of pointing gives the number of figures in the root is made evident by cubing a few numbers. Thus the cube of 1 is 1, and of 10 is 1000 ; hence the cubes of all numbers between 1 and 10 have 1, 2, or 3 (cannot have 4) figures. The cube of 100 is 1,000,000 ; hence the cube of numbers between 10 and 100 have 4, 5, or 6 figures, but can not have 7. Again, the cube of 1000 is 1,000,000,000 ; hence the cube of any number between 100 and 1000, i. e. , of any number expressed by 3 figures, contains 9, or one or two less than 9 figures. In like manner it appears that the cube of any integral number contains either three times as many figures as the root, or one or two less. Since in multiphcation of decimal fractions the number of fractional places in the product is equal to the number in both or all the factors used, the fractional part of any cube must have three times as many figures as the root. Thus (2.15)3 _- 9.938375, six fractional places. (. 612 ) ■* = . 229220928 or nine fractional places. 2nd. That the greatest cube in the left hand period is the cube of the highest order in the root, appears from the facts that the cube of any number of units between 1 and 9 falls in the 1st period ; the cube of any number of tens between 1 and 9, falls in the second ; of any number of hundreds, in the 3rd, etc. Moreover, though the left hand period usually contains more than the cube of the digit in the highest order in the root, it can not contain the cube of a unit more of that order, since all the figures that can follow this highest order in the root can not make another unit of that order. Thus the cube of 3999 can not be as great as the cube of 4000. But the cube of 4000 gives 64 in the highest period. Hence the cube of 3999 must give less than 64 in that period. 3rd. In any given case, supjoose the pointing shows that the root consists of thousands, hundreds, tens, and units. Kepresent the thousands by T, the hundreds by h, the tens by t, and the units by u. Then the number is {T -}- h -}- t -\- u^ = T' -f [3^^ + 327i + h-'-]h+ [3(7+ h)' +3(r+ h)t-i-t'-]t -f [3(r+ h -\-t)^ 4- 3(1" -j- h -\- t)u -^u^]u. Biit, having removed the cube of the thousands, T^ the next part of the power is [3 T^ -j- 3 Th -}- h^ ]h. No part of this can fall in either of the two lowest periods of the power, since its lowest order arises from h'^ which is 1,000,000, at least. Hence we need only bring down one period. For a irial, we consider this part as 31^ X h, and hence the Trial Divisor is BT'^ or 3 times the square of the root already found. Again, regarding this thousands figure as tens, makes the T, which squared and multipHed by the next figure of the root which is also hundreds, give millions, 172 POWERS AND EOOTS. tlie same order as the new dividend. But the True Divisor is 3T- -J- 337i -f A* ; hence we add to 3 T^ , 3 T/i -f h^; L e., 3 times the root pre- viously found multiplied by the last figure, and the square of this last fig- ure. In making this correction we are to remember to call the thou- sands so many tens, which reduces it to hundreds, the order of the root which we are seeking; whence the correction becomes the square of hundreds, or of the same order as the trial divisor, and can be added to it. dth. It is evident that this process is merely repeated, as we pro- ceed to obtain other figures in the root; and, as the law of notation is the same as we pass the decimal point, no special exemplification is needed in that case. EXAMPLES. 1. Extract the cube root of 99252847. MODEL SOLUTION. — OPERATION. G4 T 3(40)2= 4800 i 9.(A0\a — 790 Corrections _, ht u 9925284714 6 8 Trial Divisor 3(40)2= ^gOO 3(40)6 = 720 6^ = ^ True Divisor 5o5(i 35252 33336 Trial Divisor 3(460;=^ = G3480U 1916847 Corrections I ^(^g^l^i^^ ^'^^ True Divisor 638949 191 6847 Explanation. — As the highest order in this number is ten millions, the highest order in the root is hundreds, since the cube of a hun- dreds figure falls in millions period, while the cube of thousands falls in bilhons. Moreover, the cube of the hundreds figure is the greatest cube contained in 99, i. e. 64, the root of which is 4, which is, there- fore, the hundreds of the root. That the hundreds figure is not greater than 4, is evident, since the cube of 5 hundreds is greater than the given number. But the cube of 3 hundreds with the greatest pos- sible figures in tens and units' places, (i. e. 399) is less than the cub© of 4 hundreds. Hence the hundreds figure is not less than 4. Therefore the cube root of the given number is A -f- f -f- w, and the number itself is (A -f- f -f- w) » = K^ -f [3h^ _f- 3ht -f i^]^ -f [3(7i -j- t)^ -f- (3/i -f- ^'^ -f- w'^Jw. But having removed the h-^ by subtracting the 64 (millions), the next part of the power is [SA,* -|- 3ht -\- f^lt. Now the EVOLUTION. 173 lowest order of this is i'\ or the cube of tens, which cannot fall below thousands, so that I need only bring down thousands period, i. e. the next lower. For a Trial I now consider this part of the power (35252) as 3/1^ x t, or 3[40]2 X ^ I reduce the 4 hundreds to the same order as the root figure which I am seeking, so that the product of its square by this root figure shall be of the same order as the new dividend. Thus my new dividend is thousands, and in order that thousands divided should give tens, the divisor must be hundreds, or the square of tens. Therefore, reducing the 4 hundreds to tens it be- comes 40, whence 3(40)^ = 4800, which being hundreds, goes into the new dividend, which is thousands, tens times. This trial divisor is really contained in the dividend 7 {tens) times, but as the corrections to be made upon it for the true divisor are so great, the true divisor does not go but 6 times, as I find by trying 7 for the tens of the root. Having thus found 6 to be the tens of the root, I correct my trial divi- sor, which by the formula is 3/i,* -^3h X t -\- l^, hy adding Sh X t or 3(40) X 6, and t^ or 6*, and find the true divisor to be 5556 (hun- dreds). This multiplied by the 6 (tens) gives the second part of the power, i. e. {3h^ -f- 3ht + 1^)1 = 33336 (thousands), which I therefore subtract from the given number. [The next step is exactly like the last, and the pupil can supply the demonstration. But be sure that it is done, and the reasoning repeated in subsequent examples till it is familiar and fixed in the mind. ] 2. What is the cube root of 74088? Ans. 42. 3. What is the cube root of 12326391 ? Ans. 231. 4. What is the cube root of 122P97755681 ? Ans. 4961. 5. Wliat is the cube root of 2936.493568? Ans. 14.32. 6. What is the cube root of 12.5 ? Am. 2.321 nearly. 7. What is the cube root of .64? Ans. .8617 +. 8. What is the cube root of .08? Ans. .4308 -f. 9. What is the cube root of .008 ? Arts. .2. 10 to 12. Show that ^^2 = 1.2599 + ; ^5 = 1.7099 + ; ^9 = 2.08008 +. 13 to 15. Show that ^1 = .87 + ; '^^^ = |f ; ^34ti 174 POWEES AND ROOTS. [Note. — The author does not think it worth while for the student to consume his time learning the various shorter methods for extracting roots, the various methods of approximation and the like, inasmuch as no mathematician thinks of using them, or even those here given, but resorts at once to the table of logarithms. It is therefore thought better that the student should spend his time in becoming perfectly familiar with the demonstration of a single method, than to cumber the memory with a multiplicity of processes which he will not remember, and which if he were to remember he would never use. ] FOR REVIEW OR ADVANCED COURSE. 213, JPvob, G» To extract roots whose indices are com- posed of factors of 2 and 3. Solution. — To extract the 4th root, extract the square root of the square root. Since the 4th root is one of the 4 equal factors into which a number is conceived to be resolved, if we first resolve a num- ber into 2 equal factors (that is, extract the square root) and then re- solve one of these factors into 2 equal factors (that is, extract its square root) one of the last factors is one of the 4 equal factors which com- pose the original number, and hence the 4th root. In like manner the 6th root is the cube root of the square root, etc. EXAMPLES. 1. What is tlie 4tli root of 16a' — QQa'x -{- 21Ga"-x^ — 216ax^ -{- 81x'? Ans. ± {2a— Sx). 2. What is the 6th root of 15:^2 — 20^^ ^ x^ — 6x'- -\- 1 — Gx -f 15:^4? Ans. ± {x — 1). 3. What is the 6th root of 2985984? 4. What is the 8th root of 1679616? 214, JProb, 7. To extr^act the mth {any) root of a deci- mal number. Solution. — Any root can be extracted by a process altogether simi- lar to those given for the square and cube roots, or by a simple inspec- tion of the corresponding power of a binomial. Thus to extract the fifth root, point oif by placing a point over units and every fifth figure EVOLUTIOK. 175 therefrom, for the 7th root over every 7th figure, etc. Extract the re- quired root of the largest power of the mth degree in the left hand period for the first figure in the root. Subtract, and bring down the next pe- riod. To form the trial and true divisors, and hence to find the other figures of the root, consider the corresponding power of a binomial. Thus for the 5th root, we have {a + b)^=a^ + 5a*b + 10a3b^ + 10a'^b^ + 5ab* + b^^a^ + [5a* + 10a^b + 10aH'^ + 5ab^]b. The trial divisor is 5aS i. e. five times the 4th power of the root already found regarded as tens. The corrections are lOa^b + 10a'^b^ + 5ab^ , regarding a as the root already found and as tens, and b as the next figure, i. e. the one sought by the trial. In the 7th root the trial divisor is 7a®, and the corrections are 21a56 + Z5a*b^ + Zoa^b^ + 2'la^* + 7ab^ + b\ But in these cases, and much more in the case of higher roots, the trial divisor differs so much from, the true divisor that the process is little better than guess-work. 2 IS, I*Toh, 8, To extract the mth root of a polynomial. HULK, — Having aekanged the polynomial as for division, take the eoot of the first term, for the first term of the required root. Subtract the power from the given quantity, and divide the first term of the remainder by the first teem of the ROOT IN\"0L'VT:D to the next inferior POWER, AND MULTIPLIED by the index of the given power ; the quotient will be the next teem of the root. Subtract the power of the terms already found from the given quantity, and using the same divisor, proceed as before. Dem. — This rule demonstrates itself, as the final operation consists in involving the root to the required degree. ScH. — This rule may also be used for decimal numbers. EXAMPLES. 1. Find the fourth root of 16a* — 96a^x + 216a^x^' — 216ax^ + Six*. 2. Find the fifth root of x^ + 5x* -f lOx"^ + 10j;2 + 5j; + 1. 176 POWERS AND ROOTS. + CO I + § (^ 5^ 7 1 rO O ^r .^ t^ § 1-i ^ + o fO t- (M 1 « CD CO <4H O O -4-3 -+3 O o O o fn 'f~\ ^ I — i ^ ->J y\ lO "co 03 -M A -4J -+J o c3 T^ ^ a -^ E H ^1 '-' lO 1 t- t> t^ lO lO 10 CO CO CO o CM o: Ol t— t- Ti ZC cb (N ic t> t- -rJH •^ CM rH - 1/3 -tH 5 -> s 10 t- CO lo 2^ CO ~ CO CM -"i^ + »B g i—l t^ -* 3.+ ^ CO ^ « + 2 II 1 ^ II 1 2| , > c c •31 1 X *£• X Jo T— 1 > s rH lit ^ 1 + + i ^T II "" : 1 1 5 II - / V ^ ' 1 c (2 o a c c .2 ^1 c r 1 ^ ^ O a b^ 1 1 ^ rH CO ^^ ■^ (M CO rH 00 ■^ to tH t- 1—1 1— 1 CO c»- C ^- » ^ "^ to to CO ^ § r ^^ i? w ^^^ CO 00 CO 00 w QO CO GO O X ^ o p ^' hf^ ^ '^ ^ "x X X ^ I-* Oi CI to ^1 ^1 to C5 Cl •i^ o o o cr :n o o o o o CO to I-- r-" J2j ~ to CO ro to en h-" Ji a 7^ « c- c- C c •f ■*' 11 II II 11 to CO CO CO CO CO <-> o o o to X X X X 00 00 00 00 II II T,+ ;^^ ^^ a' ^+ a to CO , + + CO + ^ a I * a + c2 Ci ^1 CO to CC' rf- CO Ct) CO l-i hi- lO h-^ ti- Q h-' lO hU o ►i^ cc o c hf-- o o o 2 S o "• ::? t o to 4^ o + ^ § i"+ o o ^ ^ '^ -» 1— ' h-» o to to •-' CO lO^ 00 h-i ^ ^ l-» -/«■' — a^x='s/a^(a — x)'=\/d^ X v^a — x=.a\^ a — x. 180 CALCULUS OF RADICALS. 12. Simplify \/a"'+"6". Remit, abva'". 13. Simplify v{a-—b-){a-{-b). Result, (a + h)Va — b. 14. Simplify Z^'o^xK Result, Ibx^^^. Suggestion. — "When the radical has a coefficient, the factor removed from under the radical sign is to be multiphed into this coefficient. 15. Simplify {x + y) v^^< — 2x^y -\- xy'^. Remit, (jp2— ?/')v/^ 1,6. Simplify xyVx^y^ — x^y\ Result, x'^y"'\^x — y. 17. Simplify Vx^^y-''z'"^+^. Result, x-^y-'^z'"^z. 18. Reduce v^f to its simplest form. Result, ^^i>. ScH. — A surd fraction is conceived to be in its simplest form when the smallest possible whole number is left under the radical sign. The reason for this is, not only that the radical factor is thus made simpler, but, if a fraction were to be left under the radical sign the question would arise, What fraction ? Certainly not the least possible, for such i2 ( — r a fi-action can be diminished at pleasure. Thus /(.|t^=/.|-1x'« ^^ 2 Itt = ^x 'j^6 X q?" "^^v/Mfi"' ^*°'' ^^^•' "^itliout limit. Perhaps, if a fraction is to be left under a radical sign it ^\ill be proper to con- sider the expression as simplest when the fraction is nearest unity ; |2 . . IT whence pr is to be considered as simpler than 8 1^^-. 218. Cor. — The denominator of a surd fraction can alivayf< be removed from under a radical sign by multiplying both terms of the fraction by some factor ivhich will make the denominator a perfect power of the degree required. 19. Reduce >J-, >^- ,^- and ^'- to their simplest forms. REDUCTIONS. 181 ScH. — Thef root of a fraction having 1 for its numerator is equal to the same fraction into the root of the next lower power of the denom- inator 20. Reduce/^ — to its simplest form. r to its simplest form. a -\- x-^ Result, "^a^ — x\ a -\- X 22. Reduce — -, -, -v t^ and 4v --— to their simplest 11\7 aWo y bb forms. BesuUs (not in order), •— v/l5^, -Vg^, and — ^• dU 77 23. Reduce 7..v/ — — to its simplest form. a Result, -r- —j-V^. 24. Reduce /]-, and /Jj to their simplest forms. ScH. — The above simplifications can always be effected upon frac' tions, but only upon integral radicals when the integer has a factoi^ which is a perfect power of the degree of the radical. 219, J^rob, 2, To simplify a radical, or reduce it to its lowest terms, when the index is a comjMsite number, and the number under the radical sign is a joerfect power of the degree indicated by one of the factors of the index, R ULE, — Extract that root of the number which corres- ponds TO ONE OF THE FACTORS OF THE INDEX, AND WRITE THIS root as a surd of the degree of THE OTHER FACTOR OF THE GIVEN INDEX. 182 CALCULUS OF RADICAI^. Dem. — Since the 4tli root of a number is one of the four equal factors of that number, if we resolve the number into 2 equal factors, and then one of these factors into 2 other equal factors, one of the latter is one of the 4 equal factors which compose the given number. So, in general, the ?nnth root is one of the mn equal factors of a number. K, now, the number is resolved first into m equal factors, and then one of these m factors is again resolved into n other equal factors, one of the latter is the mnth root of the number. EXAMPIiES. 1. Eeduce V25a^66. MODEL SOLUTION. Opeeation. V25a^— V V'^do^P = ^Ba'^b* = ab'/'^b. Explanation. — The 4th root of 25a"* 6^ is one of the 4 equal factors of it. Hence I first resolve it into two equal factors, one of which is 5a* 6^. Then I resolve 5a- 6^ into two equal factors, or rather indicate it, as the operation cannot be fuDy performed, and have -J^a'^b^. ■v/So^is, therefore, one of the 4 equal factors of Iha^h^. ^u<^' ^y the last problem, >/5a'^b^ = ab\^5b. Hence V25a^6-* = abVUb. 2. Reduce ^'21a'¥. Result, b\^^ab. Suggestions. ^27a'b^ = ^■1/270^= ->/3a6* = b^Sab. 3. Eeduce V~ 64a^. • Besult, 2^— a. 4. Reduce v/25Ga'2j78. 5. Reduce vSln'^m^^. 6. Reduce \/x-^ — 2xy + y^. ■ - 220* JPvob, S, To reduce any number to the form of a radical of a given degree. RULE. — Raise the number to a power of the same degree AS THE RADICAL, AND PLACE THIS POWER UNDER THE RADICAL SIGN WITH THE REQUIRED INDEX, OR INDICATE THE SAME THING BY A FRACTIONAL EXPONENT. KEDUCTIOKS. 183 Dem. — That this process does not change the value of the expression h evident, since the number is first involved to a given power, and then the corresponding root of this power is indicated, the latter, or indi- cated operation, being just the reverse of the former. Thus, x = \/x^. That is, raising x to the mth power, and then indicating the mth root, le&'7es the value represented unchanged. EXAMPLES. 1. Eeduce la^x^ to a form of a radical of the 3rd degree. MODEL SOLUTION. Operation. 7a'x^ = Z^(Ja'^x'-^)* = ^34ya''x^. Explanation. — If I cube 7a* x^ and then extract the cube root of this cube, the result will evidently be the same as at first. Now (7a^ic^)^ = o-i:3a^x^. But instead of performing the operation of ex- tracting the cube root of SiSa^x^, which would evidently return it to 7a^x^, I simply indicate the operation, and have '^'d'i^a^x^. 2. Reduce 2ay — 3 to the form of a radical of the second degree. Result, ^4:a'y^ — 12a?/ + 9. 3. Reduce a — :r to the form of the cube root. |2 4. Reduce /O- to the form of the 4th root. 3 5. Reduce - to the form of the 3rd root. 5 x^ 6. Reduce - to the form of the 4th root. o Result ' ^i-i-o 221. CoR. — To introduce the coefficient of a radical under the radical sign, it is necessary to raise it to a power of the 18-1 CALCULUS OF RADICALS. same degree as the radical; for the coefficient being re- duced to the same form as the radical by the last ride, we have the product of two like roots, which is equal to the root of the product {194^ and 205). EXAMPLES. 1. Introduce the coefficient in 3^ '^2^' under the radical sign. MODEL. SOLUTION. Operation. Sxf^Sx"* = i^27aF X ^^^ = 1^27x^O<~2^ = #^5^5^ Explanation. — Cubing 3x I have 27ic^, the indicated cube root of which is F 27x"*. This is evidently the same in value as 3x. Hence ^xf^^ = t^27x"^ X ^2x*7 But f^2nx^~X^ = fWix^X 1^2x* since the former indicates one of the three equal factors of 27x^ X 2x*, or 54x^, which may be found by resolving each of its two factors 27x* and 2x^ into 3 equal factors and taking one from each group. This, however, is what is indicated by l/27x^ X 1^2x^. Hence 3x1^ 2x'^ = ^27x^ X 1^^^ = ^27x3 X 2x2 = l^six^ [Note. — Doubtless some will think that the argument contained in the above, especially in so elementary a form, is unnecessarily re- peated. But the author's experience leads him to think that such a thing is scarcely possible. There are so very few pupils who get the idea that all the processes of Evolution and Involution are but exten- sions and applications of the simple principles of factoring, that it is thought best to make use of every opportunity to fix this fundamental thought, and elucidate it. Many pupils go through with the forms of demonstration usually given in the Calculus of Radicals, and really get no meaning out of them. ] 2. Introduce the coefficient in -\/2 under the radical sign; inlv/ij;ini;j4;my9. Result., ^\, Ji, ^i, Ji REDUCTIONS. 185 3. Introduce the coefficients in the following expressions under the radical signs; ^a'^'^'lax, {a — x)'^ a + x, iv/4a, {x — y)^x__ y^ and - ^27c o Two of the results are V '^a^ — x^){a — x), and ^{^x—ij)\ 4. Introduce under the radical signs the coefficient in q j~q the following: ^^x\ o" sJi^T* 5x\/25.r-', and a -\-h \a^ a — h\a The last two are V5"» + 2^"»-2^ and \j 222, I^vob* 4. — To reduce radicals of different degrees to equivaloit ones having a common index. BULE.—Bxthb&s the numbers by means op fractional INDICES. Reduce the indices to a common denominator. Perform upon the numbers the operations represented by the numerators, and indicate the operation signified by the denominator. ^ Dem. — The only point in this rule needing further demonstration is, that multiplying numerator and denominator of a fractional index by the same number does not change the value of the expression, i. e., that a ma a^ a;6_ajOTb. Now, x'^ signifies the product of a of the b equal factors into which x is conceived to be resolved. If we now resolve each of these b equal factors into m equal factors, a of them will include ma of the mb equal factors into which x is conceived to be resolved. Hence ma of the 7nb equal factors of x equals a of the b equal factors. [The student should notice the analogy between this explanation and that usually given in Arithmetic for reducing fractions to equivalent ones having a common denominator. It is not identical.^ EXAMPLES. 1. Reduce s/ta'^x and ^J^:mHj to forms having a conamou index. 186 CALCULUS OF RADICALS. MODEL SOLUTION. ,2,A^ Opebation. \/2a^x = (2a^x)' , and '^im'^y = {4,m^yy . But {2a^xf = {2a^xf = (Sa^x^)^ = v'Sa^; and {4:m^yf= (4m *yf == s/lGm^y'^- Explanation. — Expressing the given numbers with fractional in- 1 J. dices, I have (2a^r) and {Am^T/) . These indices reduced to forms hav- L ing a common denominator are | and f. Now (2a^.v) signifies one of 3. the ^100 equal factors of 2a ^x ; while (2a^x)*' signifies three of the six J. 3. ' i equal factors of 2a^x. Hence (2a'''x) *= (2a^x) ^. In like manner {^m^y) I signifies one of the three equal factors of 4m^^ ; while (4m* ^z) signifies J. 2. two of the six equal factors of the same. Hence (4m^t/)^ = (4m*2/) • Finally, as (2a^x) '' is the same as the 6th root of the cube of 2a^x, I have\/8a^x^. In like manner {■im^y)*', meaning the 6th root of the square of A^nx^y, becomes s/l'om^y'^. [Note. — Let the pupil show that the sixth root of the square is the same as the square of the sixth root, etc. ] 2. Reduce V'^'and v^to forms having a common index. Residts, ^/^^^^ v'a 3 Reduce VW, ^/b', \/2~to forms having a common index. 4 Reduce ^^2^"^ ^^^, and ^/xyio forms having a common index. One result is V 729.'c«. 5 Reduce ^ax and 's/hocP' to forms having a common index. Results, (a^x^)^ and {b^x')^- 6 Reduce «s (5^*)^ and (3c)« to forms having a common index. Results, ''^/2f(F/\/a^, '^^/(j25b'. 7 Reduce x'"' and y" to forms having a common index. REDUCTIONS. 187 8. Eediice 2c^/x and ba\^'2y to forms having a common index. Suggestion. — The radical factors can be reduced to forms having a common index without affecting the coeflScient, since the operation does not affect the value of the radical. 9. Kednce 4:^/bx■^y, "H^^^xy, and lOaVSbx to forms having a common index. Results, "l^/^xy, \^a^%lh^x^, 4o^25x^y'^. 10. Eeduce a+c and {a—cY to forms having a common in- dex. Results, (a2 + 2ac+c2)2 and Vct—c, 223, J^vob, 5, To reduce a /inaction having a mono- mial radical denominator, or a monomial radical factor in its denominator, to a form having a rational denominator. RULE. — Multiply both teems of the fraction by the RADICAL IN THE DENOMINATOR WITH AN INDEX WHICH ADDED TO THE GIVEN INDEX MAKES IT INTEGRAL. Dem. — Since two factors consisting of the same quantity affected by the same or different exponents are multiplied by adding the ex- ponents [90), and the sum of the exponent of the denominator and the factor by which we multiply it is an integer, the product becomes rational. The value of the fraction is not altered, since both its terms are multiplied by the same number. EXAMPLES. 1. Reduce — to a form having a rational denominator. V2,x MODEL SOLUTION. «.>x..,.^^x- 5^-^ 3a(2a;)2 X (3a;)2 3a(6as2)^ .rr OPEBATION. -iznr- = '. i = .. =a\/^. ^^* (3x)* X (3a;)« '^"^ Explanation. — Using fractional exponents, I have — _— . Since (3x)^ mtdtiplying numerator and denominator by the same number does not alter the value of the fraction, and as (3x)'^ X (3x)^ makes 3a-, I can 188 CALCULUS OF BADICALS. rationalize the denominator of this fraction by multiplying both its terms by (3x)* This gives -^-; which reduces to a\/6. [K the iiX rationale of these last reductions is not perfectly familiar it should b© given. Thus (6a;-') = ^6) ^ (x^^) ^ = 6 ''x, vrhence — — = ._,^ , and Cancelling the 3x I have un/B.] 2 2. Rationalize the denominator of — -=, 2 2 J Suggestion. =r- = —^, and a^ is the factor which rationalizes ii 3v^a^ 3a^ oa- 3. Rationalize the denominator of —_. 4. Rationalize the denominator of -~ • Result, ^ . 'va a 5. Reduce -;;^ to a form having a rational denominator. Eesult, ^_^^. 12 6 ^3 6. Reduce —=, --=., -t7=^, and — to forms having ra- v3 V3 V 6i v6 3 tional denominators. One of the results is kv/16. v/3~ (3)^X(6)^ ^ig- 3^2" 1 Suggestion. _ — = ; = ^ = Ll = - ,/.J" >/6" (6r^x(6r ^ ^ ^ ScH. — This process is equally applicable to any form of radical fae< ior in the denominator, whether monomial or polynomial. 7. Rationalize the denominator of Suggestion. y/g — X {a — xy^ x (n -{- ^)^ _ s/n'^—x'^ A + iC (a -fx)^' X (a + x)^ " + '"^ REDUCTIONS. 189 8. Reduce — to a form having a rational denom- inator. Besult, ^ ^ 6c — 'Ax-^ 2 2 J:* IPvoh, 6, To rationalize the denominator of a fraction when it consists of a binomial, one or both of whose terms are radicals of the second degree. RULE. — Multiply both tkrms of the fraction by the DENOMTNATOK WITH ONE OF ITS SIGNS CHANGED. Dem. — This rationalizes the denominator, since in any case it gives the product of the sum and difference of the two terms of the denomina- tor, which being equal to the difference of their squares, frees either or both from radicals, as the square of a square root is rational EXAMPLES. n 1. Rationalize the denominator of ^b OPEEATION. MODEL SOLUTION. a—Vh t^a — ^bXai-^b) a^ — b ExpiiANATioN. — I observe that a — ^b will be rationahzed by mul- tiplying it by a -f Vb, since the product of the sum and difference of two quantities is the difference of their squares. Hence multiply- ing both terms of the fraction, so as not to alter its value, I hava a- -\- a^^b 2. nationalize the denominator of ^a—^b 190 CALCULUS OF RADICALS. 3. Reduce =: to a form having a rational denom- inator. 3(2v/2 + 3 \/3-) 6v/2 + Ov^B Result, 8 — 27 — ly 6\/2 + 9v/3 g 5\/2 4. Reduce -^ to a form having? a rational de- 3 — 2^2 nominator. Result, 4 + \/% l—\/l 1 _^ \/2 5. Rationalize the denominators of 3 + v/5 2 + V'A 3v/5 — 2^2 ,^a + x —Va — x — =, and 2n^5 — v^lS \/a + ^ + ^a— X Results, 2 — v/5, iv/2, 9 + |v/lO, and -, X 6. Rationalize the denominator of "^X- -f .77 -f 1 Vx'^ X 1 Vx- -Y X -\-l + Vx^- — X — 1 „ ,, ^2 — ^y^i — ^2 — 2^ Result, ^- + 1 FOR REVIEW OR ADVANCED COURSE. 22S* JProh, 7. — A factor may he found which will ror^ tionalize any binomial radical. Dem. — If the binomial radical is of the form /(« -f- by", or (a + &)" » n — m the factor is (a -f- 5) " , according to {223), i. I. i. If the binomial is of the form V^a* -j- Z^W or «"* -j- Z»" Let a'" = REDUCTIONS. 191 1 1 *■ a;, and b" =y; whence a'" = x' , and h" =y . Also let p be the least common multiple of m and 7i, whence x"^ and y*"^ are rational. But gp rp x'P = a"*, and y*"^ = &". If now we can find a factor which will ren- • r der xf -\- y'', x'P ± y"^, this will be a factor which wiU render a"*-)- &" , ^m _^ ^i» which is rational. To find the factor which multiphed by ^' + 2/"^ gives x^ 4- y'"'', we have only to divide the latter by the x'P 4- yp former. Now — ^ ^x^p-i) — ay(p— *)y -j-x»(p— ')y2'' — x''(p-*)y5'"-(- ± y(p— 1) (^), the -j- sign of the last term to be taken when p is odd, and the — sign when it is even {126). Therefore ic»0'—i) — x*(p-2)y'" -j- a;»(p-')y2'" — x^(p—^'>y^'' -j • 4- y''P— i), is a factor which will render v^a* + -^^ '" rational, x' being understood to be a"*, and r y = h^, and p the L. C. M. of m and n. If the binomial is ^a' — v^6'" , the factor is found in a similar man- ner, and is x»(p-^> -j- x'^^^^^y + x'^p-^^y'^'' -| f- y(»>-i). EXAMPLES. /— a/- J- 1 1. Kationalize va +'v6 or a^ + 6^. i. 1 Solution. — In this example s=l, r = 1, p = 6, x = a* andy=6^. 5. J. a 3. -L A 5. Hence formula (A.) becomes a"^ — a'-b'^ -j- a'^"* — ab -^ a* 6'* — 6"*. This factor multiphed into a^ -f- ^^ gives a^ — &-, as the rational- ized product. 2. Find a factor which will rationalize '^e^ — v u3 or 2 3 22 20^ 186. 16^ 14 12 _1_3 J_6 1_0_ 1_8 8. 2Jl 6 2_4 4 2_7 2 3_0 3_3 i> * , is the required factor, and the expression ration- alized is f?^ — t"'. 192 CALCUXiUS OF RADICALS. 226. Proh. 8, A Trinomial of the form v^a -|- v^b + v^c may he transformed into an expression with hut one radical term hy multiplying it hy itself with one of the signs changed, as v^a + ^b — v^c. The product thus arising may then he treated as a hinomial radical hy considering the sum of the rational terms as one term, and the radical term as the other. Thus, {^a+ ^b+ v^c) (^a + A — ^c) = a + 6 — c + 2^ab. Again, [{a + h — c) + 2^aF] X [(a-f 5 — c) — 2^ ah] = a-^ + 6^ + c^ — 2ab — 2bc — 2ac, a rational result. Ex. 1. Rationalize v^S — v^l _ \/3. Besult, 4. SECTIOJsf IV. COMBINATIONS OF RADICALS. ADDITION AND SUBTRACTION. 227. l?roh. 1. To add or subtract radicals. RULE. — If the badigals ake similar the rules already GIVEN {72f 77) ARE SUFFICIENT. If THEY ARE NOT SIMILAR MAKE THEM SO BY {217-222), AND COMBINE AS BEFORE. If THEY CANNOT BE MADE SIMILAR, THE COMBINATIONS CAN ONLY BE INDICATED BY CONNECTING WITH THE PROPER SIGNS. Dem. — When the radicals are similar the radical factor is a common quantity and the coefficients show how many times it is taken. Hence the sum, or difference, of the coefficients, as the case may be, indicates how many times the common quantity is to be taken to produce the required result. If the radicals are not similar, the reductions do not alter their values ; hence the sum or difference of the reduced radicals, when they can be made similar, is the sum or difference of the radicals. COMBINATIONS — ADDITION AND SUBTRACTION. 193 EXAMPLES. 1. Add v/T6 and ^212^ MODEL SOLUTION. OPEBATION. V i» = 3>/l:, and v^2'12 = ll'v/2. .-. VU -f- ^^2 = 3v/iZ -f- 11^2 = 14^2. Explanation, x/18 = x/y x '^- But the square root of the pro- duct equals the product of the square roots; hence ^^d X ^ =3V2.* In hke manner x/24:2 = x/l21 x 2 = llv/2. Therefore V'lb -f- 'v/242 = 3V'2 4- 11 v/2. But three times any quantity, as ^2, and 11 times the same are 14 times that quantity. . •. 3^2 -f- H'^^ = 14^2. 2. Add \^'24:oxy' and vj^^xy^. Sum, lly^'Sx, 3. Add ^5UU and '^lOH. 4 Add v^a-?/ and ^""c-y. ^um, (a + c)v/|/. 5. From v/6(J5"take the ViOST X>?/f., Sv^S. 6. Add ^GU5 and — v/i05. 7. Add 3^ - and 2^ J- ^N5 NflO. 8. From 3 J^ take -2^^- ^iff., |v/lO. 9. AVhat is the sum of . - and ;,. I-? ^/28., -v/3. 10. "WTiat is the difference of ^2ax- — -ku.r + 2a and ^y'lax- -f- 4a j; 4- i:a ? Ans., 2^2a. * If this reasoning is perfectly familiar it may do to omit it. But it is far better to repeat a reason when it is already clearly comprehended, than to omit it when there is the least doubt. If it is famiUar it can be given in an instant ; and if it is not fami- liar it ought to be made so by repetition and further study. 194 CALCULUS OF RADICALS. "Why should no sign be given to the last answer ? If the problem read, From v/2ttx=^ — 4ax + 2a take >/2ax^"+lax + 2o, why would the answer be — 2v/2a? 11. What is the sum of {a — xy^xy and {a + xY^xyl ^?2S.,2(a2 + x'')^x.j. 12. What is the difference of (a — x)^Vxy and {a^xy^xy ? 3 1 Ans., ^ax'^y'^. 13. Find the sum of 8 J5, Vm, —1\^Th, and A-. Sum, 4\/ij. 14. From ^ I^Z^ take ^llE Bern. , {^^x -r,) ^ ^. 15. Add «Ji+r-i* and ^Ji+rf]*. Suggestions. « ll_l-r-|-' = a>Jl+— = a\j^^ -^ = ^5Va^-f6^ lu like manner ?> jl + [-1 ' = ^' \^ J 4. ^^i .*. The sum is (a^ + 6^) V J^lTf^ = (J -{-b'){J -^b'Y^ == 16. From (a — x) v^a- — :z;' take {a — ■ x), 'Sa X Bern., (a — x — l)v/a2 — x-\ 17. Add and - X-^^^X'^ 1 X — "^x- 1 Suggestions. — Beducing the given fractions to a common denomi- nator, they become X — s/x- — ^1 and x-j-s/x- — 1. Hence the sum is 2x. 18. Add -^^^n + v^.~i ^^^ ^^^r jl_V7~y Vx'' +1 — ^x-'- — 1 Vx^ + 1 + ^^•- — 1 Sum, 2;r2. COMBINATIONS — MULTIPLICATION. 195 MULTIPLICATION. [Note. — Although the principles embraced in this section have been evolved in the preceding chaj)ters, under appropriate heads, their im- portance is so great that it is thought best to collect them here, and in connection with a careful review, extend somewhat their apphcation. ] 228, I^rox>. 1, The product of the same root of two or more quantities^ equals the like root of their product. Dem. — Thatis^yx X \/y = \/-^y- This is evident from the fact that ^xy signifies that xy is to be resolved into m equal factors. If now each factor, as x and y, be separately resolved into in equal factors and then the product of one factor from each be taken, there -wdll be m such equal factors in xy. Thus ^x is one of the m equal factors of X, and ^y is one of the m equal factors of t/. Hence [y^X 'Yy] X \_\/x X s/y] X Lv^X v^J/] etc., torn factors of ^.c X \/?/, makes up xy. Therefore \/x X s/y = \/xy. (See Arts . 205 and 202, ) 229, I^rox>, 2. Similar Radicals are multiplied by m^ultiplying the quantities under the radical sign and writing the product under the common sign ; Or by indicating the root by fractional indices, and, for the product, taking the common number unth an index equal to the sum of the indices of the factors. Dem. 1st. — Since similar radicals are the same root of the fame quantities, as X^x X v^x, this is only a partictdar case under Br op, 1, 1 1 2nd. x'" X X "' signifies that one of the m equal factors of x is to be multiplied by another of the m equal factors, or by itself. This gives 2 2 of the m equal factors of x, which is what is indicated by x"*. 230. J^rob, 2. To multiply radicals. RULE. — If the factors have not the same index, re- duce THEM TO A COMMON INDEX, AND THEN MULTIPLY THE 196 CALCULUS OF EADICALS. NUMBERS UNDER THE RADICAL SIGN AND WRITE THE PRODUCT? UNDER THE COMMON SIGN. Dem. — (This is the same as Prop, 1,) EXAMPLES. 1. What is the product of ^^1 and ^J) MODEL SOLUTION. Opekation. x/2= VbTand 1^3 = Vo. . • . •v/2 X 1^3"= Vs X t^ 9 Explanation. V 2 = ^/y, since the former is one of the two equal factors of 2, aBcl the latter is three of the six equal factors of 2. In like manner f'd=^'^. Consequently, >/2 X 1^3 = v'8 X ^9. Now since the joroduct of the same root of two numbers is equal to the lik« root of the product, -v/s X 4/9= ^W. 2. Multiply v^te by ^2aa Prod., 'y^2a^. 3. Multiply v^a — xhj Prod. Va^ — ba'x + 10a^.x-- — lUa-x'^ + ^ax^ — xK 4.Multiply Jjby JA. p^od., i 5. Multiply ya by ^^3. Prod, 3^"^^= 3"^^ = ^6561. 6. Multiply v''2a.r by v/2a^. Prod., (2ax)^^ = y4096fti2j;i2 7. Multiply ^^ by ^| Pro6?., fi - = -v^iyST \3 3 8. Multiply 3v/2a^by 2^^^. COMBINATIONS — MULTIPLICATION. 197 SuG. — Here we have the continuous product of 3, ^'lax, 2, and "^xy. But, as the order of the factors is immaterial, we may write 3 X 2 X 9. Multiply 3>g? by 6^i. Pro^., 6 v^236196. 10. Multiply 5a^ by da^ IL Multiply 2v/a6'by 3^a6! 12. Multiply 4.ah^hy5ah\ 13. Multiply Sx^y^ by 2x^y^ and express the product without fractional exponents. 14. Multiply. I- by A-^ and express the product without the use of the radical sign and in its simplest form. Prod., Jl(9000)^. 10 i. JL 15. Multiply a"* by 6". 1 1 1 Prod.,a''b'^ or, Va"Z>'", or (a"6'")"'". 16. Multiply iv^byi^iO. Frod.,^t/250. 17. Multiply a^^x, h^y, and cv/J together. Prod., abc^^x"^- y'"''- z^\ 18. Show that 2^^ X 16 = 16vl2. 19. Show that ^24 X 6V3 = 6v/l2. 20. Multiply 2v/a — Z^bhy Sv^a + 2v^. OPERATION 2v/a — 3v/6 3-v/a 4- 2 x//7 "^ 6a — 9^/06 4- 4x/a^ — 65 6a — by^ajb — 66, or 6(a — 6) — Sy oS! 198 CALCULUS OF RADICALS. 21. Multiply 3 + v/5 by 2 — ^. Prod., 1 — V^. 22. Multiply v^ + 1 by v/2 — 1. Prod., 1. 23. Multiply IW2 — WTb by v/(5 + /a _ _ Prod., 2^3 — v/lO. 24. Multiply V 12 + ^ ly by V 12 — >/l9. SuG. Since these two radicals are of the same degree we may mul- tiply the quantities under the common sign, and write the product under the same, which gives v 144 — 19 = 5. 25. Multiply a"- — av/2 4- 1 by a- + ^^^2+1. Prod., a^ + 1. 26. Expand {x^- + DC^^ — ^^3 + l){x^ + ^v^3 + 1). Prod., x^ -\- 1. 27. Multiply 3^45 — 7^^ by n/i| + 2v'ij|. Prof^., 34. 28. Multiply Va + c^6 by v/fl _ c^6. Proc?., a — d'-'^bK DIVISION. 2St., ^rop. The quotient of the same root of two quantities equals the like root of their quotient. Dem. —Let m be any integer and x and y any numbers ; we are to to/- I— m |— prove that ^yx-^^y, or ^ = l^y- Now, that ^^ ^y dent, since ~— raised to the mth power, that is yy V^ X \/x X Z^x X v/a: - - - to m factors x , .. - ^^ _ ^._ ^1-^ ^— = - ; whence it appears \/y y< \/y ^ '\/y X C^y — to m factors y isevi- X that - — is the mth root of - or equals ^ r. (See Abts. 206 and y-y 2/ -' 202,) \y COJVIBINATIONS — DIVISION. 199 232, Proh, 3. To divide Radicals. R ULE. — If the radicals are of the same degree, divide THE NUMBER UNDER THE SIGN IN THE DIVIDEND BY THAT UNDER THE SIGN IN THE DmSOR, AND AFFECT THE QUOTIENT WITH THE COMMON RADICAL SIGN. If the radicals are of different degrees, reduce THEM TO IHE SAME DEGREE BEFORE DIVIDING. Dem. — [Same as above ; or, it may be considered as the converse of the corresponding case in multiplication. ] EXAMPLES. 1. Divide ^'6a-y^ by ^2^. MODEL SOLUTION. Opeeation. v^3a-?/3 = t/27a"y\ and ^2ay = y^a-yK . v/3^ _^^27^3_ J27^B_ J27^^_le, ^,^ , , ' ' ^2^y VT^ Nj^«^r nJ ^ 2^43.a2/. Explanation. — Since \/'da-y'^ = \/27a^t/*, and ^'Aay = v^-ta'^j/S ' rrrr- '-= ' : . And since the quotient of the 6th root of two ^2ay V4a=y2 ^27a&j/9 (27aV numbers is the 6th root of their quotient, v^ =f \ .^„ ' , which, V 40=2/2 \4a-?/2 by performing the operation indicated, becomes .^ / — j — ; and by re- ducing so that the number under the radical sign shall have the inte- gral form, this becomes - 3/432a-?/7 2. Divide ^/l^Ea^x^ by \^5a^xy. QuoL, 5v^. 3. Divide v/3 by ^. Quot, ^^=2'^^2 4. Divide J? by WL QuoL, -Myii. y 5. Divide '^2a-^x3 by y2a^x^ QuoL, s/Aa'x^. 6. Divide ^72 by v^. ' QuoL, V^'S. 200 CALCULUS OF RADICALS. 2 1 11 7. Divide x'^y^ by x'^y^, and express the quotient without fractional or negative exponents. Quot, « F. 8. Divide 24v^ by Q^a^yK 6^^ ^a^ 'v'aY "S'^'y" y'^y 9. Divide 125^x'^y^ hy25'^x^y^, expressing the quotient without the radical sign. QuoL, 5x^y^. 10. Divide v^ by ^4. QuoL, v^a^. 11. Divide 20^200 by 4^. QmL, 5^5- 12. Divide ' k by 'k. 13. Divide Jv^ by v'2 + Sv^I Suggestions. s/^ -\- S^/k = ^\^k + 3\/4 = 5>/i Whence Wi ^vT i 1 ^2 + 3^4 5\/4 5 10 14. Divide ^a^ — 6-^ by ^^TITft. Quo/5., ^7:^^. 15. Divide {a^b-^c)^hj (ab)K QuoL, a'^bc. 16. Divide 200 by v/io. Q^o^., lOv'io, or (10)^. 17. Divide a^^x — ^^bx-^-a'A/ — v^by by ^x + v^t/. Suggestion. — Observe that a'^x — '>/bx'-\- a-s/y — ■\/6y" = a( Vic + Vy) — x/6 (Vx + Vj/ ) = (a — v/6 )(^x + v/y ). Whence (a — V'^ )(x/x -f- ^V) .- we have — — == a — v'6. Vx + V^/ 18. Divide a -\- b — c -\- 2^db by \/a + \/6 — ^c. COMJBINATIONS — INVOLUTION. 201 OPERATION. o + 2-^06"+ 6 — c \-y7i -[- 'Jb — j'g a -f- ^y ab — v'ac V a -\- 's/b -j- -v c >/ab -\-b -\- Vac — c V^a6 -(- ?) — V oc "/ac -|- V 6c — c -s/ac -f- ^^^^c — c 19. Divide h^a^- — 6'^ by av'(a + 6)2. 20. Divide ^a-' — ^- hy a — x. Quot, . ^ + '^. Na — X ^1 T^iwl.., -^-Iby., ,^ + 1. g.r)/./'^'^--'^ nJ^ + 1 ^ N/o;-! ^ ^K'^+1) INVOLUTION. [Note. — The principles heretofore given are sufficient to enable us to involve radicals, but it may be well to collect and review them.] 233, J* rob, 4. To raise a radical to any power. R ULE. — Involve the coefficient to the required power, AND ALSO THE QUANTITY UNDER THE RADICAL SIGN, WRITING THE LATTER UNDER THE GIVEN SIGN. Dem. — This results directly from the principles of multiphcation of radicals {230). Thus, to raise a \/b to the ?nth power, is to take m factors of a\/b, which gives a^b X a-C^b X a\/b, etc., to m factors. But as the order of the factors is immaterial {85) this may be written oxm to m factors X >7t^X \^b X v^^ to m factors. But aaa to m factors is by notation a™, and v''^ X v^^ X v'^ to m factors is by {230) v'^b'". .-. The mth power of a^'b is a^'^b'": q. e. d. EXAMPLES. 1. Raise | ^ | to the 3rd power. 202 CALCULUS OF RADICALS. MODEL, SOLUTION. Operation. {ix/Jf = iv/f • ix/^-Wl = ^ • i • i X v/f • \/| ' \/l = ^fV^^ or rizVl = ^f a v/ia Explanation. — The cube of i\/f is three factors of i\/| multiplied together; but as the order of the factors is immaterial this may be considered as 3 factors of I, or -^, and 3 factors of \/|, or f v/f . Hence (3 v^l)^ = 2V X 5 v^5 = rl^^v^l- Which by removing the denominator from under the radical sign becomes ^yjv/lO. 2. Raise 2'^8a'-^6 to the second power. 3. Raise -"^^x-y to the 5th power. 4. Raise — ^v)? ^^ ^^^ ^^^ power. 5. Square 3 — v/2. Square, 11 — 6v/2. 6. Cube v^ — \/3. Cube, llv/2 — Ov's. 7. Cube 2^x Cube, S{x — y)^^ — y,oY^{x — y)^. 234, Cor. — To raise a radical to a power v)hose index is the index of the root, is simply to drop the radical sign. Thus, the square of v/a6 is ah, the cube of "^'Ix/^y i^lxHj, the square of v^ — 2a^6 is — 1a% the 5th power of v^a* — b- is ScH. — This process of involution is a special case under (194). EVOLUTION. 23tj. I^rob. S. To extract any required root of a mono- mial radical. B ULE. — Extract the required root of the coefficient, AND OF THE QUANTITY UNDER THE RADICAL SIGN SEPARATELY, AF- COMBINATIONS — EVOLUTION. 203 FECTING THE LATTER WITH THE GIVEN RADICAL SIGN. ReDUCE THE RESULT TO ITS SIMPLEST FORM. Dem. — The nth. root of aX/b, signifies one of the n equal factors of a-^b. Hence if we resolve a into n equal factors, and v'b , into n equal factors, the product of one of each is the 71th root of aZ^b. Thus one of the n equal factors of a is -s/a, and of y^'^b, v ^]j^ for ^ ^b X v'y^ X ^ -^^ to 71 factors, is V Vb - b - b • bio n factors = 'l/b{233f 234). If now we take the product of one of each set of factors, that is \)\i V ^b, we have the 7ith root of aVb, since we have one of its n equal factors, q. e. d. EXAMPLES. 1. AVhat is the 3rd root of 4^ Ba^^? MODEL SOLUTION. Opeeation. V4v/3a3a; — ^ 1^4 V i^3a-a; = t^^ y^''Sa^x = ' Explanation. —The cube root of 4:^'da'x is one of the 3 equal fac- tors which compose it. In order to find this, I resolve each of the two factors 4 and -v 3a' x into 3 equal factors, and take the product of one of each. 4 resolved into 3 equal factors becomes F 4 X 1^4 X 1^4. (A process which in reality is only indicated.) In like manner y/da 'X, resolved into 3 equal factors becomes W V3a^ X 1^^3a^X V^ct^x or V f^3a^ X V ^Sa^ X V fSa^x since the root of the product equals the product of the roots. Now taking one factor out of each of these groups I have F 4 V ^3a^x, and as three such factors could be formed from the number, ^ 4 \/ ]7 Sa^x is the cube root of 4v/3a^x. But this expression can be reduced to a more simple form, by observing first that the square root of the cube root is the 6th root, and then reducing fT to the same radical form. Thus I have 1^4 V f 3a^= fI^y3a-x = ^I^sy-S^= v'48a^x by (233).'' * The author can hardly refrain from apologizing for so elementary a demonstra- tion at this stage of the work. But if the spirit of the treatment of radicals is ap- prehended, the reason for this will be understood. If the pupil once comes into full possession of the idea that factoring is the basis of the whole subject, he has th« key to all its difficulties. 204 CALCULUS OF RADICALS. 2. Extract the cube root of va^x^. Boot, x^ a\ 3. Extract the square root of 32'^iy2a«.'r^ SuG.— The square root of 32l^/iy2a^ = 4n/2 X v^lS^' = 4v^2 X 2a^^3a^= 8a-/ 2-^30^= 8a-^8-5'3^^=8a-^2ia^. 4. Extract the cube root of - a^^h. 8 5. Extract the 4th root of 16^2^^^. Boot, ^Va^x. 6. Extract the cube root of (a + j^) Va + x. Boot, V a ^ X. 7. Extract the cube root of ovlo- Boot, q-y's^. 8. Extract the square root of 7v Jh- Boot, -v/2. 230, ScH. — This operation is but a special case of qfeding a quan- tity with any given exponent {104), and the examples may be performed according to the rule there given. Thus, to extract the cube root of ^V^ax- ; putting it in the form 8 X (3ax,")^ and dividing exponents by 3 (multiplying by \) we have 8-*(3aic")'5 __ 2A/3ax-. The pupil may solve the following in this manner : 9. Extract the 5th root of ^^1x^^. SuG. -s/32x"^ =(32)*x5. Multiplying exponents by \ we have (32)"iV But (32)'^"' = 1(32)^ =y2. .• . V^ ^32^^= xn/'Z But the most simple way to solve this particular case is\/-v/32x"- = V ^32xi^= V2x2 = x-y% Or by the .rule given {235). 10. Extract the square root of '^49a2. 11. Extract the cube root of G4v/8a«. COMBINATIONS — EVOLUTION. 205 12. Extract the 5th root of 486a ^ia^ Boot, 3^2^. rOR REVIEW OR ADYANCED COURSE. [Note. — This article should not be taken till after Quadratic Equa- tions, and is not especially important in an ordinary course at all. ] 237* Pvoh. 0, To extract the square root of a binomial, one or both of luhose terms are radicals of the second degree. Solution. — Such binomials have either the form a + n\^b or w-v/a -4- ?i'v/6. Now observing that (x -j- y)'^ = x- + 2xy -j- y^, we see that if we can separate the first term of any such binomial surd into two parts the square root of the product of which shall be ^ the other term, these two parts may be made the first and third terms of a tri- nomial (corresponding to x- -f- 2a-?/ -f- y-), and the middle term being the second term of the given binomial, the square root will be the sum or difference of the square roots of the paxts into which the first term is separated. EXAMPLES. 1. Extract the square root of 87 — 12^^. Solution. — Let x = one of the parts into which 87 is to be sepa- rated, and 87 — x the other. Then we have v'a-,(^87 — x) = — 6x^42, or squaring, 87ic — x'^ = 1512, or ic^ _ 87x = — 1512. . • . a; = 63 or 24, and we have ^87 — 12^42" r== ^63 — 12^42^+ 24. Nom^ ^03" = 3 -v/? and 'v/24 = 2 v^6, and as the middle term of the trinomial is negative and twice the product of these roots, its square root is 3-^7 — 2^e. {123.) 2. Extract the square root of Sv'q + 2^12, SuG. 3'>/Q i^ ^^ ^^ separated into two parts. Let them be x and 3-v/6 — X. Thenx(3-v/6 — x) = 12. Whence x = 2-^ 6 or V'g^ and v^3v/6 + 2^/12" r= ^2Vg + 2^12"+ v/6 = ^^2^6 + ^^6" = t/24 -f ^Q. 206 CALCULUS OF RADICALS. 3. Extract the square root of 12 — v^l4U. Root, \/7 — ^. 4 Extract the square root of 11 + Q^2. 5. Extract the square root of 13 + 2\/30. 6. Extract the square root of ax — ^a^ax — a-. SuG.— Letting y be one of the parts into which ax is to be separated, the equation from which its value is found is y^ — axy=^ — a'{ax — a-) or — a^x -\- al Whence y = ax — a'^ ox a"^, and the parts are ax — a'^ and a-. Hence \/ax — 2a^/ax — u^ = V{ax — a^) — '■2a\/ax — a^ -|- a'^ = Vax — a-^ — -v/o^ or a — 7. Extract the square root of 2a + 2^ a- — b^. IMAGINARY QUANTITIES. 238. An Imaginary Quantity is an indicaied even root of a negative quantity, or any expression, taken as a whole, which contains such a form either as a factor or a term. Thus V — b, ^ — a:\ h^ — h\ 2+^ — 4, ^ — 6, 3 — v/ — 1^ etc., are imaginary quantities. 230, ScH. 1. — It is a mistake to suppose that such expressions are in any proper sense more unreal than other symbols. The term Impos- sible Quantities should not be applied to them : it conveys a wrong impression. The limits of this work prevent anything more than a mere explanation of the method of multiplying or dividing one imagin- ary of the second degree by another. A fuller development of the theory of these interesting and important symbols will be given in the Higher Algebra. 24:0, ScH. 2. — A curious property of these sjonbols, and one which for some time puzzled mathematicians, appears when we attempt to multi- ply -v/ — a by v' — a. Now the square root of any quantity multi- plied by itself, should, by definition, be the quantity itseH ; hence COMBINATIONS OF IMAGINARIES. 207 V — a X '^ — a = — a. But if we apply the process of multiph'- ing the quantities under the radicals we have 's/ ■ — a X ^ — « = •>/a- r= -}- a,as well as — a. What then is the product of x/ — a X -v/ — a ? Is it — a, or is it both -|- a and — a? The true product is — a ; and the explanation is, that \/a~ is, in general, -\- a or — a. But when we know what factors were multiplied together to produce a~, and the nature of our discussion limits us to these, the sign of -v/a^ is no longer ambiguous : it is the same as was its root. 241. Projy. Every imaginary term of the second de- gree {and in fact of every other degree) can he reduced to the form. mV— 1 in which m is not imaginary, m may le ra- tional or surd. Dem. — Let V — X represent any such expression. Then V — x ^^ a/x^ — 1) = -J X'J — 1, which is the required form. 24:2- ScH. 3. — By means of this proposition and the property no- ticed in Hch. 2, the multipUcation and division of imagiuaries is ef- fected. EX^VMPLES. 1. Multiply 4v^^=l5 by 2v/^^. Operation. 4^/ — o = 4^3 x V — 1. Also 2v/ — 2 = 2-^2 X -y"^^. Hence, 4n/ — 3 X 2^/ — 2! = 2 X 4x/3 X -v/^ X y — 1 X ^ — 1 = 8^ 6 X ^ - 1 X v/^ITi; =, _ 8^ ^^ since ^ — 1 X ^ — 1 = — 1. 2. Multiply 3v/ — 5by4v/ — 3. Prod, — 12\/l5. 3. Multiply ^ — X'' by v — yK Prod., — xy. 4. Multiply 2^^^^ by 3v/'^^ Prod, — 36. 5. Multiply 2v/ — 6, 5v/ — 4, 3^^ — 7 together. ProtZ., — 60 v^v' _ 1, or— 60v/— 42. 208 CALCULUS OF RADICALS. 6. Multiply 24 + v/ _ 4i) by 24 — v^ — 49. SuG.— We have here the product of the sum and difference of two quantities which equals the difference of their squar es. Hence (24 4- -v/^^l9) X (24 — y^=l9) = (24)2 _ (v^_49)2 = 575 _ ( _ 49) = 576 -I- 49 = G25. 7. Divide v/ — 16 by \/ — 4. Opeeation. ^ — iG = ■i'J — 1, and v/ — 4 = 2^ — 1. Hence ■s/ — 16 _ 4'v/^ri; 9 ScH. — A superficial view of the case might lead the pupil to think that the quotient was + 2. Thus, noticing that the radicals are simi- -v/— 16 -16 ._ lar, he might conclude that = . I = V 4 = + 2, an incorrect result. 8. Divide Gv^ _ 3 by 2v/ — 4. QwA., -^'6. 9. Divide _ v/_ 1 by — 6v/ — 8. Qiiot, — v^S. 2 1_ 18 10. Divide 1 + \/— 1 by 1 — -^ — \. 1 _|- y _ 1 . Opeeation. — Writing the example thus ' and rational- izing the denominator by multiplying both terms of the fraction by , 1 4- 2V^:ri _ 1 2x/"^=~l 1 4- V — 1 there results, (see JSc. 6) .. __ ■., = 3 The example may also be performed thus : 1 _|- ^^1~I 1^ _ V^iri -f 1 1 + y/ — 1 y — 1 Since 1 divided by — »/ — 1 gives ^ — 1, as — v' — 1 X '^ — 1 = 1. /■ SYNOPSIS. 209 Synopsis for Eeview. Power. — Degi-ee of. ] Sch. 1. P. and E. correlative. Koot. — Degree of. f Sch. 2. Square, Cube, etc. [ 1 \T 1. r Cor. Transf 'ring TT ^ T ^ J o ti f-^^' J afactorinafrac- Exponent or Index ^2.+ Fraction. ^.^^ ^^^^ ^^^ (3. -Int. or Fr. term to another. Radical Real j Rational. I Irrational. Imaginary. How indicated < ,„ Sim. Rads. Rationalization. To a£fect with Expt. Involution. Evolution. Calculus of radicals. Frob. 1. To any power. RuiiE. Bern. [Co7\ Signs. cl. +m. Prob. 2. To affect with Expt. Rule. Deyn. 4 2. 4-^. (3. — m or - T* Coi\ 1. Terminates. Cor. 2. No. Terms. Cor. 3. CoeflBcients. Cor. 4. Expt's in ea. term. Cor. 5. Rule. Cor. 6. («— Jf.- /Sc^. Signs. Cor. 1. Same as (193). Cor. 2. R. of Prod.= Prod, of Roots. Co7\ 3. R. of Quot.= Quot. of Roots. ( Sch. 1. Prob. 2. Sqr. R. of Poly. Rule. Bern. ■{ Sch. 2. ( Sch. 3. Prob. 3. Sqr. R. of Dec. No. Rule. Beiii. Prob. 3. Binomial Formula - Prob. 1. Roots of Perfect Pow- ers. Rule. Bern. i Sch. 1. I ^c^. 2. Pro5. 4. CubeR. of Poly. Rule. Bern. | |^^- 2 ( Sch. 1. Pro5. 5. CubeR. of Dec. No. Rule. Bern. ■{ Sch. 2. ( Sch. 3. Prob. 6. Any R. whose index composed of factors, 2 or 3, Prob. 7. Any Root. Solution. [Genl. Scholium. 210 SYNOPSIS. SYNOPSIS FOR -KEYIWN.— Continued. Prob. 1. Remove factor. Rule, j Sch. Simplest form. Dent. I Cor. Denom. of surd. P7^ob. 2. Index composite, etc. Rule. Dem. Prob. 3. To given index. Rule. Dem. ^ Prob. 4. To common index. Rule. Dem. Prob. 5. To rationalize Moii. Denom. Rule. Dem. Prob. 6. To rationalize Binomial Denom. Rule. Dem,. Prop. 1. To rationalize any binomial. Dem. Prop. 2. To rationalize any trinomial. Dem. Prob. 1. To add and subtract. Rule. Dem. i Prop. 1. Prod, of roots=equal root of product. Dem. \ Prop. 2. Similar Rads. , how multiplied. Rule. ( Prob. 2. To multiply Radicals. J Prop. Quot. of roots=root of quots, Dem. Dem. q I Prob. 3. To divide Radicals Prob. 4. Involution of Radicals. Prob. 5. Evolution of Radicals. VProb. 6. ^ Rule. Rule. Rule. Dem. Dem. [Sch. Dem. Def. '^b and '^^a-^'^b, etc. Sch. 1. Called Impossible. Sch. 2. Reason for name. Prob. 1. To add or subtract. Rule. Devi. Prob. 2. To multiply. Rule. Dem. Prob. 3. To divide. Rule. Dem. Test Questions. —By what must numerator and denominator of be multiphed to reduce it to a simple fraction ? Give the ^' . /- ^- various significations of an exponent. Perform the operation x/2 X 1^ 3, and explain the process. Repeat the Binomial Formula, and by means of it expand (1 — x^)^. Demonstrate the rules for square and cube en ^^ ^ a -\- hx. — '>/ (fi A-h'^x^ ^rri _i_ M-r,"- —a root. Show that — X ' __ n^ a -f- t> .c a ^j^^^ a — hx-{- y ^2 _j_ ^.^2 5^ sign is given to a square root? Why? To a cube root? Why? What is the value of ic" ? [Note. — Here ends the subject of Literal Arithmetic. The student is now prepared for the study of Algebra, properly so-called ; i e. , The Science of the Equation. ] PART II. ALGEBRA. CHAPTER I. SIMPLE EQUATIONS. SECTION L Equations with One Unknown Quantity. DEFINITIONS. 1, Ati Eqiiatiori^ is an expression in mathematical symbols, of equality between two numbers or sets of numbers. III. 3x — 2a"y = — '— is an equation because it is an expres- sion of equality between 3ic — 2a'^?/ and 2, Algehl'Ct is that branch of Pure Mathematics which treats of the nature and properties of the Equation and of * Do not pronounce this word " Equazion." For this common error there is no authority. " Equashun " is the correct pronunciation. 212 SIMPLE EQUATIONS its use as an instrument for conducting mathematical in- vestigations. 3, The First If ember of an equation is the part on the left hand of the sign of equality. The Second Jlf ember is the part on the right. 4, A JV^ufnerical JEquatiou is one in which the known quantities are r-epresented by decimal numbers ; as 12^2 _ 3^ = 48. ScH. — This is another instance of the unfortunate, because incor- rect use of the word number. 25 is no more a number, or representa- tive of a number, than is a or x. The former is a number expressed in the decimal notation, and the latter in the Uteral. But perhaps we must retain the term. 5, A. Literal Equation is one in which some or all of the known quantities are represented by letters ; as 4cj;2 — 2 ax ■ — c -f 'Sby = 8 G, The I>egree of an Equation is determined by the highest number of unknown factors occurring in any term. I1.L. ax — tx' = c -|- £c^ is of the 3rd degree; a'^x — 4a; = 12 is of the 1st degree ; x'^y"- = 18 is of the 4th degree, etc. 7. A Simx^le Equation is an equation of the first degree. III. y ^=: ax-\-h is a simple equation, as also is — f- 4a; = J-. + 5. S, A Quadratic Equation is an equation of the second degree. 9. A Cubic Equation is an equation of the third degree. A IMquadratic is one of the fourth degree. WITH ONE UNKNOWN QUANTITY. 213 10. Equations above the second degree are called mgher Equations, TRANSFORMATION OF EQUATIONS. 11, To Transform an Equation is to change its form without destroying the equality of the members. 12, There are ybwr principal transformations of simple Equations containing one unknown quantity, viz : Clearing of Fractions, Transposition, Collecting Terms, and Dividing by the coefficient of the unknown quantity. 13, These transformations are based upon the fol- lowing AXIOMS. Axiom 1. Any operation may be performed iipon any term or upon either member, which does not affect the value of that term or member, without destroying the Equation. Axiom 2. If both membei^s of an Equation are increased or diminished alike, the equality is not destroyed. 11, JProh, To Clear an Equation x-\-l ^ ' = 6x -j- 12. (The term 7;,. ]^3 ~ — 7—— is multiphed by 3 by dividing the denominator, and by the 6x -f- 3 other factor, 3, of 9, by multip>lying the numerator.) Now dropping 21a*, 39 6cc from both members and subtracting 7, we get — ^ — -— ~ = 5. Ax — j— 1 Whence x = 4. ,n n- 4.r+3 7.r — 29 8^ + 19, . , 41. Given -^- + ^^^—3- = -^ to find x. 4x4-3 SuG. — Multiply numerator and denominator of the term — - — by 2, WITH ONE UNKNOWN QUANTITY. 227 transpose and unite it to the second member, and there results Ix — 29 13 ^^^^ — = -— . Whence x = 6. 5x — 12 18 .. _. 9.r: + 5 , 8a;— 7 36.^; + 15 ,10^, . , 42. Gwen-^ + g;^^=-^^— + — , to find x, and verify the result. .or.- 6^ + 7 2^ — 2 2^ + 1 •., , .. 43. Given — -— = — - — , to find x, and verify 15 ix — b 6 the value. .. ^. 6a: + 1 2^ — 4 2^ — 1 . _ , ... 44. Given — -— — , = — , to nnd x, and verity 15 Ix — 16 5 the result. Result, x = — 2. 2-f-l —4 — 4 _ — 4 — 1 _ii_i 15 ~_14 — 16~ 5 ' ^^ 15 15 — 12-4-1 —4 — 4 —4 — 1 11 4 Verification. — -! — —z = = , or — -— 45. Given —^ — '- — -H- H ; = 8, to find x and verify the X + 3 ^+1 ^ value. XX Qi 46. Solve for x the equation - + = •; . a — a -\-a SuG. — Clearing this of fractions and solving in the ordinary way, we have h'-x — a'^x -j- ohx -\- a'^x = a-b — a^, or {b^-]-ab)x = a"b — a\ Whence a-b — a^ a-(b — a) g. —^ or — — ■ . 5^ _|- ab' b{b-\-a)' A more elegant solution is obtained by first uniting the two terms « , , , . , , ,■■ ,. bx — ax A- ax a of the first member, which makes the equation ; =-— — , a{b — a) b-\- a a-{b — a) Whence x = a(b — a) b -\- a b{b -j- a) 47. Solve -{x — -) _ -(^ _ -) + - (^ _ -) == 0. SuG. — Performing the multiplications and transposing, we have X x , X a a a 5 2 2-^ + 4= 6--12+^'^'l2^ = -15"- ^^"^"" ^^'^^ ^y r Q j-j- we have x = --r=a. The pupil should solve this in the ordinary way and compare the processes. A" 228 SIMPLE EQUATIONS 48. Solve ^ - ^- = *— ^. . Result, y = % a a c ad 49. Solve ^^ = 3+ ^ a — 26 2a — 6 2b^ Result. X = 2a — 5b -\ . a X SuG. — Transpose r to the first member and unite the terms, 2a — 6 giving -— X = 3. . • . ic = 2a — 56 + — -. 2a2 — dab -{- 26^ a 50. Solve 3^ — a = X — . ^(a — 6) a6 j: 51. Solve -- — — - = a + . ax — hx , X , ah 3a — 36 4- 2 4a + a6 Sue. _^_ + - = <,+ -,or x=-^-. ia+ab 6 _ 3a(4 -f 6) »c =^ X 52. Solve 4 3a— 36-f-2 6^a— 6)4-4 3.V — a y+ 2b ly a c 4 8&2 — 4ac + abc Kesult, y - 12(26 — c) 53. Solve \{x — a) _l(2a; — 36) — l(a — .r) =6 — fa. o 4 M Result, X = j6. 3 ' that ^ = 0. 55. Solve -^— - bx , ax SuG. — First divide both members by a, and then write the first mem- ber in two parts, reducing each fraction to its lowest terms. This gives 1 = c A-—. Drop V from each member and - = c. . • . x := -. jc 6 6 6 X c The pupil should also perform this by the ordinary process. WITH ONE UNKNOWN QUANTITY. 229 56. Solve ^- ~ -L -— = X -^1. SuG. — This example is readily solved by first clearing of fractions, if the pupil notices that the terms involving x^ and x^ then destroy each other. But a much more elegant solution is obtained by noticing that (2x + 3)a; 2.T24.3X , , 1 . , ., ,• -u ■ — — = ! z= X -4- 1 r ' whence the equation be- 2a: -fl 2.1- +1 ^ 2x + 1 ^ comes X -j- 1 — ^ ■ -\- —~ = x -\- 1. Dropping x -\-l from both Ax — j— 1 oX members and transposing and changing signs ; — - = —-, or 3x = Ax -j- 1 oX 20! + 1. . • . x=l. S„a. ^4^ = 1 + 1. .-. 4^ = i. and. = 2. 2x 2 ' X X -f 2 X 58. Solve -j-^ ___-=_, and verify. 2x SuG.— First destroy the term — — 5 rncii ^ — 1 X 2 X 5 X 6 69. Solve = -' X — 2 07 — 3 X — b X — 7 SuG. — Reducing each term to a mixed number we have 1 -\- ^ 1 L- = 1-1 ?— — 1 1— . Whence ^ X— 2 X— 3 ' X — 6 X — 7 X — 2 = • Reducing the terms in each member sepa- X— 3 X— 6 X— 7 ^ ^ rately to common denominators and adding, we get ;r- = (x — ^ ) (X — oj Whence (x — 2)(x — 3) = (x — 6)(x — 7), or x-^ (x — 6nx — 7)" — 13x -f 42 = x2 — 5x 4- 6, and x = ^\. 24:, ScH. 4. — It often happens that an equation which involves the second or even higher powers of the unknown quantity is still, virtu- ally, a simple equation, since these terms destroy each other in the re- daction. Several of the preceding examples afiford illustrations of this fact if solved in the regular way. 230 SIMPLE EQUATIONS 60. What value for x satisfies 3 — x — 2 {x — l)(^ + 2) = (^ _ 3)(5 _ ^x) ? Ans., X = If 61. Solve the equation (^ — 5) (^ — 2) — {x — 5) (2^7 — 5) + '(^ + 7)(^ — 2) = 0? Ans., X = 2^^ 62. Solve iL±^ - 2(3. - 4) + (3^ - 2)(2. - 3)^ do o x^ — — , and verify. 15 63. Solve {x + ^){x — f ) _ (^ _^ 5)(^ __ 3) ^_ | = 0, and verify the result. 64 Solve {a -^ x){h + x) = (c + x){d + x). cd — ah Result X b — c 65. Solve ^il^=-^- + X — c X — a X — h ab{a + 6 — 2c) Residt, X a'^ + ¥ — ac — be 66. Solve (a + x){b ^ x) — a(6 + c) = — + xK Result, X = -r- SuG. — Perform the multiplication indicated in the first member, and write the terms without clearing of fractions; this gives (a + b)x — ac = -— • Whence (a -f h)x = r = ^ — -r-^ — Dividing by ac (a + b) we have x = — • SIMPLE EQUATIONS CONTAINING RADICALS. 2S, Many equations containing radicals which involve the unknown quantity, though not primarily appearing as simple equations, become so after being freed of such radi- cals. INVOLVING RADICALS. 231 20, I*roh. 2, To free an equation of Radicals. RULE. — The common method is to so transpose the TERMS THAT THE RADICAL, IF THERE IS BUT ONE, OR THE MORE COMPLEX RADICAL, IF THERE ARE SE\'ERAL, SHALL CONSTITUTE ONE ME:MBER, and THEN INYOLVE K\CH MEMBER OF THE EQUA- TION TO A POWER OF THE SAME DEGREE AS THE RADICAL, If A RADICAL STILL REMAINS REPEAT THE PROCESS, BEING CAREFUL TO KEEP THE MEilBERS IN THE MOST CONDENSED FORM AND LOWEST TERMS. Dem. — That this process frees the equation of the radical which constitutes one of its members is evident from the fact that a radical quantity is involved to a power of the same degree as its indicated root b}^ drojjping the root sign. That the process does not destroy the equaUty of the members is evident from the fact that the like powers of equal quantities are equaL Both members are increased or decreased alike. EXAMPLES. 1. Find the value of x in V^^x + 16 =12. MODEL SOIiUTipN. OPERATION. -v/^C -)- 16 = 12. 4a; 4- 16 = 144. 4x = 128. X = 32. Explanation. — I first square each member of the equation. The first member, ^-ix -j- 16, is squared by dropping its radical sign, since the square of a square root is the number itself. The square of the second member, 12, is 14 4. This process is equivalent to multiplying the fii-st member by \/'Lv-\- 16 and the second by 12, hence as ^^4 X + 16 is equal to 12, both members have been increased ahke. [The equation being freed from radicals the explanation becomes the same as before. ] 2. Solve \^5x + 6=4, and v6rify. 232 SIMPLE EQUATIONS 3. Solve \/lO^+~a = 7, and verify. z> ;. 23 5 Veblfication. VlU - Y -f 3 = 7, or ^/iB + 3 = 7, or %/49 = 7. 4. Solve v^2a; + 3 + 4=7, and verify. I SuG. — First transpose tlie 4 and unite it with the 7 ; otherwise, squaring will not free the equation of radicals. 5. Solve 8 + V'Sx + 6 = 14, and verify. 6. Solve and verify d\/2x + 6 + 3 = 15. Result, X = 5. 7. Solve Vax -^ 'lab — a = b. a^ A- b^ Besult, X = — — — a 1 1 8. Solve Vx -{- x-^ — x — ^= {). Besult, x = ^■ 9. Solve ax -\- aV'Aax -\- x;^ = ab. SuG. — Before squaring put the equation in the form N/2aa; + x^=< b — X. Besult, X = — -;-• 2(a + b) 10. Given v 12 + ?/ — vy = 2 to find the value of y. Queries. — If both members are squared as the equation stands, will it be freed from radicals ? Is the first member a binomial or a trino- mial ? What is its square ? Which will give the most simple result, to square it as it stands or to transpose one of the radicals ? Which one is it best to transpose ? Will once squaring free it from radicals ? 12 + 2/ = 4 + 4:Vy + y, or 'i*/y = 8 .•.?/= 4. 11. Given \/x — 16 = 8 — Vx to solve and Yerifj. SuG. — Solve this and the five following like the preceding by squar- ing twice. INVOLVING RADICALS. 233 12. Solve \/a -\- x -r Va — x =■ Vax. Result, X = a- + 4 13. Solve \^x — a = \/ x — g^^- lo 14. Solve v^5 X v^o; + 2 = ^Vx + 2. 9 Result^ ^ := — . JO 15. Solve a + a; = ^a^ + xVb^ + ^ Result, 57= ;; 4a 16. Solve "iVb + j; — v/4a + a; == \/^. (5 — a)2 Result, X = -T 7— , 2 a — b Query. — Why is it best to transpose one of tlie terms of the first member to the second before squaring ? , a 17.Given^+ Va — x = . to find x. V a — X ScH. 1.— It is frequently best to clear the equation of fractions first, even when it involves radicals, especially if it is noticed that the de- nominator is of such a form as will not make the equation comph- cated. In this case, clearing of fractions we have jcv/a — x-j-^ — x = a. Whence x^a — x = x, or ^/a — ic = 1. Finally x = a — 1. X — ax v X 18. Solve Vx ^ SuG. — This may be solved in several ways. The pupil should exer- cise his ingenuity in seeing in what different ways he can solve such examples, and notice the most elegant methods. For example, com- pare the following methods. 1st Method. — Clearing of fractions x- — ax'^ = x. Dividing by x, 1 x — ax = l, .' . x = - . 234 SIMPLE EQUATIONS 2nd Method. —Multiplying both members by ^x we have x — ax = - = 1, etc. a; 3rd Method. — Dividing numerator and denominator of the second member by ^x, we have ^r- = — =• Whence multiplying both members by ^/x, x — ax = 1, etc. 4.th Method.— Divide the numerator of the first member by its de- nominator and x/ic — a^x = — -. Dividing both members by x/a;, X 1 — a = -, or X = :; . X 1 — a ^^ ^ , \/ax — b Svax — 26 19. Solve -—=: = — —=z . vax + b dvax + 56 SuG. — Clearing of fractions Sax -\- 26'v/a« — 5&2 = Sax -j- b\/ax — 2&2. Transposing, uniting terms, and dividing by b, "J ax = 3b. . • , x = 96- — . A much more elegant solution is to reduce each member to a mixed a number, obtaining 1 — = — =1 — = . Dropping the 1, V ax -j- & 3 V ax -|- 5b 2 7 anddividingby— 6, — r=: = — 7== • Clearing of fractions V ax -j- > 3 V ax -j- 56 Gx/^-f 106 = 7v/ax"+ 76. . • . n/^= 36, etc. In a similar manner solve the two following : 20. Solve — = ——^ — . Itemlt, ar = 4. V ^ + 4 V ^ -f 6 0-1 a 1 v/^ + 2a y^ + 4a ^, ,, /_«LV 21. Solve — 73: = -71= . llemlt, x = ( r ) ^x^b v/a; + 36 Va— 6/- 00 a 1 3^ — 1 . , s/^x — 1 '22. Solve -— == = 1 H ^r ^/^x + 1 2 Sua. — Observing that — = = v^3x — 1, we have -vSx — 1 =s _ -JSx + 1 l_f_2:^_^, or V3^=U, or>v/3^=3. .•.x = 3. 23. Solve 4^ = 5v/5 - 8 + ^. INTOLVING RADICALS. 235 SuG. — In this and the two following use the same expedient as in the 22nd. Thus V^x — 2 = 5^x — 8-\- i^x, or 54^x = 6, or ^/^ = 12 _144 24. Solve -^^^ =. ^^^:^ ^ 2v/a. SxjG.— We have y/x — \/a = iv^4- Ifv^oi or |n/x = 2|ya. .• . X = 16a. ^^ c^ 1 «'^ — ^'^ , ^«^ — ^ 25. Solve — =: = c H Vax + 6 c Result, x= -(h A ) . a\ c — 1/ va -\-x -\- va — X , 26. Solve "7= 7==^ = v ^• V a-j-o; — V a — x Sua. — Eationalize the denominator {224-), and after reducing some- what \/a-3 — x'^ = x\/7ii — a. Squaring a- — x^^ mx'^ — ^aVtnx -j- a^ or — X ^= mx — 2a'v/m. . • . x = 2a's/'in l-\-m — \/,^ + a + y^r _, ,^ a((? — l)^ 27. Solve , 7= = c. Result, x=-^ — '-. ^^ „ , v/i^Tfl + "iVx ^ „ ,, 4 28. Solve Trr = 9. Result, x=-. V4.x-\-i — 'ivx y 29. Solve v^^ + v/^ — v^^ _ v/^ = tvr~^- Sua. — ^Multiply both members by ^x4- v'^x, and after some reduo* tion \/x2 — x=x — Wx. Squaring, transposing, and combining, /- 5 25 30. Solve Vv'^r + S— Vv^^r— 3 = ^"^^x. Result, j; = 9. 236 SIMPLE EQUATIONS 31. SolveJ^ + J^=J- — x^ 46c Sue. — Squaring each member, notioe that the term 6 — c 4 X a ^a-^ yia^x^ x* J a2 — x'i occurs on both sides. 82. X a ^a^ ■ \a^x^ x* Result^ X = 2a. 33. Solve .15^7 + 1.575 — .875a; = .0625x. Besulf, X = 2. 34. Solve 1.2x — , = Ax + 8.9. .5 Besult, X = 20. 35. Solve 4.8^ — — , = 1.6a7 + 8.9. .5 Besulf, X = 5, SUMMARY OF PRACTICAL SUGGESTIONS. 27* — In attempting to solve a simple equation which does not con- tain radicals, consider, 1. Whether it is best to clear of fractions first. 2. Look out for compound negative terms. 3. If the numerators are polynomials and the denominators mono- mials, it is often better to separate the fractions into parts. III. — Ex. 22, Art. 22, may be written thus, by this expedient : A X +i-^\x + ^ = \x-l- 2f. Whence (-,\ - ^ - \)x = -3|-i-i or-iix = -f|. .-. 05 = 35. 4. It is often expedient, when some of the denominators are mono- mial or simple, and others polynomial or more complex, to clear of the APPLICATIONS. 237 most simple first, and after each step see that by transposition, uniting terms, etc., the equation is kept in as simple a form as possible. (See Examples 40 to 44. ) 5. It is sometimes best to transpose and unite some of the terms be- fore clearing of fractions. (See examples 46 to 54.) 6. Be constantly on the lookout for a factor which can be divided out of both members of the equation {Ejc. 55), or for terms which de- stroy each other {Ex. 60 to 66). 7. It sometimes happens that by reducing fractions to mixed num- bers the terms wiU unite or destroy each other, especially when there are several polynomial denominators. (See Ex. 59. ) 28. When the Equation contains Radicals, considee, 1. If there is but one radical, by causing it to constitute one mem- ber and the rational terms the other, the equation can be freed by in- volving both members to the power denoted by the index of the radi- cal. (See Exs. 1 to 9.) 2. If there are two radicals and other terms, make the more com- plex radical constitute one member, alone, before squaring. Such cases usually require two involutions. (See Exs. 10 to 16.) 3. K there is a radical denominator and radicals of a similar form in the numerators or constituting other terms, it may be best to clear of fractions first, either in whole or part. (See Exs. 17, 18, 29.) 4. Frequently a fraction may be reduced to a whole or mixed num- ber -vs-ith advantage. (See Exs. 19 to 25.) 5. It is sometimes best to rationahze a radical denominator. (See Exs. 26 to 28.) 6. Always watch for an opportunity to divide out a factor, or drop terms which destroy each other. APPLICATIONS. 29. According to the definition {2), Algebra treats of, 1st, The nature and properties of the Equation ; and 2nd, the method of using it as an instrument for mathematical investigation. The Simple Equation has been so thoroughly discussed 238 SIMPLE EQUATIONS : in the preceding part of the section, that the pupil will now be able to use it in solving problems. 50, The Algebraic Solutiofi of a problem con- sists of two parts : 1st. The Statement, which consists in expressing by one or more equations the conditions of the problem. 2nd. The Solution of these equations so as to find the values of the unknown quantities in known ones. This process has been explained, in the case of Simple Equa- tions, in the preceding articles. 51, The Statement of a problem requires some knowledge of the subject about which the question is asked. Often it requires a great deal of this kind of know- ledge in order to " state a problem." This is not Algebra; but it is knowledge which it is more or less important to have according to the nature of the subject. 52, Directions to guide the student in the State- ment of Problems : 1st. Study the meaning of the problem, so that, if you had the answer given, you could prove it, noticing carefully just what operations you would have to perform upon the answer in proving. This is called, Discovering the relatioris between the quantities involved. 2nd. Represent the unknown (required) quantities (the answer) by some one or more of the final letters of the alphabet, as x, y, z, or to, and the known quantities by the other letters, or, as given in the problem. 3rd. Lastly, by combining the quantities involved, hoth known and unknown, according to the conditions given in the problem (as you would to prove it, if the answer were known) express these relations in the form of an equation. PROBLEMS. 1 . What number is that to the double of which, if 18 be added, the sum is 82 ? APPLICATIONS. 239 MODEL SOLUTION. Statement. — Let ic represent the unknown number sought. Then the problem says that double this number, that is 2a*, with 18 added, that is 2x -j- 18, is 82. Hence 2x + 18 = 82 is the statement. Resolution or the Equation. — [With this the pupil is famihar.] Solving 2x + 18 = 82, we find x = 32. 32 is, therefore, the number Bought. , Veeitication. 2 X 32 -}- 18 = 82. [Note. — This is a very simple question ; but the pupil will do well to study it carefully. There are three numbers involved, the 18, 82, and the one to be found, which we call x. Having noticed this, we see that the relations between these numbers are explicitly stated in the problem, and the statement is easily made. This is not always so. These relations are often the most difficult thing to discover, and their discovery requires a knowledge of other subjects than Algebra. Sup- pose the problem to be : Given three masses of metal of equal weight, one of pure gold, one of pure silver, and one part gold and part silver. When they are immersed in water, the gold displaces 5 ounces, the sil- ver 9 ounces, and the compound 6 ounces. ^\'Tiat part of the last was gold and what part silver ? Now in this problem the relations between the quantities are not explicitly stated. Yet by a knowledge of Natu- ral Philosophy and a httle of Mensuration, they can be found out, and the statement of the problem made in an equation. ] 2. What number is that to the double of which, if 44 be added, the sum is 4 times the required number? SuG. — What are the numbers involved in this problem ? Ans. Only two, 44, and the number sought. Suppose I guess that the number sought is 30, how will you tell whether I am right or not ? You say : " Double 30 and add 44 to it, and, if you are right, the sum will equal 4 times 30." But 2 X 30 -f 44 does not = 4 X 30 ; so I am wrong. Now, call the number sought x, and use it as 30 was used in proving that it is not the answer, and the statement, 2x -f- 44 = 4x is obtained. Whence x = 22. Verify it. 3. What number is that, the double of which exceeds its half by 6? Ans., 4. 4. The property of two persons is $16000, and one owns three times as much as the other. How much has each? 240 SIMPLE equations: Statement. Let x = the less amount. Then 3a; = the greater amount. And3x4-a; = 16000. .-. x = 4000 and 3x = 12000. 5. A man being asked his age replied: "If to my age you add its half, and third, and then deduct 10, the result is 100." What was his age? Ans., 60. 6. After paying j of a bill and -J of it, $92 still remained due. What was the bill at first ? Statement. — Let x = the amount of the bill; then \x and ^x had been paid. Taking these amounts from the bill we have x — \x — ^x. But this, by the problem, was $92. Hence x — \x — -^x = 92. . • . x = 140. 7. The sum of 2 numbers is 180 and the difference 10. What are the numbers ? Statement. — Let x = the less number: then x -f- 10 is the greater, and X -{-x -{- 10 = 180. . • . x = 85, and x -{-10 ■■= 95, and the two numbers 85 and 95. 8. The sum of two numbers is 5760, and the difference is ^ the greater. What are they ? Statement. — Let x = the greater; then 5760 — x is the less, and X — (5760 — x) = |. . • . X = 3456, and the less is 2304. o 9. A man divided 80 cents among four beggars; to the first two he gave equal amounts, to the third twice as much as to one of these, and to the fourth twice as much as to the third. How much did he give to each ? Ans. , The equation is x -\- x -{- 2x -\- 4:X = 80. Whence he gave the first and second 10c. each, the third 20c., and the fourth 40 cents. 10. A, B, C, and D, invest |4755 in a speculation. B fur- nishes 3 times as much as A, C as much as A and APPLICATIONS. 241 B together, and D as much as C and B. How much does each invest ? Ans., A, $317; B, $951; C,$1268; and D, $2219. THE SAME PROBLEMS WITH THE LITERAL NOTATION. 11. What niynbef is that to n times which, if m be added, the sum is s ? — m The equation is nx -}- m = s. Ans., n ScH. — To make this conform to Pbob. 1, we call s = 82,m = 18, - - s — m 82—18 oo-Di-i-v, ^ — ■^ and n =z2. . • . = = 32. But the answer ap- ji 2 n plies just as well to any other problem of the kind, no matter what numbers are involved. Thus, let the problem be, — What number is that to 5 times which, if 20 be added, the sum is 100 ? Now s = 100, m = 20, and n = 5. Whence = = = 16. We, there- fore, see that is a general answer to all such problems. 12. What number is that to a times which, if 6 be added, the sum is c times the number ? Ans.f The equation is ax -^ b == ex, and the num- ber c — a Queries.— How is this adapted to Prob. 2 ? What other problems can you state to which this value of x affords an answer ? 13. What number is that, a times which exceeds - times it by m? MiQuation, ax — - = m. .* . Ans., x = -; 7 ab — 1 is the number. QtrEBiES. — How is this adapted to Prob. 3? What other problems can*S'ou state to which — affords an answer ? Repeat tiie same ab — 1 inquiries after each of the seven following examples. 242 SIMPLE equations: 14. The property of two persons is $m, and one owns n times as much as the other. How much has each ? Ans., and 1 + ^ 1 + n 15. What number is that to which if th and -th of it- m n self be added, and then a deducted, the result is 5 ? (a + b)mn A71S., ran + 77i + n 16. After paying — th and -th of a bilL a remained due. *- m n What was the bill ? , amn Ans., ' mn — n — m 17. The sum of two numbers is s, and their difference d. What are the numbers ? , s -\- d , s — d Ans. 33, Cor. — Observe that the solution of this problem proves the very useful theorem : Having the sum and the difference of two quantities, the greater is found by taking half the sum of the sum and differ- ence, and the less by taking half the difference of the sum and difference. Or, since the above results may be written - + 77 and - — -5 the theorem may be stated: 2 2 2 2 ^ I Half the sum plus half the difference of two quantities is the greater of the quantities, and half the sum minus half the dif- ference is the less. Thus, the sum of two numbers is 20 and s d their difference 12. What are the numbers ? ^ + - = 10 + 6 = 16, the greater; and | — ^ = 10 — 6 = 4, the less. APPLICATIONS. 243 18. The sum of two numbers is s and the difference is -th of the greater. "What are the numbers ? „ . . . X . ms s(m — 1) Equation, x — (s — x) = — . An^., -, — — -. ^ ^ ' m 2m — 1 2m — 1 19. A man divided $a among 4 beggars ; to the first two he gave equal amounts, to the third m times as much as to each of these, and to the fourth m times as much as to the third. How much did he give to each ? a Ans., To each of the first two $-r-^ ; -, to the third am aiw^ %-T—. ■. -, and to the fourth %-r-, ■. ;• ^2 -f m + m-' ^2 + m + m* 20. A, B, C and D invest %s in a speculation. B furnishes m times as much as A ; C as much as A and B together, and D as much as C and B together. How much does each invest ? S 771.9 8(1 + ^^) Alls. J A furnishes o , a — ; B, ^ , a — ; C, » , ~< » ' 3 + 4m ' ' d + 4m ' * 3 + 4?7i s(l-i-2m) and D, .> . -; 3 + 4?7i 21. At a certain election 943 men voted, and the candidate chosen had a majority of 65 votes. How many voted for each? * ^7is., 439 and 504. 22. A farmer has two flocks of sheep, each containing the same number. From one he sells 39, and from the other 93, and then finds just twice as many in one flock as in the other. How many did each flock originally . contain? Ans., 147. 23. A person spends -^ of his income in board and lodging, ■|- in clothing, and yV ^^ charity, and has $318 left. What is his income ? Ans., $720. 214 SIMPLE EQUATIONS l 24. From a bin of wheat -^ was taken, and 20 bushels were added. After this i of what was then in the bin was sold, and i as much as then remained + 30 bushels was put in, when it was found that the bin contained just j^ as much as at first. How much did it contain at first? SuG. — After taking out 2, there remains ^x, calling x the amount at first in the bin. To this add 20 bushels and there is in the bin ix -f- 20. After selling i of this f remains, or f (gx -j- 20). i of this is ^{ix + 20). . • . f (ix + 20) 4- |(ix + 20) 4- 30 = hx. Whence X = 560, the answer. 25. A person has a hours at his disposal ; how far may he ride in a coach which travels b miles per hour, and yet have time to return on foot walking at c miles per hour ? . abc Ans. 6 + c 26. After paying -th of my money, and then -th of what remained, I had $a left. How much had I at first? Equation, x — \x — {^x — \x)^ = a. . amn Ans., — — -. ran — m — n + 1 27. A boy, being asked his age, replied, 11 years are 7 years more than -| of my age. How old was he ? Statement. — Let x = his age. Then |x -j- 7 = 11. . • . x = 10. [Note. — This example is taken from Stoddard's Intellectual Arith- metic, page 118, Ex. 27. The whole series of exajiples in that work styled "Algebraic Questions," on pages 116-140, will afford good ex- ercise for the pupil at this stage of his progress in Algebra. It wiU be well for him to solve them, both by the processes given in the Arith- metic and by the use of the equation. He will thus see how the equa- tion is an instrument of great simplicity and power for the solution of mathematical problems. ] 28. A boy, being asked how many sheep his father had, replied, that 40 were 5 less than | of his father's flock. How many sheep had his father ? Ans., 60. APPLICATIONS. 245 29. If A can perform a piece of work in 10 days, and B in 8 days, in what time will they perform it together ? Statement. — Let x = the number of days. Then since A can do the work m 10 days, in 1 day he will do ^^J, and in x days — of it. In like manner B will do -. Hence 77: + o = (^11 ^^^ work) or 1. This fl. sometimes troubles pupils. But let them consider that if two of us do a piece of work, one doing f and the other j^, if the work is all done, the sum of the parts done is 1. 30. There is a certain piece of work which A and B can do in 8 days ; but A and C can do it in 6 days, or B and C in 10 days. How long would it take any one of them to do it alone ? How long if all work ? 1 1 SuG. - — 5 is the difference between what B and C can do in a day ; and — is the sum of what they can do in a day. (See 33,) Ans., A in lOif days, C in 14^2-^, B in 34f , and aU in 5^V 31. A, B, and C can do a piece of work in 4 days, which A alone can do in 12, and B alone in 8 days. C begins the work alone, and is joined after 2 days by A, and they work together 2 days more. A and C being then called o£f, how long will it take B to finish the work ? Ans., 5i days. 32. A performs | of a piece of work in 4 days ; he then receives the assistance of B, and the two together finish it in 6 days. Required the time in which each could have done it alone ? SuG. — How much does A do in 1 day? Let x = the time B would 6,6 5 require to do it all. The equation is — 4- - = 7;. A could do it in 14 X 7 14, and B in 21 days. 33. A vessel can be emptied by 3 taj)s ; by the first alone it could be emptied in 80 minutes, by the second in 246 SIMPLE EQUATIONS : 200 minutes, and by the third in 5 hours. In what time will it be emptied if all the taps are opened at once? ^ns., 48 minutes. 34, A can do a piece of work in a days ; B in 6 days, and C the same piece in c days. In how many days will they finish it, when all work together ? , abc Ans., — ; . aO'i-oc-\- ac 35. Three men A, B, and C are employed on a certain piece of work. A and B can do it in « days ; A and C in ^ days, and B and C in c days. How long would it take each to do it alone ? How long would it take them to do the work together ? . , . 2ahc ^ ^ . 2abc Ans., A m 7 days, 13 m bo -\- ac — ab be -\- ab — ac C in — '■ — r-> and aU together in ab -\- ac — be 2abc ab -}- ac -{- be Sua. — Let the student notice the singular symmetry of these an- swers. Such symmetry is common, and sometimes becomes very use- ful in complicated processes. See Stoddard's Intellectual Arithmetic, Lesson LI., for other exam- ples of this kind. 34-, ScH. — It is not always expedient to use x to represent the num- ber sought. The solution is often simplified by letting x be taken for some number from which the one sought is readily found, or by let- ting 2x, 3x, or some multiple of x stand for the unknown quantity. The latter expedient is often used to avoid fractions. 36. There is a fish whose head is 9 inches long; the tail is as long as the head and half the body, and the body is as long as the head and tail together. "What is the length of the fish ? Ans., 72 in. APPLICATIONS. 247 SuG. —Let X = the length of the body ; then % -{- \x= the length of the tail. The equation is x= 18 -\-\x. K x = the whole length the equation is9-f-4-9-|-j=a^- 37. A general whose cavalry was ^ of his infantry, after a defeat found that before the battle -^ of his infantry less 120, and y^ c>f his cavah-y plus 120 had deserted. After the battle he found \ of his whole army in gar- rison, f on the field, and that the rest of those engaged were either taken prisoners or slain. Now 300 plus the number slain was \ the infantry he had at first. Of how many did his army consist originally ? SuG. — Let X == the number of cavalry; then 3x = the infantry, and 4x = the whole army. Alls., Whole army 3600 men; viz., 900 cavalry, and 2700 infantry. 38. A shepherd, in time of war, was plundered by a party of soldiers, who took i of his flock and ^ of a sheep more; another party took ^ of the remainder and ^ of a sheep ; and a third party took ^ of the last remain- der and ^ of a sheep, when he had but 25 sheep left. How many had he at first ? SuG.— Letting 12x = the number in his flock at first, 9x — ^ is what remained after the first plundering, 6x — ^ after the second, and 3x — f after the third. The equation is 3x — f = 25. .-. 12x = 103, the number in his flock at first. The pupil should notice that he wants the value of 12x instead of x. 39. A cask. A, contains 12 gallons of wine mixed with 18 gallons of water ; another, B, contains 9 gallons of wine mixed with 3 gallons of water. How many gal- lons must be drawn from each to make a mixture of 7 gallons of wine and 7 of water ? SuG. f of the mixture in A is wine, and f water; and in B f is wine and \ water. Let x = the number of gallons to be drawn from A ; then 14 — x represents the number to be drawn from B. Of the first, fx is wine, and |-x is water; of the second |(14 — x) is wine, and ^(14 — x) is water. But in this new mixture the wine and water are equal. •.•.^x + a(14 — x) = |x4-4(U — «). 248 SIMPLE EQUATIONS : 40. In the composition of a quantity of gunpowder the ni- tre was 10 lbs. more than f of the whole, the sulphur was 4|- lbs. less than ^ of the whole, and the charcoal 21bs. less than i of the nitre. "What was the amount of the gunpowder ? Ans., 69 lbs. SuG. X being the whole, the nitre is |ic + IO5 the sulphur ^a; — 4:|-, and the charcoal |(|x + 10) — 2. 41. Several detachments of artillery divided a certain num- ber of cannon balls. The first took 72, and 9 of the remainder; the next 144, and 9 of the remainder; the third 216, and i of the remainder; the fourth 288, and 9 of those that were left ; and so on ; when it was found that the balls had been equally divided. "What was the number of balls and detachments ? Ans., 4608 balls, and 8 detachments. 42. A gentleman bequeathed his property as follows: To his eldest child he left $1800, and ^ of the rest of his property; to the second, twice that sum and ^ of what then remained; to the third, three times the same sum and -^ of the remainder, and so on; and by this ar- rangement his property was divided equally among his children. How many children were there, and what was the fortune of each ? Ans., 5, and $9000 the fortune of each. 43. A brandy merchant has two kinds of brandy ; the one cost y dollars per gallon, the other o". He wishes to mix both sorts together in such quantities that he may have 50 gallons, and each gallon, without profit or loss, may be sold for 8 dollars. How much must he take of each sort to make up this mixture ? Ans., 37^ gallons of the best, 12|^ of the other. /- APPLICATIOi;iS. 249 44. Let the price of the best brandy in the preceding prob- lem = a dollars, the price of the poorest = b dollars, the number of gallons in the mixture = n, and the price of the mixture = c. How many gallons of each kind must he use ? Ans., 7— gallons of the poorest, and -^ : — of the other. a — b 45. A father, who has three children, bequeaths his prop- erty by will in the following manner: To the eldest son he leaves $1,000, together with the 4th part of what remains; to the second he leaves $2,000, together with the 4th part of what remains after the portion of the eldest and $2,000 have been subtracted from the estate; to the third he leaves $3,000, together with the 4th part of what remains after the portions of the two other sons and $3,000 have been subtracted. The property is found to be entirely disposed of by this ar- rangement. What was the amount of the property ? Ans., $9,000. 46. A father, who has three children, bequeaths his prop- erty by will in the following manner : To the eldest son he leaves a sum a, together with the nth part of what remains ; to the second he leaves a sum 2a, to- gether with the nth part of what remains after the portion of the eldest and 2a have been subtracted from the estate; to the third he leaves a sum 3a, to- gether with the nth part of what remains after the portions of the two other sons and Sa have been sub- tracted. The property is found to be entirely disposed of by this arrangement. What was the amount of the property? (6>i2 — 4n + l)a. ^nS., ; ITT — ■ (n — iy 250 , SIMPLE EQUATIONS : 47. A general arranging his troops in the form of a solid square, finds he has 21 men over, but attempting to add 1 man to each side of the square, finds he wants 200 men to fill up the square; required the number of men on a side at first, and the whole number of troops. Remit, 110 and 12121. QuEEY. — Can the pnpil show that this problem is the same as the following ? The diflference between the squares of two consecutive numbers is 221. What are the numbers ? 48. Gold is 19^ times as heavy as water, and silver 10^ times. A mixed mass weighs 4160 ounces, and dis- places 250 ounces of water. What proportions of gold and silver does it contain? Ana., Gold, 3377 ounces; silver, 783 ounces. THE TRANSLATION OF EQUATIONS INTO PRACTICAL PROBLEMS. SS. Cor. — Since the statement of a problem is expressing the conditions of the problem in an equation, or in equations (30), it follows, conversely, that an equation may be considered as the enunciation in algebraic language of the conditions of a practical problem. EXAMPLES. rrj /v» 1. Form problems of which - — - = 6 is the statement, o 5 SuG. — Any number of problems may be stated which will meet the requirements. Thus, ' ' What number is that whose third part exceeds its fifth by 6 ?" Again ; " A man being asked his age said, ' When I was 3 as old as I now am, I was 6 years older than when I was -t as old as I am now;' what was his age ?" Or, again ; "A man being engaged to work a certain number of days, continued in the service i of the whole time, but during this time was sick, and lost a number of days equal to ^ of the whole time, when finally he had to break the engage- ment, having actually worked but G days. What was the whole period engaged for ?" APPLICATIONS. 251 2. Form problems of whicli x + lO^r = 66 is the statement. SuG. — The pupil may say, "What number is that to which if 10 times itself be added the sum is 66, " and meet the requisition. But it is urged that he exercise his ingenuity to frame more indirect enunci- ations. Thus, for this example, the problem may be, "Charles has his money in cents and dimes, of each an equal number. The whole of his money is 66 cents. How manj^ copper and how many silver pieces has he ?" ,. X X ^ X X ^ . X ^x tions: -+-4-2 = ^; 4-:^l; 2x — - 2 4 3 b 2 5 3. Give similar translations of each of the following equa- -, X ^x 2"-2-T = 7; a:+i^ — 760 + 600=2000; 15(fL±il+5 X -\- X -{- 4: b SuG. — One of the above may be enunciated thus : "The tens digit of a certain number exceeds its units digit by 4 ; and when the num- ber is divided by the sum of its digits the quotient is 7. What is the number ?" X X 4. Translate the equations 3^ + ^ = 16000 ; ^ + n + q — 2 o 10 = 100 ; x — ^—^ = 92; x-{-x + 2x + 4:X= 80, 5 7 [Note. — See examples 4, 5, 6, and 9, {32).'\ 5. Enunciate problems which will give rise to the following 6^ + 7 2^-— 2 2^—1 2(4j7 + 3) equations: — — -. =: — - — ; -^ — ^ + ^ 15 7a; — 6 5 x-\-3 ^ 3 [Note. — The following examples give rise to equations found under Art, 26, The pupil should look them up, and make similar enun- ciations of other examples.] 6. What number is that which gives the same quotient when 28 is added to its square root, and the sum di- vided by its squaie root -f 4, as when 38 is added to the same root, and this sum divided by the square root of the number -f 6 ? 252 SIMPLE EQUATIONS 7. A man has a certain number of square rods of land ly- ing in a square ; if 12 rods be added, the whole being kept in the form of a square, his plat is increased by 2 rods on a side. How much land has he ? 8. What number is that to which if 12 be added its square root is increased by 2 ? 9. If 4 times a certain number be increased by 1, the square root of this sum + twice the square root of the number itself, divided by the difference of the same quantities is 9. What is the number ? 10. There is a certain number to which if its own square root be added, and also subtracted, the difference be- tween the square roots of the results is 1^ times the square root of the quotient of the number divided by the number -f its square root. What is the number? 11. What number is that from which if 32 be subtracted, the square root of the difference is equal to the square root of the number — ^ the square root of 32 ? 12. What number is that from which if a be subtracted, the square root of the difference is equal to the square root of the number — ^ the square root of a ? Ans. 25a 16 13. What number is that whose square root + 2a, divided by its square root -f- b, equals its square root + 4a, divided by its square root + 36? .^.j^f WITH TWO UNKNOWN QUANTITIES. 263 SECTION 11. Simple, Simultaneous Equations with two Unknown Quantities. DEFINITIONS. S6, The preceding problems have all been solved by a single equation containing only one unknown quantity. In some of them several quantities have been sought, it is true, but we have managed to express these quantities by the use of a single unknown quantity, x. There are, however, many problems in which this is not practicable. In such problems there are two or more quantities sought, and the conditions are such as to give rise to two or more equa- tions. III. — To make this latter statement clear, consider the following problem : A says to B, " If ^ of my age were added to f of yours, the sum would be 19^ years." But, says B to A, " Iff of mine were subtracted from \ of yours, the remainder would be IS-j- years. " Eequired their ages. Here are two distinct quantities sought; viz., A's age and B's age. Suppose we represent A's age by x, and B's by y. Now notice that there are also two sets of conditions. 1st, the statement which A makes to B ; and, 2nd, the statement which B makes to A. Accord- ing to the 1st, we have the equation 5X + f !/ = 19^; and according to the 2nd, ^x — ly = 18^. 37 > Tndependent Equations are such as express different conditions, and neither can be reduced to the other. 38. Simultaneous Equations are those which express different conditions of the same problem, and con- sequently the letters representing the unknown quantities signify the same things in each. Any grouj) of such equa- tions are all satisfied by the same values of the unknown quantities. 254 SIMPLE EQUATIONS 111. — Thus in the example above the two equations ^x -{- ^y = 19^ and ^x — ^y =:18\ are indepemdent equations, since they express dif- fereni conditions, and neither can be produced from the other. But, since these conditions are of the same problem, so that x in the first equation means the same as a; in the second, and y in the first, the same as y in the second, they are simultaneous equations. It is evi- dent that the true values of x and y must satisfy, or verify, both equa- tions. If, however, we were to write one equation from one problem, and one from another, while they would be independent, they would not be simultaneous ; x and y would not mean the same things in the first equation as in the second. In fact, the equations would be so inde- pendent, that they would have nothing to do with each other. 39, JEjlhninatiofi is the process of producing from a given set of simultaneous equations containing two or more unknown quantities, a new set of equations in which one, at least, of the unknown quantities shall not appear. The quantity thus disappearing is said to be eliminated. (The word literally means 2yuUmg out of doors. We use it as meaning causing to disajDpear. ) 40. There are Three 3Iet1iods of Blimination in most common use; viz., by Compaiuson, by Substitution, and by Addition or Subtraction. There is also a very elegant method, by Undetermined Multipliers, which is worthy of more attention than it generally receives, but which will be reserved for the advanced course. ScH. — ^Any one of these methods will solve all problems; but some problems are more readily worked by one method than by an- other, while it is often convenient to use several of the methods in the same problem, especially when there are more than two unknown quantities. ELIMINATION BY COMPARISON. 4:1, J^rob. 1, Having given two independent, simul- taneous, simple equations between two imbnoivn quantities, to i^educe fherefrom by comjMrison a neiu equation containi'ng only one of the unknown quantities. ^ A WITH TWO UNKNOWN QUANTITIES. 255 [Note. — This Prob. might be stated simply — To eliminate by com- parison ; but it is very important that the statement be made in full, as above.] HULE. 1st. — Find expressions for the value of the SAME UNKNOWN QUANTITY FROM EACH EQUATION, IN TERMS OF THE OTHER UNKNOWN QUANTITY AND KNOWN QUANTITIES. • 2nd. Place these two values equal to each other, and THE result WILL BE THE EQUATION SOUGHT. Dem. — The first operations being performed according to the rules for simple equations with one unknown quantity, need no further dem- onstration. 2nd. Having formed expressions for the value of the same unknown quantity in both equations, since the equations are simultaneous this unkno^-n quantity means the same thing in the two equations, and hence the two expressions for its value are equal, q. e. d. ScH. — The resulting equation can be solved by the rules already given. EXAMPLES Of Independent Simultaneous Equations. 1. Given 4a7 + ?/ = 34 and Aij -\- x = 16 ; to find x and y and verify the values. MODEL SOLUTION. 34 — t/ ••• ^^— r- (2) 4?/ -f X = 16 . • . X = 16 — 4?/ ^1^^ = 16-4, ... y = 2 4x+2=34 .'. x = 8 Explanation. — Transposing y and dividing by 4 I have from the 1st equation, x = '-. Transposing 4?/ in the 2nd equation I have X = 16 — -iy. Now, since these equations are assumed to be simulta- neous, X means the same thing in both ; and since things that are equal to the same thing are equal to each other, — - — - = 16 — 4i/. From this equation I find y = 2. Finally, since y is iound to be 2, putting 2 for y in the first eotr^tion, I havo -Ix -f- ^ = 34 ; whonce x = 8. OPEEATION. (1) 4x -f- y = 34 256 SIMPLE EQUATIONS VEEincATioN. — Substituting for x, 8, and for y, 2, in the 1st equation I have 32 -f 2 = 34; and in the 2nd, 8 -f- 8 = 16, both of which equa- tions are satisfied. 2. Given 5^ + 4?/ ^ 58, and 3^7 + 7^= 67, to find x and y, eliminating by comparison. Verify the results. Besults, -j „' 3. Given 11^ + 3i/ = 100 and 4:X — ly = 4, to find x and y, eliminating by comparison. Verify the results. x = 8, 4. Results, \ \y . r. , . r. ^ + '"5 „ 3^ — 2?/ , 4. Same as above, given 8 -j— =7 p and 8 — ?/ _.j 2.^ + 1 4^- -^=241 —. SuG. — Observe the effect of the — sign be^bre the compound quan- 5 -I- 7x tities. The value of y from the 1st is y = —^ — , and from the 160 — 30x ^^ 5 4- Ix 160 — 30x , _ ^ 2nd, V = 7 • Hence — \ = ■ , andx = 5, and ' ^ 2 8 2 5. Given ax -^ hy ^= m, and ex -\- dy= 7i, to find x and y, eliminating by comparison. Sua— From the 1st, y = — ; and from the 2nd, y = — . m — ax n — ex ^ hn — dm ^^ • . -, ^ ■. i.-j. l and a; = -^ ;-. Now, instead of substitut- ' ' b d be — ad ing this value of x for x in one of the given equations, it will bo found more expeditious to eliminate x as we did y. Thus, from the 1st, x = TO — by , ^ ^i r> 1 ^ — (^y '^ — ^v ^^ — (^y J -: and from the 2nd, x = -. . • . = -, and a c a c (try CTTl y = — — . To find this value of y by substituting for x m the ad — be !> J. i.- -i. 1 bn — dm , bn — dm- first equation its value, -. , we have a- 4- by = m, or ^ be — ad bc — ad'^ abn — adm , , _ . . , abn — adm -\-by = m. By transposition, by = m be — ad be — ad bem — adm — abn -J- adm bem — abn ^. . -,. , , cm —an T 7 = —r 5-. Dividing by b,y = —. be — ad . be — ad ° ^ 6c — ad Let the pupil consider whether this is the same as the former value of y ; and if it is, how we might have made the results alike in form. WITH TWO UNKNOWN QUANTITIES. 257 ELIMINATION BY SUBSTITUTION. 4:2, J*rob, 2, Having given two independent, simulta neous, simple equations, between two unhnoivn quantities, to de duce therefrom by substitution a single equation with but on> of the unknown quantities. RULE. — 1st. Find from one of the equations the value OF THE UNKNOWN QUANTITY TO BE ELIMINATED, IN TERMS OF THE OTHER UNKNOWN QUANTITY AND KNOWN QUANTITIES. 2nd. Substitute this value for the same unknown quan- tity IN THE OTHER EQUATION. Dem. — The first process consists in the solution of a simple equation, and is demonstrated in the same way. The second process is self evident, since, the equations being simul- taneous, the letters mean the same thing in both, and it does not de- stroy the equality of the members to replace any quantity by its equal. Q. E. D. EXAMPLES. X + y 1. Given the independent, simultaneous equations 2 ^ ^ = 8, and "^^^ + "^--^ = 11, to find x and 3 ' 8 ' 4 t/, eliminating by substitution. Verify the results. MODEL SOLUTION. /-iN^ + v ^ — y o OPEBATION. (1) ^ — ^ = 8 3x -f 3?/ — 2x + 2y = 48 jc -f- 5?/ = 48 a; = 48 — 5t/ (2) ^4-^-^ = 11 ,o^ 48 - 5.V 4- y , 48 - Sy - y (3) + I = 11 48-4.V 24-3y _ 3 "^ 2 96 — 8i/ + 72 — 9y = 66 — 17y = — 102 .-. 2/ = 6 258 SIMPLE EQUATIONS XX .-. ic = 18 ErpiANATiON. — Taking equation (1) I clear it of fractions and solve it for X, finding that x = 48 — 5y. Now I observe that if I take the 2d equation and substitute this value of x for x, I shall have a simple equation with only the unknown quantity y in it. This substitution does not destroy the equality, since, the equations being simultaneous, X has the same value in both. Making the substitution I have 48 — 5v+ y , 48-— 5?/ — V ,., -o ^ • .r,- .. , .^, ^ — ■ — - -| Y '~ = II- lieducmg this equation by the method for simple equations with one unknown quantity, I find 2/ =6. Finally, resuming (1) I substitute this value of y for y, which evi- X -i-6 X 6 dently does not destroy the equation, and have —J- — = 8. Solv- 2 o ing (4), which is now a simple equation with one unknown quantity, I find X = 18. (Instead of taking (1) in its first form it would be better, because so much shorter, to take its reduced form, tc = 48 — 5y. .'. ic = 18.) [Note. — The student should keep a sharp lookout for opportunities to effect such reductions of terms as are made in the equations follow- ing (3) and (4). In the latter the process consists in observing that -— — is - -|- 3, and — is — q -|- 2, hence the first member be- comes J5 + 3 — o ~f~ 2' ^^^ transposing the 3 and 2, we have - — x = 3, a O 2 3 aU of which can readily be effected mentally, ] Veeification. — Substituting in (1) 18 for x, and 6 for y, I have -1 Q I n 1 Q /» — ^ — =8, or 12 — 4 = 8. Also, in Uke manner from (2), 1 Q I n -f Q n I have .7" H — ■ =11, or 8 -f 3 = 11. Whence I see that 3 ' 4 ' X =: 18 and y = 6 satisfies both equations. 2. Given 3^ — 2ij = 1 and 3y — 4:X = 1, to solve as above. Result, x = ^ and'i/ = 7. WITH T^^O UNKNOWT^ QUANTITIES. 259 SxjG. — From the second, y = —\: — , hence the first becomes 3x — o "^ =1 . • . X = 5. Taking the reduced form of the second, y = — — , 3 .. ■ ' + ' - "■™ la 56ar + 13.421^ = 763.4 58.54 Result, i^'^^ rro^/ ' I" nearly. APPLICATIONS. 1. There are two numbers, such, that three times the greater added to one-third the less is 36; and if twice 266 SIMPLE EQUATIONS the greater be subtracted from 6 times the less, and the remainder divided by 8, the quotient will be 4. "What are the numbers ? OPEEATION. Then MODEL SOLUTION. Let X = the greater number, and y = the less number. (1) Sx-\-y = 36 (2) 6y — 2x 8 4 (3) 6y + 54.x = 648 (4) 6y — 2x = 32 5&X = 616 X = 11 (5) 6y — 22 = 32 y = 9 Explanation. — As there are two unknown quantities involved in this example, L e. , the two numbers sought, I let x represent the greater and y the less. There are also two sets of conditions stated in the problem : 1st, 3 times the greater added to ^- the less is 36. This, ac- cording to the notation, is 3.i; -\- ^y z= 36, which is the first equation. The 2d set of conditions is, that tmce the greater is to be subtracted from 6 times the less, which is 6y — 2.r, and this difference divided by 8, i. e. Qy 2x - — -— ^ . This quotient is equal to 4. Hence the second equation, 8 6v — 2x , [Note. — The explanation of the resolution of the equations can be given as heretofore. But as the attention is now to be directed chiefly to the statement of questions, if the former subject has been thoroughly- mastered, little need be said about it here. Nevertheless the pupil should not he allowed to run over, his solution simply reading the successive equa- tions. If he does anything more than simply give, as above, the ex- planation of the statement then saying ' ' Solving which equations I find K = 11 and t/ = 9," he should at least tell what he does, thus : "Multiplying (1) by 18, and (2) by 8, I have 6y -f- 54a; = 648. and Qy — 2x = 32. Subtracting (4) from (3), member by member, I elimi- nate y and find 56x = 616. Dividing by 56, x = 11, &c."] WITH TWO UNKNOWN QUANTITIES. 267 2. Find two numbers, such, that if the first be increased by a, it will be m times the second; and if the second be increased by h, it will be n times the first ? a -\- hm b -\- an EesuU, -, and -. mn — 1 mn — 1 3. What two numbers are those, to i of the sum of which if I add 13, the result will be 17 ; and if from ^ their . difference I subtract 1, the remainder will be 2? Verify. Ans., 9 and 3. [Note. — In verifying the results in such examples as these, no attention should be paid to the equations ; but the results should be tested directly by the statement. Thus, in this example, ^ of the sum of 9 and 3 is 4. Adding 13 the result is, as the example requires, 17. Again ^ the difference of 9 and 3 is 3. Subtracting 1, the remainder is 2, as required. ] 4. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes f; but the denominator being doubled, and the numerator increased by 2, the value becomes f ? Ans., f . SuG. — The pupil should ask himself, "How many sets of con- ditions in this problem ?" " What are they ?" " How many unknown (required) quantities ?" " W^hat are they ?" There must always he as many of one as of the other. The unknown (required) quantities here are the numerator and the denominator of the fraction. K these are X called respectively x and y, the fraction is — . Now, by the first set of 2a: a: -}- 2 conditions, j — - = |, and, by the second set ' 1/ 4- 7 - — ' 2y 5. What fraction is that which becomes f when its nu- merator is increased by 6, and i when its denominator is diminished by 2 ? Ans., sV 6. If 1 be added to the numerator of a certain fraction, its value is ^; but if 1 be added to its denominator, its value is \. What is the fraction ? Verify. Ans., y4^ 268 SIMPLE EQUATIONS 7. There is a certain number, to the sum of whose digits if you add 7, the result will be three times the left hand digit ; and if from the number itself you subtract 18, the digits will have changed places. AVhat is the number? Verify. Ans., 53. SuG. — The two numbers sought are the two digits. Hence let y =3. the units digit, and x = the tens digit. The number tiien is lOx -f- y. (Just as when 6 is the units digit of a number and 5 the tens, the num- ber is 10 • 5 + C. Of course the number would not be represented by xy, for this would indicate the product of the digits. (See Part I., 30, Second Law, Scholium 1st. ) The first conditions give 2x — y = 7, and the second lOx + .V — 18 = lOy -j- x, i. e. the units becomes the tens figure and t&© tens becomes the units. 8. A certain number of two digits contains the sum of its digits four times and their product twice. What is the number ? Ans., 36. 9. There is a number consisting of two digits; the num- ber is equal to 3 times the sum of its digits, but if the number be multiplied by 3, the product equals the square of the sum of its digits. What is the number ? Verify. SuG. — The equations are 10a:; -f V = ■^{^ -\- ?/\ and 3(10x -}- y) = (x -\- yY: Multiplying the 1st by 3, we have 3(10.c -\-y) = 9(x -^ y). Equating (placing equal) the second members, 9 x -f- y- = (^c -f- y'^. Dividing by a; -{- y, 9 = a: -f- ?/• This equation and the first are easily combined and solved. 10. .A number consisting of 2 digits, when divided by 4 gives a certain quotient and a remainder of 3; when divided by 9, gives another quotient and a remainder of 8. Now, the value of the digit on the left hand is equal to the quotient which was obtained when the number was divided by 9; and the other digit is equal to yV ^^ ^^^ quotient obtained when the number was divided by 4. What is the number ? Verify. SuG. — Letting x represent the tens figure, and 7/ the units, the equa- tions are — '—- — - = lly -]-\, and — ' = x -f- f- Th^ pupil should give the explanation. WITH TWO UNKNOWN QUANTITIES. 269 11. A farmer parting with his stock, sells to one person 9 horses and 7 cows for 300 dollars; and to another, at the same prices, 6 horses and 13 cows for the same sum. What was the price of each ? Ans., $24 and $12. 12. A son asked his father how old he was. His father answered him thus : If you take away 5 from my years, and divide the remainder by 8, the quotient will be ^ of your age ; but if you add 2 to your age, and multi- ply the whole by 3, and then subtract 7 from the pro- duct, you will have the number of the years of my age. What was the age of the father and son ? Ans., 53 and 18. 13. A farmer purchased 100 acres of land for $2450; for a part of the land he paid $20 an acre, and for the other part $30 an acre. How many acres were there in each part ? Verify. ScH. — Very many such problems can be solved equally weU by means of one or of two unknown quantities. The pupil should do such in both ways. 14. At a certain election 946 men voted for two candidates, and the successful one had a majority of 558. How many votes were given for each candidate ? Verify. 15. A jockey has two horses and two saddles. The saddles are worth 15 and 10 dollars, respectively. Now if the better saddle be put on the better horse, the value of the better horse and saddle will be worth ^ of the other horse and saddle. But if the better saddle be put on the poorer horse, and the poorer saddle on the bet- ter horse, the value of the better horse and saddle will be worth once and -^^ the value of the other. Required the worth of each horse ? Besult, 65 and 50 dollars. 16. A sum of money was divided equally among a certain number of persons ; had there been four more, each 270 SIMPLE EQUATIONS would have received one dollar less, and had there been four fewer, each would have received two dollars more than he did : required the number of persons, and what each received ? Verify. SuG. -= — -— -+ljand-= 2. Hence xy 4- 4x = xy 4- y y+^ y y-^ ^^ ^^ y- -f- 4?/, and xy — 4x = xy — 2?/' -f- 8?/, or 4ic = ?/- -f- 4?/, and — 2x = — ?/2 -j- 4i/. Adding 2x = 8y. 17. A farmer hired a laborer for ten days, and agreed to pay him $12 for every day he labored, and he was to forfeit $8 for every day he was absent. He received at the end of his time $40. How many days did he labor, and how many days was he absent ? Verify. 18. A- boatman can row down stream a distance of 20 miles, and back again, in 10 hours, the current being uniform all the time ; and he finds that he can row 2 miles against the current in the same time that he rows 3 miles with it. Required the time in going and return ing. Result, 4 and 6 hours. SuG. — If £c and ?/ are the times of rowing down and up, respectively, at what rate does he row down ? At what rate up ? Twice one of these rates equals 3 times the other. 19. A and B together could have completed a piece of work in 15 days, but after laboring together 6 days, A was left to finish it alone, which he did in 30 days. In how many days could each have performed the work alone ? An^., 50, and 21f days. SuG. — If a; represent the number of days A would require to do it alone, and y the number B would require, how much would each do in a day? How much both? How much would they do in 6 days? How much would remain to be done by A alone ? How much would A do in 30 days ? In resolving these equations do not clear of fractions. 20. Two pipes, the water flowing in each uniformly, filled a cistern containing 330 gallons, the one running during 5 hours, and the other during 4 ; the same two pipes, the first running during two hours, and the second WITH TWO UNKNOWN QUANTITIES. 271 three, filled another cistern containing 195 gallons. The discharge of each pipe is required. Verify. 21. If I were to enlarge my field by making it 5 rods longer and 4 rods wider, its area would be increased 240 square rods ; but if I were to make its length 4 rods less, and its width 5 rods less, its area would be dimin- ished 210 square rods. Required the present length, width, and area. Verify. 22. A farmer sells a horses, and h cows for $7?i ; and at the same prices a^ horses, and h^ cows for %mi; what is the price of each ? Apply the results to Ex. 11. ^- -, J>\^n — ftm, - .o,m — an\ Ans., Of a horse $ — ; — ; of a cow $ — 7 r— . a6i — a^h ajb — 061 [Note. — Observe the symmetry of such results. Thus, in these nu- muerators the a and b change places and in the denominators the sub- scripts change letters.] 23. A man bought s acres of land for $m. For a part he paid %a per acre, and for the rest %a^ per acre. How many acres in each part? Deduce from the general answer obtained in this case the particular answers to ^ ,^ . m — a,s _ m — as Ex. 13. Ans., and acres. a — fli fli — a 24. A waterman rows a given distance a and back again in h hours, and finds that he can row c miles with the cur- rent for d miles against it : required the times of row- ing down and up the stream, also the rate of the cur- rent and the rate of rowing ? rr,. -. hd ,. be , - Ans., Time down, ; time up, — — ; ; rate of c-^-d c-{-d a{cr- — d-^) , . . a(e-hd)^ current, -—z — ; rate of rowmg, 2bed ' *' '2bcd Deduce from these answers those of Ex. 18. 272 SIMPLE EQUATIONS [Note.— Several varieties of examples usually given in this place are reserved for their proper places under the discussions of the general principles or problems of which they are special examples. Such are, examples involving ratio, proportion, percentage, aUigation, etc. ] SECTION III Simple, Simultaneous, Independent Equations with more than Two Unknown Quantities. 40* JProh. Having given seiwral simple, simultayieoiis, in- dependent equations, involving as many unhioivn qiianti- ties as there are equations, to find the values of tlie unknoiun quantities. RULE.. — Combine the equations two and two by either OF THE METHODS OF ELIMINATION, ELIMINATING BY EACH COMBINA- TION THE SAME UNKNOWTSf QUANTITY, THUS PRODUCING A NEW SET OF EQUATIONS, ONE LESS IN NUMBER, AND CONTAINING AT LEAST ONE LESS UNKNO^TSf QUANTITY. CoMBINE THIS NEW SET TWO AND TWO IN LIKE MANNER, ELIMINATING ANOTHER OF THE UNKNOWN QUAN- TITIES. Repeat the process until a single equation is found with but one unknown quantity. solve this equation and then substitute the value of this unknown quantity in one of the next preceding set of equations, of which there will be but two, with two unknown quantities, and there w^ll result an equation containing only one and that another of the unknown quantities, the value of which can therefore be found from it. substitute the two values now found in one of the next preceding set, and find the value of the remaining unknown quantity in this equation. Continue this process till all the unknown quantities are determined. ScH.l. — If any equation of any set does not contain the quantity you are seeking to eUminate from the following set, this equation WITH MORE THAN TWO UNKNOWN QUANTITIES. 273 cau be written at once in that set and the remaining equations com- bined, ScH. 2. — In ehminating any unknown quantity from a particular set of equations, any one of the equations may be combined with each of the others, and the new set thus formed. But some other order may be preferable as giving simpler results, ScH. 3. — It is sometimes better to find the values of all the unknown quantities in the same way as the first is found, rather than by substi- tution. Dem. 1. — The combinations of the equations give true equations be- cause they are all made upon the methods of eUmination akeady de- monstrated, 2, That the number of equations can always be reduced to one by this process, is evident, since, if we have n equations and combine any one of them with each of the others, there will be ?i — 1 new equations. Combining one of these n — 1 new equations with all the rest there will result n — 2. Hence n — 1 such combinations will produce a sin- gle equation; and as one unknown quantity, at least, has disappeared from each set there will be but one left. q. e. d. EXAMPT,FS. 1. Given (10 1X — 2Z + du = 17, (20 ^ + 4?/ — 2z = 11, (30 5y — Sx — 2iL= 8, (40 - - 3w ^2t + 4ty= 9, (50 Sz + 8a = 33, to find the value of X, y, z, t, and u. MODEL SOLUTION. OPEBATION. (1*) t^iy-2z=ll- (2,) 2« -f- 4?/ — 3u = 9 2nd set , from which (3*) 3z-f-8u= 33 rx is absent. (4,) 35y — 62 — 5u =107 (I3) (23) (33) Sz-{-8u= 33-] 35r/ - 6s - 5u = 107 L ^rd set, from which 4y - 4. 4- 3u = 13 J ^ ^^^ ^ ^^^ ^^'^^^• (I4) 3r 4- 8it = 33 ) 4th set, fi-om which (2,) 116z — 125 a = — 27 fa;, t, and y, are absent. 274: SIMPLE EQUATIONS (I5) lS03u 3909 iu) 3z + 21 = 33 .-. z = 3 (33) 47/ _ 12 + 9 = 13 ••• 2/ = 4 (12) i_f_ 16 — 6 = 11 .-. f = l (3x) 20 — 3a3 — 6= 8 .-. x = 2 Explanation. — I notice tliat I have 5 equations with 5 unknown quantities. From these I wish to produce a new set of 4 equations from which one at least of the unknown quantities shall be eliminated. I observe that x does not appear in (2i), (4,), and (Sj), hence I write these as three of the 2d set of equations. Then eUminating x between (li) and (3i) I have (42), and thus obtain the 2nd set of 4 equations containing only 4 unknown quantities. (If desirable the pupil may be asked how he knows that (I2) is a true equation.) Again, as t is contained in a less number of this set of equations than any one of the other unknown quantities, leUminate it next; i. e., I produce a 3rd set which does not contain it. As (3,j) and (4j) do not contain t, I transfer them at once to the 3rd set; and then ehminating t between (Ig) and (23) this set is complete, having 3 equations with 3 unknown quantities. Now eliminating y from this set by combining (23) and (83), and transferring (I3), I have the 4th set of two equations with only 2 un- known quantities. Combining these two so as to eliminate z I find M = 3. Finally, substituting 3 for u in (I4), I find z = 3. Substituting 3 for u and 3 for z in (33), I find y =:^ 4,. Substituting 4 for y and 3 for z in (I2), I find t == 1. Substituting the values oiy and u in (3j ), I find x == 2. K 2. Given -^ ^ -f z = 25, V Values, (y+ z = 15.) / 8j; — 4?/ = 24 — z,\ 3. Given -| 6^ + y ^ 2; + 84, V Values, [ a; + 80 = 3?/ + 42. j f 3u + a: + 2?/ 4 Given -\ z = 22, ] 4.x— 2/ + 3z = 35, ! . Y Values, -\ ^u + 3:r — 2?/ 19, L 2u -^ 4y -\-2z= 46. J WITH MORE THAN TWO UNKNOWN QUANTITIES. 275 5. Given X 2 + i-^ 2 X a + !+ 5 X 4 + \- z = 124, = 94, = " 76. Results, - 48, 120, 240. SuG. — Such examples afford opportunity for the exercise of no Httle ingenuity, m order to avoid large numbers and inconvenient fractions. For example, this may be solved in the ordinary way by clearing of fractions, etc. , but the following is far more elegant : Dividing the 1st by 3 and the 2nd by 2 and subtracting, we have 11 ) "3~ y_ A.^ 72 "^ GO Subtracting ^ the 1st from the 3rd 30"^ 2r 2nd set. 14 Or, 'dm "*" 5-60 15 360 ^24- 12 Subtracting 24-1^.^^- = 30 ^^ 240-^' 2 5 1 >h-l r.= -^ 6. Given - - Values, -< = 1, and z = 240. a + 6 — c 2 6 + c — a SuG. — Do not clear of fractions. Having found the value of one un- known quantity, do not get the others by substitution, but return to the original equations and get each in the same manner. — 7. Given 2 X j3 3 4 y z 4 5 6. X y x= 6, Values,] 2/ = 12, 276 SIMPLE EQUATIONS 8. Given J 9. Given, y -\- a :^ 2x -{- 2z, [ Values, ^ y 9^+6y + 4z + 3u + 2w + 1 = 0, 16^ + 4i; + 1 = 0, 25^ — ^y -\- z-\-bv — u-\- 1=0, X -\- 1y + ^z — V — 2u + 1 = 0, 4:X — 2y-\-z — 2v + u + 1 = 0. j Values, - zi + V + ^ + ?/ = 10, If + u + ^' + 2 = 11, 10. Given, \ u -\- v -^ y + z =12, \ u + ^ + ?/ + z = 13, I (; + :^ -f 2/ + 2 = 14. J f 3.f/ — 1 62 j; 12/ Values, \ z I 11. Given, 4 5 2 5.r 42 5 1 = 1/ -1- _i_ 14 T^ -'■5J T=y+& - Fa/i^e. 3^ + 1 2^ , y + ^ = 777 + '*■> 14 ' 6 21 ' 3 J 12,v — 11^ 11^ — lOy 12. Given, \ x-\-z — 2y z — y — 1 Values, - 3 Zx 2/ + 2 + 7. y = a il' 5a II' la ir 169 ~924' 220 924' _89_ 924' 4^5 924' 113 924* = 3, = 4, = 5, = 2, = 1. = 2, = 3, = 1. 10, 11, 12. WITH MOKE THAN TWO UNKNOWN QUANTITIES. 277 13. Given, ' Values, ^=9, !/-7, z = 3. 14. Given, ■{ }j(x -\- z) = d — ij, i(^_z) = 2^-7. Values, -J ?/ = 4, z =3. APPLICATIONS. 1. The sum of three numbers is 9. The sum of the first, twice the second, and three times the third is 22. The sum of the first, four times the second, and nine times the third is 58. What are the numbers ? Ans., 1, 3, and 5. SuG. — How many unknown quantities? How many sets of condi- tions? What are they? Express the first in an equation, — the second, — the third. The form of explanation is the same as in the last section. 2. Five persons, A, B, C, T>, and E played at cards ; after A had won half of B's money, B one-third of C's, C one- fourth of D's, and D one-sixth of E's, they each had $7.50. How much had each to begin with? Ans., A,$2.75; B $9.50, C $8.25; D,$8, and E,$9. 3. There are 4 men. A, B, C, and D, the value of whose estates is $14,000 ; twice A's, three times B's, half of C's, and one-fifth of D's, is $16,000 ; A's, twice B's, twice C's, and two-fifths of D's, is $18,000 ; and half of 278 SIMPLE EQUATIONS A's, with one- third of B's, one-fourth of C's, and one- fifth of D's, is $4,000. Kequired the property of each, Ans., A's,$2,000; B's,$3,000; C's,$4,000; r>'s^$5,000. 4. A number is expressed by three figures ; the sum of these is 11 ; the figure in the place of units is double that in the place of hundreds, and when 297 is added to this number, the sum obtained is expressed by the figures of this number reversed. What is the number? Ans., 326. SuG. — Letting x represent the hundreds figure, y the tens, and z the Units, the number is represented by lOOx -f- lOy -f- z. The number with the digits reversed is lOOz -f lOy -f a;. 5. A man worked for a person ton days, having his wife with him 8 days, and his son 6 daj^s, and he received $10.30 as compensation for all three ; at another time he wrought 12 days, his wife 10 days, and son 4 days, and he received $13.20 ; at another time he wrought 15 days, his wife 10 days, and his son 12 days, at the same rates as before, and he received $13.85. What were the daily wages of each ? Ans., The husband 75 cts.; wife, 50 cts. The son, 20 cts. expense per day., SuG. — Solving the equations which express the conditions, the value of the quantity representing the son's wages is found to be negative. But as negative quantities are such as are opposed in their nature to those called positive in the same problem, the son produced the oppo- site effect from wages. 6. Three masons, A, B, C, are to build a wall. A and B, jointly, can build the wall in 12 days ; B and C can accomplish it in 20 days, and A and C in 15 days. How many days would each require to build the wall, and in what time will they finish it, if all three work ' together ? Ans., A requires 20 days; B,30;and C,60 ; and all three require 10 days. WITH MORE THAN TWO UNKNOWN QUANTITIES. 279 7. Three laborers are employed on a certain work. A and B, jointly, can complete the work in a days ; A and C require h days, B and C require c days. What time does each one, Avorking alone, require to accom- plish the work, on the condition that each one, undet all circumstances, does the same quantity of work? And in what time would they finish it, if they all three worked together ? Ans.. A requires , days, B r — ■ — 7 he -\- ac — ab be -\- ab — ac days, and C -- ; r~ days. -^ ' ab ^ ac —be "^ Jointly, they require ^^ ^ ^^ ^ ^^ days. Deduce from these results those of the preceding ex- ample. 8. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C to- gether in 10 days, in how many days can each alone perform the same -svork ? Ans., A in 14f * days, B in 11 1^ days, and C in 233V days. 9. A gentleman left a sum of money to be diyided among his four sons, so that the share of the oldest was -^ of the sum of the shares of the other three, the share of the second ^ of the sum of the other three, and the share of the thu'd \ of the sum of the other three; and it was found that the share of the oldest exceeded that of the youngest by $14, What was the whole sum, and what was the share of each person ? Ans., Whole sum, $120 ; oldest son's share., $40 ; second son's, $30; third son's, $24 ; youngest son's, $26. 280 SIMPLE EQUATIONS. Synopsis. r Algebra. Equation. Members. Defs. \ Independent equations. Simultaneous. I Transposition. Elimination. [ Statement. Solution. rr- J f { Simple, ) With one unknown quantity. Kinds of jQ^jadratic, V j^quauon^. j higher. ) With more than one nnk'n quant. Axioms. r 1. Clearing of frac's. Kule. Dem. 111. m /. ^- 2. Transposition. Rule. Dem. 111. Transformations. ^ -^^.^.^^ ^^^,^^^ [ 4. Dividing by coeff. of unk'n quant. Proh. 1. To solve simple equations. ) „ p .„„^p^tioTm Rule. Dem. \ ' ^^^^- suggestions. Proh. 2. To free an equation of radicals ( ^ -r. i.- Rule. Dem. [ 6 Prac. suggestions. !^ f Number of methods. Reason for several. H [ Prob. 1. By comparison. Rule. Dem. < J Prob. 2. By substitution. Rule. Dem. M I Py^ob. 3. By addition and subtraction. Rule. Dem. M Prob. 4. With several unknown quan- ) q^, i o q i titles. Rule. Dem. \ ^^^- ■^' ^^ ^' Test Questions. — Upon what principle is an equation cleared of fractious ? How is it done ? Upon what principle is elimination by addition and subtraction performed ? What comparison ? Substitu- tion ? Give the seven Practical Suggestions upon solving Simple Equations. The six upon freeing of Radicals. Give the reason for changing the signs of the terms of a fraction having a poljmomial nu- merator, preceded by a minus sign, when clearing of fractions. What is the general method of procedure in stating a problem ? Does the statement involve a knowledge of anything but algebra ? Illustrate. Upon what principle may all the signs of an equation be changed ? (This may be explained in at least four different ways.) Having given the sum and difference of two quantities, how are the quantities found ? Prove it. RATIO, PROPORTION, AND PROGRESSION. 281 CHAPTEE 11. RATIO, PROPORTION AND PROGRESSION, SUCTION I. Eatio. 47. Mcitio is the relative magnitude of one quantity as compared with another of the same kind, and is ex- pressed by the quotient arising from dividing the first by the second. The first quantity named is called the Antece- dent, and the second the Consequent. Taken together they are called the Tei^ms of the ratio, or a Couplet. 48, The Sign of ratio is the colon, : , the common sign of division, -i- or the fractional form of indicating division. Q III. — The ratio of 8 to 4 is expressed 8 : 4, 8 -^ 4, or _, any one of 4 > whicli may be read " 8 is to 4," or, " ratio of 8 to 4." The aniecedeni is 8, and the consequent 4. The sign : is an exact equivalent for -j- , and by many writers, especially the Germans, the former is used ex- clusively. The sign : is, probably, a mere modification of -f-, made by dropping the horizontal line, as unnecessary. Possibly the sign ~ finds its analogy to the fractional form of expressing division, by con- sidering the upper dot as symbohzing a dividend, and the lower a divisor. 40, CoR. — A ratio being merely a fraction, or an unexe- cuted problem in Division, of which the antecedent is the nu- merator, or dividend, and the consequent the denominator, or divisor, any changes made upon the terms of a ratio produce 282 RATIO, PROPORTION, AND PROGRESSION. the same effect upon its value, as the like changes do upon the value of a fraction, lohen made upon its corresponding terms. The principal of these are, 1st. If both terms are multiplied, or both divided by the same number, the value of the ratio is not changed. Thus, the ra- tio 16 : 8 is 2. So, also, 4 x 16 : 4 x 8, or 64 : 32 is 2 ; 16 8 or — : -, i. e., 8 : 4 is 2. Ji A 2nd. A ratio is multiplied by multiplying the antecedent (i. e., the numerator or dividend), or by dividing the con- sequent (i, e., the denominator, or divisor). Thus, 32 : 8 is 4, but 2 X 32 : 8, i. e., 64 : 8 is 2 x 4, or 8. So, also, 32 : o -^, i. e., 32 : 4, is 2 X 4, or 8. 3rd. A ratio is divided by dividing the antecedent, (i. e., the numerator, or dividend), or by multiplying the consequent, ({. e., the denominator, or divisor). Thus 24 : 6 is 4, but 24 4 — : 6, i. e., 12 : 6 is -, or 2. So, also, 24 : 2 x 6, i. e., Z A 24 : 12 is -, or 2. A 50. A Direct Ratio is the quotient of the antece- dent divided by the consequent, as explained above. {4t ,) An Indirect or Meciprocal Ratio is the quotient of the consequent divided by the antecedent, i. e., the re- ciprocal of the direct ratio. Thus, the direct riitio of 6 to 3 is 2, but the inverse ratio is f or ^. When the word ra- tio is used without qualification it means direct ratio. The in^ verse or reciprocal, it will be seen, is the ratio of the reciprocals. Thus the inverse ratio of 8 to 4 is the ratio of \ to \, or ^. 51, A ratio is always written as a direct ratio. Thus, if the inverse ratio of 6 to 2 is required, we write 2 : 6. The inverse ratio of aiobi^b -. a, or - : -, the latter beincr ex- a b pressed as the direct ratio of the reciprocals. RATIO. 283 52. A ratio of Greater Inequality is a ratio which is greater than unity, as 4 : 3. A ratio of Less Iiie^ quality is a ratio which is less than unity, as 3 : 4. 53. A Compound Ratio is the product of the cor- responding terms of several simple ratios. Thus, the com- pound ratio a : h, c \ d, m : n, \?> acm : bdn. This term cor- responds to compound fraction. A compound ratio is the same in effect as a compound fi'action. 54. A Duplicate Ratio is the ratio of the squares, a trijMcatef of the cubes, a suhduplicate^ of the square roots, and a suhtriplicate, of the cube roots of two num- bers. Thus, a' : b\ a^ : b^, Va : Vb, and "^a : '^b. EXAMPLES. 1. What is the ratio of 3am3 to Qam" ? Model Solution. — Since the ratio of two quantities is the quotient of the antecedent divided by the consequent, the ratio Zarn? : Gam'^ is 3am^ 1 2. What is the inverse ratio of a — - 6 to a2 — 6-^? Ans., a + 6. 3, What is the ratio of 2 1 Of 5 , 6 ' ■ 3' a- a — x'> Of*^-='"to 2cm ^ Of*'' -y- . X — -y db^y 'db'^y- x^ -\- y^ x- — xy-\-y^ Answers to three, 1, 1\, and 1^. 4. What is the triplicate ratio of 6 to 2. Ans., 27. 5. What is the subduplicate ratio of 64 to 16 ? Ans., 2. 6. What is the compound ratio of 3 to 4, 8 to 9, 2 to G 4 and 4 to 2 ? Ans., 4 : 9, or -. y 7. Reduce 360 : 315 to its lowest terms. SuG.--This is the same as reducing a fraction to its lowest terms. The result is 8 : 7. 284 KATIO, PEOPORTION, AND PROGRESSION. 8. Eeduce 1595 : 667, and a^ + 2a"x : a^ to their lowest terms. Result of the last, a -\- 2x : 1 . 9. Which is the greater, 16 : 15, or 17 to 14? SuG. To compare two fractions, reduce them to a common denom- inator. On the same principle these ratios become 224 : 210, and 255 : 210. 10. Which is greater, a^ — b^ : a — 6, or a^ -j- 2ah -\- b^ : . a + bl 11. Which is least of the ratios 20 : 17, 22 : 18, and 25 : 23? Which is greatest, 8 : 7, 6 : 5, or 10 : 9 ? 12. Which is greater, a + 2 : ^a + 4, or a + 4 : ^a + 5 ? Alls., a + 4:^a + 5>>a + 2 :ia + 4. 13. What is the compound ratio of 15 : 12, 6 : 7, and 9:4? Ans., 135 : 56. 14. Compound the ratios a2 — ^2 • a'i^a-^-x : b, and 6 : a — x. Result, The duplicate ratio of a -\- x io a. 15. Show that the compound ratio oi x -\- y \ a, x — y '. b, and : IS 1. a 16. Is the compound ratio of 3^ + 2 : 6a + 1, and 2a + 3 : a + 2, a ratio of greater or of less inequality, if a is — f ? If a is 2,? If a is — 2? Ans., — Zero. Greater. Infinity. 17. Compound the following : 7 I 5, the duplicate of 4 : 9, and the triplicate of 3 : 2. Result, 14 : 15. 2 Suggestion. - X ^'4 X ^-^, -- 14 : 15. 5 0.0 t-t-t 3 18. Compound the sub-duplicate of x'^ : y^, and the triph- cate of '^J7 : "^y. Result, x^ : yK PROPORTION. 285 19. Compound the inverse ratio of vx + \/y to :c — y, and the direct ratio x + 2V xy -\- y : Vx + Vy. Result, X — y. 20. Which is the greater, the inverse subtriphcate ratio of 8 to 64j or the direct dupUcate ratio of 2 to 3 ? SECTION II. Proportion. do, I*t*opoVtion is an equahty of ratios, the terms of the ratios bemg expressed. The equahty is indicated by the ordinary sign of equahty, ^, or by the double colon : : . Thus, 8 : 4 = 6 : 3, or 8 : 4 ; ; 6 : 3, or 8^ 4 = 6 -^- 3, or - = -; all mean precisely the same thing. A propor- tion is usually read thus : " as 8 is to 4 so is 6 to 3." ScH. — The pupil should practice writing a proportion in the form a c — =: — » still reading it " a is to 6 as c is to d." One form should be as familiar as the other. He must accustom himself to the thought that a : h \: c '.d means -r = -j and nothing more. It will be seen that the o ^ . — 1 = . From the given proportion, by Prop. 1, ilh^mx = 30a-6m"a;- or 2&2 = Sa-ma;. Therefore the answer is, Yes. 11. If lax^ : \hy :: a^x : ¥ij, show that h : ^ \: a^ '. ^x. COMPOSITION AND DIVISION. 12. If (2 : 6 : : 7?i : ??, show that a -\- h \ a '. : m -{- n : m. Solution. — I am to prove that = , knowing that a m a :b : : m :n. The two ratios to be tested, when reduced to a C. D. am -{-bm ^ am -\- nn . . , , , , . . ,, are and , which are seen to be equal, since from the am am given proportion bm = an. 13. If a : 6 : : J7 : ?/, show that a — h : b '.'. x — y ' y- 14. li m : n : : X : y, show that m -{- n : m — n ;: x -\- y : x — y. 15. If ^a — X : ^a -{- X w h ■ — y : b -{- y, show that 2x : y '.'. a : b. 2a; a Solution.— I am to compare — '- and -, which reduced to a C. D. y b are — and r^^— . But from the given proportion I have, by Prop. 1, ^ab -{-bx — ^ay — xy = ^ab — bx -f- k^iy — xy, which reduced gives 2x a 2bx = ay. . • . -^ = -, which was to be proved. 16. If a : 6 : : ^ : y, does it follow that a — y : b — x :: a : X? Ans., No. 17. If four quantities are in proportion, does it follow that they are in proportion by composition, or by division, separately, or by both at once? That is if a \ b '.'. m : n, is a + ^ \ a :: m -\- n \ m, or a — b : a :: m — n : m, or a + 6 : a — b ; : m-{- n : m — n? Ans., Yes. 292 RATIO, PROPORTION, AND PROGRESSION. 18. li a : b :: X : ^j isa^ : b^ :: x-^ \ y-l Is a" : 6" : : X* : 2/"? Is v/a : v/^ : : v/^j : v/^ ? Is a^ : 6"^ : : x?^ \ y'-^l Is a"* : 6"* : : a:"' : 7/"* whether m is integral or fractional, positive or negative? AVhy is it that the ratios remain equal in each case? How are they changed? MISCELLANEOUS. 19. If a : & : : c : (Z, show that ma -{- nh \ pa -]- qb : : mc -\- nd '. pc -\- qd. SuG. — The ratios to be compared when reduced to a C. D. are, acmp -\- bcnp -f- ndmq -\- hdnq acmp -f- bcmq -\- adnp -\- hdnq (Up -\-bqi{cp -^dq^ {ap -}- bq}{cp -\- dq) Now from the given proportion we learn that ad = be. Therefore, ex- changing them in the two middle terms of the first ratio, the ratios become identical. This may also be shown as follows : Multiplying antecedents by m and consequents by n, ma : 7ib : : mc : nd. By composition ma -j- nb : ma :: mc -{- nd : mc, or multiplying both ratios by m, ma -f- >'^ : a : : mc -{- nd : c. By changing the places of the means ma -f- nb : mc -\- nd : : a :c. In like manner it may be shown that pa-\-qb '■ pc -\- qd : : a :c. . • . ma -{- nb : mc-\-nd : : pa-\- qb :pc-\- qd, or ma -{- nb : pa -\- qb : : mc -\- nd \ \pc -{- qd. The student should give the reason why each step does not vitiate the proportion, according to {09). 20. If {a-\-b-^c + d){a — b — c+d) = (a — b-\-c — d) (a -f 6 — c — d) prove that a : b : : c : d. ^ ^ ,, . ^. , a + ?)-j-c-f-d a — b-j-c—d buG. — From the given equation we have — -——^ ■ — -, = ; — ■ ; — •,. a-^b — c — d a — b — c-^d Clearing of fractions and reducing 2ad — 2bc = — 2ad -f- 26c, or ad = be. Whence - = -. b d This may also be proved by writing according to Prop. 2, a -f- ^ + c-{-d : a — b -f-c — d :: a + 6 — c — d : a — 6— c-fd. Comparing the sum of each antecedent and its consequent with their difference, 2a 4- 2c : 26 + 2d : : 2a — 2c : 26 — 2d Whence a -}- c : a — c : : b-\-d : 6 — d. Repeating the same processes we have a : 6 : : c : d 21. If — : = b, show that a — x : 2a : : 2b : a -\- x. Produce other forms of proportion from the given re- lation. How many can be produced ? PROPORTION. 293 22. If r=^^s show that r : s : : 1 : ^2. 23. If (a 4- j:)^ : (a — x)^ : : x + y : x — y, show that a : X : : ^2a — y : ^^. Solution. a2 -f- 2ax + x^ : a^ — 2ax -{- x^ : : x + y : x — y, a2 4- 2ax + x2 : a2 — 2ax 4- x2 : 2a2 + 2x2 : 4ax : : 2x : 2y, a^ + x2 : x2 : : 2a : y. a2 : x2 : : 2a — y '■ 2/. .-. a : X : : ^/2^ —y : v/y. Let the student give the reasons. 24. If a : 6 : : c : d :: e :f :: g : h :: i : k, etc., show that {a-\-c-\-e + g + i+, etc.) : (6 + cZ -h/4- h + k-^, etc.) : : a : i*), or c : c?, or e : /*, etc. That is, in a series of equal ratios, the sum of all the antecedents is to the sum of all the consequents, as any antecedent is to its consequent. Solution. a a , - - = - OT ah = 6a, b b a c , , — =: - or aa = he, a - = - OT af = h%, a q , ^ - _ = ^ or a/. = 6gr. — = - or a/c = hi, b k etc. Adding, a(h -\- d 4-.f + h-\^k-\-, etc. ) = h{a -\- c ^ e -\- g -{-i ■\; etc.) : whence {a-\-c-\-e-{-g-\-i-\-, etc.) : (6 + d -j-/ -f- ^ + ^ +» etc. ) : : a : 6 or (since a -.b = c -.d, etc. ), c : d : : e :/, etc. 25. Four given numbers are represented by a, b,c,d; what quantity added to each will make them proportionals ? . be — ad Ans,, — •. a — 6 — c + a 26. If four numbers are proportionals, show that there is no number which, being added to each, wiU leave the resulting four numbers proportionals. 294 RATIO, PROPORTION, AND PROGRESSION. SECTION III Progressions. 70, A Progression is a series of terms which in- crease or decrease by a common difference, or by a com- mon multiplier.* The former is called an Arithmeiical, and the latter a Geometrical ProgreHsion. A Progression is in- creasing or Decreasing according as the terms increase or decrease in passing to the right. The terms Ascending and Descending are used in the same sense as increasing and decreasing, respectively. In an increasing Arithmetical Progression the common difference is added to any one term to produce the next term to the right ; and in a de- creasing progression it is subtracted. In an increasing Geometrical Progression the constant multiplier by which each succeeding term to the right is produced from the preceding is more than unity; and in a decreasing progres- sion it is less than unity. This constant multiplier in a Geometrical Progression is called the Ratio of the series.f 7-1. The character, • • , is used to separate the terms of an Arithmetical Progression, and the colon, : , for a like pur- pose in a Geometrical Progression. * This is the common use of the term. It is also used to include what is called a Harmonical Progression. But our limits do not allow, neither does the importance of the subject demand the treatment of the latter topic in this volume. t This is an unfortunate use of the term Ratio : it were better to use the term Rate, as some French writers do. To harmonize the use of the term in proportion, with this vise, may have led some writers to define ratio, as used in proportion, as the quotient of the cousi'quent divided by the antecedent. But the definition has neither logic nor the common usage of autliors, English or c'oiitinental, to support it. How the notion that this definition of ratio is the common French definition, has gained currency, it is not easy to say. So far as the author's observation has oxteni'^d. it certxinlv i^ not the fact. ARITHMETICAL PROGRESSION. 295 LLLUSTEATIONS. 1 .. 3 .. 5 .. 7, etc., etc., is an Increasing Arithmetical Progression with a common difference 2, or -j- 2. 15 •• 10 •• 5 •• •• — 5, etc., etc., is a Decreasing Arithmetical Pro- gression with a conamon difference — 5. a"a + d"a + 2d"a-\- 3d, etc. , etc. , is the general form of an Arithmetical Progression, d being the com- mon difference. 2:4:8 : 16, etc., etc., is an increasing Geometrical Progression with ratio 2. 12 : 4 : ^ : I : ^V e^-. etc., is a Decreasing Geometrical Progression with ratio ^. a : ar : ar'^ : ar^ : ar*, etc., etc., is the general form of a Geometrical Progression, r being the ratio, and greater or less than unity, according as the series is increasing or decreasing. 72, There are Five Things to be considered in any pro- gi'ession ; viz., the first term, the last term, the common difference or the ratio, the number of terms, and the sum of the series, either three of which being given the other two can be found, as will appear from the subsequent dis- cussion. ARITHMETICAL PROGRESSION. 75. JPro2>» 1» The formula for Jinding the nth, or last term of an Arithmetical Progression; or, more properly, the formula expressing the relation between the first term, the nth term, the common difference, and the number of terms of such a series ts / = a -f (71 — l)d, in which a is the first term, d the common difference, n the number of terms, and I the nth or last term, d being posi- tive or negative according as the series is increasing or de- creasing. RATIO, PROPORTION, AND PROGEESSIOIT. Dem. — According to the notation, the series is a • • a -f- d . • a -f- 2d • • a -|- 3d . . a -f- 4d • • a -f- 5d, etc., etc. Hence we observe that as each succeeding term is produced by adding the common difference to the preceding, when we have reached the nth term, we shall have added the common diflference to the first term n — 1 times ; that is, the nth term, or / = a -f- (n — l)d. q. e. d. ScH. — As this formula is a simple equation in terms of a, I, n, and d, any one of them may be found in terms of the other three. 74. JProp, 2. The formula for the sum of an Arith- m^etical Progression, or expressing the relation between the sum of the series, the first term, last term, and number of terms is ■a + /- n^y s representing the sum of the series, a the first term, I the last term, and n the number of terms. Dem. — If I is the last term of the progression, the term before it is I — d, and the one before that I — 2d, etc. Hence, as a • • a + d • • a -f- 2d •• a-\-^d I, represents the series, 1--1 — d--l — 2d"i! — 3d a, represents the same series reversed. Now, the sum of the first series is s = a -f (a -l-d) +(a+ 2d)-\ (Z _ 2d) + (? — d) + Z; and reversed s = I -f (I — d) -j- (^ — 2d) H (a + 2d) -{- (a -)- d) -f- a. Adding, 2s = (a -|- /) -|- (a -\- 1} -]- {a -{- 1)-\ [a-\- 1) -\-ya -\-l]-{-{a-\-l}. If the number of terms in the series is n, there will be n terms in this sum, each of which is {a-\-l); hence 2s = (a -|- hn, or s = 7" p- Q. E. D. ScH. — This formula being a simple equation in terms of s, a, I, and n, any one of the four can be found in terms of the other three. 75, Cor. 1. — Formulas (1) 1 = a + (n — l)d, and (2) s= — TT" r^' being two equations between the Jim quantities, a, 1, n, d, and s, any two of these five can he found in terins of the other three. ARITHMETICAL PROGRESSION. 297 [Note. — It is not considered worth while to make separate cases out of the different problems which arise in the progressions, or to cumber the memory with the multipHcity of formulas which can ba deduced from the two fundamental ones, but rather that these should be fixed in memory, and their use clearly understood. ] EXAMPLES. 1. The first term of an A. P. is 2, and the common differ- ence 3, what is the 11th term ? "What the sum of the series ? Solution. — In the first case there are under consideration the first term, a = 2, the common difference, d = 3, the number of terms, n = 11, and the last term, which is the thing required. The relation between these is given in Z = a -f- (n — l)(i ; in which by substituting the given values there results / = 2 -f (11 — 1)3 = 32. In the second case the formula s = \ J~ n gives the relation, in which by substituting the given values there results s = — — ^ 11 = 187. 2. The first term of an A. P. is 8, the last term 203, and the common difference 5, what is the number of terms ? What the sum of the series ? Ans., n = 40, s = 4220. 3. The first term of an A. P. 8, the last term 203, and the number of terms 40, what are the common difference and the sum ? 4. The last term is 1, the sum 1717, and the number of terms 34, what are the first term and the common difference ? SuG.— The equations are 1 = a + (34 — l)d, and 1717 = (^-^~-^3iy from which to find a and d. a = 100, and d = — 3. 5. What is the sum of the numbers 1, 2, 3, 4, etc., to 1000? Suggestion. — The common difference is 1, and the last term 1000. 6. The first term of an arithmetical progression is 1, and the number of terms 23, what must be the common dif- ference that the sum of all the terms may be 100? What the last term? 298 EATIO, PROPORTION, AND PROGRESSION. 7. If the first term of an arithmetical progression is 100, and the number of terms 21, what must the common difference be that the sum of the series may be 1260 ? What the last term ? Ans., — 4. 8. Two persons, A and B, start from the same place together, and travel in the same direction. A goes 40 miles per day; B goes 20 miles the first day, and increases his rate of travel | of a mile per day. How far will they be apart at the end of 40 days, and which will be in advance ? Ans. A, 215 miles. 9. The first term of an arithmetical progression is — 7, the common difference — 7, and the number of terms 101, what is the sum of the series? Ans., — 36057. 10. Insert 8 arithmetical means between 3 and 21. Series., 5- • 7- -O- -11- -13. -15. -17- -19. 11. Insert 3 arithmetical means between i and -i-. Series, -3. . . _5^ . . li. 12. What is the nth term of the series l««3-'5««7", etc. ? Ans., 2n — 1. 13. What is the sum of n terms of the series l-«3-'5--7'*, etc. ? Ans.^ n^. 7G, Cor. 2. — The formula for inserting a given number of \ a arithmetical meaiis between two given extremes is d= . ^ m -h 1' in which m represents the number of means. From this d, the common difference, being found, the terms can readily be written. Dem. — If a is the first term and I the last, and there are m terms between, or m means, there are in all m -f- 2 terms. Hence substituting in the formula I = a-\-[n — l)d for n, m-{-2, we have Z ■= a -f- ;m -)- l)d. From this d = — — -. q. e. d, m-{-l GEOMETRICAL PROGRESSION. 299 14. If a body falling to the earth descends a feet the first second, 3a the second, 5a the third, and so on, how far will it fall during the ^th second ? Ans., {2t—l)a. 15. If a body falling to the earth descends a feet the first second, 3a the second, 5a the third, and, so on, how far will it fall in t seconds ? Ans., at\ 16. A debt can be discharged in a year by paying $1 the first week, $3 the second, $5 the third, and so on ; re- quired the last payment and the amount of the debt. Ans.^ Last payment, $103 ; amount, $2704 17. A person saves $270 the first year, $210 the second, and so on. In how many years will a person who saves every year $180 have saved as much as he ? Ans., 4. 18. A board, 2| inches wide at the narrow end, and 10 feet long, increases in width 1\ inches for every foot in length ; what is the width of the wide end ? Ans., 17| in. 19. If 100 oranges are placed in a line, exactly 2 yards from each other, and the first 2 yards from a basket ; what distance must a boy travel, starting from the basket, to gather them up singly, and return with each to the basket? Ans., 11 mi. 3 fur. 32 rd. 4 yd. [Note. — For other examples involving the principles of Arithmetical Progi-ession, see Problems after Quadratics, and also the subject of Interest.] GEOMETRICAL PROGRESSION. 77. I^rop, 1. The formula for finding the nth, or last term, of a Geometrical Progression; or, more properly, the formula expressing the relation between the first term, the nth 300 RATIO, PROPORTION, AND PROGRESSION. term, the ratio, and the numher of terms of such a series is l=ar"~*, in which 1 is the last, or nth term, a the first term, r the ratio, and n the numler of terms. Dem. — Letting a represent the first term and r the ratio, the series is a : ar : ar^ : ar'^ : ar^ : etc. Whence it appears that any term con- sists of the first term multiplied into the ratio raised to a power whose exponent is one less than the number of the term. Therefore the nth term, or I = ar''-'^. q. e. d. 7S, I^vop* 2, The formula for the sum of a Geometrical Progression, or expressing the relation between the sum of the series, the first term, the ratio, and the number of terms is ar" — a in which s represents the sum, a the first term, r the ratio, and n the number of terms. Dem. — The sum of the series being found by adding all its terms, we have, s=a -|-ar-j-ar2-j-ar3 -j ar"~^-\-ar^ -^-f-ar"— i, and multiplying by r, rs = ar-\-ar^-\-ar^ -| ar"-^-\-ar''-^-\-ar''-^-\-ar". Subtracting, rs — s = ar" — a, or (r — l)s = ar" — a, and s = — q. e. d. SuG. —The student will notice that multiplying the first series by r, and placing the terms of the product under the like terms of the se- ries, simply moves each term, when multipUed, one place to the right, so that however many terms there may be in the series, each will have a similar one in the product except the first term, a ; and each term in the product wiU have a similar one in the series, except the last one, 70, Cor. 1. — Formulas (1) 1 = ar— S and (2) s = — ; -— being two equations between the five quantities, a, 1, r, n, and s, are sufficient to de- termine any two of them when the others are given. GEOMETRICAL PROGRESSION. SOI S0» Cor. 2. — Since 1 = ar"~\ Ir = ar", which substituted in (2) tfives s = ; which formula is often convenient. EXAMPLES. 1. The first term of a geometrical progression is 2, the ra- tio 3, and the number of terms 6. What are the last term and the sum of the series ? r — 1 2. The last term of a geometrical progression is 62500, the ratio 5, and the number of terms 7. What are the first term and the sum of the series ? SuG. — From I = ar"-^, tliere results by substitution 62500 = a-5^ or 15625a = 62500. . • . a = 4. From s = ~ "^ we find s = r — 1 78124. 3. By saving 1 cent the first week, 2 cents the second week, 4 cents the third week, and so on, doubhng the amount every week, how much is saved the last week of the year ? A7is., $22,517,998,136,852.48. SuG.— This problem requires us to raise 2 to the 51st power. This is readily effected thus : the 3rd power of 2 is 8. The 3rd power mul- tiplied by the 3rd power gives the 6th power; hence 8 X 8 = 64 is the 6th power. In like manner 64 X 64 = 4096 is the 12th power, and 4096 X 4096 = 16777216 is the 24th power; and, finally, 16777216 X 16777216 X 8 is the 51st power. 4. What is the sum of 10 terms of the series 8:4:2: 1 : i, etc. ? 1) 'ih-') a(r--l) "V2'" / ^, 2'" - 1 1023 SUG. s= —-^ = __^ = 2. -^^^ =-^ =. 2 ^ = 15tJ. Also, Z^saf-i = -==- =-^4-. In such cases the ingenious student will avoid unnecessary multipUcations by suppress- ing faetors. 302 RATIO, PROPORTION, AND PROGRESSION. 5. If 4 is the first term, 324 the last, and 5 the number of terms, what is the ratio ? SuG. — Since I = ar''-^, we have 324 = 4r*, or r = ^^81 = 3. 6. Insert 5 geometrical means between 3 and 192. The ratio is 2. 7. Find 4 geometrical means between -^ and ^. (2\ ^ - ) , whence the series becomes 11 1 1 11 mtans between a and 1 is 9' 4 1'3 2*2 3'1 4''^ "^ 2^-3^ 2^ • 3"^ 2^ • 3"^ 2^ • 3'^ *^ 81* CoR. 3. — The formula for inserting m geometrical fi IS r = "^+ M--. \a ScH. — This and many other problems in Geometrical Progression, are more readily solved by means of logarithms. Many also require a knowledge of quadratic equations, and even of the higher equations. Some farther illustrations will be given in their proper place, espe- cially in treating the subject of Interest. 82, CoR. 4. — The formula for the sum of an Infinite De- creasing Geometrical Progression is s = ;. Dem. — Since in a decreasing progression the ratio is less than unity, the last term, ar'^"^, is also less than the first term, and numerator and denominator of the value of s, •- , become negative. Hence it is well enough to write the formula for the sum of such a series s = , that is, change the signs of both terms of the fraction. 1 — r Now, if the terms of a series are constantly decreasing, and the num- ber of terms is infinite, we can fix no value, however small, which will not be greater than the last, or than some term which may be reached and passed. Hence we are compelled to call the last term of such a series 0, which makes the formula s = — — . q. e. d. GEOMETRICAL PROGRESSION. 303 ScH.— Decimal Repetends afford illustrations of such series. Thus .333 +, is ,% + r^u + TJlo +, etc., to infinity. Again, .5434343 -f- is, fj -f the series t^^o + To^gotr +Ttrtr^^ot7 +> etc., to infinity. 8. Find the sum of the series 1 : ^ : i : etc., to infinity. Sum, 2. 9. Required the sum of the series 1, i, -^j to infinity. Sum, li. 10. Find the value of .1212 to infinity. Sum, ^V 11. Find the value of .2333, etc., to infinity. Sum, ^V 12. Find the value of .3411111, etc., to infinity. Sum, ^U- 13. Find the value of .323232, etc., to infinity. Sum, ^i. 14 Find the value of .20414141, etc., to infinity. Sum, ^U- 15. Suppose a body to move eternally in this manner; viz., 20 miles the first minute, 19 miles the second minute, 18^ the third, and so on in geometrical progression. What is the utmost distance it can reach ? Ans., 400 miles. 16. What is the distance passed through by a ball, before it comes to rest, which falls from the height of 50 feet, and at every fall rebounds half the distance ? Ans., 150. 17. In the preceding problem, suppose the body falls 16yL. feet the first second, 3 times as far the next second, and 5 times as far the third second, and so on, how long will it be before it comes to rest? Ans., -h%- v/579(4 + 3^2) = 10.27657+ seconds. 304 RATIO, PROPORTION, AND PROGRESSION. Synopsis. o o I o t {Ratio. — Terms of. — Antecedent. — Consequent.—' Couplet. Direct. — Inverse. — Gr. inequality, less. Comp. ratio. — Duplicate, sub-duplicate, etc., etc. Sign of. Cor. — Changes in terms of. r Proportion. — Extremes. — Means. J Mean proportional. — Third proportional. 1 Inversion. — Alternation. — Composition, — Division, [inverse or reciprocal proportion. — Continued. {Pro]). l.—Cor. l.—Cor. 2. J Prop. 2. 1 Prop. 3. — Principles on which transformations are [ made. Equi-multiples. — Why proportion not destroyed, dig. in order of terms, — " " " 'Composition, or division. — " *' " Involution, or evolution. — " " " Progression. — Arithmetical. — Geometrical. Increasing, or ascending, — Decreasing, or descend- ing. Common difference, positive, negative. , Ratio, greater than 1, less than 1. Sign of Ar. Prog. — Of Geometrical. Pive things. — Given, required. Produce them. 2 Fund, formulas means. -To insert 2 Fund, formulas. — Pro- ) To insert means. duce them. s Sum of infinite series. Test Questions. — Give the various changes which can be made upon the terms of a ratio and tell how the ratio is aflfected, and Why ? State the various transformations which can be made upon a propor- tion without destroying it, and Give the reason in each case. Produce the two fundamental formulas of Arithmetical Progression. Also of Geometrical. APPLICATIONS. 305 APPLICATIONS. [Note. — ^Teacher and pupil should bear in mind that the object of this section is to teach the properties of Ratio and Proportion ; hence all the operations should be performed upon the proportion. The pro- portion should be kept in the form of a proportion, and not reduced to an equation.] 1. Divide 60 into two parts which are to each other as 2 : 3. SuG. — Letting x and 60 — a; be the parts, ic : 60 — x : : 2 : 3. Hence x : 60 : : 2 : 5, or X : 24 : : 2 : 2 ; and x = 24. The pupil should give the reason for each transformation. What is the first transformation ? Composition. Why does it not destroy the proportion? What the second transformation ? Why does it not destroy the proportion ? 2. A boy being asked his age said : John, who is 18, is older than I ; but, if you add to my age ^ of it, and from this sum subtract ^ of my age, the result will be to John's age as 10 : 9. How old was the boy ? Verify. OPEEATioN. X -\- ix — 4 X : 18 : : 10 : 9, f X : 18 : : 10 : 9, X : 18 : : 8 : 9, X : 18 : : 16 : 18, .-. x = 16. Let the pupil tell, in each instance, just what the transformation is, and why, according to (60), the proportion is not destroyed. Vebification. 16 -|- -•/ — ^£- = 20, which is to 18 as 10 is to-9, the ratio in case being ^^. 3. Two brothers being asked their ages, the younger re- plied, my age is to my brother's as 2 to 3 ; and if you add 18 to mine and 2 to his, the sums will be as 3 to 2. What were their ages ? SuG. — To solve with one unknoASTi quantity we may represent the younger brother's age by 2x and the older's by 3x. Then 2x + 18 : 3x -f 2 : : 3 : 2 ; Whence 2x -f 18 : 27x + 18 : : 3 : 18, 2x + 18 : 25x : : 3 : 15 : : 1 : 5, 2x -f 18 : X : : 5 : 1, 2x + 18 : 2x : : 5 : 2, 18 : 2x : : 3 : 2, 9 : X : : 9 : 6, . • . X = 6, 2x = 12 and 3x = 18. 306 PROPORTION. [Note. — True, it gives a somewhat shorter solution of this example to put the first proportion immediately into the equation 4x -j- 36 = 9x + 6, whence 5x = 30 and tc = 6. But the object is to become famiUar with the properties of a proportion. ] 4. A man's age when lie was married was to his wife's as 3 to 2 ; but after 4 years, his age was to hers as 7 to 5. What were their ages when they were married ? SuG. — This may be solved with one unknown qiiantity, like the pre- ceding, and that is the more elegant way. We may also use two. Thus, X :y : :3 :2, and x + 4 : j/ + 4 : : 7 : 5. From the former X : f i/ : : 3 : 3 . • . x = fy. Substituting in the latter f y -|- 4 : r/ -}- 4 : : 7 : 5. Whence 32/ + 8 •.2y-\-8 : : 7 : 5, y:2y-i-S : : 2 : 5. 2^/ : 2y + 8 : : 4 : 5, 2?/ :8 : :4 : 1, andy : 1 : :16 :1. .-. y=16, andx = fy=24. 5. A man is now 25 years old and his brother is 15. How many years before their ages will be as 5 to 4 ? Verify. 6. A man has two flocks of sheep, each containing the same number; from one he sold 80 and from the other 20, when the numbers in the flocks were as 2 to 3 How many were there in each flock in the first place ? Verify. 7. It is known to every one that a small body near the eye hides a large one farther off; and it is a principle in optics so nearly axiomatic that we will take it for granted, that, in order to have the smaller body just cover the larger their distances from the eye must be in proportion to their breadths, or lengths. From this some very pleasing calculations can be made. The pupil may make the following : 1st. The breadth of a man's thumb is about 1 inch, and he can readily hold it at 2 feet from his eye ; how far off is the man who is 5 feet 8 inches high, when the breadth of the man's thumb at 2 feet from his eye just covers the man? Arts., 136 feet APPLICATIONS. 307 2nd. Wishing to know approximately the height of the top of a steeple from the ground, I found that my hand, which is 4 inches wide near the thumb, when held 2 feet from my eye, just covered the height of the steeple at a distance of 240 paces of 3 feet each. What was the height of the top of the steeple from the ground? Ans., 120 feet. 8. What number is that to which if 1, 5, and 13 be seve- rally added, the second sum will be a mean propor- tional between the other two ? Verify. 9. What number is that whose ^ increased by 2 is to its 10. i diminished by as 6 is to 2^. Ans., 30. The number of acres a farmer planted with corn is to the number he planted with potatoes, as f to 1 ; but if he had planted 6 acres less of corn, and ^ as many po- tatoes + 15^ acres, the ratio would have been as f to I". How many acres of each did he plant ? [Note. — The five following examples are designed to be solved by using two or more unknown quantities. ] 11. Find two numbers, the greater of which shall be to the less, as their sum to 42 ; and as their difference is to 6. SuG. Let X = one and y the oth Then, (1) X : y : : X -j- y : 4:2, 02) X : y :: X — y : 6, By equality of ratios X + y :A2::x~ y and X -\- y -x — y :: 7 2x :2y :: 8 : 6, X :2y ::4: :6, x : f y : : 4 : 4, .-. X = ^ y. Substituting in (2) 1?/ ■ y •■ 3.y — y Or, 1 : 1 : : iV : 6, 4 : 1 :: y : 6 1 : 1 : : y : 24: ■■• 2/ = 24, and x=^y = 32. : 6, 308 PKOPORTION. 12. Two numbers have such a relation to each other, that if 4 be added to each, they will be in proportion as 3 to 4; and if 4 be subtracted from each, they will be to each other as 1 to 4. What are the numbers ? Ans.j 5 and 8. SuG. X -{- 4: : y -{- 4: ::3:4 and x — 4:1/ — 4:: 1:4. Whence -f 1 : 3 : : 1/ : 4, or —~ : 1 : : y:l. .-. y= — ^. Substitut- o o ing,x — 4 : — J^-— 4 : : 1 :4, a — 4 : 2a; — 4 : : ] :6, x — 4 : x : : 1 :5, 4 : X : : 4 : 5. . •. x = 5. Substituting, 6 : 3 : : t/ : 4, or 3 : 3 : : y : 8. .-.2/ = 8. 13. Find two numbers in the ratio of 2|- to 2, such that, when each is diminished by 5, they shall be in the ratio of 1^ to 1. Numbers,25 and 20. 14. There are two numbers, which are to each other, as 16 to 9, and 24 is a mean proportional between them. What are the numbers ? Ans.y 32 and 18. [Note. — The following examples may be solved by converting the proportion into an equation, at whatever stage of the solution it is found expedient. ] 15. Find two numbers in the ratio of 5 to 7, to which two other required numbers in the ratio of 3 to 5 being re- spectively added, the sums shall be in the ratio of 9 t,3 13; and the difference of those sums = 16. Numbers, 30 and 42, and 6 and 10. 16. A farmer hires a farm for $245 per annum; the arable land being valued at $2 an acre, and the pasture at $1.40; now the number of acres of arable is to half the excess of the arable above the pasture as 28 : 9. How many acres are there of each ? Ans., 98 acres of arable, and 35 of pasture. 17. The quantity of water which flows from an orifice is proportioned to the area of the orifice, and the velocity APPLICATIONS. 309 of the water. Now there are two orifices in a reservoir, the areas being as 5 to 13, and the velocities as 8 to 7, and from one there issued in a certain time 561 cubic feet more than from the other. How much water did each orifice discharge in this time ? Ans., 440 and 1001 cubic feet. X8. At an election for two members of parliament, three men offer themselves as candidates, and all the elect- ors give single votes. The number of voters for the two successful ones are in the ratio of 9 to 8; and if the first had had seven more, his majority over the second would have been to the majority of the second over the third as 12 ; 7. Now if the first and third had formed a coahtion, and had one more voter, they would each have succeeded by a majority of 7. How many voted for each ? Ans., 369, 328, and 300, respectively. 19. A man, driving a flock of geese and turkeys to market, in order to distinguish his own fi'om any he might meet on the road, pulled 5 feathers out of the tail of each turkey, and 2 out of the tail of each goose, and upon counting them, found that the number of turkeys' feathers lacked 15 of being twice those of the geese. Having bought 20 geese and sold 15 turkeys by the way, he found that the number of geese was to the number of turkeys as 8 to 3. What was the number of each at first? Ans.y 45 turkeys, and 60 geese. PROBLEMS OF PURSUIT. 20. A fox starts up 120 feet ahead of a hound at exactly ■i- past 2 o'clock P. M.; the hound gives chase and gains 5 feet every 2 minutes. At what time will he overtake the fox ? 310 PEOPORTION. Statement. — Lotting x be the time which will elapse before the hound overtakes the fox, the problem becomes; If a hound gain 5 feet in 2 minutes, how long will it take him to gain 120 feet? That is 5 : 120 : : 2 : a;. . •. x = 48, and the hound overtakes the fox at 3 o'clock and 18 minutes. 21. A privateer espies a merchantman 10 miles to lee- ward at 11.45 A. M., and, there being a good breeze, bears down upon her at 11 miles per hour, while the merchantman can only make 8 miles per hour in her attempt to escape. After 2 hours chase the topsail of the privateer being carried away, she can only make 17 miles while the merchantman makes 15. At what time will the privateer overtake the merchantman ? Ans., At 5.30 P. M. 22. A hare, 50 of her leaps before a greyhound, takes 4 leaps to the greyhound's 3; but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to overtake the hare? SuG. Let 3x = the number of the hound's leaps, whence 4a; = " " " hare's " in the same time. Then 2:3 : : 3x : 4iC -J- 50. .•. x = 100. ; and the hound takes 300 leaps. 23. The hour and minute hands of a clock are exactly to- gether at 12 M. When are they next together ? SuG. — Measuring the distance around the dial by the hour spaces, the whole distance around is 12 spaces. Now, when the hour hand gets to 1, the minute hand has gone clear around, or over 12 spaces. But as the hour hand has gone one space, the minute hand has gained only 11 spaces. Now as the minute hand must gain an entire round, or 12 spaces, to overtake the hour hand, we have the question: If the minute hand gains 11 spaces in 1 hour, how long will it take to gain 12 spaces? .-. 11 : I'i : : 1 hour : x hours ; and x ' minutes. APPLICATIONS. 311 ScH. — 1. It is evident that the hands are together every l-j^,- hours ; hence to find at what time they are together between any two hours on the dial, we have only to multiply l-,J-i- by the number of the whole hours past 12 o'clock. Thus between 7 and 8 they pass each, other at 7-n-, or 7 o'clock and 38-i\ minutes. Between 10 and 11 they pass each other at 10 o'clock o4-i^f minutes. 24. At what time between 6 and 7 o'clock is the minute hand just i of the circle in advance of the hour hand? SuG, — The question is : If the minute hand gains 11 spaces in one hour how long will it take it to gain 64 rounds, or 75 spaces ? Or, if it gains 1 round in l/j- hours, how long will it take it to gain 64 rounds? Ans., At 49-jL minutes past 6. 25. At what time between 4 and 5 is the hour hand of a watch just 20 minutes in advance of the minute hand ? Ans., At no time between these hours. The minute hand is within 20 minutes of the hour hand at 4 o'clock, and at 5y\- minutes past 5. 26. Before noon, a clock which is too fast, and points to afternoon time, is put back 5 hours and 40 minutes ; and it is observed that the time before shown is to the true time as 29 to 105. Required the true time. Si'G. — Letting x = the time pointed to, x : x + C)^ : : 29 : 105. Observe that to turn the hands back 6h. 40m. is the same as to turn them for- ward 6h. 20m. X = 2h. 25m., and the true time was 2h. 26m. + 6h. 20m., or 15m. before nine o'clock in the morning. 27. Two bodies move uniformly around the circumference of the same circle, which measures s feet. When they start, one is a feet before the other ; but the first moves m and the second 31 feet in a second. When will these bodies pass each other the 1st time, when the 2nd, when the 3d, etc., supposing that they do not disturb each other's motion, and go around the same way ? SuG. — 1st. If J/>> m, the second gains M — m feet a second, and having a feet to gain, overtakes the first, and does it in sec- M — m onds. The problem is then like the preceding, as the second gains a 312 PROPORTION. whole round every — seconds. Hence the second passing is at from the starting, the third at — , the fourth at M — m ° M a-f3s M etc. 2nd. If Jf << m, the second is over- taken by the first after the first has gained s — a feet, or in — seconds ; and in m — M g every — seconds thereafter ; that 2 o n is, from the time of starting, in — . TO — M 3s — a etc. m — M 28. When will they pass if the 1st starts t seconds before the second, and M ^ m? When if M <^m? Ans., If M^m, in — , — ^— , —^ — , etc., seconds. If Jf <" m, m ^^j^, — -, 3s — a — TYit . T -— -, etc., seconds. [Note. — Observe that the results which are most symmetrical in the two results do not correspond to the same meetings. Thus the num- ber of the time of meeting in the first, is one ahead of the coefficient of s. The time of the third meeting has 2s, etc. But in the second the times of meeting and the coefficients of s are the same. 29. When will they pass if the first starts t seconds later than the second and M'^ m? When if M <^m? ^ a~Mt s-}- a—Mt 2s-j-a—Mt ^ Ans., In -T-p , — r— , — , etc., seconds, . s— a + 3ft 2.S' — a +3fl; ds — a-^3ft . ^^ ^^ -^^:z: M ' m — 31 ' ~^^r=: jf * ^^"^-^ seconds. 30. When will they meet if they start at the same time and move towards each other, or over the distance a, first ? APPLICATIONS. 313 If they move from each other, or over the ares — a first ? Ans., In -77— , —p , -^rp- — , etc., seconds, or in s — a 2s — a 3s — a , ^ etc., seconds. M -\-m' M-\- rn M + m 31. When will they meet if the first starts t seconds before the other, and they move toward each other, or over the distance a first ? If they move from each other, or over the arc s — a first ? _ s — a — mt 2s — a — m^ 3s — a — mt , Ans., In — TT , rp— , — iT^- , etc., seconds after the second starts, if they move over the arc s — a first. 32. If they move in opposite directions, and the first starts t seconds later than the second : when they move over the arc a first ? When they move over the arc s — a first? a — 3It s + a — 3It 2s + a ~ 3It Ss+a—M/ ^ Ans., 1st. ^rp- , — ^rr— , -— , — 5rj-- , etc. seconds from the time the first starts, or a -{- mt s -\- a -{- mt ^s -}- a -}- mt M-\-m M -\- m M^m from the time the second starts. 2nd. [Let the pupil determine.] etc., ScH, — These problems are of little value for the purpose of illustrat- ing the use of the equation, but yet will be found well adapted to ini- tiate the pupil, into the method of reasoning by means of general sym- bols. The pupil should exercise his ingenuity in ascertaining how many different cases these problems give rise to. Thus, 1st Class of cases, when both move in the same direction. This will be subdivided into (A), when they start together, and {B) when they do not; the lat- ter of which will comprise two cases, (a) when the first starts first, ( h) when the second starts first. Finally, each of these will have iwo varie- ties depending upon whether 3f ^ m, or if << m. 314 BUSINESS RULES. CHAPTER m. BUSIJ^ESS RULES [OF ARITHMETIC^* SECTION L Percentage. 83. According to our definition, the Equation, of which it is the special province of Algebra to treat, is the grand instrument for investigating the relations of quantities. Now, in simple Percentage, there are four quantities to be compared ; viz., the Base, the Rate Per Cent., the Percentage, and the Amount, and the problem is. To discover and ex- press in equations the relations between these four quan- tities so that if a sufficient number of them are given the others may be found. [Note. — For Definitions see Elements of Arithmetic, 235 et seq.l^ 84, I^TOh, 1, To express the relation between base, rate per cent., and percentage. Solution. — Let b represent the base, r the %, and p the percentage. Now, if we divide the base, h, by 100, we get 1 of every 100, or — . But r% means r of every hundred of the base. Hence r% of 6 is r h rb rb times J- , or _. ■■ . p = ^. 86. I*rob. 2. To express the relation between rate per cent., amount or difference, and base. Solution. — Let s represent the sum or difference of the base and percentage. Thens = &4. ,^ = ]m~~' *^® + ^^^° *° ^^ ^^^^ PERCENTAGE. 315 when the base is increased by the percentage, and the — sign when the base is diminished by the percentage. ScH. — The two formulce (1) p = j^ and ._100i^) "^ ' lUU expressing the relation between the four quantities h, r, p, and 5, two of which must always be given to find the others, are in themselves suflScient for the solution of all problems in Simple Percentage. EXAMPLES. 1. Bought a borse for $840, and sold it for $560. How much did I lose per cent. ? Ans., 33J%. SuG. — Here h and s are given to find r. Hence formula (2) is to be used. And as there is loss involved, the — sign is to be taken. Sub- stituting in this formula, 560 =: — — — . Solving for r we have 5^ = 100-,., or?i° =100 - r; whence .= 100 - ?55 = "»= 33i. «-iO ' 3 3 3 2. A n amber being increased by 2 equals 14. Kequired the increase per cent. ? Ans., 16|%. Bug.— Formula (1) gives 2 = — ; and (2) gives 14 = ttttt--- 'Jrh From which we are to find r. Multiplying (1) by 7, 14 = — - . Whence — = ■ or 7r = 100 4-r, orr= -7- =165. 100 100 ^ ' 6 3. A piece of cloth sold for $779, cash, which was 5% off. Kequired the price of the cloth. Price, $820. 4. Sold 40%" of my wheat, and had remaining 981 bushels. How much had I at first ? Ans., 1635 bushels. 5. A man sold two horses at $420 each ; for one he re- ceived 25% more, and for the other 25% less than its value. Required his loss. Loss, $56. SuG. — Letting b and bi be the values of the horses we have 420=3 316 BUSINESS RULES. 5(100-25) .... 6,(1004-25) „, ..„ . 5^ ^ - — , and 420= -— . . • . 75ft = 1255 1, or 5 = -5^ and 100 J.UU o o b-\-bi = -5i, the value of the horses. Now 5 1, found from the 2nd, o o is 336, and as -5i — 840 = the loss, we have 896 — 840 = 56 = the o loss. Those who are not famihar with algebraic reasoning will prefer to find the values of 5 and 5i from the two formulas, and add them together. 6. A man sold 72 turkeys, which was 32% of the number he had remaining. How many had he at first ? Ans., 297. 7. A farmer saved annually $145^, which was 33^% of his annual income. Required his income ? -4/i.s.,$436^. 8. A merchant having 400 barrels of cider, sold at one time 45% of it; at another time 20% of the remainder. How many barrels did he sell in all? ^?is.,224 bbls. rh 45 X 400 ... OPEEATION. P =J^ = —^^^ = 180. _r,6, _ 20 X 220 _ ^'~I00"~ 100 ~ ... p-\-p^= 224. 9. A housekeeper gave to her neighbor ^ of a pound of tea, and had f of a pound remaining. What per cent. of her tea had she remaining? Ans., 85f%. 3 H ^ 7r 600 .^, OPEBATION. - = _-,6=-- .-.r^ — = 85f 10. John has -| of a dollar, and Henry has -| of a dollar. What per cent, of John's money equals Henry's? What of Henry's equals John's ? Ans., 120%, 83^%. r = 120. [Note. — For other examples in percentage, see Stoddard's Com- plete Arithmetic, 173-177 pp.] OPERATION. 3 _ ^-i 6 - ^ 5 100' 20 1 r| 3r 2~" 100' 50 SIMPLE INTEREST, AND COMMON DISCOUNT. 317 SUCTION 11. Simple Interest, and Common Discount. [Note. — For definitions see Elements of Arithmetic] 8G, I^rob, 1. To express the relation between principal, rate per cent, time^ and interest. Solution. — Letp, r, t, and i represent respectively the principal, rate percent, time and interest; i being in the denomination for which the rate per cent, is estimated. Thus, if the rate per cent, is rate per cent, per year, t is to be understood as years; if the rate per cent, is per month, i is months, etc. ITD Then as p is the base, the percentage for a unit of time is ■—- {84); and for t units of time it is -—• .-. i = — -—• lUU 100 87* I*rob, 2, To express the relation between amount, principal, rate per cent, and time. Solution. — Since the amount is the sum of principal and in- terest, representing the amount by a, we have a = p -{- i. But trp „ , trp 100 4- fr 100a t = — -—' Hence a = r> -4 i— , or a = o ■ — • . •. 0= 100 ^ ^ 100 ^ 100 -^ 100+^r. 88, ScH. — This problem embraces the common rule for Discount (see Arithmetic, 268 et sq.). The pupil should be careful to under- stand the reasonableness of discount. For example, if I hold a note against Mr. B. , which note is payable at any future time, if the note is drawing interest for all money is worth, the Present Worth of the note at the time it icas given was its face. If the rate at which it is drawing interest is less than money is worth (or if it draws no interest) the Present Worth at its date is less than the face of the note. (If it draws no in- terest, its Present Worth at any time is less than its face. ) Finally, if the rate which the note draws is greater than the market rate, the Present Worth of the note at its date is more than its face. ScH. — The two formulce = ^ and 100 100 + tr (1) i=^ and (2) a=p-\-i=p- 100 318 BUSINESS RULES. are sufficient to solve all problems in Simple Interest and Common Dis- count. EXAMPLES. 1. What is the interest on $250 for 1 yr. 10 mo. 15 da., at 6% per annum? Ans., $28.12^. Solution. — In this example I am to consider principal, time, rate per cent, and interest, the latter of which is the unknown quantity. Formula (1) expresses the relation between these quantities. The rate per cent, being per annum, the time must be in years. 15 days = ^g = 10 5 .5 of a month. 10.5 months = — ^ :=.875 of a year. . • . 1 yr. 10 mo. 3 5 15 da. = 1.875 years. Now (1) gives i = -^ = ^ = 28.124. 2. What is the interest on $47.25 for 1 tjr. and 6 mo., at 6% per annum ? Ans., $4.2525. 4.721 .3 3 -^A^ OPERATION. I = — — <^ = 4.2525. -20- 4- -^■ 3. "What is the interest on $145.50 for 1 yr. 9 mo. 24 da., at 6% per annum f Ans., $15.86 nearly. . 1.816X6X145.50 ,^ __ OPERATION. I = = 15.86 nearly. 4. What is the interest on $123.75 for 2 yr. 8 mo. 12 da. at ()% per annum? Ans., $20.0475. 5. What is the interest on $475 for 2 yr. 7 mo. and 20 da., at 6% per annum ? Ans., $75.208|. 6. What is the interest on $340.60 for 4 yr. and 5 mo., at 6% per annum f Ans., $90,259. . 7. ¥/hat is the interest on $50.40 for 1 yr. and 10 mo., at 7% per annum f Ans., $6,468. SIMPLE INTEREST, AND COMMON DISCOUNT. 319 SxjG. — In solving this example the operation upon paper consists simply in multiplying 12.833 by .504. The pupil should always reduce the written details of his arithmetical work to the minimum. Thus, in this case, he sees mentally that 10 mo. are . 833 of a year. Hence, ^^00.40x1-833x7^ But 1.833 X 7 he produces mentaUy, 12.833 100 504 and writes 12.833. And cancelling the 100 from 50.40 makes it . 504. Hence the written work should only be as f^/^p? in the margin. In practice, nothing but this multiphcation ""^^"^ should be written down. 6.4G7832 Solve the following by thus reducing the written work to a minimum. The answers are given as in practice in busi- ness. 8. "What is the interest on $49.80 for 2 yr. and 11 mo., at 7% per annum ? Ans., $10.17. Opekation. i = ^^-^^^J^^^'^^^^ = 2.9166 X 7 X .498 = 10.17. 9. What is the interest on $95.40 for 3 yj\ 9 mo., at 8% per annum ? Opekation, i = — '■ '- — = 28.62. The operation should be performed mentaUy. Thus 8 X 3. 75 = 30. Dropping the 0, and one from the denominator, and for the 10 remaining in the denomi- nator removing the decimal point in 95.40 so as to make it 9.54, we have simply to multiply 9.54 by 3. All of this should be done at a glance, without writing more than is given above. 10. What is the interest on $196 for 5 yr. 7 mo., at 9% per annum ? Ans., $98.49. 11. What is the interest on $471.11 for 4 yr. 8 mo., at 1\% per annum? Ans., $164.89. 12. What is the interest on $18.60 for 3 mo. 12 da., at 8% per mo. ? Ans., $1.90. Opekation, { = ?:l2ii^l2:^ = 1.7 x 3 X .372 = .372 X 5.1 = $L90. 13. What is the interest on $400 for 150 days, at 2h% per month? Ans., $50. 320 BUSINESS RULES. - 2 5 400 X 5 X^ Opebation. i = ^ = 50. loe- 14 What is the interest on $1,000 for 2 mo. 12 da., at 1^% per month f Ans., $36. 15. What is the amount of $432.10 for 5 yr. 4 mo. 24 da., at 7% per annum f Ana., $595 .43. Opebaxiok. a =pl5^#: ^432.10 ^2^^^^ = 4.321x137.8 = 595.4338. •^' lUU ■" 100 16. What is the amount of $325.25 for 2 yr. 9 mo. 12 da., at 6i% per annum ? Ans., $384.09. Opebation. a = 325.25^+4^^^21^ =. 384.09. 100 17. In what time will $13, at 6% per annum, give $0,975 interest? Ans., 1 yr. 3 mo. SoiiUnoN. — Since principal, rate per cent., time and interest are compared i — » ■ gives the relation. As time is required, I solve this equation for t, and have i = . Substituting the given values, t = rp .025 50 .075- 18. In what time will $45.25, at 6% per annum, give $1.81 interest? Ans., 8 mo. 19. In what time will $70.50, at 9% per annum, give $31,725 interest? Ans., 5 yr. 20. In what time "will $140, at 1% per annum, give $10.861|- interest? Ans., 1 yr. 1 mo. 9 da. SIMPLE INTEKEST, AND COMMON DISCOUNT. 321 21. In what time will $48.50, at G% per annum, amount to $56,187-4? Ans., 2 yr. 7 mo. 21 da. SuG. —Subtract the principal from the amount to find the in- terest. 22. In what time will $248, at 6% per annum, amount to $282,224 ? Ans., 2 yr. 3 mo. 18 da. 23. In what time will $700, at 9% per annum, amount to $712.35? Ans., 2 mo. 10.5 + da. 24. At what per cent, will $325 produce $3.25 interest in 2 months? Ans., 6%. trp Solution. — Same as above, finding r from the equation i = -— ;. __ lOOt 100 X 3.25 _ ** "~ ip ~ i X 325 ~ • 25. At what per cent. wiU $40 produce $13.36 interest in 2 yr. 9 mo. 12 da. ? Ans., 12%. 26. At what per cent, will $125 produce $32,375 interest in 3 ?/r. 6 mo. ? Ans., 7|%. 27. At what per cent, will $124 produce $29.17 a interest in 4 \jr. 3 mo. 10 da.'i Ans., U%. 28. At what per cent, will $2,360.25 amount to $2,470,395 in 7 months ? Ans., 8%. Suggestion. — Find the interest and then proceed as before. 29. At what per cent, will $230 amount to $249. 83f in 11 mo. 15 da.? Ans., 9%. 30. What principal will in 3 yr. 8 mo. 15 da., at 6% per awniwi, give $76,095 interest ? Ans., $342. 322 BUSINESS KULES. Solution. — Solving i = —~ forp, I have lOOi 100X76.095 100 X 76.095 _ ^ tr 6 X 3.7-iV ^^-^5 31. What principal will in 4 yr. 9 mo. 18 6?a., at 9% pjr annum, give $65,016 interest? J/i.s., $150.50. 32. "What principal will in 8 y7\ 8 mo. 12 cZa., at 5% per annM?7i; give $147.9435 interest? Ans., $340.10. 33. How long will it take for $200, at simple interest, at 6% per annum, to amount to $500 ? Ans., 25 years. Solution. — I have under consideration p, r, a, and i, the latter of which is the unknown quantity. The relation between these is a=p T-rp Solving this for U I find _ 100(a — V) _ 100(500 — 200) _ 300 _ ~ pr ~ 6 X 200 ~ 12 ~ ' 34. How long will it take $1, at simple interest, at 10% per cent, jjer annum, to amount to $100 ? 35. How long will it take $75, at interest at 5% per an- num, to amount to $100 ? 80, CoR. — To find the time required for a principal to double, triple, or become n tim.es itself at any rate of simple inter- est, ive have a = 2p, 3p or np. Hence the last formula becomes t = = for the time required to double. pr r -^ ^ rp , • 1 L , 100 X 2 . ^ ^ , ^ 100 X 3 10 triple we have t = ; to quadruple, t = ; to become n times itself t = ■ -. ScH, — It appears that the time in such a case is independent of the principal, as might have been anticipated. 36. How long will it take $100 to become $500 at S% per annum ? Ans., 50 years. SIMPLE INTEREST, AND COMMON DISCOUNT. 323 37. How long does it take a principal to double at 6% jjer annum ; at 7% ; at 5%? 38. Sold property amounting to $3,000, on a credit of 12 mo. without interest. Money being worth 8% per annum, what sum in hand is equivalent to the $3,000 under the contract? Verify it. Ans., %2,111.11 +. SoiiUTioN. — I am to find a sum which put at interest for 1 yr. at 8% will amount to $3,000. I therefore have given a, r, and t to find p» The relation of these is given in the formula a= p — -^ — from 100 100a ,. ,^. , 300000 ^^^^ ^^ , which p = -j^jjj-TT^ = (m this case) -^^^ = 2777. 77 +. 39. A debt of $500 will be due in 3 yrs. without interest. "What is its present worth if money commands 6% per annum '^ Ans., $423.73. 40. What discount should be allowed for the present pay- ment of a note of $400, due 3 yrs. 5 mo. hence, the note not bearing interest, though money is worth 6% per annum ? Ans., $68.05. 41. A man buys a piece of property to-day for $5,000, giving his note with security at 12% per annum, pay- able 2 yrs. 9 mo. hence. What is the present worth of this note if money brings in market 6% per annum ? Ans., $5,708.13 — . Operation. a=p ^^^^ = 50(100 + 12 X U) = 50 X 133 =, 100« 665000 ^^^„ ^ _ ««50- P' =m> + lF, = mOr = '">'■'' - 42. A merchant buys $2,000 worth of goods on 90 days time, at 2% a month. He could have borrowed money at 1^% a month. How many more goods could he have bought for the same money if he had borrowed ? Alls., $28.71 worth. 324 BUSINESS RULES. 43. July 20tli, 1869, I hold a note for $500 dated April 6th, 1867, due Jan. 1st, 1872, and bearing interest at 1% l^er annum. What is the present worth of this note, money being worth 10% per annum? Ans., $534.86+. * 00, I*VOhm To find what each payment must he in order 'to discharge a given principal and interest in a given number of equal payments at equal intervals of time. Solution. — Letp represent the principal, r the rate per cent., t one of the equal intervals of time, n the number of payments, (i. e., nt is the whole time), and x one of the payments. There will be as many solutions as there are different methods of computing interest on notes upon which partial payments have been made. 1st. By the United States Court Rule. — As the payments must exceed the interest in order to discharge the principal, this rule requires that we find the amount of p, for time t, at r per cent. This is done by multiplying by 1 + -—- , and gives p( 1 -f -tht )• From this subtract- ing the payment x, the new principal is p( 1 -j- --— ) — x. Again find- \ 100 / ing the amount of this for another period of time, t, and subtracting the second payment • In like manner, after the third payment there remains After the 4th payment, the remainder is Finally, after the ?ith payment, we have SIMPLE INTEREST, — PARTIAL PAYMENTS. 325 Whence ' i+o+w)+o+w)V(i+4y-o+4)""'' This denominator being the snm of a geometrical progression whose first term is 1, ratio (1 -\- V and number of terms n, its sum is + 4)" Hence x rt /. . rt ^ lUO " ■ ' 2nd. By the Vermont Rule. — The amount of the principal for the whole time is pfl-\- ^ttt- ]• The amount of the 1st payment is a-l+ ttjttC^ — 1) > - - " 2nd " ^[l+T50^'^-4 " " " 3rd - a-[l + -^^{n - 3)], Etc., etc., etc. (( (( ____-__■•_ " The nth payment (with no interest) is x. The sum of the amounts of these payments is nx + ^x[(n-l) + (>i-2) + (n- 3) 1]. The series in the brackets being an arithmetical progression whose first term is (n — 1), common difference — 1, last term 1, and number of terms (w — 1), its sum is ( — - — )n. Hence the sum of the payments is nrt nic4--Yn77^( — o — )'^' ^^' ^L'^"' 9 J' -^^^^ by the condition this sum equals the amount of the principal; consequently nrt -{11 — !)• x = . , 326 BUSINESS RULES. ScH. — If the payments are made annually, t = 1. And letting 7' r' = — -. i. e. , letting the rate per cent, be expressed decimally, the formulas become pr'il -j- r')" , By the U. S. Rule, (1 4- r')'' - 1 2p(l 4- r'n) By the Vermont Fade, x = - zn-\^rwji — 1) 44. AVhat must be the annual payment in order to dis- charge a note of $5,000, bearing interest at 10% per annum, in 5 equal payments ? Ans., By the U. S. Rule, $1,318.99 within a half cent. By the Vermont Rule, $1,250. Query. — What occasions the great disparity between the payments required by the different rules ? 45. What annual payment is required to discharge a note of $300, bearing interest at 7 per cent, per annum, in 4 equal payments ? 46. The sum of $200 is to be applied in jDart towards the payment of a debt of $300, and in part to paying the interest, at 6% in advance, for 12 months, on the re- mainder of the debt. What is the amount of the pay- ment that can be made on the debt ? SuG. — Let X represent the payment; then (300 — ic) X t'oo is the in- terest on the remainder of the debt; and we have therefore the equa- tion, X + (300 — X) X T&o = 200. Am., $193.62. 47. A is indebted to B $1,000, and is able to raise but $600. With this sum A proposes to j^ay a part of the debt, and the mterest, at 8% in advance, on his note at 2 years for the remainder. For what sum should the note be drawn? Ani^., $476.19. PARTNERSHIP. 'd'27 SECTION III. Partnership. [Note. — For definitions, see Elements of Arithmetic] 91. JBrinclple. In Simple Partnership, i.e., ivhen all the capital is employed the same length of time, the fundamental principle is that the shares of the gains shall bear the same ratio as the shares of the stock. EXAMPLES. 1. A and B enter into partnership. A puts in $1,200 and B $1,800. They gain $900. What is each one's share of the gain ? 2 : 3. 2. A, B, and C enter into partnership. A puts in $340, B $460, and C $500. They gain $390. What is the gain of each ? Ans., A's $102, B's $138, and C's $150. Opekation. ic + Xi -{-Xi^^ 390. x : ic, : x^ : : 340 : 460 : 500, or jc _^ a-i 4- X2 : X : : 1300 : 340, or 390 : x : : 130 : 34, or 3 : x : : 1 : 34. . • . X = 102. X + Xi 4- X2 : Xi : : 1300 : 460, or 390 : x, ; : 130 : 46. or 3 : Xi : : 1 : 46. . • . Xi = 138. x -f- x, -f x, : x^ : : 1300 : 500, or 390 : X2 : : 130 : 50, or 3 : X2 : : 1 : 50. . • . x^ = 150. 3. Any number of persons as A, B, C, D, etc., unite in a part- nership, putting in respectively a, b, c, d, etc., dollars each. The gain or loss is s. What portion falls to each ? Solution. - -Let X and Xi be their respeeti Then x + Xi = 900 and X : Xi : : 1200 : 1800 . . Whence x + Xi : X : : 5 : 2 or 900 : X : : 5 : 2 . •. X = 360, andxi = 540. 328 BUSINESS EULES. Solution. — Let Xa, Xh, Xc, xa, etc., be the respective shares of the gaio or loss. Then Xa -\- Xb -^ Xc -\- Xd -\-, etc. — s, . and Xa : Xb : Xc : Xd etc. : : a :h : c : d etc. Whence Xa -\- Xb -{- Xc -\- Xd -\- etc. :Xa : ■ a -\-b -\- c -\- d -\- etc. : a or s : Xa : : a -\- b -^ c -{- d -\-, etc. : a. ■ _ ^'^ "~a-f-6 + c-f-t/-|-, etc.* In like manner Xb = Xc a-\-b-\-c-^d-\-, etc. cs a -\- b -{- c -^ d -{-, etc' ds Xd = . a-\-b-{-c-\-d-{-, etc. ScH. — This is the common mle for Simple Partnership, viz. : Multi- ply the gain or loss by each partner's stock and divide the products by the whole stock. [Note, — In such examples the solution is so simple that the equa- tion scarcely renders any assistance. In the following its advantages will appear.] 4. Two men commenced trade together. The first put in $40 more than the second ; and the stock of the first was to that of the second as 5 to 4. What was the stock of each? Ans., $200, and $160. Opebation. — Let x = what the first put in. Then x — 40 = what the second put in. And we have x : x — 40 : : 5 : 4 or x : 40 : : 5 : 1. .-. x = 200. 5. Three men trading in company gained $780, which was to be divided in proportion to their stock. A's stock was to B's as 2 to 3, and A's to C's as 2 to 5. What part of the gain should each have received ? Ans., A, $156 ; B, $234 ; C, $390. 6. Three men trading in company, put in money in the following proportion : the first, 3 dollars as often as the second 7, and the third 5. They gain $960. What is each man's share of the gain ? Ans., $192 ; $448 ; $320. PARTNERSHIP. 329 7. A, B, and C found a purse of money ; and it was mu- tually agi'eed that A should receive $15 less than one half, that B should have $13 more than one quarter, and that C should have the remainder, which was $27. How many dollars did the purse contain ? A?is., $100. 92* J^rinciple. In Compound Partnership ^ i. q., part- nership in which the several partners' shares of the capital are in for different lengths of time, the gain or loss is divided in the ratio of the products of the several amounts of stock into the times which they respectively remained in the business. This is assuming that the use of $a for time t in business is equal to $atyb?' time 1. 8. A and B enter into partnership. A furnishes $240 for 8 months, and B $560 for 5 months. They lose $118. How much does each man lose ? Ans., A 48, and B $70. Solution. — Let Xa = A's share of the loss, and Xb, B's. Then Xa -\- a;o = 118, and x^ : Xb : :8 X210 :5x 560 : : 24 :35. Whence Xa + X(, :Xa : : 59 : 24 or 118 : iCa : : 59 : 24 or 2 : iCa : : 2 : 48. . • . Xa =48, and Xu = 118 — 48 = 70. 9. A, B, and C entered into partnership. A put in $100 for 4 months, B $300 for 2 months, and C $500 for 3 months. They gained $250. How much was each man's gain ? Opebation. Xa -\- Xb -\- Xc =^ 250, and Xa : Xft : Xc : : 4 X 100 : 2 X 300 : 3 X 500 : : 4 : 6 : 15. Whence 250 : Xa : : 25 : 4, or 10 : Xa : : 1 : 4. . • . Xa = 40. 250 : X6 : : 25 : 6, or 10 : Xft : : 1 : 6. .-. Xb = 60. 250 : Xc : : 25 : 15, or 10 : Xc : : 1 : 15. . • . x^ = 150. 10. A, B, and C hire a pasture for $180. A puts in 8 cows for 10 weeks, B 20 for 5 weeks, and C 30 for 9 weeks. How much ought each to ipaj ? Ans., A $32, B $40, and C $108. 330 BUSINESS EULES. 11. To gather a field of wheat, A furnished 8 laborers for 5 days, B 12 for 3 days, and C 6 for 4 days. For the whole work A, B, and C received $45.50. How much should each have received ? Ans., A $18.20, B $16.88, and C $10.92. 12. Any number of persons, as A, B, C, D, etc., unite in partnership, A putting in $a for time ^ ; B, $6 for time th ; C, $c for time t, ; D, $d for time t^, etc., etc. The gain or loss is s. How is it to be shared by the part- ners? Solution. — Letting Xa, Xb, Xc, Xa, etc., represent the respective shares of the gain or loss, we have Xa -\- Xh -^ Xc -\- Xd -f-, etc. , = s, and Xa :Xb :Xc :xci etc., : : ata : hU : cle : dta etc. Whence s:Xa : : atu-\-Ut,-{-ctc-\-dtd - -etc., :ata. .'. Xa= , ^^ s. s :Xb : : ata-{-htb-\-dc-]-dtd - - etc. : Ub. .' . Xb= ^^^ , ,^ , ^^ ,\,^ rrr*- And in Hke manner for the others. ala-\-btb-\-ctc-\-dtd- - etc, Ub ata-\-btb-\-ctc-\-dtd- - etc' ScH. — This is the common rule for Compound Partnership, viz. : Take the produ.ct of each partner's share of the stock into its time in trade. For any partner's share of the gain or loss multiply the whole gain or loss by the product of his stock into its time, and divide by the sum of the several shares of the stock into their respective times. SECTION IV. Alligation. 1. A farmer mixes together 10 bushels of oats, at 40 cents a bushel, 15 bushels of corn at 50 cents a bushel, and 25 bushels of rye at 70 cents a bushel? What is the value of a bushel of the mixture ? Ans., 58 cents. ALLIGATION. 331 Solution. — Let a be the value o\ a bushel of the mixture of which there are 10 + 15 -f- 25 = 50 bushels. Hence 50x represents the value of the whole mixture, and 50a; = 10 X 40 + 15 X 50 + 25 X 70 = 2900. .-. x=:58. 2. A grocer mixes 120 pounds of sugar at 5 cents a pound, 150 pounds at 6 cents, and 130 pounds at 10 cents. What is the value of a pound of the mixtui'e ? Ans., $0.07. 8. A liquor dealer mixes 8 gallons of alcohol 100%, 12 gallons 80%, 25 gallons 60%, 40 gallons 40%, and 60 gallons 20% strong. What is the strength of the mix- tui-e? Ans., 41fi%. 4. One kind of wine is 40 cents a quart, and another 24. How much of each must be taken to make a quart worth 28 cents ? Statement, x -\- y = 1. 40x -f '^4?/ = 28. 5. Three kinds of sugar are worth respectively 6, 8, and 10 cents a pound. How much must be taken of each to make a mixture worth 7 cents a pound ? Solution. — Let x, y, and z be the amounts of each required, a; pounds at 6 cents a pound are 6a: ; t/ at 8, 8?/ ; z at 10, lOz. The whole amount is x -\- y -{- z, and the whole value Q,x -{- ^y -[• lOz. Hence 6x -f- 8?/ + lOz = 7(.x -{-y -\- z). But here are three unknown quanti- ties, and the example gives but one set of conditions. The problem is therefore indeterminate ; that is, there are not conditions enough given to fix the values of the unknown quantities. Suppose we add the two conditions : To make a mixture of 48 lbs. ; and that twice as much of the 6 cent sugar shall be used as of both the others. We then have the two additional equations x -\-y -\- z =» 48, and x =2(y -\- z). These equations, together vrith the former, 6x -f 8?/ 4- lOz = 7[x-\-y -\- z) readily give x = 32, y=8, and z = 8. 93. ScH. — The last example is a case in Alligaiion Alternate, as it is , denominated in our Arithmetics. Such examples, as they are usually stated, are simply problems in which there are more unknown quan- tities than equations, and are hence Indeterminate. The Indetermi- 332 BUSINESS BULES. nate Analysis can not be treated in this volume, but a few further illus- trations of such examples will be given. All the various methods of treating this subject usually given in our Arithmetics are exceedingly cumbrous and perplexing to the pupil, and, after all, fail to give a full view of it. The propriety of puzzhng pupils with any of them is exceedingly questionable. They are very clumsy and incomplete efforts at doing a thing which becomes very simple when the proper principles are developed, which principles cannot be brought forward in Common Arithmetic. 6. How mucli of each sort of grain, at 48, 50, and 68 cents a bushel, must be mixed together, so that the com- pound will be worth 60 cents a bushel ? Statement, x, y, and z being the amounts of each kind respective- ly, we have 48a; + 50y -f 68z = 60(ic -\- y -\- z) ov &£ -\- 5y = 4z. Now any real, positive values may be assigned to either two of these unknown quantities, which will give a positive value for the other. Thus if z = 4, and y = 2, 6x = 16 — 10, or x = 1. Verify these re- sults. Again, if z = 5, and y = 1, 6x = 20 — 5, or x = 2^. Verify these results. Again, if z = 6, and t/ = 3, 6x = 24 — 15, or x = 1^. Verify these results. In like maner an unlimited number of sets of answers can be ob- tained. J£, however, we try z = 2, and y = 3, we have x = — 1^; The neg- ative sign in this case shows an impossibility. The cause of this is evident when we notice that 2 bushels, at 68 cents, and 3 at 50, make 5 bushels, worth 286 cents, or 57i cents per bushel. This mixture can- not be made worth 60 cents per bushel by patting in grain worth only 48 cents. The — 1^ of the 48 cent grain indicated by this result, may be un- derstood to mean that we are to take out of the 5 bushels, worth 286 cents, 1^ bushel, worth 48 cents per bushel. This leaves 3| bushels, worth 230 cents, or just 60 cents per bushel. 7. A merchant has two kinds of wine. The first kind is worth 12 shillings per gallon, and the second is worth 7 shillings per gallon. How many gallons of each kind must he use in order to form a mixture worth 9 shil- lings per gallon ? ALLIGATION. 333 Alls., Letting x and y represent the quantities re- spectively, he may take any quantity he pleases of either, so that he takes such an amount of the other as to preserve the relation 3x == 2y. That is, he must take 1-^ times as much of the latter as of the former. If he takes 2 of the former, he must take 3 of the lat- ter. If he takes 6 of the former, he must take 9 of the latter, etc., etc. 8. How much corn at 48 cents, barley at 36 cents, and oats at 24 cents per bushel, must be taken to make a compound worth 30 cents per bushel ? SuG. — Show that he can take any amount he pleases of either one, if he takes proper amounts of the other two respectively. Show what he may take of the second and third kinds if he take 1 bushel of the first. Is there any limit to the number of ways he can make up the mixture when he takes 1 bushel of the first kind ? Can he take 3 bush- els of the first kind and 2 of the second ? What value would this give to z (representing the amount of the third kind) ? 9. A merchant wishes to mix 32 pounds of tea at 36 cents per po^ind, with some at 48 cents, and some at 72 cents. How many pounds of each kind must he take to form a mixture worth 56 cents per pound ? SuG. — The relation is 2?/ — x = 80. Any value for either x or y may be taken which gives a positive value for the other; and any positive value of X will give in this case a positive value for the y, for 2^ =i 80 -|- a;. Kx = 4, i/ = 42. If ic = 10, y = 45. lix = \,y = 40|^, etc. , etc. If we add to this example the condition that the whole amount of the mixture shall be 102 pounds, the problem becomes determinate ; as there are then two equations with two unknown quantities. The equations, when reduced, are 2?/ — x = 80, and y -{• x = 70. "Whence jc = 20, and y = 50. 10. A man bought horses at .$50 each, oxen at $40, cows at $25, calves at $10, so that the average price per head was $30. How many were bought of each ? SuG. — Of course fractional values as well as negative are excluded by the nature of this example. The conditions to be met are 4.C + 2y = z -f- 4io. negative and fractional values being excluded by the nature of the problem. Can the conditions be met by taking 3 cowa and 5 calves ? Why ? Can they by taking 2 cows and 5 calves ? How ? Can they by taking 4 horses and 6 cows V How? 334 BUSINESS RULES. 11. A bought 240 barrels of molasses for $4,320 ; worth, respectively, $10, $14, $20, and $22 ; how many barrels of each did he buy? Ans., 40, 20, 160, and 20. SuG. — The conditions are x-\-y-\-z-\-w= 240, and 4:X-{-2y:=z-{- 2w, negative and fractional values being excluded by the nature of the exam- ple. Here are two equations with four unknown quantities, hence any other two conditions maybe imposed which do not conflict with the na- ture of the example. Can x = GO and ?/ = 20 ? Yes, This reduces the equations to z-\-w = 160, and z-j- 2m5 = 280; from which to = 120, and z = 40. Can the condition that half the quantity shall be of the first two kinds be met ? No. This gives z -\- w = 120, which subtracted from z -^ 2io = 4:X -{- 2y, makes lo = 2(2iK -f- y) — 120. But since x-{-y = 120, 2 (2.x -{-y)— 120 > 120, . • . ro > 120, which would make z negative. 12. I have two kinds of molasses which cost me 20 and 30 cents per gallon ; I wish to fill a hogshead, that will hold 80 gallons, with these two kinds. How much of each kind must be taken, that I may sell a gallon of the mixture at 25 cents per gallon and make 10 per cent, on my purchase ? Ans., 58-^\ of 20 cents, and 21^\ of 30 cents. 13. A lumber merchant has several qualities of boards ; and it is required to ascertain how many, at $10 and $15 per thousand feet, each, shall be sold on an order for 60 thousand feet, that the price for both qualities shall be $12 per thousand feet. Ans., 36 thousand at $10, and 24 thousand at $15. 14. How many ounces of gold 23 carats fi.ne, and how many 20 carats fine, must be compouuclod with 8 ounces 18 carats fine, that the alloy of tb'j three diiferent quah- ties may be 22 carats fine ? Ans., 48 oz. of the first, and 8 oz. of the second. [Note, — These applications might be extended to much greater length, did space permit. The equation renders important aid in many problems in Compound Interest, but their discussion usually requires a knowledge of Quadratics, and some of them ot* liogaritnms. Thejf must, therefore, be reserved for the future. ] PURE QUADRATICS. 335 OHAPTEE lY. QUADBATIC EQUATIONS, sucTiojsr I, Pure Quadratics. 94, A Quadratic Equation is an Equation of the second degree {0, 8). 95, Quadratic Equations are distinguished as Pure (called also //icompfefe), o^ndi Affected (called also Complete.) 96, A I^ure Quadratic Equation is an equation which contains no power of the unknown quantity but the second ; as a^^ _j_ ^ := ^^^ ^2 — 2>h=^ 102. 97, An Affected Quadratic Equation is an equation which contains terms of the second degree and also of the first, with respect to the unknown quantity or quantities ; as x'^ — 4^ = 12, hxy — x — y^= 16a, imxy -\-y:x=b. 98, A Moot of an equation is a quantity which sub- stituted for the unknown quantity satisfies the equation. 99, I*roh, To solve a Pure Quadratic Equation. R ULE. — Transpose all the terms containing the unknown QUANTITY INTO THE FIRST MEMBER, AND UNITE THEM INTO ONE, CLEARING OF FRACTIONS IF NECESSARY. TRANSPOSE THE KNOWN TERMS INTO THE SECOND MEMBER. DiVIDE BY THE COEFFICIENT OF THE UNKNOWN QUANTITY. FiNALLY, EXTRACT THE SQUARE ROOT OF BOTH MEMBERS. Dem. — According to the definition of a Pure Quadratic, all the terms C9ntaining the unknown quantity contain its square. Hence they can 336 QUADRATIC EQUATIONS. bo transposed and united into one by adding with reference to tlie Bquare of the unknown quantity. That transposition, and division of both members by the same quantity do not destroy the equahty has aheady been proved. [Let the student repeat the reasoning. ] Ex- tracting the square root of the first member gives the first power of the unknown quantity, i. e. the quantity itself. And taking the square root of both members does not destroy the equation, since like roots of equal quantities are equal. 100, Cor. 1. — Every Pure Quadratic Equation has two roots numerically equal but loith 02J2^osite signs. For every such equation, as the process of solution shows, can be reduced to the form x^^=a {a representing any quantity whatever). Whence, extracting the root, we have x = -\- \/a ; as the square root of a quantity is both +, and — (203,) ScH. — The question naturally arises, Why not put the ambiguous sign (the -f ) before the x, as well as before the second member ? It is proper to; but there is no advantage gained by it. Thus, if we write 4- icz= 4- s/a, we have -|- a;= + \/a, or — ic =+ Va. But the former is ic = 4- Va, and the latter becomes so by changing the signs of both members. So that all we learn in either case, is that x =-{- \/a, and X = — \/a. 101, Cor. 2. — The roots of a Pure Quadratic Equation may both be imaginary, arul both will be if one is. For if after having transposed and reduced to the form x'^=^a, the second member is negative, as ^^ :::= — a, extracting the square root gives x = -^ V — a, and x = — v — a, both imaginary. EXAMPLES. 1. Given 8^2 — lo — x^ = 12 + 4tx^ — 54 to find the vahie of X. MODEL SOLUTION. Operation. 3x2 — 10 — x^ = 12 -^ 4x2 — 54, (2) — 2x2 = — 32. (3) x2 = 16, X = -I- 4. PURE QUADRATICS. 337 Explanation. — Transposing and uniting terms, I ha.ve — 2x- = — 32. [K necessarj% let the pupil show why the equality is not destroyed. ] Dividing both members by — 2, I have x^ = 16. Extracting the square root of both members, I find x = 4. 4 (read, ' ' x equals plus and minus 4"). [Be careful and not omit showing why the equation is not de- stroyed by the processes, as long as there is a possibility of a doubt that it is imderstood. ] Veetfication. — Substituting -f- 4 for x, the equation becomes 48 — 10 — 16 = 12 -f- 64 — 54, or 22 = 22. Substituting — 4 gives just the same since — 4 squared, ( — 4)2, is the same as -}- 4 squared. 2. Given x-^ -{- 1 = -— -{- 4:, to find x. Hoots, x = ^2. 4z — _ ^. x(9-^2x) 3^-f 6 . . , 6. Criven -— = , to nnd x. 10 5 Boots, a: = 4- 3. 4. Given [- = -, to find the values of x. 4:-\- X 4: X 6 Boots, a: = + 1. 5. Given x'^ — ah = d, to find the values of x. Boots, X = -\^ vd + ab. 6. Given 3 + t^ =- 777 — ^^ + ^A to find the values of x. Boots, x = 4-3. 7. Given 13 — v'dx-^ -f 16 = 5, to find the values of x. Boots, 0* = -4- 4. 2a 8. Given x -f \/x-^ -f- a = — " to find the values of x. V x-^a Boots, X = -^ I V 8a. SuG. — Clear of fractions. Transpose and condense. Square both members. 9. Given —^== = a: + ^x^-{-d. Boots, x= ±2. v/a:2-f 5 10. Given \/x + a=- ^ Boots, x = \/2a{a-{-b)+bK v^x — a '6'6t i QUADRATIC EQUATIONS. 11. Given - - a a X ^1 — a-^ 12. Given — - X - X 2^77111 n -^ , , ,— /- X X 13. Given a- ^ a^ Boot^, X = 4- V a2 — 52, 14. Given 12 — ;r^ : ^x^ : : 100 : 25. Boots, x=±2. OPEEATION. 12 — ic2 : ^x'^ : : 100 : 25 12 — ic2 : a;2 : : 50 : 25 : : 2 : 1 12 : x2 : : 3 : 1 4:x2: :1 :1 .-. x=4: 2. [Note. — Use the principles of proportion in solving these.] 15. Given ^x^- + ij;^ — 3 : ^x^ — ^x^ -]- 3 : : 9 : 3. Boots, X = ^ 4v^2. 16. Given ^{x'^ — 5)^ : j;^ — 5 : : 2 : 1. Boots, ^= + 3. OPERATION. ^(x2 — 5)2 : x2 — 5 : : 2 : 1 x2 — 5 : 1 : : 4 : 1 x2 - 5 : 5 : : 4 : 5 x2 : 5 : : 9 : 5 . • . x = + 3. 17. Given ^(11 + ^') : 1(4^^ — 2) : : 5 : 2. Boots, ;r= 4- 2. 18. Given x^ + 'i : x:^ — 11 : : 100 : 40. _ Boots. x=^ + v/21. 19. Given ^ , =2 — . Boots, x= + 2. ^3 + 4 x^ — 4 — /r4-4 r 4 10 20. Given — ^- -f =■ = ^. Boots, x= + 8. X — 4 x-^4c 6 — 21. Given (x + 2)2 = Ax + 5. Boots, x=±l. 22. Given -^(.r^ _ 12) = i:r' — 1. Boots, x== ±6. x+1 x—1 7 23. Given 072 — 7^ ^2 _[_ 7^7 072 — 73* BootSf X PURE QUADRATICS. 339 24. Given x -\- ^ x'- — ^2 — ix = 1. Eooi^, x ■■ ±h Query. — If the members are squared as the equation stands will it be simplified? (See Abt. 26, and examples under it for hints as to methods of freeing of radicals. ) 25. Given Va + x = '^ x+ ^'x- Boots, x-^^ -^ v/flj — b-. 26. Given ^1 -\- x-^ + ^^ 1 + ^^ + v^l — ^- = v^l — x'-. Verify. Boots, ar== + fv — 6. 27. Given Vb^ + x^ =^ \/a-' + x'-. Boots, x= ± ^ v^{b' — a^). 28. Given h — v/a^ + x'^ a- — x^ a + Va-^ _|- x-^ Boots, x= 4- -/wM •• — a — 6 N a __ ^. \/a''-\-x"~a , _ ^ 2av/6 29. Given := h. Boots, a; = + -r V a-^ -{- x^ -{- a J- — ^ 30. Given ^'^'+^ + 2 ^ ^^ ^^^^^^ x = -^^. \/3j7^ + 4— 2 "" 31. Given -=== ^. Boots, =^^^5- VX-^+1 + VJ7--2 — 1 32. Given 1 -— -— - = x. 33. Given ^ H- V 2 — x-^ X — v^2 — X- Boots, a; = -f ^. 1 la Boots, j: = + ^a^. 34. Given ==-. i?ooifs, ^= + _\ V a- + .r- — jr ^ -* v^ftc 34C 1 QUADRATIC EQUATIONS. 35. Given 1 + 1 1 v/1- -x^-^1 VI — j;2- _1 ^'* iioo^s, a: = = +iv/3. 36. Given 1 + 1+37 — n ■ ■ 1 — X X -{- ^l+x-^ 1 — ^ + ^1 + x-^ Boots, X = ±^{a~ -2y-l. 37. Given ' + i = ax. J7 + V 2 — x-^ X — v/2 — x^ Moots, X = -\- j^l . — y a 38. Given — ^= + , = a. Boots, 4 I 4 — a\ a^ APPLICATIONS. 1. What two numbers are those whose «um is to the great- er as 10 : 7, and whose sum multipHed by the less pro- duces 270? yl/is., 21 and 9. SuG. — Let lOx = the sum of the numbers, and 7x the greater. ScH. — It is customary to omit the negative roots in giving answers to examples, the nature of which renders such answers impossible. In this case the question is about pure number, and hence the answers should be given without signs. 2. There are two numbers whose ratio is that of 4 to 5, and the difference of whose squares is 81. What are the numbers ? Ans., 12 and 15. 3. What two numbers are those whose difference is to the greater as 2 to 9, and the difference of whose squares is 128 ? Verify. PURE QUADRATICS. 341 4. Find three numbers which bear the same ratio to each other as ^, f , and f do to each other, and the sum of whose squares is 724, Numbers, 12, 16, 18. 5. Find three numbers in the ratio of m, n, and p, the sum of whose squares is equal to a. „ 7 I am' I an-' Numbers, +k\ ; — , +v > and 4. ap2 6. Divide 14 into two psirts so that the greater part di- vided by the less shall be to the less divided by the greater as 16 to 9- SuG. — Having '■ : : : 16 : 9, it follows that x* : ^14 — X X (14 — a;)2 : : 16 : 9, and ic : 14 — ic : : 4 : 3, and* ; 14 : : 4 : 7. . •. x = 8, and 14 — a; = 6. 7. Divide a into two parts so that the greater part divid- ed by the less shall be to the less divided by the greater as m to n. „ , a^^m- , ttv^n Parts, -—z — —, and — = — — . ScH. — Example 7 is example 6 generalized. The pupil should de- duce the results in the former from thesa Thus,- substituting « = 14 m = 16, and n = 9, — — — : = 8, etc. 8. What two numbers are they, whose product is 126, and the quotient of the greater divided by the less, 3^ ? Generalize this. Ans., 6 and 21. 9. The sum of the squares of two numbers is 370, and the difference of their squares 208. Required the numbers. Numbers, 9 and 17. SuG. — If X represents the first, 370 — x- is the square of the second number. 342 QUADRATIC EQUATIONS. 10. Generalize the 9th, and show that ^\/2(8 + d) and -^x/2(s — d) are general results. J.1. For comparatively small distances above the earth's surface the distances through which bodies fall under the influence of gravity are as the squares of the times. Thus, if one body is falling 2 seconds and another 3, the distances fallen through are as 4 : 9. A body falls 4 times as far in 2 seconds as in 1, and 9 times as far in 3 seconds. These facts are learned both by obser- vation and theoretically. It is also observed that a body falls IB^^- feet in 1 second. How long is a body in falling 500 feet? One mile (5280 ft. ) ? Five miles ? Ans., To fall 500 ft. requires 5.58 seconds. To fall 5 miles requires 40.51 seconds. 1 1. A and B lay out some money in a speculation. A dis- poses of his bargain for $11, and gains as much per cent, as B lays out. B succeeds in gaining $86 ; and it appears that A gains four times as much per cent, as B. Eequired the capital of each. Results, $5 = A's capital, and $120 == B's. SuG.— To avoid fractions let 4« represent B's capital, and conse- quently A's gain per cent. 13. A money safe contains a certain number of drawers. In each drawer there are as many divisions as there are drawers, and in each division there are four times as many dollars as there are drawers. The whole sum in the safe is $5,324 ; what is the number of drawers ? Ans., 11. SuG. — This example gives rise to a cubic equation, 4x3 ^^ 5324. 14. Two travelers, A and B, set out to meet each other ; A leaving the town C at the same time that B left D. They travelled the direct road from C to D, and on meeting it appeared that A had travelled 18 miles more PUKE QUADKATICS. 343 than B ; and that A could have gone B's journey in 15f days, but B would have been 28 days in perform- ing A's journey. What is the distance between C andD? Ans., 126 miles. SUG. M D J 1 I I I A • > < — B If X := CM = the distance A travelled, then x — 18 = MD = the dis- a; ig 3. tance B travelled. - ■.-, — = distance A travelled a day ; and -^ lof '' 28 = distance B travelled a day. N'otice that the times are equal. 15. From two places at an unknown distance, two bodies, A and B, move toward each other till they meet, A going a miles more than B. A would have described B's distance in n hours, and B would have described A's distance in m hours. What was the distance of the two places from each other ? \/^ -f- \/n Ans., a X -7=- T^. vm — V n 16. A and B engaged to work for a certain number of days. A lost 4 days of the time and received $18.75. B lost 7 days and received $12. Now had A lost 7 and B 4 days, the amounts received would have been equal. How long did they engage to work and at what rates? Ans., Whole time, 19 days. Sua. — If X = the whole time, what represents A's daily wages? What B's ? Alter the equation is formed, see if you cannot strike out a numerical factor from both members, and extract the root without expanding. 17. A vintner drew a certain quantity of wine out of a full vessel that held 256 gallons ; and then filled the vessel with water, and drew off the same number of gal- lons as before, and so on for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time ? Ans., 64, 48, 36, and 27 gallons. 344 QUADRATIC EQUATIONS. SuG. — If he drew out - part of the contents of the cask each time, X 1 there remained after the first drawing? th of the wine : after the X second X or , and after the fourth X X X- x^ (x — l)^ _, x — 1 ••• -^r-=-^V..or— -=^. 18. A number a is diminished by the nth part of itself, this remainder is diminished by the ntli part of itself, and so on to the fourth remainder, which is equal to b. Eequired the value of n. y ~ 19. There is a number such that, if the square root of three times its square + 4, be taken, the quotient of this root increased by 2, divided by the root diminished by 2, is 3. What is the number ? QuEKY. — Which of the equations in the preceding part of this sec- tion does this give rise to ? 20. If the square root of the difference between the square of a certain number and 2, be both added to and sub- tracted from the number itself, the sum of the recip- rocals of the result is j'- of the number itself. What is the number ? Query. — Which of the equations in the preceding part of this sec- tion does this give rise to? With what modifications? SECTION IL Affected Quadratics. 102. An Affected Quadratic equation is an equation which contains terms of the second degree and also of the first with respect to the unknown quantity. AFFECTED QUADRATICS. 345 .^ . « 2ax4-Sbx-^ ^a-^x^ a:2 — 3j: = 12, 4 j: + ^ax'^ = , and ^ax + 36'^ = are affected quadratic equations. 103. Proh, To solve an Affected Quadratic Equaiion. RULE. — 1. Keduce the equation to the form x--\-ax^=h, the characteristics of which are, that the first member consists of two terms, the first of which is positive and simply the square of the unknown quantity, its coefficient being unity, while the second has the first power of the unknown quantity, with any coefficient (fl) positive or nega- tive, integral or fractional ; and the second member con- sists of known terms (6). 2. Add the square of half the coefficient of the second term to both members of the equation. 3. Extract the square root of each member, thus produc- ing A SIMPLE equation FROM WHICH THE VALUE OF THE UNKNOWN QUANTITY IS FOUND BY SIMPLE TRANSPOSITION. Dem. — By definition an Affected Quadratic Equation contains but tliree kinds of terms, viz. : terms containing the square of the unknown quantity, terms containing the first power of the unknown quantity, and known terms. Hence each of the three kinds of terms may. by clearing of fractions, transposition, and uniting, as the particular example may require, be united into one, and the results arranged in the order given. If, then, the first term, L e. the one containing the square of the unknown quantity, has a coefficient other than unity, or is nega- tive, its coefficient can be rendered unity or positive without destroy- ing the equation by dividing both the members by whatever coefficient this term may chance to have after the first reductions. The equation will then take the form x- ±(ix = ± h. Now adding (-) to the first member makes it a perfect square ( the square of x -f - j, since a trino- mial is a perfect square when one of its terms (the middle one, ax, in this case) is + twice the product of the square roots of the other two. these two being both positive (123 f Pakt I. ) But, if we add the square of half the coefficient of the second term to the first member to make 346 QUADRATIC EQUATIONS. it a complete square, we must add it to the second member to preserve the equality of the members. Having extracted the square root of each member, these roots are equal, since like roots of equals are equal. Now, since the first term of the trinomial square is x-, and the last (I) does not contain x, its square root is a binomial consisting of x 4- the square root of its third term, or half the coefl&cient of the middle term, and hence a known quantity. The square root of the second member can be taken exactly, approximately, or indicated, as the case may be. Finally, as the first term of this resulting equation is simply the unknown quantity, its value is found by transposing the second term. EXAMPLES. 1. Given 6 — x tt =" ^ — 2^+— -to find the value of o Xj and verify. MODEL SOLUTION— OPERATION. fi — x—^' — a- — 2-1- 4- — 48 —8x— x2 = 8x — 18 -J- x2 — 2x2 _ i6x = — 66 x2 -f- 8x = 33 xi -f 8x-|-16=33 + 16=49 X + 4 = + 7, x= + 7 — 4=3, and — 11. Explanation. — Clearing the equation of fractions, transposing and uniting, and dividing both members by — 2, I have x^ -|- 8x = 33. [Give the reasons why these processes do not destroy the equation often enough to keep them fresh in memory. ] Now since 8x contains the square root of x- as one of its factors, the other factor, 8, is twice the square root of the other term of a trinomial square {123 f Pakt I). Hence ^ of 8, or 4, squared, 16, is the third term. Adding this term I have x2 _j_ 8x -f- 16, which is a perfect square. But as I have added 16 to the first member to make it a perfect square, I must add it to the second member to preserve the equality. This gives x^ -f 8x -f 16 = 49. Extracting the square root of both members, I have x -j- ^ = -f 7. [Notice that (x -f- 4)-' = x^ -|- 8x + 16, and tell why this does not destroy the equation. ] Finally, transposing the 4, I have x = — 4 4- 7, t. c, X =: 3, and — 11, 3 if I take the square root of 49 as -f 7, and — 11 if I take the square root of 49 as — 7. Both are correct. Hencn there are two roots (values of or) of this equation, 3, and — 11. AFFECTED QUADRATICS. 347 Verification. — To verify the value x = 3, I substitute 3 for a; in the given equation, and have 6 — 3 — t = 3 — 2:^-|-8»oJ^ 3 — | = f-f- I, or ^f = -^if-. To verify the value x = — ^11, 1 substitute for x, — 11, in the given equation and have 6 -f- H — "^1^ = — H — 2^ -f- ''"a^. or 17 — -HJ^ = - 13i -f H^ or ^i- = \^. 2. Given x-^ — 8.r + 5 = 14, to find the values of x, and verify. Result, j; = 9, and — 1. 3. Find the roots of the equation x- — 12^ + 30 = 3, and verify. Roots, 9 and 3. 4j; 9 4. Find the roots of x — 2 = — . SuG. — This reduces to x^ — 6x = — 9. Whence, completing the square, x"^ — 6x + 9 = 9 — 9 = 0, and x — 3 = 0, or x = 3. In this case it appears that the equation has but one root. 5. Given - = — '■ to find the vahies of x. 5 X 6. Find the roots of S{x — 4) = ^- —. Sua— This reduces to x'^ — 8x = — 20. Whence, completing the square x- — 8x -)-16 = 16 — 20 ==—- 4, and extracting the root X 4 = -f 2^/"!, or X = 4 4- 2^^ — 1, two imaginary roots. 7. Find the values of x m 5 X — 5 * Results, a; =-- 5 + 2^—5, and 5 — 2^—5. 300 „„ o ^ — 300 8. Find the roots oi x -] -f 73 = 3 . X X Roots, — 30, and — 40. „4(2j^ + 6) 8 9. Find the roots of - — ^r = ^. 2x X Only one root, — 2. 348 QUADBATIC EQUATIONS. 10. Find the roots of l{x +1) + Slf-Z 1 = Q, Boots, 5(— 1 + v/— 1), tncl 5(— 1 — v/_l), or as the same may be written, — 5(1 — v^ — 1)^ and — 5(1 + v'^T). 11. Find the roots of |^-^ — ^.x + 20^ = 42f, and verify. Boots, 7, and — 6^. ScH. 1. — This process of adding the square of half the coefficient of the first power of the unknown quantity to the first member, in order to make it a perfect square, is called Completing the Squake. There are a variety of other ways of completing the square of an affected quadratic, some of which will be given as we proceed ; but this is the most important. This method will solve all cases : others are mere matters of convenience, in special cases. 12. Given x^ — ^ + 3 = 45 to find its roots, and verify. 13. Given 2x^ + 8^ — 20 = 70 to find its roots, and verify. IOj; _, 5.^2 — 65 and verify. 35 3j. 15. Given Gx + = 4:4: to find the values of x. X SuG. — Clearing of fractions, Qx^ 4-35 — 3x = 44a;, Transposing and uniting, 6x^ — 47a; = — 35, Dividing by 6, x^ — ^^x = — ^^, Completing the square, x^ — ^^x + (H)^ = (ff )^ — Y = ^W, Extracting the root, x — f |- = 4. f f. Transposing, x = ^ ± f J = 7, or f . 16. Find the roots of 5x '■ = 2x -\ '—- . X — 3 2 SuG. — Notice the compound negative term. Cleared of fractions and reduced, the equation becomes x^ — 3a; = 4. . • . x = 4, and — 1» 17 -c.- ^ ^1. . 4^ 16 100—9^ 17. Fmd the roots of = 3. X 4^2 Suggestion. — Multiply by 4x- to clear of fractions. 14. Given 5 ^ = 13^^ '■ to find its roots AFFECTED QUADKATICS. 349 18. Find the roots of 3x^ — 20j: — 62 = Ix — 2x^ + 100. Boots, 9, and — 3f . 19. Find the roots of 2a; — 2 = 2 + -. X Boots, 3, and — 1. 20. Find the roots of -^^^ — ^a; + 7 = 8 — |. Boots, f , and — f. 48 21. Find the values of x in the equation h 5 = ^ ^ +3 165 ^ , — — TTT- Besult, X = b% and 5. 8j7 20 22. Find the values of x in the equation — — =6. x-\-'A "dx Besult, X == 10, and — f . 23. Given ^{x + 4) — l-I— ^ = i (4^ + 7) _ l, to find X O the values of x. Besult, x = 21, and 5. 24. Given ^ 1 — — = 5^-^, to find the roots and X .r + 12 verify. Find t] and verify. 25. Find the roots of ^(8 — x) = U^ — 2), oa T?- A ^i. ^ « 2x + 9 , 4^—3 3:r — 16 26. Find the roots of — - — + - — -- = 3 + , y 4:X -\- 6 10 and verify. 27. Find the values of x in the equation Sx^ — 2ax = b. Besult, X = — = — ~ ■ 350 QUADRATIC EQUATIONS. OPERATION. 3a;2 — 2ax = 6; Dividing by 3, cc'^ —x = -, ^ 1 .. .1 , 2« , /ay- a2 5 a2_j_3& Completing the square, x^ —x + (o) =n +0= — z ^ Extracting the square root, a; — — = 4- -v^a- -\- '6b, ^ . a "/ a!^ -\- oh « 4- '^«' + <^^ Transposing, a; = — 4- » or — = 00 o ScH. — The form — ^ ; — ~ — signifies that there are two values o of ic ; i. e., that each of the signs 4" and — i^iay be used. Thus the Values in this case are a -J- >/a2 + 36 , or. — ^/as -(- 36 — ' and X = -; 3 *28. Find the roots of 3x'^ -\- hax = m. — 5a -i- ^25a- + 12m Boots, ■ — =^ -^ — b 29. Find the roots of da'^b^x^ — Ga^b'-x = bK a 4- v/a2 _^ 52 Boots, 3a'6-^ X a 2 30. Find the values of x in the equation ■- -\ — = -. a X a 31. Find the roots of BesuU, X = 1 -j- v^l — a-\ 2x(a — x) a '6a — 2x 4* Boots, fa, and ^a. 104, CoR. 1. — An affected quadratic equation has two roots. These roots may both be positive, both be negative, or one posi- tive and the other negative. They are both real, or both imaginary. Dem. — Let X- -j-px = q be any affected quadratic equation reduced to the form for completing the square. In this form p and q may be either positive or negative, integral or fractional. Solving this equa- AFFECTED QUADRATICS. 351 tion we have x = — ^ + kJj, — I" 9- ^^ ^^^^ ^^^ observe what dif- ferent forms this expression can take, depending upon the signs and relative vahies of p and q. 1st. Wheyi p and q are both positive. The signs will then stand as given; i. e., x = — -■ ± \/ "T" + 5- Now, it is evident that .1^ — h 5 > "T"' ^^^ \l ^" '? ^^ ^^® square root of something p2 p I p" p p- than -^. Fence, as — < ^— -{- q, — ^ + \Jx "^ ^^ ^^ more positive; but — ^ — \/~T — h ? i^ negative, for both parts are nega- tive. Moreover the negative root is numerically greater than the posi- tive, since the former is the numerical sum of the two parts, and the latter the numerical difference. .*. When p and q are both -|- in the given form, one root is positive and the other negative, and the nega- tive root is numerically greater than the positive one. See Example 1, above. 2nd. When p is negative and q positive. We then have x= — — ^ ± I — T ^ 5 = 7 ± I 4- ^- I^ "^^ take the plus sign of the radical, x is positive; but if we take the — sign, x is negative, since J 1 "I" ? ^ o- Moreover, the positive root is numerically the greater. Wh'en p is negative and q positive, one root is positive and the other negative ; but the positive root is numerically greater than the nega- tive- See Example 2, above. — P 8rd. When p and q are hoth negative. We then have x :=^ ^r^ 4- real, and as it is less than — , both values are positive. See Example 3. li J =:q, \— — g=0 and there is but one value of x, and this is positive. (It is customary to call this two equal positive roots, for the sake of analogy, and for other reasons which cannot now be appreoi- 352 QUADRATIC EQUATIONS. -f<^.J ated by the pupil.) See Examples 4 and 5. I^ i" ^ |x — ? ^®- comes the square root of a negative quantity and hence imaginary. See Examples 6 and 7. 4th. When p is positive and q negative. We then have x z= -^ ~ -1- ' q. As before, this gives two real roots when q • 2xV )ns in the first member, and Whence 2x^ = 26x — ab SuG. — Add the fractions in the first member, and ■ = /-• 2x — a V X FOR REVIEW OR ADVANCED COURSE. 100, Cor. 3. — UjDon the principle that the middle term of a trinomial square is twice the product of the square roots of the other two {94^ OS, Part l.),we can often complete the square more advantageously than by the regular rule. [Note.— A few examples will be given to illustrate this corollary, and if more are needed the pupil can use those which precede. The methods which follow are rather such as experts will use than such as are important in an elementary course of training. The use of them might be left optional with the student. Those who have special ap- titude for the science will learn them with profit, while to perplex others with them may be an injury rather than a benefit. ] 46. Solve 4^2 4. \Qx = 33. Solution. — Dividing 16a; by twice the square root of 4x2, i. e., by 4x, and adding the square of the quotient, (4)-, to both members, 4x2 -f. 16x + 16 = 49. Extracting the root 2x -f 4 = + 7. . • . x = §, and — J-/. 47. Solve 8^2— 12a: = 36. SuG. — Divide the equation by 2, and proceed as above. 4x2 — 6x-f- (1)2 = fii, 2x — f — 4- I, and x = 3, and — l^-. In this example the regular method is better. 356 QUADRATIC EQUATIONS. 48. Solve 3^2 ^ 2x = 5. SuG. —Multiply by 3 and 9x2 -f 6a; = 15. Whence 9x2 -f- 6x -{- 1 = 16, x=l, and — f . 49. Solve 110^2 — 21^ = — 1. Besidt, x = ■^^, and -Jy- SuG.— Multiply by 110, and (110)2a;2 — 21 • 110x = — 110. Whence 1102x2 — 21 . 110x-)-(^/)2= i, and llOx — V= ± h- 50. Solve 3^2 ^ 5^ = 2. SuG. (3)2x2 4- 3 . 5x + (I )2 = 6 + \i = Aa. Whence 3x + f = + f . . • . X = I, and — 2. ScH It appears that by this method the term to be added to com- plete the square is the square of i the coefficient of the first power of .r. Therefore when this coefficient is odd, fractions arise. These can always be avoided by doubling the equation when this coefficient is odd, before completing the square. 51. Solve 3:r2 _ 7^ = 40. Solution. — Multiplying by 2 to avoid fractions, 6x2 — 14x = 80. (6)2x2 — 6 . lix + (7)2 = 49 + 480 = 529. 6x — 7 = ± 23. x = 5, and -2|. 52. Solve ^x{x + 5) = ^6 + 4:x{l — x), and verify. SuG. — Perform as few multiplications as possible. When the square is completed, this stands (14)2x2 + 14 • 22x -f- (11)2 = 14 . 192 -|- 121 = 2809. 53. Solve 4-^2 4- i^ -f y7_3_ = 0. 1 2 S 54. Solve - x — 2 X + 2 5 55. Solve {ax — b)(bx — a) = c\ SuG. a&x2 — (a2_f_&-2)a; = c2 — ah. 2abx^—{m) =■= 2c2— 2ah, letting (m) stand for the term which becomes the middle term of the trino- mial square and which disappears in the subsequent process. Then (2a6)2x2 — ( m) + (a2 -(- 62)2 ^ (^2 4. j^iyz _j_ 4a6c2 — 4a262, . • . x = a2 -j- &2 4. v/(a- — 6^)-4-4a6c2 56. Solve , = — '■ — . Boots, x = a, and la. a + V2ax — x-^ «— ^ Suggestion. — Clear of fractions and condense. EQUATIONS SOLVED AS QUADBATICS. 357 SECTION III, Equations of Other Degrees which may be solved as Quadratics. 107, JProp, 1, Any Pure Equation {i.e., one containing the unknown quantity affected with hut one exponent) can he solved in a manner similar to a Pure Quadratic. Dem. — In any such equation we can find the value of the unknown quantity affected by its exponent, as if it were a simple equation. If then the unknown quantity is affected with a positive integral expo- nent it can be freed of it by evolution ; if its exponent be a positive fraction it can be freed of it by extracting the root indicated by the numerator of the exponent, and involving this root to the power indi- cated by the denominator. If the exponent of the unknown quantity is negative it can be rendered positive by multiplying the equation by it with a numerically equal positive exponent, q. e. d. EXAMPLES. 1. Solve y — • — = ttV. SuG. 2/3 = 8. . • . y = 2. Why not put the + sign before the 2? ~ /> q 2- Solve ^j^ = - gj^rr^. BesuU, y = -4. Query. — Can the root of a Pure Cubic Equation be imaginary? Why? 3. Solve 3a;^ — 5 = 2A Result, x = 125. 4. Solve ^ + 1 = 5(^ 1). Eesult, x = 32. — 2. 5 2 25 5. Solve 12^ 3 _|_ = _j_ Eesult, x = 27. o -4 if 358 EQUATIONS OF OTHER DEGREES 6. Given {x + {Ux + lla^)^^ =x^ + a^ to find the value of X. Result, x = 16a. 7. Solve 3aa;» = 2aa7'» + 56. Result, -"Nl(?y- 8. Solve a — l=6a;» — a;™ . a — I Result, -(^ef)--"- 9. Solve x^ = 27. Also x^ = 4. Also ?/^ = 32. Roots, 81, + 32, and 8. QuEKY. — Why the -f sign in one case and not in the others ? 108, J*rop. 2. Any equation containing one unknoum quantity affected with only two different exponents, one of which is twice the other, can he solved as an Affected Quadratic. Dem. — Let m represent any number, positive or negative, integral or fractional ; then the two exponents will be represented by m and 2m ; and the equation can be reduced to the form x-"" + px'^ = q. Now let y = x^, and y" = x"^^, whatever m may be. Substituting we have y2 ^ py = q^ whence ?/=— |- -+. .|^+^- ^^^ V = a;'" ; hence EXAMPLES. 1. Solve Sx^ + A2x^ = 3321. 3. Solution. — Let y = x^, whence ?/- = ic^ ; and Sy- -f- 42y = 3321. From this y = 27, and — 41. Taking the first value, x^ = 27. . • . a = 9. Taking the second a;^ == — 41 . • . x = 1^16817 We therefore find that aj = 9, and ^1681. SOLVED AS QUADRATICS. 359 2. Solve x^ + Ix^ = 44. ^ = + 8, or + (— 11 )i QuEKY. — How many values ? "Which are imaginary ? 3. Solve Ax^ + j;« = 39. x= 729, and (^)'* 4. Solve 3^6 _|_ 42^3 = 3321. x = d, and — '^il. 5. Solve — + 2 = ^ x=4., and ^A^i ScH. —It is not necessary to substitute another letter for the unknown quantity as given in such examples. Thus, in Ex. 3, doubling, to avoid fractions, Sx'^ -f 2x^ = 78. Completing the square 8"x^ -f- (m) + 1 = 8-78 + 1 = 625. Extracting root8x^ + 1= + 25. a:« = 3, 13 /13 \6 and — ^. . • . X = 729, and ( — - j . [Note. —Solve the next six without substituting.] 6. Solve x^o _j_ six^ =32. or = 1, and — 2. 1 1 7. Solve a;" + 13^'^ = 14. ^ = 1^ and (— 14)'". 8. Solve 3^2 _ 4^1 ^ 7. 9. Solve 3x + 2v/^ = 1. /7\^ Suggestions. (3)-x -|- (?n) + 1=4. . • . -/x = -^ and — 1. X = ^, and 1 . 10. Solve ^2x — Ix = — 52. .x = 8, and ^%K 11. Solve X + 2v/a^ + c = 0. X = { — s/a -4- v^a — c}2. jg _ 1 + ^2 12. Solve x^ X — — 7=— • yx V X X =.± v/l -f 1^27 360 EQUATIONS OF OTHER DEGREES SuG. — The first member may be written — . Hence drop- ping the denominator, x/x, and squaring, 6ic"^ ■ — x^ = 1 -\- 2x^ -{- x*. FOR REVIEW OR ADVAINCED COURSE. 100, J^VOJ}* 3> Equations may frequently he put in the form of a quadratic by a judicious gi^ouping of terms contain- ing the unknoum quantity, so that one group shall be the square root of the other. Dem. — This proposition will be established by a few examples, as it is not a general truth, but only points out a sjiecial method. EXAMPLES. 1. Solve 2x'- + 3^ — 5 \/2x'^-\-'6x -{- d + 3 = 0. Solution. — Add 6 to each member and arrange thus, (2x2 -j- 3x -f- 9) — 5 (2x2 + 3x -I- 9 ) ^ = 6. Put (2x-2 -{•3x-\-9f =y, and the equation be- comes t/'^ ^ 5^/ = 6. Whence t/ = 6, and — 1. Taking j/=*6, 2x--j- 3x -f 9 = 36. Whence x = 3, and — 4i Taking y = — 1, 2x2 -f 3x -|- 9 = 1. Whence x = = • 4 2. Given {2x + 6)i + (2a; + 6)4 = 6, to find the values of x. SuG.— Put 2/ = (2x + 6) ; whence y^- -{- y = 6, y = 2 and — 3. .-. 2x-f6 = 16, andalso2x + 6 = 81. x = 5, and37i QuEBY. —Will the value x = 37 ^ verify ? Why ? Ans. — Since (2x -J- 6) and(2x-j-6) are even roots, their signs are strictly ambiguous, though not so expressed in the example. Substi- tuting for X, 372, the equation becomes (81) + (81)^ =6. If now we regard (81) = — 3, as it is, as really as it is -|-3, the value verifies. Such cases are frequent. 3. Given (x -j- 12)^ == 6 — {x -\- 12)^ to find the values of X, and verify both values. SOLVED AS QUADRATICS. 361 1 2 4. Given -— = i + — —- to find the values of X. Values, x = 3, and 1. SUG.— Put {2x — 4)2 = y. 5. Solve x-\-5 — vx +5 = 6. J7 = 4, and — 1. 6. Solve 2^x-' — 'Sx+ll==x^ — 3x SuG. xi — 3x-{-8 — 2 Vx^ — 3a; -f 11 = 0. Add 3 to each member id x2 — 3x -f 11 — 2v/x^- — 3x -Pli =3. , Put v ! — 2t/ = 3. .-. x = 2, 1, or i(3 + v/— 31.) 7. Solve (x-^ _ 9)2 = 3 + ll(a;2 _ 2). :c = 4- 5, and + 2. 8. Solve (x + -V + X = 42 — -. \ ^/ X 07 = 4, 2, and ^(— 7 + ^/vf). / 1 \2 9. Solve x4l + _ j _ (3;r2 + a;) = 70. SuG. x'(l + —)~ = i(3x2 -j-a;)2. Hence the equation may be written (3x2 _|_ ic)2 — 9(3xi + .X) = 630. Besults, X = S, — 8-^, and ^(— 1 + ^— 251). 10. Solve — = —— — . Boots, x = 4, and 1, X — v^j? ^ / — 1 X — V X SuG. — Divide by x + vx, and = — . Hence 4 = X — \/x ^ (x — >/x)2, or X — \/x = -f 2. 110 • CoR. — The foem of the compound term may some- times be found by transposing all the terms to the first member, arranging them with reference to the unknown quantity, and extracting the square root. In trying this expedient, if the highest exponent is not even it must be made so by multiplying the equation by the unknown quantity. In like manner the coefficient of this term is to be made a perfect square. When 3B2 EQUATIONS OF OTHER DEGREES the process of extracting the root terminates, if the root found can be detected as a part, or factor, or factor of a part of the 'emainder, the root may be the polynomial term. .-, c . ^ 30 12 + ^^ 7 ,, 11. Solve--- + -^^=-- + H. Solution. — Cleared of fractions, transposed and arranged, this be- comes Sx-i — 42x^5 -}-■ 168x2 — 147x — 180 = 0. Dividing by3, x^ — 14x3 -f 56x2 — 49x — 60 = 0. Extracting square root x" — 14x^ -j- 56x2 — 49x — 60 1x2 — 7x. x" 2x2 _ 7a. | _ i4a;3 _^ ^q^^ — 14x3 -{- 49x2 7x2"-- 49x _ 60 7(x2 — 7x)— 60 After obtaining two terms of the root we observe the root itself as a fac- tor in part of the remainder. The polynomial is therefore (x2 — 7x)2 -f- 7(x2 — 7x) — 60 = 0. This is solved as before, x = 4, 3, and ^(7 + Vm). 12. Solve x^ — 12x^ + 4Ax^ — 48^ = 9009. One value of x is IS. 13. Solve J73 — 6j:2 4- lla: _ 6 = 0. x = 1, 2, and 3. 14 Solve 4:X* + ^ = 4^3 ^ 33. x = 2,—^, andi(l + \/ — 43)- 15. Solve X* — 2x3 + ^ = a. ^ ^ |. _|_ V| ^ v/a + i 16. Solve X* -\- x^ — 4:X^ -{- X -{- 1 = 0. J7 = 1, and = . Solution. — Dividing by x2, x2 -f- x — 4 -j }- - = 0, which may be written x2 + 2 -f - -f x -f - = 6, or (x -f -) -|- f'^^ + - ) = f>- Whence the compound term appears. SOLVED AS QUADRATICS. 363 [Note. — The discovery of such devices as are required for the solu- tion of this equation, is quite beyond the skill of any but expert alge- braists. ] 111, JPvop, 4. When an equation is reduced to the form x" + Ax«-' + Bx"-' + Cx"-^ +L = 0, the roots, with their signs changed, are factors of the absolute (Jcnown) term, L. Dem. — 1st. The equation being in this form, if a is a root, the equa- tion is divisible by x — a. For, suppose upon trial ic — a goes into the polynomial x" \- A.t" — ^ -(-, etc., Q times voiih a remainder E,. (Q rep- resents any series of terms which may arise from such a division, and K, any remainder. ) Now, since the quotient multipHed by the divisor, -}- the remainder equals the dividend, we have (x — a) Q -}- 11= x" 4" Aic"~^ -\- Bx"-2 -f- Cx"— ^ \-L. But this polynomial = 0. Hence (a; — a)Q -f- E = 0. Now, by hypothesis a is a root, and consequently X — a = 0, Whence E, = 0, or there is no remainder. 2nd. If now x — a exactly divides x" -|- Ax" - ^ -|-^^" ~ ^ + Ca;" ~ ^ |-L, a must exactly divide L, as readily appears from considering the pro- cess of division. Hence —a is a factor of L, a being a root of the equation. Q. E. D. [Note. — In most of our better text books there will be found special methods for reducing some forms of higher equations (usually cubics, it will be observed, ) to quadratics, by separating and arranging the terms, adding some term, multiplying or dividing by some number, BO as to make some binomial factor appear which can be divided out, or so that both members shall become perfect squares. This method is always difficult of application, depending, as it does, solely upon the ingenuity of the student. The binomial factor sought by these vieans is readily found by the principle above given. The difficulty in always discovering a root, on this principle, arises from the fact that the num- ber of factors of L, including fractional and imaginary ones, is infinite. So that this method will be of service only when some one or more of the roots is an easily recognized factor (usually an integral one) of L. But, in this respect, the method is equally as good as the cumbrous ones usually given, since, in cases where we could not recognize the root in this way, it would be little less than hopeless to attempt to do it in the other. In fact, the examples ordinarily given to illustrate the expedi- ents referred to, are here given, and solved by a simple and uniform method.] 364 EQUATIONS OF OTHER DEGREES EXAMPLES. 1. Solve 1x^ — 68a: = 32 — IT^^. 17x3 Solution. — This may be written x^ -| 34a; — 16 = 0. By 2 Pkop. 4 the roots of this equation are factors of 16. We therefore try- in order ±\ ±% ±^, ±'^, ± 16, (all the integral factors of 16) till we find whether there is an integral root. We see at once that neither -f- 1 nor — 1 satisfies the equation. Trying -f- 2 we find it is a root. 17a;3 Hence x — 2 is the divisor sought. Dividing x^ -J — 34a; — 16=0, 21 by X — 2, we have x^ -f — x^ -f- 21x + 8 = 0. But as — 2 satisfies the a 21 equation as well as -j- 2, x + 2 is a divisor. Dividing x^ -f- — x^ -f- 21x 17 4-8 = 0byx-j-2we have x'^ + — x + 4 = 0. From this equation a x = — 8, and — h- The roots therefore are 2, — 2, — 8, — i. 2. Solve ;r—l= 2+-?=. \^x SuG. — Put y = \/x, and clearing of fractions and transposing, y3 — 3y — 2 = 0. If there are integral roots of this equation they are 4- 1, or 4- 2. -f-1 cloes not satisfy the equation and hence is not a root. But — 1 does. Hence y -j-1 is a. divisor. Dividing, we get t/2 — y — 2 = 0. Whence j/ = 2, or — 1. ". •. The roots of the equa- tion y^ — ^y — 2^0 are — 1,2, and — 1, there being two equal roots. Now as 2/ = \/x, X = 1, and 4. 3. Solve x^ — 3x^-}-x + 2 = 0. The irrational roots are ^{1 -f n^5). 4. Solve x^ = Qx -\- 9. Imaginary roots, ^{ — 3 -f- y — 3). [Note. — The simpler roots are not given in these results, since they would at once indicate the factor to be divided out of the equation. ] 5. Solve 2x^ — x^ = 1, or x^ — ^x'^ — i = 0. ( + 1 is a factor of — -1-). Imaginary roots, ^( — 1 -f v^ — 7). 2 6. Solve x^ — 3- = H'y or x^ — -\^-x — f -= 0. ( + 1 and ox 4- f are factors of the absolute term.) Surd roots, ^{1 ± v^). 365 SOLVED AS QUADRATICS. 7. Solve X -{-Ix^ == 22. (Put x'^ = y.) Imaginary Boots, 29 + 1^ — 10. 8. Solve x^ + -i/a;3 — 39a; = 81. Imaginary Boots, ^{ — 13 -f V — 155). ^ ^ . 12 + 8x/x 9. Solve X = — . X — 5 Imaginary Boots, ^( — 3 + v — 7). Suggestion. — See Ex. 2 above, ,^ ^ , 49^2 48 ,^ ^6 10. Solve — 49 = 9 + -. 4: X'^ X Surd Boots, }{— 3 + v^93). 232 24: 192 Sua — This put in the required form is x* — x^ — —a; + 377 = 0' Examine for integral roots by trying 4- 1, and + 2, etc. The /or?n of the absolute term suggests also a fractional root. To test for this we would try fractions with 7 for denominator. These processes may be a Httle tedious, but there is method in them. But suppose this ques- 232 24 192 tion given in the form x^ — Iq""*'^' = To"'^ — Tq~' -^^ what possible means could the pupil discover that it must be put in the form given in the example, and then the square of both members completed by 1 49x- 49 adding — ^ to each member, and writing the result, — ^ 49 -^ XX- \. c -, nrr 841 17 232 1 , r 11. Solve27.2---+-=— -— +5. 2 232 840 SuG. — From the equation we have x^ -{-ZTr""^^ 31"^ oi~ =^- ol ol ol / Searching for integral roots we readily find 2, and dividing out the 326 420 factor x — 2, have cc^ -f- 2x-i -| — -x + --— = 0. To find a root of 81 ol 14 14 this by inspection would be a hopeless task. It is — , and x -| — — is the divisor which reduces the equation to a quadratic whose roots are 9 (— 2 + v'- 266> 366 EQUATIONS SOLVED AS QUADRATICS. The method of solving this equation by completing the square of both members is, to multiply by 3, transpose — j- and — (not uniting them), and add one to each member, thus deducing 81ic2 -^ 18 -f- 1 841 232 ~ = — ~ -j- — — -f- 16. Extracting the square root of both members 9x-\ — = ± ( ^ ^r ,^ ^ , 18 81 — X-' ^2_65 12. Solve h — t: • = 7r\ ' SuG. — Putting this in the form required the absolute term is — 1296 ; the integral factors of which are very numerous. But trying ± 1' ± 2, ± 3, 4- 4, + 6, 4- 8, + 9, we find — 4 and 9 to be two of the roots. The method of solving this by putting it in form so as to complete 8t 81 the square of both members, is to multiply by 2, and add -— -f- ^^ 36 So 36 18 9 x^ 2x 4 to both members, obtaining ^ -| -\- — = .^ -\- — — f- — . Ex- 6 3 / X 2 \ tracting root --j-=4-(-4--). Whence x = 9, — 4, — 4, and — 9, there being two roots — 4. 3 _^ 4v/^ 13. Given x — 3 = to find the roots. SuG. — The roots being all surds and imaginaries in this equation cannot be found by the principle in the proposition. The following special expedient may be resorted to : Clear of fractions and add x -f- 1 to each member and xfi — 1x -\-\ = 4 -j- 4v/x + X. 14. Solve the following by the principle in the proposition: jr2 — 1 3 x — \ _ 3 15 J7 + 1 ^ + 5 X X X + 5' ^4 4. ^3 _ 4^2 _4_ ^ ^ 1 = 0. (See Ex. 16, Prop. 3.) [Note. — Many of the examples unde*- Prop. 3 can be readily solved in this manner. ] SIMULTANEOUS EQUATIONS. 367 SECTION IV. Sunultaneous Equations of the Second Degree between two Unknown Quantities. 112, I^VOjy. 1, Two equations^ between two unknown quantities, one of the second degree and the other of the first, may always be solved as a quadratic. Dem. — The general form of a Quadratic Equation between two un- known quantities is ax^ 4- hxy + cy^ + cZx + ey + / = 0, since in every such equation all the terms in a;- can be collected into one, and its coefficient represented by a ; all those in xy can also be collected into one, and its coefficient represented by b, etc ., etc. The general form of an equation of the First Degree between two variables is a'x -f- h'y -\- c' = 0. J)'lJ (;' Now, from the latter x = , which substituted in the for- a mer gives no term containing a higher power of y than the second, and hence the resulting equation is a quadratic, q. e. d. EXAMPLES. X — y ^ + 3i/ 1. Griven X ■ — ~ = 4, and y — r = 1- SuG. — From the first a; = 8 — y. Substituting this value of x in 8 — V + 3y 8 4- 2v the second, we have y — — — = — ■; — - = 1, or v — tt. =^ 1 ; whence ^ 8— v-f-2 ^ KJ — 2/ yz — 9y = — 18, and y = 6 and 3. 2. Given x + y = 1, and x^ + 2?/2 = 34. Verify. 3. Given — | — = 2, and x -}- y = 2. Verify. X y 4. Given x -\- y = 100, and xy = 2400. 1 1 14- 5. Given 2^ + 3v = 37, and - + - = — . X y 4:5 6. Given 2x^ -r xy — 5y^ = 20, and 2x — Sy = 1. 368 SIMULTANEOUS QUADRATIC EQUATIONS 7. Given x -^\ = — ^-^, and —-L-'l ==, —J, -^ 3 X 2 8. Given .ly + .125x = y — x, and ?/ — .5x ^= . 15xy — ■ Sx. Besults, X ^ 0, and 4; and ^ = 0, and 5. 9. Given .3^ + .125y = 3x — y, and dx — .5y = 2.25xy + 3i/. Results, a; = 0, and — 1; and 3/ = 0, and — 2f. 113 • ^Top* 2, In general, the solution of two quadrat- ics between two unknown quantities, requires the solution of a biquadratic. Dem. — Two General Equations between two unknown quantities have the forms (1) ax2 4- hxy -f- ct/2 _|_ dx + ey + / = 0, and (2) a'a;2 _|_ h'xy + cY + d'x + e'y +/ = 0. From (1), .=_^-^ +^ l(!i^)l_ 2^1 + ^05. Now, to substitute this value of cc in equation (2), it must be squared, and also, in another term, multiplied by y, either of which operations produce rational terms containing j/^, and a radical of the second degree. Then, to free the resulting equation of radicals will require the squaring of terms containing y^, which will give terms in y*, as well as other terms. Q. e. d, [Note. — Since it is not the purpose of this treatise to embrace the resolution of the higher equations, only such special cases of Simulta- neous Quadratics with two unknown quantities, will be introduced, as can be resolved by the methods of quadratics .] 114. I*voj)» 3, Two Homogeneous Quadratic Equations between two unknown quantities can always be solved by the method of quadratics, by substituting far one of the unknown quantities the product of a new unknown quantity into the other. Dee. — A Homogeneous Equation is one in which each term contain? the same number of factors of the unknown quantities. 2x2 — 3^;^ — j/2 = 16 is homogeneous, dx- — 2?/ -\- y- = 10 is not homogeneous. WITH TWO UNKNOWN QUANTITIES. 369 Dem. — The truth of this proposition will be more readily appre- hended by means of a particular example. Taking the two homoge- neous equations x'^ — ^V -\- V' = 21, and y"^ — ^xy -|- 15 = 0. Let X =1 vy, V being a new unknown quantity, called an auxiliary, whose value is to be determined. Substituting in the given equations, we have v^y- — vy^ -}- t/2 = 21, and y^ — 2y?/' = -^ 15. From these we 21 15 find y- = -— , and y- = r , . Equating these values ^ v^ — v-\-l ^ 2u — 1 ^ ^ 21 15 of y2, — — ,-— = -; ; whence 42t; — 21 = 15u2 _ ISu + 15. u^ — V -j- 1 2i; — 1 This latter equation is an aflfected quadratic, which solved for v, gives V = 3, and |, Knowing the values of v we readily determine those of 15 — y from y- = -, and find y = 4. v^3 when v = 3, and y = + 5 when u = f . Finally as x = vy, its values are x = 4- 3v/3, and 4- 4. By observing the substitution of vy for x in this solution it is seen that it brings the square of y in every term containing the unlcnown quantities, in each equation, and hence enables us to find two values of t/2 in terms of v. It is easy to see that this will be the case in any homogeneous quadratic with two unknown quantities, for we have in fact, in the first of the given equations, all the variety of terms which such an equation can contain. Again, that the equation in v vnll not be higher than the second degree is evident, since the values of 2/2 con- sist of known quantities for numerators, and can have denominators of only the second, or second and first degrees with reference to v. Whence v can always be determined by the method of quadratics ; and being determined, the value of y is obtained from a pure quadratic {y- = -, in this case), and that of x from a simple equation (» = vy in this case). EXAMPLES. 1. Given Sx- + ^y = 18, and 4?/^ + dxy = 54. Boots, X = ^ 2, and + 2 v^S ; and y = _}_ 3, and + 3v/3. 2. Given x-^ -\- xy + 2?/2 = 74, and 2x^ -f 2xy + y' = 73- Hoots, ^ = -j- 3, and + 8 ; and y = + 5. 3. Given x'^ + Sxy = 54, and xy + 4y2 = 115. Boots, j7 = 4- 3, and + 36 ; and2/= -f 5, and -f 11-^. 370 SIMULTANEOUS QUADRATIC EQUATIONS 4. Given 2x^ + Sxy = 26, and dy^ + ^^y = 39. Boots, X ^ -^ 2, and y = ± ^' The other roots areoo- 6. Given x^ — 4?/2 = 9, and xy + 2^/2 = 3. j?oo/s, ^ == i --== ; and y = ± -y==. The other roots are oo- 6. Given Sx^^-i-xy — 9 = 9, and 4?/2 + 3xy — 4 = 50. (See Ex 1.) Boots, x= ±2, and y = + 3. 7. Given x"^ — ^y = 70, and xy — y~ = 12. 8. Given x^ -{- xy = 84, and x'^ — y- = 24. FOR REVIEW OR ADVANCED COURSE. US* J^TOp, 4. When the unknown quantities are simi- larly involved in two quadratic, or even higher equations, the solution can often he effected as a quadratic, by substituting for one of the unknown quantities the sum of two others, and for the other unknown quantity the difference of these new quan- tities. [Note. — As this and the following are merely special expedients, they need no demonstrations other than is furnished by applying them to examples. ] EXAMPLES. 1. Given x^ -{- y^ = 52, and x -}- y -{- xy = M. Solution. — In these equations x and y are similarly involved, and hence I try the expedient of putting a; = m -)- n, and y = m — n, whence X- 4- y^ = 2m2 -f- 2?z'- = 52, and x -\- y -\- xy = 2m -{- m^ — n^ = 34. Now, from the two equations m2 -f- n- = 26, and 2m -f- wi^ — n^ = 34, by adding I have 2m -f- 2m2 = 60, whence I find m = 5, and — 6. Substituting these values in m^ -j- n'^ = 26, n = -i- 1, and 4- s/ — 10. Whence the real values of x are found to be 6, and 4 ; and of y, 4, and 6. 2. Given x^ -\- x -{- y ^= 18 — y^, and xy = 6. Bational roots, x = 3, and 2; and y = 2, and 3. WITH TWO UNKNOWN QUANTITIES* 371 3. Given —^•- + ^ = — , and x'^ + y^ = 45. X — y X -{- y o SuG. — Using the same notation as above, 1 = -rrr ^^^ ^m'^ 4- ^ n m 3 2n2 = 45 ; whence 3m- -j- 3n- = lOmw, and we have imn = 27, or 27 9 3 3 9 m = — . m= + -, and + - ; 7i = + -, and + .. The roots are 4n ~2 ~"2 ~2 ~ ^ x=± 6; and 2/ = + 3. 4 Given 4(^ -|- y) = 3a;t/, and x -\- y + x^ -]- y"- = 26. Boots, x = 4,, and 2 ; 2/ == 2, and 4. Also x = — i^- + ^v/377, and 7/ = — -y. =F iv/377: Def. and Sch. — It will be observed that the above equations are of the second degree, and that they have the unknown quantities simi- ^ larly involved ; that is, in the last, for example, in the first member there is -f- ^^, and also -f- % ; in the second member x and y are mul- tiplied together ; in the second equation there is -j- x, also -|- y ; there is -f- x^, and also -f- 2/^- Such equations can usually be readily solved in 5 1 this manner. But the equations x- -\- y- = -xy, and x — y = -xy have the unknown quantities similarly involved in the first but dissimilarly in the second. There is -f- ^ in the second, but no -|- y, hence they are not similarly involved. Whether the solution of such equations will be facilitated by this expedient can only be ascertained by trial. In this case the expedient will be found successful. 5. Given x'^ -]- y^ = ^xy, and x — y = \xy. Boots, or = 0, 4, and — 2 ; y = 0, 2, and — 4. 6. Given x^ + xy-i- Ay'^ = 6, and S^-^ + 8if = 14. Boots, 07= 4- 2, and + |yiO ; and ?/ = + ^, and ± 1^10. 7. Given x-^ — 4?/2 = 9, and xy + 2rj^ ^d. SuG. — The student will find by experiment that the above expedient is of no service in this example. The example is readily solved by finding the value of x from the first equation and substituting it in the . second, thus obtaining ys/Ay^-^-d =3 — 2f, or iy^ + dy^ =d — 12y^ -\-iy*. Whence 21y* = 9, and y = + I-- ^^ ^^^ ecpuations can be treated as in [^114:), they being homogeneous. 8. Given - -\- K' = 18, and j; + y == 12. V X 372 SIMULTANEOUS QUADRATIC EQUATIONS SuG. — In these equations the unknown quantities are similarly in- volved, and although the first is of the third degree the expedient of the proposition is successful. x = 8, and 4 ; and y = 4:, and 8. ScH. — In all symmetrical equations the value of the unknovm quan- tities must, of course, be the same numerically, but taken in the re- verse order, since the letters can change places in the equation without altering the equations. When, therefore, in such equations the values of one of the unknown quantities are found the values of the other are known. 9. Given x^ — x-y^ -\- y^ = ±2, and x — xy -{- y = L SuG. — Putting x = m-^n, and y^=m — n, x^ -\-y^ = 1m^ -\- 2n^, and x^y^ ={m -f- n)^{m — n)^ = {m -}- n){m — n) X {m -\- n){m — n) - = (rn'^ _ n^)2. Hence 2m^ -|- 2n2 — (m^ —n^f =19. From the second equation, n^ z= 4 -j- wi^ — 2m. Boots, X = -1 (9 + \/73) ; and ?/ = ^(9 + ^73). 10. Given x + y = 11, and x^ + y^ = 407. Boots, X = 1, and 4 ; and ?/ = 4, and 7. 11. Given x — ?/ = 3, and x"^ -{- y^ = 641. Boots, X = 5, and — 2 ; y = 2, and — 5. SPECIAL EXPEDIENTS. no* Many equations of other degrees than the second, and which do not fall under the preceding cases, may still be solved as quadratics by means of special artifices. For these artifices the student must depend upon his own inge- nuity, after having studied some examples as specimens. These methods are so restricted and special that it is not expedient to classify them ; in fact, every expert algebraist is constantly developing new ones. 1. Given x^ -{- y'^ =^ 5, and x^ -\- y^ = 13. SuG. — In case of fractional exponents, it will usually be found expe- dient for the learner to put the unknown quantities with the lowest exponents, equal to the first powers of new unknown quantities, and thus make the exponents integral. Thus, putting x* = 7n, and y* = n, 1 2, we have x^ = m^, and ?/' = n'. Whence the equations becom(, T^^TH TWO UNKNOWN QUANTITIES. 373 tn -^ n = 5, and m- -j- n- = 13. These equations are readily solved by methods already learned, and we find m = 3, and 2, and n = 2, and 3. Hence x"* = 3, gives a; «= 81 ; and x* =2, gives x =16, Also 1/=* = 2, gives ?/ = 8 ; and ^■' = 3, gives y = 27. 2. Given ^^ + y ^ = 6^ a^d j;* + y^ = 126, i^oo^s, it; = 625, and 1, and y = 1, and 3125. 3. Given x^y^ = 2y\ and 8^^ — y'^ = 14. 4. Given x — y = v/^ 4- v/j^^ and x^ — 2/^ = 37. SuG. — Observe that both members of the first are divisible by -s/x + Vy, giving V'x — Vy = I. x = 16, and 9 ; i/= 9, and 16. 5. Given x^ -\- x -{- y ^ IS — t/2, and xy = 6. SuG. — From the first, by adding twice the second, we may write x2+2xy4-y2 4_x-fy = 30, or(x + 2/)* + (ic4-2/) = 30. .-. x + y = 5, and — 6. 6. Given x^ -\- y^ ^ 52, and x -\- y -\- xy = 34. Boots, X = 6, 4:, and — 6 -f y/ — 10 ; ?/ = 4, 6, and — 6 + v/— 10- 7. Given x^ -{■ y^ + 4= >/x^ + y2 = 45, and x^ _|_ ^4= 337. SuG, — In the first, put -^x* -(- y* = u, whence v* -f- 4i; = 45 ; and y ^=: 5, and — 9. x = 3, and 4 ; y = 4, and 3, X* x^ 8. Given f- 2— = 9|S and j;^ + v' = 65. X* 39 SuG. — In the first, put = v, whence v^ -f- 2u = 9 — , y 49 Real and rational roots, x == 4, y = 7. 9. Given x'' + 2xy -\- y^- -^ 2x = 120 — 2y, and xy — if = 8. i^oo^s, y = 1, 4, — 3 — v/5 a^^l — 3 + v/5'; x= 9, 6, — 9 + v/5, and — 9 — v^ 5. 374 SIMULTANEOUS EQUATIONS. 10. Given ^72^2 ^= I8O — 8xy, and x -\- dy = 11. Boots, X = 5, and 6 ; y = 2, and f . 11. Given x-^ -\- 3x -{- y = 7^ — 2xy, and y^ -]- 3y -{- x = 44 SuG. — Add the two equations together, and proceed as before, a; = 4, 16, and - 13 +^58" ; and ?/ = 5, — 7, and — 1 + n/58. 12. Given xy + xy^ = 12, and x + ^y^ =18. 12 18 SuG.— From the first, x = — - — ; ; and from the Becond, x = r— — . 12 18 .'. = . Dividing denominators by 1 -4- V» and nu- ya-hy) 1 + 2/' y -ry^ 2 3 merators by 6, we have - = ; whence 2 — 2v 4- 2v2 = y 1 — .V + y- y ^ y 3y. Hence x = 2, and 16 ; and y ==2, and ^. 13. Given x"^ -}- xy -\- y^ = 26, and x^ + x^j^ -\- y^ = 364. SuG. — From the first, x^ -\-y^ =2G — xy ; and from the second, by adding x'-y- to both members and extracting the square root, x^ -f- y^ = -v/y64: + x'^- Equating these values of x'^ -\- y^, and squaring, we have 676 — 52xy -j- ^^y'^ = 364 -j- x'-y', whence xy = 6. Squaring this and adding it to the second, and extracting the square root, x^ -\- y'^ = 20. .Also subtracting Sx'^y^ = 108 from the second, and extracting the square root, x- — y'^ = 16. Whence x^ = 18, and y^ = 2. Another Solution. — Dividing the second by the first, we have x* — xy -\- y'^ = 14. Subtracting this result from the first, we have 2xy = 12. Whence the solution proceeds as above. [Note. — Though this field is illimitable, it is not thought necessary for the learner to pursue special methods farther, inasmuch as what is given will enable him to catch the spirit of such solutions, and no writer in discussing a problem involving processes even as complex as Bome given above, would fail to give hints at his methods of solution.] APPLICATIONS. [Note. — One of the most important things to be learned from the following examples is several devices frequently found serviceable in stating a problem, wfhich make the equations arising more simple and easy of solution. These devices are of special necessity in examples involving progressions.] OF THE SECOND DEGREE. 375 1. What number is that which being divided by the pro- duct of its two digits, the quotient is 2, and if 27 is added to it the digits are reversed ? 2. There are three numbers, the difference of whose dif- ferences is 8 ; then- sum is 41 ; and the sum of their squaft'es is 699. What are the numbers ? Notation. — Let x be the second number, and y the difference be- tween the second and first, so that x- — y represents the first. Now as the difference of their differences is 8, and as the second is y more than the first, the third is ?/ -f- 8 more than the second, and hence is x -f- 2/ -f- 8. The first equation is 3x -|- 8 = 41, and the second (11 — y)'^ -j- 121 -f- (19 -|- y)'-^ = 699. Having solved the problem in this manner let the student solve it by letting x, y, and z represent the numbers, and then compare the solutions. 3. There are three numbers, the difference of whose dif- ferences is 5 ; their sum is 44 ; and their product is 1950. What are the numbers? 4. A grocer sold 80 lbs. of mace and 100 lbs. of cloves for $65 ; but he sold 60 lbs. more of cloves for $20, than he did of mace for $10. "What was the price of a pound of each? 5. A and p have each a small field, in the shape of an ex- act square, and it requires 200 rods of fence to enclose both. The contents of these fields are 1300 square rods. What is the value of each, at $2.25 per square rod? Ans., One, $900 ; other, $2,025. 6. Find two numbers, such that the sum of their squares being subtracted from tliree times their product, 11 re- main ; and the difference of their squares being sub- tracted from twice their product, the remainder is 14. SuG.^This example gives rise to homogeneous equations (114:). 7. What two numbers are those whose difference multi- phed by the difference of their squares is 32, and whose sum multiphed by the sum of their squares is 272 ? 376 SIMULTANEOUS EQUATIONS 8. The difference of two numbers is 2, and the square of their quotient added to four times their quotient is 9*. What are the numbers? Ans., 5, and 3. 9. There are two numbers, whose sum multipHed by the less, is equal to 4 times the greater, but whose sum multiplied by the greater is equal to 9 times 'the less. What are numbers? 10. Find two numbers, such, that their product added to their sum shall be 47, and their sum taken from the sum of their squares shall leave 62. Ans., 5 and 7. 11. Find two numbers, such, that their sum, their product, and the difference of their squares shall be all equal to each other. Ans., | -f ^^5, and ^ -\- ^v^S. 12. Find two numbers whose product is equal to the dif- ference of their squares, and the sum of their squares equal to the difference of their cubes. Am., iv/5, and J(5 + ^5). 13. A person has $1,300, which he divides into two portions, and loans at different rates of interest, so that the two portions produce equal returns. If the first portion had been loaned at the second rate of interest, it would have produced $36, and if the second portion had been loaned at the first rate of interest, it would have pro- duced $49. Eequired the rates of interest. Ans., 7 and 6 per cent. 14. The fore wheel of a wagon makes 6 revolutions more than the hind wheel in going 120 yards ; but if the periphery of each wheel be increased 1 yard, the fore wheel will make only 4 revolutions more than the hind wheel in going the same distance. What is the circum- ference of each wheel ? Ans., 4 and 5. 15. The sum of two numbers is 8 and the sum of their cubes is 152, what are the numbers? OF THE SECOND DEGREE. 377 SuG. — Let X -}- y he one of the numbers, and x ■ — y the other. Thena; == 4, and 2^3 +6xy^ = 152 [111). 16. The sum of two numbers is 7, and the sum of their 4th powers is 641. What are the numbers ? 17. The sum of two numbers is 6, and the sum of their 5th powers is 1056. "What are the numbers ? 18. The product of two numbers is 24, and their sum mul- tiphed by their difference is 20 ; find them. Ans., 4 and 6. 19. What two numbers are those whose sum multipHed by the greater is 120, and whose difference multiphed by the less is 16 ? Ans., 2 and 10. 20. What two numbers are those whose sum added to the sum of their squares is 42, and whose product is 15 ? A)is., 3 and 5. 21. A's and B's shares in a speculation altogether amount to $500 ; they sell out at pa7\ A at the end of 2 years, B of 8, and each receives in capital and profits $297. How much did each embark ? Ans., A, $275 ; B, $225. SuG. — Letting x be A's capital, and y B's, A gained 297 — x, and B, 297 T— y. And as the gains are proportioned to the products of the respective times into the capitals, 2x : 8y : : 297 — x : 297 — y. 22. What three numbers are those in A. P., whose sum is 120, and the sum of whose squares is 5600 ? Ans., 20, 40, 60. SuG. — La solving examples involving several quantities in arithmet- ical progression, it is usually expedient to represent the middle one of the series, when the number of terms is odd, by x, and let y be the common difference. If the number of terms is even, represent the two middle terms by x — y, and x -^ y, making the common difference 2y. Thus the statement of the above problem becomes x — y -\- ^ -{z^'^hU r= 120, or 3x = 120 ; and {x — 2/)^ + ic^ -f (x -4- vj^ =? 5600, pr 3x^ -f 2t/2=5600. '^ 378 SIMULTANEOUS EQUATIONS 23. What four numbers are those in A. P., the sum of whose squares is 84, and their product 105 ? SuG, — Call tbie numbers x — Sy,x — y,x -\- y, and x + 3?/. 24. The sum of five numbers in A. P. is 35, and the sum of their squares 285 ; find the numbers. 25. What three numbers are those in A. P., the sum of whose squares is 1232, and the square of the mean greater than the product of the two extremes, by 16 ? 26. Find four numbers in A. P. such, that the sum of the squares of the extremes is 4500, and the sum of the squares of the means is 4100. 27. The product of five numbers in A. P. is 945 ; and their sum is 25. What are the numbers ? Ans., 1, 3, 5, 7, 9. 28. The product of four numbers in A. P. is 280, and the sum of their squares 166 ; find them. 29. The sum of nine numbers in A. P. is 45, and the sum of their squares 285 ; find them. Ans., 1, 2, 3, etc., to 9. 30. The sum of seven numbers in A. P. is 35, and the sum of their cubes 1295 ; find them. Ans., 2, 3, etc., to 8. i 31. There are three numbers in geometrical progression, whose sum is 52, and the sum of the extremes is to the mean as 10 to 3. What are the numbers? Ans., 4, 12, and 36. SuG. — Let X be the first term and y the ratio. Then x-}-xy-{- xy^ = 52, and x + xy^ :xy ::10 :d, orl-\-y^ :y ::10 : 3. 32. The sum of three numbers in geometrical progression OF THE SECOND DEGREE. 379 is 13, and the product of the mean and sum of the ex- tremes is 30. What are the numbers ? Ans., 1, 3, and 9. SuG. — We have x -\- xij -{- xy^ = 13, and (x -\- xy-) xy = 30. From 30 the first x -f- ^y'^ = Vi — xy, and from the second, a; -f- ^- = — '■> xy 30 whence 13 — xy=^ — , or x-y- — 13a:?/ = — 30. 33. If the seventh and tenth terms of a geometrical pro- gression are 6 and 750 respectively, what are the inter- mediate terms ? SuG. — The equations area;?/6=: 6, and a'^/^ = 750. Divide the second by the first. 34. If the third and fifth terms of a geometrical j^rogTession be 75 and 300 respectively, what will the fourth term be? Ans., 150. 35. If the first and fourth terms of a geometrical progres- sion are 3 and 24 respectively, what are the two inter- mediate terms? 36. There are four numbers in geometrical progi'ession. The sum of the means is 30, and the product of the extremes 200, what are the numbers ? SuG. — Using the notation given above, we have for the equations xy-^-xy^-z^ 30, and x'^y^ = 200. But a more elegant and simple solu- x^ v^ tion is obtained by representing the numbers by — , x, y, and — , in y X which - is the ratio. This gives for the equations x -\- y = SO, and xy = 200. When this form of notation is used for an odd number of terms, it is expedient to make xy the middle term. Thus for five terms, we have 0-3 ^ y^ — , X-, xy, y-, — . 37. The sum of three numbers in G. P. is 26, and the sum of their squares is 364 ; required the numbers. SuG. — The equations are x^ + xj/ + ^^ = 26, and x* + .r^yz + y* = 364, which have already been solved. 380 SYNOPSIS. £ j Quadratic Equation \ P^i^e. -Incomplete. o i ^ ^ i Affected. — Complete. Root of Equation. Frob. To Solve. Bern. Cor. 1. Number of Boots. Dem. Cor. 2. Imaginary Roots. r Prob. To Solve. Rule. Bern. Completing Square. What ? | <^^^-. ^lethod. ( Special methods. Cor. 1. Roots of an Aff. Quad. Dem. Cor. 2. To -write the root of x- + px = q directly. Pure. Prop. 1. r Prop. 2. Dem. When solved as Quad. eS in S j Expedients. ] ^^^i^- 3- Cor. [ ( Prop. 4:. How applied. Bern. Bern. Bern. Sch. Affected. Prop. 1. Prop. 2. Prop. 3. Prop. 4. Expedients. Enumerate the 7 given. r Common method. ^ A. P. -{ j Even number of terms. ! Special methods, i _ , , [ A (Odd " " " r Common method. A. G. P. ] j Even number of terms. Special methods. S ^^^ I ( Odd " " " Test Questions. — If one of two equations between two unknown quantities is of the first degree, and the other of the second, what will be the degree of the resulting equation after ehminating one of the unknown quantities? Prove it. How may such equations be solved? In general, what is the degree of the equation arising from eliminating one unknown quantity from two equations, each of the second degree ? Prove it. Mention the several cases g'ven in which such equations can be solved by quadratics, and state the process in eaclt c^se. LOGARITHMS. 381 CHAPTER Y. logahithms. [Note. — It is the purpose of this chapter to give a simple. arithmet« ical view of the nature of logarithms, with some illustrations of their practical utility. For the production of the Logarithmic Series and its use in computing Logarithms, see Appendix II. Enough, how- ever, is here given for practical use in trigonometry, as usually studied, and to enable the student to understand the use of loga- rithms in ordinary operations.] 117 » A Logarithm is the exponent by which a fixed number is to be aJBfected in order to produce any required number. The fixed number is called the Base of the System. III. — Let the Base be 3 : then the logarithm of 9 is 2 ; of 27, 3 ; of 81, 4; of 19683, 9; for 32 = 9; 3^=27; 3^ = 81; and 33 = 19683. Again, if 64 is the base, the logarithm of 8 is i, or .5, since 64^, or 64-"' = 8 ; {. e., i, or .5 is the exponent by which 64, the base, is to be affected in order to produce the number 8. So also , 64 being the base, i, or .333 -f- is the logarithm of 4, since 64* , or 64-333+ = 4 ; i. e., •^, or .333 -}- is the exponent by which 64, the base, is to be affected in order to produce the number 4. Once more, since 64^, or 64-ece+ = 16, §, _ i or .666+ is the logarithm of 16, if the base is 64. Finally, 64 ^, or 64. — .5 __ i^ Qj. j^25 ; hence — i. or — .5, is the logarithm of g, or .125, when the base is 64. In like manner, with the same base, — i, or — .333 -f- is the logarithm of 4, or .25. EXAMPLES. 1. If 2 is the base what is the logarithm of 4 ? of 8 ? of 82? of 128? of 1024? MODEL SOLUTION. 7 is the logarithm of 128, if 2 is the base, since 7 is the exponent by which 2 is to be affected in order to produce the number 128. 382 LOGARITHMS. 2. If 5 is the base what is the logarithm of 625 ? of 15625? of 125? of 25? 3. If 10 is the base what is the logarithm of 100? of 1,000? of 10,000? of 10,000,000? If 2 is the base what is the -1-, or .125? of ^\, or .03125? 4. If 2 is the base what is the logarithm of ^, or .25 ? cf Ans. to the last, — 5. 5. If 8 is the base, of what number is f , or .666 + the logarithm? of what number is f or 1.333+ the loga- rithm ? of what number is 2 the logarithm ? of what number is 2^, or 2.333 + ? of what number 3f, or 3.666 + ? Ans. to the last, 2048. . ScH. — Since any number with for its exponent is 1, the logarithm of 1 is 0, in all systems. Thus 10^ = 1, whence is the logarithm of 1, in a system in which the base is 10. 118, A St/stein of Logarithms is a scheme by which all numbers can be represented, either exactly or approximately, by exponents by which a fixed number (the base) can be affected. Negative numbers can have no logarithms. 110, There are Two Systems of Logarithms in common use, called, respectively, the Briggeaiv or Common System, and the Najnerian or Hyperbolic System. The base of the former is 10, and of the latter 2.71828 +. 120 • One of the most important uses of logarithms is to facilitate the multiplication, division, involution, and the extraction of roots of large numbers. These processes are performed upon the following principles : 121, JProp, 1, TJie sum of the logarithms of two num-^ hers is the logarithm of their p?vduct LOGAKITHMS. • 38? Dem. — Let a be the base of the system. Let m and n be any two numbers whose logarithms are x and y respectively. Then by defini- tion a^ = m, and a^ r= n. Multiplying these equations together we have a'^+y = mn. Whence x-\- y is the logarithm of mn. q. e. d. 122. ^rop, 2. The logarithm of the quotient of two numbers is the logarithm of the dividend minus the logarithm of the divisor. Dem. — Let a be the base of the system, and m and n any two num- bers whose logarithms are, respectively, a*, and y. Then by definition we have ai^=m. and a" = n. Dividing, we have a* — " = — . "Whence n X — w is the logarithm of — . q. e . d. 71 123. J^rop. 3. The logarithm of a jwwer of a number is the logarithm of the number multiplied bij the index of the power. Dem. — Let a be the base, and x the logarithm of m. Then a''^=^m', and raising both to any power, as the 2th, we have a^^ = m~. Whence xz is the logarithm of the zth power of m. q. e. d. 124, Prop. 4:. The logarithm of any root of a number is the logarithm of the number divided by the number express- ing the degree of the root. Dem. — Let a be the base, 9.nd x the logarithm of m. Then a^=m. X Extracting the zth root we have a^ = ym. Whence - is the loga- rithm of v/?u. Q. e. d. 12 o. In order to apply these principles practically, we need what is called a Table of Logarithms. That is, a table from which we can readily obtain the logarithm of any number, or the number corresponding to any logarithm. We will, therefore, proceed to show how such a table can 384 - LOGARITHMS. be computed. Though the method about to be given is not the most expeditious now known, it is, nevertheless, the one used when our tables were first computed. jL20» JProb, To compute the common logarithm, of any decimal number. Dem. — 1st. It is evident that it is only necessary to compute the log- arithms of prime numbers, since the logarithm of a composite number is the sum of the logarithms of its factors {121). 2nd. To compute the logarithms of the series of prime numbers. In the first place we know that the logarithm of 1 is 0, since 10^ = 1. Also the logarithm of 10 is 1, since 10' = 10. Now if we find the log- arithm of 5, we can get the logarithm of 2 by subtracting the logarithm of 5 from log. 10. * If there be any number which is the logarithm of 5 it is evident it must lie between 0, which is log. 1, and 1, which is log. 10. Therefore starting with 10^ = 1 and 10' =10 multiplying them together we have 10' = lO Extracting the square root, 10-5 = v/lO = 3.162277 +. Again, as 5 lies between 10 and 3 .162277 + its logarithm lies between 1 and .5, Multiplying the last two equations we have 10' '5 = 31. 62277+. Extracting the square root, 10-75= v/^i. ^-A'zn + = 5. 623413 -f-. Again, 10- =3.162277 4- and 10-" =5.62 3413 4- Multiplying 10'-- = 17.78^7895914 4- Extracting square root lO-csa = v/17 .7827895914 -f = 4.2169644- Again, multiplying this last by 10 -^^ =: 5.623413 -{-, as 5 lies between these numbers, and extracting the square root, we have 10 •^'^■'■5==: 4.869674 4-- I^ each case the exponent of 10 is the logarithm of the number ; thus .6875 is log. 4.869674-)-. Continuing this process to 22 operations (!) we have log. 5. 000000 -|-= . 698970 -{-, which i3 suffi- ciently accurate for ordinary purposes. Now log. 10 — log. 5 = log. 2 = 1 — .698970 = .301030. To find log. 3, we would take lO-^ = 3.162277 4-, and 10° = l,and proceed as before. Were it our purpose to find log. 11, the computation would be as follows : * This is the common abbreviation of "logarithm of 10," and should be read "logarithm of 10," not " log ten," which is grossly inelegant. LOGABITHMS. 385 101 = 10 102 ^ 100 10^ = 1000 10- = 101-5 = v/lOoi) = 31.62277 -f 101 = 10 10^ 101 101 = 102-5 = 316.2277 + •25=^316.2277+ = 17.78278 + 10 10^ 101 lOi ••Zb = • 125 - -. 177.827« + = v/l77.8278 + == 13.33521 + 10 102 101 101 .125 = •0625 : = 133. 3521 + = v/l33.3621 + = = = 11.54782 — 10 102 101 101 •0025 : •03125 •0625 : = 115.4782 + = >/ll5.4782 + = = 10.74607 + 11.54782 — 102- 09375 101.04687; 101-03125 = 124.09368 + * = V'124. 09358 + . = 11.13973 + 10.74607 + 102.07S12O = 119.70845 + 101 •0390625 = 10.94113 + Whence 1.0390625 is log. 10.94113; and proceeding with the computa- tion, the logarithm of 11 may be found -with suflacient accuracy. Example. — Let the pupil compute the logarithm of 23, and compare his result with the logarithm as found in the table following. ScH. — The pupil will not fail to be impressed with an idea of the im- mense labor involved in computing a tabic of logarithms. The com- mon tables give the logarithms of numbers from 1 to 10,000, with provision, as mil be seen hereafter, for using them to find the loga- rithms of much larger numbers, with sufficient accuracy for practical purposes. One page of such a table is given. (Page 386.) 127* JProb, To find the logarithm of a number from \he table. 386 (a page of) a TABLE OF LOGARITHMS. N. 1 2 3 4 5 6 7 8 9 I*. 280 447158 7313 7468 7623 7778 7933 8088 8242 8397 8552 155 281 8706 8861 9015 9170 9324 9478 9633 9787 9941 ••95 154 282 450249 0403 0557 0711 0865 1018 1172 1320 1479 1633 154 283 1786 1940 2093 2247 2400 2553 2700 2859 3012 3165 153 281 3318 3471 3624 3777 3930 4082 4235 4387 4540 4692 153 285 4845 4997 5150 5302 5454 5606 5758 5910 6062 6214 152 286 6366 6518 6670 6821 6973 7125 7276 7428 7579 7731 152 287 7882 8033 8184 8330 8487 8638 8789 8940 9091 9242 151 288 9392 9543 9694 9845 9995 •146 •296 •447 •597 •748 151 289 460898 1048 1198 1348 1499 1649 1799 1948 2098 2248 150 290 2398 2548 2697 2847 2997 3146 3296 3445 3594 3744 150 291 3893 4042 4191 4340 4490 4639 4788 4936 5085 5234 149 292 5383 5532 5080 5829 5977 6126 6274 6423 6571 6719 149 293 6868 7016 7164 7312 7460 7608 7756 7904 8052 8200 148 294 8347 8495 8643 8790 8938 9085 9233 9380 9527 9675 148 295 9822 9969 •116 •263 •410 •557 •704 •851 •998 1145 147 296 471292 1438 1585 1732 1878 2025 2171 2318 2464 2610 146 297 2756 2903 3049 3195 3341 3487 3633 3779 3925 4071 146 298 4216 4362 4508 4653 4799 4944 5090 5235 5381 5526 146 299 5671 5816 5962 6107 6252 6397 6542 6687 6832 6976 145 300 7121 7266 7411 7555 7700 7844 7989 8133 8278 8422 145 301 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 302 480007 0151 0294 0438 0582 0725 0869 1012 1156 1299 144 303 1443 1586 1729 1872 2016 2159 2302 2445 2588 2731 143 304 2874 3016 3159 3302 3445 3587 3730 3872 4015 4157 143 305 4300 4442 4585 4727 4869 5011 5153 5295 5437 5579 142 306 5721 5863 6005 6147 6289 6430 6572 6714 6855 6997 142 307 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 141 308 8551 8692 8833 8974 9114 9255 9396 9537 9677 9818 141 309 9958 -99 •239 •380 •520 •661 •801 •941 1081 1222 140 310 491362 1502 1642 1782 1922 2062 2201 2341 2481 2621 140 311 2760 2900 3040 3179 3319 3453 3597 3737 3876 4015 139 312 4155 4294 4433 4572 4711 4850 4989 5128 5267 5406 139 313 5544 5683 5822 5960 6099 6238 6376 6515 6653 6791 139 314 6930 7063 7206 7344 7483 7621 7759 7897 8035 8173 138 315 8311 8148 8586 8724 8862 8999 9137 9275 9412 9550 138 316 9687 9824 9962 -99 •236 •374 •511 •S48 •785 •922 137 317 501059 1196 1333 1470 1607 1744 1880 2017 2154 2291 137 318 2427 2564 2700 2837 2973 3109 3246 3382 3518 3655 136 319 3791 3927 4063 4199 4335 4471 4607 4743 4878 5014 136 320 5150 5286 5421 5557 5693 5828 5964 6099 6234 6370 136 321 6505 6640 6776 6911 7046 7181 7316 7451 7588 7721 135 322 7856 7991 8126 8260 8395 8530 8664 8799 8934 9068 135 323 9203 9337 9471 9006 9740 9874 •••9 •143 •277 •411 134 324 510545 0679 0813 0947 1081 1215 1349 1482 1616 1750 134 325 1883 2017 2151 2284 2418 2551 2684 2818 2951 3084 133 326, 3218 3351 3484 3617 3750 3883 4010 4149 4282 4414 133 327 4548 4681 4813 4946 5079 5211 5344 5476 5609 5741 133 328 8874 6006 6139 6271 6403 6535 6608 6800 6932 7064 132 329 7196 7328 7460 7592 7724 7855 7987 8119 8251 8382 132 330 N. 8514 8646 8777 8909 9040 9171 9303 9434 9566 9697 131 1 2 3 4 5 6 7 8 9 D. LOGARITHMS. 387 Solution. — Page 386 is one page of a table of logarithms giving the logarithms of numbers from 1 to 10,000 directly, and from which the logarithms of other numbers can also be found with little trouble. Thus, let it be required to find the logarithm of 325. Now, as the loga- rithm of 100 is 2, and of 1000 is 3, the logarithm of 325 must be between 2 and 3, i e. 2 and a fraction. The fractional part is all that is given in the table, as the integral can be kno-«Ti by simple inspection. Look- ing in the table down the column marked N (numbers), we find 325, ^ and opposite it in the column headed 0, we find 1883, but just above this we observe 51 , which belongs to this logarithm and which is simply omitted to save space in the table, since it really belongs as a prefix to all the logarithms clear do-vvTi to the number 332 where it is replaced by 52. Prefixing the 51 to the 1883, we have .511883 as the decimal part of the logarithm. Hence log. 325 is 2.511883. In like manner the logarithm of any number consisting of three figures is found from the table. To find the logarithm of a number consisting of four figures. Let it be required to find the logarithm of 2936. Looking for 293 (the first three figures) in the column of numbers, and then passing to the right until reaching the column headed 6, the fourth figure, we find 7756, to which prefixing the figures 46, which belong to all the logarithms following them till some others are indicated, we have for the decimal part of the logarithm of 2936, .467756. But, as 3 is the logarithm of 1000, and 4 of 10,000, log. 2936 is 3 and this decimal, or log. 2936 = 3.467756. To find the logarithm of a number consisting of more than 4 figures. Let it be required to find the logarithm of 2845672. Finding the deci- mal part of logarithm of the first 4 figures 2845, as before, we find it to be .454082. Now the logarithm of 2846 is 153 (millionths, really) more than that of 2845. Hence, assuming that if an increase of the number by 1000 makes an increase in its logarithm of 153, an increase of 672 in the number, wiUmake an increase in the logarithm of -]^o~i.ro o^ .672 of 153, or 103, omitting lower orders, and adding this to .454082, we have .454185 as the decimal part of log. 2845672. The integral part is 6, since 2845672 lies between the 6th and 7th powers of 10. Hence log. 284672 = 6.454185. q. e. d. ScH. 1. — If in seeking the logarithm of any number any of the heavy dots noticed in the table are passed, their places are to be filled with O's, and the first two figures of the decimal of the logarithm taken from the column in the line below. Thus log. 3166 is 3.500511. This ar- rangement of the table is a mere matter of convenience to save space. 388 LOGARITHMS. ScH. 2, — The column marked D is called the column of Tabular Differences ; and any number in it is the difference between the loga- rithms found in columns 4 and 5, which is usually the same as be- tween any two consecutive logarithms in the same horizontal line. The assumption made in using this difference ; viz., that the logarithms increase in the same ratio as the numbers , is only approximately true, but still is accurate enough for ordinary use. 128^ The Integral Part of a logarithm is called the Characteristic^ and the decimal part the Man^ tissa. 129. I*rop, The Mantissa of a decimal fraction, or of a mixed number, is the same as the mantissa of the number considered as integral. Dem.— Above it was found that log. 2845672=6.454185. Now this means that lO^-* « 4 1 8 s =:= 2845672. Dividing by 10 successively we have 105.464185 ^ 284567.2, or log. 284567.2 = 5.454185, lQ4.464i85^ 28456.72, or log. 28456.72 =4.454185, 103.464186^ 2845.672, or log. 2845.672 =3.454185, 102-464185^ 284.5672, or log. 284.5672 =2.454185, 101.454186^ 28.45672, or log. 28.45672 =1.454185, 100-464186 _ 2.845672, or log. 2.845672 = 0.454185. Now if we continue the operation of division, only writing 0. 454185 — X 1,454185, meaning by this that the characteristic is negative and the mantissa positive, and the subtraction not performed, we have 10T.4o4i8.5 _ .2845672, or log. .2845672 —1.454185, 10iT454i85 =- .02845672, or log. .02845672 =2.454185, IOT.454185 = .002845672, or log. .002845672=3.454185, etc., etc. Q. E. D. ScH. — The characteristic of an integral number, or of a mixed integral number and decimal, is one less than the number of integral places, as will appear by comparing such numbers with the powers of 10, as is done in demonstrating {12S). The characteristic of a number en- tirely decimal fractional, is negative, and one greater than the number «of O's immediately followinier the decimal point, as appears from the last demonstration, or as appears from the fact that 10~^ = xV = .1: 10-2 = T^o = .01 ; 10-2 ^ ^i__ ^ ^001 ; etc., etc. LOGAKITHMS. 389 EXAMPLES. Find the logarithms of the following numbers : 285 ; 3145 ; 2905624 ; 30942716 ; 298.026 ; 32.5614 ; 2.8641 ; .3205 ; .00317 ; 00000328. Results, log. 298.026 = 2.474254 ; log. .00317 = 3.501059. 130, PtoJ), — To find a number corresponding to a given logarithm. Solution. — Let it be required lo find the number corresponding to tbe logarithm 5.515264. Looking in the table for the next less mantissa, we find .515211, the number corresponding to which is 3275 (no account now being taken as to whether it is integral, fractional or mixed ; as in any case the figures will be the same). Now, from the tabular differ- ence, in column D, we find that an increase of 133 (millionths, really) upon this logarithm (.515211), vrould make an increase of 1 in the number, making it 3276. But the given logarithm is only 53 greater than this, hence it is assumed (though only approximately correct) that the increase of the number is -^^ of 1, or 53 -j- 133 = .398-4 +. This added (the figures annexed) to 3275, gives 32753984+. The characteristic, being 5, indicates that the number hes between the 5th and 6th powers of 10, and hence has 6 integral places. . • . -5.515264 = log. 327539.84+. EXAMPLES. Find the numbers corresponding to the following loga- rithms : 3.467521; 2.467521; 0.467521; 4.500281; 1.520281; 4.520281; 0.52081; 1.520281; 2.490160; 2.490160; and 0.490160. Results, 2.490160 = log. 309.1435 + ; 2.490160 = log. .030914 + ; 0.490160 = log. 3.091435 +. 131, As logarithms are largely used to facilitate numer- ical computations, it is important that the student be able to take any formula representing such operations and write at once the equivalent logarithmic operations. 390 LOGARITHMS. EXAMPLES. 1. If 28.035 : 3.2781 : : 3114.27 : x, what logarithmic op- erations will find X ? SvG. — The logarithm of the product of the means is the sum of their logarithms ; and the logarithm of the quotient of this product divided by the first extreme, is the logarithm of said product minus the logarithm of the other extreme. . • . log.iC = log. 3. 2781 -}- log. 3114.27 — log. 28.035 =0.515622 + 3.493356—1.447700 = 2.561278. Having a table sufficiently extended, the number corresponding to this logarithm could be found, and would be the value of x. 2. Find the product of 23.14, by 5.062, knowing that log. 23.14 is 1.364363, log. 5.062 is 0.704322, and log. 117.1347 is 2.068685. 3. How is 287 raised to the 5th power by means of loga- rithms ? How is the 5th root extracted ? 4. Extract the 5th root of 31152784.1 by means of loga- rithms. SuG.— Log. >y31 152784.1 = i log. 31152784.1 = 1.498699. The num- ber, therefore, is 31.52 -J-. 5. What is the cube root of 30? Ans., 3.107 +. 6. What is the cube root of .03 ? SuG.— Log. .03 = 2.477121. Now to divide this by 3, we have to bear in mind that the characteristic alone is negative, i. e., 2.477121 =— 2-1-.477121, or— 1.522879. This divided by 3 gives — .507626, or — . 507626 =1. 492374. But a more convenient method of effecting this division is to write for the — 2, — 3-f-l, whence we have foi 2.477121, —3 4-1.477121, which divided by 3 gives~l. 492374, nearly. 7. Divide 3.261453 by 2, by 4, by 5. Last quotient, 1.4522906. APPENDIX SECTION I. DIFFERENTIATION. [This subject is inserted as the best metliod of reaching the demonstration of the Binomial Formula and the production of the Logarithmic Series. While it is equally simple, to say the least, ■with the old method, it is more direct, and gives the student nothing but what is of fundamental importance in subsequent mathematical work.] 132, In certain classes of problems and discussions the quantities involved are distinguished as Constant and Variable. 133, A Constant quantity is one which maintains the same value throughout the same discussion, and is represented in the notation by one of the leading letters of the alphabet. 134, Variable quantities are such as may assume in the same discussion any value within certain limits deter- mined by the nature of the problem, and are represented by the final letters of the alphabet. III. — If X is the radius of a circle and y is its area, y = -nx^, as we learn from Geometry, tt being about 3.1416. Now if x, the radius, * The preceding part of this volume famishes a course in Algebra quite as full as will be found practicable or desirable in most high schools and academies, and is an adequate preparation for college. This appendix, selected from Olnet's Uni- versity Algebra, is inserted for such of tht- above schools as desire a fuller conrse, and as adapting the book to the needs of many of our colleges which do not tod it expedient to give as much time to this subject as is required to master the University Algebra. There is nothing in the ordinary college course wuiclx requireB more Algebra than is found in this volume. 392 DIFFERENTIATION. varies, y, the area, will vary ; but tt remains tlie same for all values of a; and y. In this case x and y are the variables, and it is a constant. Again, if y is the distance a body falls in time x, it is evident that the greater x is, the greater is y, i. e., that as x varies y varies. We learn from Physics that y = 16 j^^^. for comparatively small distances above the surface of the earth. In the expression y = lQ^^a^\ X and y are the variables, and IGyi^ is a constant. Once more, suppose we have y^ — 25iC^— 3a;'^— 5, as an expressed relation between x and y, and that this is the only relation which is required to exist between them ; it is evident that we may give values to x at pleasure, and thus obtain corresponding values for y. Thus if 2! = 1, 2^ = ± -v/l^» \ix = 2,y= ±/v/l83, etc., etc. In such a case X and y are called variables. But we notice that if we give to x such a value as to make 3a;2 + 5>25a;3 (as, for example, |, \, etc.), y will be imaginary. This is the kind of limitation referred to in our definition of variables. 135, ScH. — The pupil needs to guard against the notion that the terms constant and variable are synonyms for known and unknown, and the more so as the notation might lead him into this error. The quantities he has been accustomed to consider in Arithmetic and Elementary Algebra have all been constant. The distinction here made is a new one to him, and pertains to a new class of problems and discussions. 136, A Function is a quantity, or a mathematical expression, conceived as depending for its yalue upon some other quantity or quantities. III. — A man's wages for a given time is a function of the amount received per day, or, in general, his wages is a function of the time he works and the amount he receives per day. In the expression y = 1Qy2X^ {134), second illustration, y is a function of x, i. e., the space fallen through is a function of the time. The expression 2ax^—dx-\-5b, or any expression containing x, may be spoken of as a function of x. J37, When wc wish to indicate that one variable, as y, is a function of another, as x, and do not care to be more specific, we write y=f{x)y and read "?/ equals (or is) a DIFFEREI^-TIATION. 393 function of a:." This means nothing more than that y is equal to some expression containing the variable x, and which may contain any constants. If we wish to indicate several different expressions each of which contains x, we write/(a;), go [x), or/' (re), etc., and read " the/ function of a:," " the (f> function of x" or "the/' function of x." III. — The expression /(a?) may stand for a^— 2a? + 5, or for Z{a^—x'^), or for any expression containing x combined in any way with itself or with constants. But in the same discussion f{x) will mean the same thing throughout. So again, if in a particular discussion we have a certain expression containing x{e. g., dx^—ax + 2ah). it may be repre- sented by /(a-), while some other function of x, e. g., 5 {a^—x^) + 2x'^, might be represented by/' (x), or (p (x). 138. In equations expressing the relation betweeen two variables, as in y"^ = daa^—x^, it is customary to speak of one of the variables, as y, as a function of the other, x. Moreover, it is convenient to think of x as varying and thus producing change in y. When so considered, x is called the Independent and y the Dependeyit variable. Or we may speak of «/ as a function of the variable x. 139. An Infinitesimal is a quantity conceived under such a form, or law, as to be necessarily less than any assignable quantity. Infinitesimals are the increments by which continuous number, or quantity (8), may be conceived to change value, or grow. III. — Time affords a good illustration of continuous quantity, or number. Thus a period of time, as 5 hours, increases, or grows, to another period, as 7 hours, by infinitesimal increments, i. e., not by hours, minutes, or even seconds, but by elements which are less than any assignable quantity. 140. Consecutive Values of a variable are values which differ from each other by less than any assignable quantity, i. e., by an infinitesimal. Consecutive values of a 394 DIFFERENTIATIOlf. function are values which correspond to consecutive values of its variable. 141. A Differeiitictl of a function, or variable, is the difference between two consecutive states of the function, or variable. It is the same as an infinitesimal. III. — Resuming the illustration y — IQ^^x^ (134), let x be thought of as some particular period of time (as 5 seconds), and y as the dis- tance through which the body falls in that time. Also, let x' represent a period of time infinitesimally greater than x, and y' the distance through which the body falls in time x'. Then x and x' are consecu- tive values of x, and y and y' are consecutive values of y. Again, the difference between x and of, as x'—x, is a differential of the variable X, and y' —y is a differential of the function y. 142. Notation. — A differential of x is represented by writing the letter d before x, thus dx. Also, dy means, and is read "differential 2/." Caution. — Do not read dx by naming the letters as you do ax ; but read it " differential x." The d is not a factor, but an abbreviation for the word differential. 143. To Differentiate a function is to find an expression for the increment of the function due to an infinitesimal increment of the variable ; or it is the process of finding the relation between the infinitesimal increment of the variable and the corresponding increment of tlie function. RULES FOR DIFFERENTIATING. 144. Rule I. — To differentiate a single valuable, simply write the letter d before it. This is merely doing what the notation requires. Thus, if x and x' are consecutive states of the variable x, i. e., if x' is what x becomes when it has taken an infinitesimal increment, x'—x is the differential of X, and is to be written dx. In like manner, y' —y is to be written dy, y' and y being consecutive values. DIFFEREIJTIATIOK. 395 145, Rule U, — Constant factors or divisors ap- pear in the differential the same as in the function. Dem. — Let us take tlie function y = ax, in wliich a is any constant, integral or fractional. Let x take an infinitesimal increment dx, becoming x + dx; and let dy be the corresponding * increment of y, so that when x becomes x + dx^ y becomes y + dy. We then have 1st state of the function . . y =: ax; 2d, or consecutive state . . y + dy = a {x+dx) = ax+adx. Subtracting the 1st from the 2d dy = adx, which result being the diflference between two consecutive states of the function, is its differential {14:1). Now a appears in the differ- ential just as it was in the function. This would evidently be the same if a were a fraction, as — . We should then have, in like man- ner, dy = —dx2^ the differential ofy=—x. o. e. d. 146. Rule m. — Constant terms disappear in dif- ferentiating ; or the differential of a constant is 0. Dem. — Let us take the function y = ax + h, in. which a and h are constant. Let x take an infinitesimal increment and become x + dx ; and let dy be the increment which y takes in consequence of this change in x, so that when x becomes x+dx, y becomes y + dy. We then have 1st state of the function . . y = ax + h; 2d, or consecutive state . . y + dy = a{x+dx) + b = ax+ ado + h. Subtracting the 1st from the 2d dy = adx, which being the difference between two consecutive states of the function, is its differential {141). Now from this differential the constant b has disappeared. We may also say that as a constant retains the same value, there is no difference between its consecutive states (properly it has no con- secutive states). Hence the differential of a constant may be spoken of (though with some latitude) as 0. Q, E, d. * The word " contemporaneous " is often used in this connection. 396 DIFFERENT! ATIOlir. 14:7, Rule rV. — To differentiate the algebraic sum of several variables, diff^ereutlate each term sep- arately and connect the differentials with the same signs as the terms. Dem. — Let u = x + y—2, u representing the algebraic sum of the variables x, y, and —z. Then is du = dx + dy—dz. For let dx, dy, and dz be infinitesimal increments of x, y, and z ; and let du be the increment which u takes in consequence of the infinitesimal changes in X, y, and z. We then have 1st state of the function. . * . u = x-vy—z ; 2d, or consecutive state . . . . u + du = x + dx + y + dy—{z + dz). Or u + du = X + dx + y + dy —z—dz. Subtracting the 1st state from the 2d du = dx+dy—dz. Q. e. d. 148. Rule V. — Tlie differential of the product of two variables is the differential of the first into the second, plus the differential of the second into the first. Dem. — Let u ■= xy be the first state of the function. The consecu- tive state is u-\-dii = {x + dx){y+ dy) — xy+ ydx + xdy + dx ■ dy. Sub- tracting the 1st state from the consecutive state we have the difler- ential, i. e.,du = ydx + xdy + dx • dy. But, as dx • dy is the product of two infinitesimals, it is infinitely less than the other terms {ydx and xdy), and hence, having no value as compared with them, is to be dropped.* Therefore, du = ydx + xdy. Q. E. D. * It will doubtless appear to the pupil, at first, as if this j^ave a result only approximately correct. Such is not the fact. The result is absolutely correct. Ko error is introduced hy dropping dx ■ dy. In fact this term must be dropped accord- ins: to the nature of infinitesimals. Notice that by definition a quantity which is infinitesimal with respect to another is one which has no assic^able magnitude •with reference to that other. Hence we must so treat it in our reasoning. Now dxdy is an infinitesimal of an infinitesimal (i.e., two infinitesimals multiplied together), and hence is infinitesimal with reference to ydx and xdy, and must ba treated as having no assignable value with respect to them; that is, it must be dropped. DIFFEKEKTIATIOK. 397 149. Rule VI. — TJie differential of the product of several variables is the sum of the products of the differential of each into the product of all the others. Dem. — Let u = xyz; then du ■= yzdx + a:zdy-\-xydz. For the 1st state of the function is w = xyz, and the 2d, or consecutive state, u + du = (x + dx) iy^dy) iz + dz), or u-\-du = xyz-iryzdx + xzdy^-xydz + xdydz+ydxdz+zdxdy + dxdydz. Subtracting, and dropping all infini- tesimals of infinitesimals (see preceding rule and foot-note), we have du = yzdx + xzdy + xydz. In a similar manner the rule can be demonstrated for any number of variables. Q. E. d. ISO. Rule VII. — TJie differential of a fraction having a variable numerator and denominator is the differential of the numerator multiplied by the denominator, minus the differential of the denom- inator multiplied by the numerator, divided by the square of the denoTninator. Dem. — Let u = -; then is du = ~ -„ — - . For, clearing of frac- tions, yu = x. Differentiating this by Rule 5th, we have udy+ydu = X xdv dx. Substituting for u its value -, this becomes — ~ + ydu = dx. y y Finding the value of du, we have du = - — —^ — -. q. e. d. lol. Cor. — Tlie differential of a fraction having a con- stant numerator and a variaUe denominator is the jjroduct of the numerator ivith its sign changed into the differential f the denominator, divided by the square of the denom- inator. Let u = -. Differentiating this by the rule and calling the dif- n 398 DIFFEEEN^TIATION. f erential of the constant {a) 0, we have du = -^ = f- . Q. E. D. 152, SCH. — If the numerator is variable and the denominator constant, it falls under Rule 2. 158. Rule VIII.— I7ie differential of a variable affected with an exponent is the continued product of the exponent, the variahle with its exponent diminished by 1, and the diff^erential of the variable. Dem.— 1st. When the exponent is a positive integer. Let y = x"", m being a positive integer ; then dy—mx"'-^dx. For y=x'^=x .x.x .x. to m factors. Now, differentiating this by Rule 6, we have dy = (xxx . . to m—1 factors) dx+{xxx . . to m— 1 factors) 'a^+ etc., which is true for any value of x. Dividing by x, we obtain B+Cx + Dx^+ etc. = B' + Ox + D'x'^ + etc., likewise true for any value of x. Making a; — 0, B ■= B', as before. In this manner we may proceed, and show that =C',I) = B', etc. Q. E. D. 1S7. Con.— If A + Bx-|-Cx2 + Dx3-f- etc. = 0, is true for all values ofx, each of the coefficients A, B, C, etc., is 0. For we may write A + Bx + Cx^ + Dx^ + Ex^ + Fx"" + etc. = + 0.r + 0a;2 4-0.r3-f-0a;^-l-0x=+ etc. Whence by the proposition ^ = 0, B = Q, C=0, etc. 402 DEMONSTRATION OF THE BINOMIAL FORMULA. SECTION II. DEMONSTRATION OF THE BINOMIAL FORMULA. lo8. Theorem, — Letting x and y represent any quan- tities tvhatever (i. e. he variables) and m a7i2/ coiistant, / , n™ « . ml. mim—l) „, „ „ m(m—l){m — 2) Dem. — We may write {x + y)"" == x'"\\+ \ . Now put - = z and assume (l + z)"' = A + Bz+ Cz'' + I)z^ + Ez* + Fz'> + etc., (1) in which A, B, C, etc., are indeterminate coefficients independent of z [i. e. constants), and are to be determined. To determine these co- efficients we proceed as follows : Differentiating (1), we have m{\+ zy-Hz = Bdzi'2Czdz + ZDzHz + 4J]]z'clz + ^Fz^dz + etc. Dividing by dz, we have m{\ 4- z^-' = B + 2Cz + SBz^ + 4:Ez^ + 5Fz*+ etc. (3) Differentiating (2) and dividing by dz, we have m(7?j-l)(l+2)™-2 =2(7+3 • dBz + d - 4Ez' + 4: . 5Fz^ + etc. (3) Differentiating (3) and dividing by dz, we have w(m- l)(m-2)(l + 2)"^-3 = 2 . 3i) + 2 . 3 . 4^s + 3 . 4 . 5M^ + etc. (4) Differentiating (4) and dividing by dz, we have m(m-l){m-2){m-d)[l +2)'"-^ = 2.3. 4^+2 .3.4. 5^2+ etc. (5) Differentiating (5) and dividing by dz, we have m{m-l)(m-2){m-d)(m-i){l + z)"'-' = 2.S.4:.5F+ etc. (6) * This form is read " factorial S," " factorial 4," etc. ; and signifies the product of the natural numbers from 1 to 3, 1 to 4, etc. DEMONSTRATION^ OF THE BINOMIAL THEOREM. 403 We liave now gone far enough to enable us to determine the co- ^ efficients A, B, G, B, E, and F, and doubtless to determine the law of the series. As all the above equations are to be true for all values of 2, and as the coefficients A, B, G, etc., are constants, i. e., have the same values for one value of z as for another, if we can determine their values for one value of z, these will be their values in all cases. Now, making 2 = 0, we have from (I) J. = 1 ; from {^),B — m; from (3), (7 = — ^ - — - (the factor 1 being introduced into the denominator for the Bake of symmetry) ; from (4), B = — ^ J^^~~^ ; from (5), E = m{m- l)(m-2)(m-S) . ... ^ m(m-l)(m-2)(m-Sym-4) \i ' "^ ^^' ^^ j5 • These values substituted in (1) give m{m—l) 'm(m—l)(m—2) .. (1+2)"'= l+mz + 12 ' [3 m{m—l)(m—2)(m—S) ^ m {m-l) { m-2 ) (m-d){m -A) ^ + rj 2 + ._- 2 + etc. |4 |5 Finally, replacing 2 by its value - , we have X I A y\^ S -. y ^(^ — 1) y^ {x + yY—a^\\ + -\ =a;"M 1+w- + -^.3 ^—, \ X/ \ X I w X' m(m — l)im-2) y^ m(m—l )(m — 2)(m-S) 1/ "^ ^ x^"" |4 x^ m{m — 1) (77* — 2) {m — 3) (m — A)y^ ■^ [5 ¥ + ^*^- , m(m — 1) - w(77i — 1) (m — 2) , . l£ E ^ m{m-l){m-2){m-Z) ^^_,^ ^ m{m-l){m-2 ) (m - 8) (m - 4) ^^^^^ ^ ^^^ 159, CoR. 1. — TJie nth, or general term of the scries is m(m-l)(m-2) (^-^ + 2) ^_+, _., "T3i *^ y • 404 THE LOGAKITHMIC SERIES. For we observe that tlie last factor in tlie numerator of tlie co- efficient of any particular term is m — the number of the term less 2, i. e., for the ?ith term, m — {71 — 2), or m — n + 2; and the last factor in the denominator is the number of the term — 1, i. e., for the nth term, n— 1. The exponent of x in any particular term is m — the number of the term less 1, i. e., for the nth. term, m — (n — 1), or m — n + 1; and the exponent of y in any term is one less than the number of the term, i. e., for the 71th. term, n — 1. 160, Def. — In a series the Scale of Relation is the relation which exists between any term or set of terms and the next term or set of terms. 161, Cor. 2. — 77ie scale of relation in the hinomial series is I ^ ) ^' ^^^^^ t^^ ^i^ ^^^^^^ muUiplied by this produces the (n + l)th term. This is readily seen by inspecting the series, or by writing the (n + l)th term and dividing it by the nth. Thus, substituting in the general term as given above, n + 1 for n, we have in{m — 1) (m — 2) ----- (m — w -i- 1) Iji . as the (n -f- l)th term. This divided by the nth, or preceding term, — n + ly /m + 1 n X \ n J gives X SECT ION III. THE LOGARITHMIC SERIES. 162, The Modulus of a system of logarithms is a constant factor which depends upon the base of the system and characterizes the system. 163* ^rop, — The differential of the logarithm of a THE LOGARITHMIC SERIES. 405 number is the differential of the numler multiplied by the onodulus of the system, divided by the number ; ' Or, in the Napierian system, the modulus being 1, the differefitial of the logarithm of a number is the diff'erential of the number divided by the number. Dem. — Let X represent any number, i. e. he a variable, and nhe a, constant sucb that y = x\ Then log y = n log x {123). Differen- tiating y = nf; we liave dy = nx'^~'^dx ; whence x^'-^dx x^ ^ «/ , dx' W — dx -dx — Again, whatever the differentials of log y and log x are, n being a constant factor we shall have the differential of log y equal to 11 times the differential of log x, which may be written d (log y) = ,i'd (log x), whence n = j^~r^ (3.) Now equating the values of n as represented in (1) and (2), we have dy T?r- — T = I" • Whence d^(logw) bears the same ratio to — , as fZ(loga;) dx v & i// ^' X d (log x) does to — . Let m be this ratio. Then d (log y) = , and X y ,■ . mdx d{logx) = -^-. We are now to show that m is constant and depends on the base of the system. To do this, take y = zn' , from which we can find as above n' = ,,. ^ , = —■ . Now as m is the ratio of d (log v) to — , it is also the d(log z) dz \ ^ if/ ^ ^ 2 ratio of d{\og z) to — ; and d(\og z) = — - , Thus we see that in any z z case in the same system tbe same ratio exists between the differen- tial of the logaritlim of a number and the differential of the number divided by the number. Therefore wz is a constant factor. 406 THE LOGARITHMIC SERIES. Aerain — = w - indicates the relative rate of change of a loga- ^ dx X ritlim and its number. Now it is evident that the larger the base the slower the logarithm will change with reference to the number. (See examples under Art 117.) But the factor - varies inversely in the number ; hence m must vary with, or be a function of the base.* .7(>4. Frob. — To produce the logarithmic series. Solution. — The logarithmic series, which is the foundation of th^ usual method of computing logarithms, and of much of the theory of logarithms, is the development of log {1+x). To develop log (l + x), assume log{l + x) = A + Bx+Cx^ + I)x^ + m!^ + Fx^+ etc., (1) in which a; is a variable, and A, B, C, etc., are constants. Differentiating (1), we have ?^ = Bdx + 2Gxdx + ZDx'dx + 4Mi?dx + hFx'^dx + etc. 1 + a; Dividing by dx, ^ = B+2Ckc+^Di^+4^xf'-^^F^+ etc. (2) 1+x Differentiating (2), and dividing by dx, we have -m— J— -=2C+3.3Da;+3.4JS^2^.4.5j«7Tj.3+ etc. (3) (1 + xf Differentiating (3), and dividing by 2 and by dx, we have m ^ — rr3D + 3.4£'ic + 2-3.5i^a;2 + 'etc. (4) Differentiating (4), and dividing by 3 and dx, we have ~m - ^ — = 4F+ 4 • 5Fx + etc. (5) (1 + a;)4 * What the relation of the modulus to the base is, we are not now concerned to know ; it will be determined hereafter. t The number is 1+a; ; hence the differential is m times the differential of l+a? divided by the number 1+x. THE LOGAEITHMIC SEKIES. 407 Differentiating (5), and dividing by 4 and dx, we have m ^— , = 5i^+ etc * (6) (1 + xf We have now gone far enough to enable us to determine the coef- ficients A, B, C, D, E, and F, and these will probably reveal the law of the series. As all the above equations are to be true for all values of x, and as the coefficients A, B, C, etc., are constant, i. e., have the same values for one value of x as for another, if we can determine their values for one value of x, these will be their values in all cases. Now, making aj = 0, we have, from (1), ^ = log 1 = ; from (2), B = ni; from (3), G = -im ; from (4), I) = lm; from (5), E = —\m ; from (6), F=\m. These values substituted in (1) give /jj2 rr& qA nA log (1 + ic) = m (^ - - + - - ^ + ^ - etc), the law of which is evident. This is the Logarithmic Series, and should be fixed in memory. ScH. — The Napierian system of logarithms is characterized by the modulus being l{m= 1). Hence the Napierian logarithmic series is ,, ^ x^ xfi x^ a? ^ log(l + a") = ir-- + --^ + -- etc. 165, Cor. 1. — Tlie logarithms of tlie same oiumler 171 different systems are to each other as the moduli of those systems. This is evident from the general logarithmic series. Thus the logarithm of 1 + a; in a system whose modulus is m, is expressed />^2 /pZ /jA /*>5 log™ {\ + x)\ = m{x-^^ + --'^ + '-- etc.) ; and the logarithm of the same number in a system whose modulus is m! is expressed * Of course the student will o'bserve what forms the succeeding terms in this and the other similar cases would have. Thus here we should have bF+bQGx + 3 -5. 75a;' + etc. t The subscripts m and m' are used to distinguish between the systems, as log (1 + cc) is not the same in one system as in the other. Read logm (1 +«), " logarithm of 1 +aj in a system whose modulus is m," etc. 408 THE LOGARITHMIC SERIES. flj^ 0^ iC^ 0^ log^/(l + 2;) = m'(a;-- + --^ + -- etc.). Now, as the number (1 + x) is, by hypothesis, the same in both cases, X is the same. Hence, dividing one equation by the other, we have \o^m (1 + a;) _ m log,„/(l ■\-x)~ m' ' 166, Cor. 2. — Havifig the logarithm of a nwnber in the Napierian system, we have hut to multi2Jly it by the modulus of any other system to obtain the logarithm of the same nu?n- her in the latter system. Or, the logarithm of a number in any system divided by the logarithm of the same number in the Napierian system, gives the modulus of the former system. 167. JProb, — To ado2ot the Napierian logarithmic series to numerical computation so that it can be conveniently used for computing the logarithms of numbers, /y»2 /vi3 /v»4 /Tfl5 Dem. — That log (1 +a?) = a! — ^ + ^ — 7" + ^+ ®tc., is not in a .4 o 4 o practicable form for computing the logarithms of numbers will be evident if we make the attempt. Thus, suppose we wish to compute the logarithm of 3. Making X — 2, we have log (1 +2) = log 3 = 2— 22 2^ 2* 2^ , , , — + 5 — 7" + ^ — ®*^* ^ series m which the terms are growing ^ o 4 O larger and larger (a diverging series). We wish a series in which the terms will grow smaller as we extend it (a converging series). Then the farther we extend the series, the more nearly shall we approximate the logarithm sought. To obtain such a series, substitute —x for x in the Napierian loga- rithmic series, and we have x^ x^ x^ x^ log{l-x) = -x---------eto. Subtracting this from the former series, we have log (l + a;)-log {1—x) = log (r^) = % {x + l^? + \x^ + ^x'' -\- etc.). THE LOGAEITHMIC SEKIES. 409 Now put X = - — -. , whence 1 + x 2z + l 22 + 1 23 22 ^ , and 1±^ = l±f . Hence, as log (—-] = log (1 + s) - log g, 4-1 1— a; z \ z / substituting, and transposing, log(l + .)=Iog. + 2(^j + g^3 + 5^, + ^^, + etc, )(A) This series converges quite rapidly, especially for large values of s, and is convenient for use in computing logarithms. 168. I^rob. — To compute the Napierian logarithms of the natural mmiiers 1, 2, 3, 4, etc., ad liiitum. Solution. — In the first place we remark that it is only necessary to compute the logarithms of prime numbers, since the logarithm of a composite number is equal to the sum of the logarithms of its factors (121). Therefore beginning with 1, we know that log 1=0 (ScH., p. 382). To compute the logarithm of 2, make s = 1, in series (A), and we havelog(l + l)-logl=log2 = 2(l + -^-3+J-, + J:- + -'3,+ 1 1 1 + + etc. 11.811 13.313 15.81^ The numerical operations are conveniently performed as follows : 3 2.00000000 .66666667 1 .07407407 3 .00828045 5 .00091449 7 .00010161 9 .00001129 11 .00000125 13 .00000014 15 •. log 2 .66666667* .02469136 .00164609 .00013064 .00001129 .00000103 .00000009 .000000 01 .69314718 * * Though the decimal part of a logarithm is generally not exact, it is not cus- tomary to annex the + sign. 410 THE LOGAKITHMIC SERIES. Second. To find log 3, make z = 2, -whence log S = log 3 + 3 (1 + g^j.3 + 5^5 + Jg, + Jg5 + etc.). COMPUTATION. 5 2.00000000 25 .40000000 1 .40000000 25 .01600000 3 .00533333 25 .00064000 5 .00012800 25 .00002560 7 .00000366 .00000102 9 .00000011 .40546510 log 2 = .69314718 .-. log 3 = 1.09861228 Third. To find log 4. Log 4 = 2 log 2 = 2 X .69314718 = 1.38629436. FouKTH. To find log 5. Let cc = 4, whence log 5 = log 4 + 3 (i + gij5 + g-^g, + ^g-, t 6^.) . 9 COMPUTATION. 2.00000000 81 81 81 .22222222 .00274348 .00003387 .00000042 1 3 5 7 .22222222 .00091449 .00000677 .00000006 log 4 = .22314354 1.38629436 .-. logs = 1.60943790 Li like manner we may proceed to compute the logarithms of the prime numbers from the formula, and obtain those of the composite numbers on the principle that the logarithm of the product equals the sum of the logarithms of the factors. Thus, the Napierian logarithm of the base of the common system, 10, = log 5 + log 3 = 2.30358508. HIGHER EQUATIONS. 411 169, I^rop, — Tlie modulus of the common system is .43429448 + . Dem. — Since the logarithm of a number, in stoy system, divided bv the Napierian logarithm of the same number is equal to the modulus of that system {166), we have Com. log 10 , , , — r-^-T7^ = modulus of common system. Nap. log 10 But com. log 10 = 1, and Nap. log 10 = 2.30258508, as found above. Hence, Modulus of common system = — = .43429448 170. Prop.— The NajneriaJi base is 2.718281828. Dem.— Let e represent the base of the Napierian system. Then by (163) com. log e : Nap. log e : : .43429448 : 1. But the logarithm of the base of a system, taken in that system is 1, since a^ = a. Hence, Nap. log e = l, and com. log e = .43429448. Now finding from a table of common logarithms the number corres- ponding to the logarithm .43429448, we have e = 2.718281828. SECTION IV. HIGHER EQUATIONS. 171, Since every equation with one unknown quantity, and real and rational coefficients, can be transformed into one of the form X- + Ax"-^ 4- Bx-~^ + Cx^-^ L = 0, (1) this will be taken as tlie typical numerical equation whose solution we shall seek in this and the succeeding sections ; 412 HIGHER EQUATION-S. and we shall frequently represent it by/(^) = 0, read "function x equals 0." The notation /"(a;) signifies in general, as has been before explained {187), simply any expression inyolving x. Here we use it for this particular form of expression. We shall also use/' {x) as the symbol for the first differential coefficient of this function. 172. JProp, — When an equation is reduced to the form x"4-Ax"~^ + Bx"~'^ + Cx"~^ . . . .\i = ^,the roots, ivith their signs changed, are factors of the absolute {known) term, L. [For demonstration see p. 363-] Its. CoE. — If a is a root of f (s) = 0, f (x) is divisible by x—a] and, conversely, ift{x) is divisible by x—a, a is a rootofi{x)^0. Dem. — The first statement is demonstrated in tlie proposition, and tlie second is evident, since as /(;«) is divisible by x—n, let the quotient be ^(a?); whence (,»—«)(* (a-) - 0. Now x=a will satisfy this equation, since it renders x—a=:0, and does not render (p{x) infinity, since by hypothesis x does not occur in the denominator.* I 174. I*rop.—If the coefficie7its and absolute term in x" + Ax»-' + Bx"-' + Cx°-^ 1^ = 0, are all integers, the equation can have no fractional root. Dem. — Suppose in this equation a? = - ; .- being a simple fraction * t in its lowest terms. Substituting this value of x, we have i» ^—1 fn—l ^n— 3 * Could there be a term of the form — - in 6 {x\ x — a would render it oc, and x—a (X—a) <}> (x) would he X 00, which is indeterminate, since OxQo = Oxi = §. HIGHER EQUATIONS. 413 Multiplying by t"^^ we obtain on Now, by liypotliesis, all tlie terms except the first are integral, and the first is a simple fraction in its lowest terms, as by hypothesis s and t are prime to each other. But the sum of a simple fraction in its low- est terms and a series of integers cannot be 0. Therefore x cannot equal - , a fraction. 17i>» ScH. — This proposition does not preclude the possibility of surd roots in this form of equation. These are possible. Its. Prop.— An equation i{x) = {171) of the nth degi^ee, has n roots {if it has any), and no rnore. Dem. — Let a be a root of / (x) = 0, which is of the nth degree. Dividing /(.t) hjx — a {17 S), we have (p {x) = 0, an equation of the {n — l)th degree. Let 6 be a root of {x) = 0, and divide (p{x)'bj x — b {173). Call the quotient 0' {x), whence 0' {x) = 0, an equation of the {n — 2)th degree. In this way the degree of the equation can be diminished by division until, after n — 1 divisions, there results 0" {x) of the first degree, and the equation is x — I = 0. Therefore, f{x) = {x - a) (p{x) = (x — a){x — h) cp' {x) = (x — a) {x — b) (x — c) (f," (,r) = {x — a){x — b){x — c) {x — I) = ; i. e.,f{x) is resolvable into n factors, of the form x — m. Now, 2iS X = a, ov X = b, or X = any one of the quantities a, b, c . . . . I, will render /(a?) equal to 0, each one of these will satisfy the equation /(a;) = 0. Therefore this equation has n roots. Again, since it is evident that we have resolved /(.r) into its 2^rime factors with respect to x, there can be no other factor of the form x — m in/(.i'), hence no other root oif{x) = 0, and this whether m is equal to one or more of the roots a,b, c . . . . n, or not. Therefore f{x) = has only n roots. 1 77. Cor. l.—TIie polynomial x° + Ax"-' + Bx"-" + Cx"-^ L, or f (x), = (x — a) (x — b) (x — c) . . . . (x — 1), in which a, b, c .... 1 are the roots of i (x) = 0. 414^ HIGHER EQUATIONS. 178, Cor. 2. — The equation i{x) = may have 2, 3, or even n equal roots, as there is no inconsistency in su^oposing a = b, a = b = c, or a = b = c == . . . . 1, i^ the above demonstration. 179, Cor. 3. — Imaginary roots enter into equations having only real coefficients, in conjugate pairs ; that is, if f (x) = has only real coefficients, if it has one root of the form « 4- ^V—'^^it has another of the form a — §\/ — 1 ; or, if it has one of the form ^V—h it has another of the form —§ V— 1. This is evident, since only thus can / {x) = {x — a) {x — b) {x — c) .... (a? — n); that is, if one root, a for example, is a — 13^ — 1> there must be another of the form a + (3'\/ — 1, in order that the pro- duct of these two factors shall not involve an imaginary. Thus, [aj_(a + /3y'iri] X [x — (a — ^^^^l] = x' — ^ax + (a^ + /32),a real quantity. So also (a? — /3y^ — \){x + ii V — 1) = ^^ + /32, a real quan- tity. But if the assumed imaginary roots be not in conjugate pairs, the product of the factors {x — a) {x — b) {x — c) . . . .{x — I) will in- volve imaginaries. 180, Cor. 4. — Hence an equation of an odd degree must have at least one real root ; but an equation of an even degree does not necessarily have any real root. Cor. 5. — If an equation has a pair of imaginary roots, the Tcnoimi quantities entering into the equation may be so varied that the two imaginary roots shall first give place to tiuo equal roots, and then these to two unequal roots. As shown above, imaginary roots arise from real quadratic factors in/(^). Let x^—2ax + b be such a quadratic factor, whence x'^ — 2ax + b = satisfies /(;r) = 0, and a ± '\/a? — h are the corresponding roots otf{x) — 0. Now, if & < a^, these roots are imaginary. If, however, b diminishes or a increases (or both change thus together), when b =. a? the two imaginary roots disappear and we have in their place HIGHER EQUATIOITS. 415 two real roots, each a. If tlie same change in a and h continues, so that a? becomes greater than 6, the two real, equal roots in turn give place to two real, unequal roots. Now as a and 6 are functions of the known quantities of the equation /(.r) = 0, such changes are evidently- possible. 1S1» By means of the property exhibited in (187), produce the equations whose roots are given in the follow- ing examples : 1. Roots 1, — 3, 4. 2. Roots a/2, — V2, — 1, 3. 3. Roots 1, 2, 2, — 3, 4. 4. Roots — 3, 2 + V^^l, 2 — V^^' 6. Roots 3, — 2, — 2, — 2, 1. 6. Roots f , i - |. 7. Roots 1 ± V^^, 2 ± V^^. 8. Roots 1J-, 2, V3, — V^- 9. Roots V— 2, — V— 2, VS, — a/5. 10. Roots 10, — 13, I, 1. 11. Roots 3 — 2 a/3, 3 + 2^/3, 2 — 3 a/^^, 2 + 3 V~^l, 1, - 1. 182. Prop, — If the equation f (x) =0 has equal roots, tlie Mgliest common divisor o/f (x) a}id its differential coeffi- cient,"^ f (x), being ptU equal to 0, constitutes an equation luMch has for its roots these equal roots, and mo other roots.\ * The differential coefficient of a function is sometimes called its first derived polynomial, The student must not suppose that the roots off (or) = 0, and its first diflJeren- tial coefficient/' (a?) = 0, are necessarily alike. f'{x) — a series of terms some of which may be + and some — , and which may destroy each other, so as to render fix) 3= 0, for other values of x than such as render/ (a:) — 0, and not necessarily for any which r; {x — &)'•-'" = 0, when r > m ; or (x — a) (x — b) = 0, when m = r. From any one of these forms we can readily determine a root. HIGHER EQUATIOJ^S. 417 184, ^vop, — In an equation f (x) =0, f(x) ivill change sign luhen x passes through any real root, if there is iut one such root, or if there is an odd number of such roots ; lut if there is an eveis" numler of such roots, f (x) will not change sign. Let a, 1), c .... e be tlie roots of f{x) = 0, so that f((c) = {x—a) (x—h) {x—c) .... (x—e) = {It 7)' Ck)iiceive x to start with some value less tlian tlie least root, and continuously increase till it becomes greater tlian tlie greatest root. As long as x is less tlian tbe least root, all the factors x—a, x—b, etc., are negative; but when x passes the value of the least root, the sign of the factor containing that root will become +,* and if there is no other equal root, this factor will be the only one which will change sign. Hence the pro- duct of the factors will change sign. But if there is an even number of roots, each equal to this, an even number of factors will change sign ; whence there will be no change in the sign of the function. If, however, there is an odd number of equal roots, the passage of X through the value of this root will cause a change of sign in an odd number of factors, and hence will change the sign of the function. Finally, as it is evident that the signs of the factors, and hence of the function, will remain the same while x passes from one root to another, and in all cases changes or does not change as above when x passes through a root, the proposition is established. The following example will be found very instructive : x^ + 4x*— 14t;"-'— 17a;— 6 = 0. The least root of this equation is — 3. When x< —d,f{x) is — ; when x = —S,f{x) = 0; when x passes —3, increas- ing, /(a*) changes from — to +, and remains + till x = —1, when it becomes 0, and changes sign as x passes —1, iioticitJiStanding they are equal roots. But there is an odd number of such roots, viz., three. But in a^— 14c2 + 64a;— 96 = 0, two equal roots of which are 4, if we substitute 2 we get f{x) = —16, and substituting 5, f{x) = — 1, the function not changing sign, although a root has been passed. * Suppose c be the least root, and that c' is the next state of x greater than c ; then = 0. This is satisfied hy x = —a,ii the former is by a? = a. For, substituting —a for x, we have a^—Aa'^ + Ba^—Ca + D = 0, the same as in the first instance. 186. Cor. — Changing the signs of the terms containing the even potvers will ansiver equally ivell, since it amounts to the same thing ; and if we are careful to put the equation i7i the complete form, changing the signs of the alternate terms ivill accomplish the purpose. III. — The negative roots oi x^—lx + Q =0, are the positive roots of — a;^ + 7:c + 6 = 0, or of x^—lx—Q = (0 being considered an even exponent) ; or, writing the equation x^±{)x''-—lx + Q — 0, changing the signs of alternate terms, and then dropping the term with its coeffi- cient 0, we obtain the same result. in, the negative roots of x^ — *lx^ —^x^->rQx^—\^2x- + ^0?>x-- 240 = 0, are the positive roots of x^ + 1lx^-^x^-%x^-\Z2x:^—b^^x-2^<(i = 0, or of —x^-7x^ + 5x^ + 8x^ + 132.c- + 508^ + 240 = 0. 187, Frob. — To evaluate"^ f (x) for any particular value of X, as X ^ a, more expeditiously than hy direct sub- stitution. Solution.— As /(:?•) is of the form x'^ + Ax''^^ + Bx""-"^ + Cx''-^ . . L, let it be required to evaluate x'^ + Ax^ + Bx^ + Cx + I> for x = a. Write the detached coefficients as below, with a at the right in the fonn of a divisor ; thus * This means to find the value of. Thus, suppose we want to find the vahie of a?*— 5cK° + 2cc*— 3aj* +6a;^— a;~12, for x = 5. We might substitute 5 for x, of course, and accomplish the end. But there is a more expeditious way, as the solution of this problem will show. HIGHER EQUATIONS. 419 1 +A +B +G +D \a__ a a^ + Aa a^ + Aa^ + Ba a^ + Aa^+Ba'^ + Ca a + A a' + Aa + B a^ + Aa- + Ba + C a'^ + Aa^ + Ba- + Ca + B Having written tlie detaclied coefficients, and the quantity a for which f{.i') is to be evaluated as directed, multiply the first coefficient 1 hy a, write the result under the second, and add, giving a + A. Multiply this sum by a, write the product under the third coefficient B, and add, giving a^ + Aa + B. In like manner continue till all the coeffi- cients (including the absolute term, which is the coefficient of x^) have been used, and we obtain a* + Aa^ + Ba^ +Ca + B, which is the value otf{x) for x = a. Illustration.— To evaluate x^ —5x^ + 2x^ — dx^ + 6^ —a;— 12, for X = iy: 1 -5 + 2 -3 + 6 -1 -12 15 '5 10 35 205 1020 2 7 41 204 1008 Now 1008 is the value of x^—5x^ + 2x^—Sx^ + ex^—x—12,foTX = 5; and it is easy to see that much labor is saved by this process. We are now prepared for the solution of the following important practical problem : 188. ^Proh, — To find the commensuraUe roots of numer- ical higher equations. The solution of this problem we will illustrate by practical examples. EXAM PLES. 1. Find the commensurable roots of x^—2o!:^^16x^-{-^x^-{- 68.T + 48 = 0, if it has any. Solution. — By (17 d), if this equation has any commensurable roots they are integral ; — it can have no fractional roots. Again, by {172), the roots of this equation with their signs changed are factors of 48. Now, the integral factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Hence, if the equation has commensurable 420 HIGHER EQUATIONS. roots, tliey are some of these numbers, with either the + or — sign. We will, therefore, proceed to evaluate f{x){i. e., in this case x^ — 2«4-15«3 + 8a;-^ + 68a; + 48), f or a? = +1, x = -l,x = +2, x = -2, etc., by (i^ 7), as follows: 1 -2 -15 + 8 +68 +48 I +1 _i ni Zli? IL^ _J2 -1 -16 - 8 60 108 Hence we see that f or cc = + 1, f{x) = 108, and + 1 is not a root of f[x) = 0. Trying a; = — 1, we have -2 -15 + 8 + 68 + 48 -1 3 13 -20 -48 -3 -13 20 48 -15 + 8 + 68 + 48 1 + 3 -30 -44 + 48 -15 -23 24 96 Thus we see that for x = —l,f(x) = 0, and hence that —1 is a root of our equation. We might now divide /(.r) by x + 1 (173) and reduce the degree of the equation by unity. But it will be more expeditious to proceed with our trial. Let us therefore evaluate /(a^) for x = +2. Thus : 1 -3 _2 Hence for x = +2,f{x) = 96, and +3 is not a root. Trying a; — —3, we have 1 -3 -15 + 8 +68 +48 |_-3 -3 _8 _14 -44 -48 -4-7 23 24 Hence for x = —2, f{x) = 0, and —3 is a root. Trying x = +3, we have 1 -3 -15 + 8 +68 +48 |_+3 _3 _3 -36 -84 -48 1 -13 -28 -16 * Of course it is not necessary to retain the + sign, as we have done in the pre- ceding operations : it has been done simply for emphasis. HIGHER EQUATIONS. 421. Hence for x = +S, f(x) — 0, and +3 is a root. Trying x = —3, we have 1 -3 -15 +8 +68 + 48 |_--3 -3 15 -24 -133 -5 8 44-84 Hence for x = —S,f(x) = —84, and —3 is not a root. Trying a; = 4, we have 1 -3 -15 + 8 +68 +48 1^ 4 _8 -28 -80 -48 3 - 7 -30 -13 Hence for x = 4,/(x') = 0, and 4 is a root. We have now found four of the roots, viz., —1, —3, 3, and 4. Their product with their signs changed is 24. Hence, by {172) 48-f- 24 = 3 is the other root with its sign changed, i. e., there are tico roots —3. That our equation had equal roots could have been ascertained by the principle in {182) ; but as the process of finding the H. C. D. is tedious, it is generally best to avoid it in practice. to 11. Find the roots of the following : 2. a^—a^—39x^-{-24:X -{-180 = 0; 3. x^-{-5x^—9x—4:6 = 0; 4. a^-\-2x^—2dx—60 = 0; 5. a^—dx^—Ux^ + ^8x—32 = 0; 6. x^—Sx^-\-ldx—6 = 0; 7. cf^—llx^-\-lSx—S = 0;^ 8. x^—dx^-{-6a^—3x^—dx + 2 = 0; 9. a^—lSx^-\-6W—171x^-i-216x— 108 = 0; 10. a^—4:6x^—4.0x-\-84: = 0; 11. a^—3x^—da^ + 21x^—10x + 2i = 0. * In order to apply the process of evaluation, the coefficients of the missing powers must be supplied. Thus we have 1 + 0—11 + 18—8. 422 HIGHER EQUATIONS. 12 to 17. Apply the process for finding equal roots (192, 183) to the following: 12. a;3 + 8a;2 + 20a; + 16 = 0; 13. x^—x^—^x + VZ = 0; 14. a;3_5^2_8ic + 48=:0; 15. a;4_ii^2_|.i8^_8 z=0; 16. ic4_j_i3^3_f_33^2_j_31a;4.io=: 0; 17. aP—ldx^ + QW—l'ilx^+^lQx—1^^ = 0. 18 to 24. Having found all but two of the roots of each of the following by (187), reduce the equation to a quad- ratic by (Its), and from this quadratic Qnd the remaining roots : 18. x^—Qx^ + lOx—% = 0; 19. a;4_4a:3_8a;-}-32 = 0; 20. a^—Sx^-\-x-{-2 = 0; 21. x^—Qx^ + Ux— !& = ()', 22. Q^—l^T^^bW—Ux + A:^ = ;* 23. a;4_9^ + i7a;2_^27a:— 60 = 0; 24. 0:5— 42;^— 16:^3 _|_ii2a;2— 2082; + 128 — O; 25. 2x^—dxi-\-2x—^ = 0; t 26. 3a:3— 22;2— 62; + 4 = 0; 27. 8^:3— 262;2+lla; + 10 = 0; 28. x^—^x-{-^ = ; (Look out for equal roots.) 29. c^—6x^-\-9ix^—dx-]-4:i^ = 0. * Apply the method for finding equal roots. + We have a;'— '^je' + a;— - = 0. Put » = |i '^^^"^^ |^ ~ 2^ ^'^ '^k^~i ~^' or. or, 3 or y^—~-^y^ + Vy—-Y = 0- If now ^ = 2, we have y^—Sy^' + Ay—n = 0, which can be solved as before, for one value of y, and the equation then reduced to a quad- ratic and solved for the ( the values of x required. ratic and solved for the other values. Finally, remembering that x = ^y, we have INTERPRETATION OE EQUATIONS. 4^ SECTION V. DISCUSSION, OR INTERPRETATION, OF EQUATIONS. ISO. To jyiscusSf or Interpret, an Equation or an Algebraic Expression, is to determine its significance for the various values, absolute or relative, wbich may be attributed to the quantities entering into it, with special reference to noting any changes of value which give changes in the general significance. Such discussions may be divided into two classes: 1st. The discussion of equations or expressions with reference to their constants ; and 2d. The discussion of equations or expressions with reference to their variables. The following principles are of constant use in such dis- cussions : 190. Prop. — A fraction, when compared with a finite quantity, becomes : 1. Equal to 0, tvhen its numerator is and its denomina- tor finite, and ivhen its numerator is finite and its denomi- nator 00 . 2. Equal to oo , luhen its numerator is finite and its de- nominator 0, and when its numerator is oo and its denomi- nator finite, 3. It assumes an indeterminate form when numerator and denominator are loth 0, and when they are both oo .* Dem. — These facts appear when we consider that the value of a fraction depends upon the relative magnitudes of numerator and de- nominator. * By this it is meant that - and — may have a variety of values, not that they necessarily do have. 424 IKTERPEETATIOK OF EQUATIOKS. 1. Let a be any constant and x a variable, then tlie fraction - '' a diminishes as x diminishes, and becomes when x is 0. Again, the fraction ~ diminishes as x increases, and when x becomes oo, i. e., a greater than any assignable magnitude, - becomes less than any assignable magnitude or infinitesimal, and is to be regarded as in comparison with finite quantities. (See 139 and 14:8, Dem., and foot-note.) x 2. As X increases, the fraction - increases, and hence when x be- CL comes infinite, the value of the fraction is infinite. Also, as x dimin- a ishes, the value of - increases; hence, when x becomes infinitely small, or 0, the value of the fraction exceeds any assignable limits, and is therefore oo . X 3. Finally, if x and y are variables, - diminishes as x diminishes, and increases as y diminishes. What then does it become when x=0, and 2/ = 0? i. e., what is the value of x? Simple arithmetic would lead us to suppose that t, was absolutely indeterminate, i. e. , that it might have any value whatever assigned to it, for n = 5, since = 5x0 = 0; Q = ''', since = 7x0 = 0, etc. But a closer inspection will enable us to see that the symboi x is not necessarily indetermi- nate, or rather that the expression which takes this form for parti cu-' lar values of its components, has not necessarily an indefinite number of values for these values of its components. Thus, what the value X of - will be when x and y each diminish to will evidently depend y upon the relative values of x and y at first, and which diminishes the X faster. Suppose, for example, that y = ^x\ then = ^ . Now, y ox suppose X to diminish ; the denominator will diminish 5 times as fast as the numerator, and whatever the value of x, the value of the frac* tfc^TEEPIlETATION" OF EQUATIONS. 425 tion will be \. So if y = 7x, - = =- , wMch is | for any value of x. if •'*' X X ^ Hence, when x = 0, and v = 0, we liave - = ^ = -- = - ^ or ■^ ' y {) ox X X ^ x - = Tr = ^^ = - ^ or - = 7, = any other value depending upon the rel- X x> ative values of x and ?/. So, also, if a; = oo , and 2/ = oo , - — — ; but J > if ' y (X) ' a? 00 a; 1 if y = 6x, we have - = — z= ^r^ = ^. And so if y — lOa", we have a; 00 x 1 = — = ZTYT = T7i • Thus we see that the mere fact that numerator y 00 lOa* 10 and denominator become 0, or become oo , does not determine the value of the fraction, i. e., gives it an indeterminate form. 191, A Ileal J^iiniber or* Quantity is one which may be conceived as lying somewhere in the series of num- bers or quantities between — oo and + oo inclusive. III. — Thus, if we conceive a series of numbers varying both ways from 0, i. e., positively and negatively to oc , we have - 00 .... -4, -3, -2, -1, 0, +1, +2, +3, +4, . . . . + oo. Now a real number is one which may be conceived as situated somewhere within these limits ; it may be + , — , integral, fractional, commensurable, or incommensurable. Thus, +15624 and —15624 will evidently be found in this series. +^^ may be conceived as somewhere between + 5 and + 6, though its exact locality could not be fixed by the arithmetical conception of discontinuous number. So, also, —y- is somewhere between —5 and —6. Again, + ^^5 is some- where between + 2 and + 3, though, as above, we cannot locate it exactly by the arithmetical conception. The following Geometrical Illustration is more complete than the arithmetical. Thus, let two indefinite lines, as CD and AB, intersect (cross) each other, as at 0. Now let parallel, equidistant lines be drawn between them. Call the one at and + oo , in- clusive.* 192, An Imaginary JV^uniber or Quantity is one which cannot be conceived as lying anywhere between the limits of — oo and + oo , as explained above. The algebraic form of such a quantity is an expression involving an even root of a negative quantity.f EXAM PLES. 1. "What are the values of x and y in the expressions 1)' — 1) aV — a'l) , 7 ,, T , , X = , , if = -r- , when o = o and a and a are a — a a a; * For example, the student who is acquainted with the elements of geometry knows how to construct a line which is exactly equal to V^ (Geom., Part 1, 11<¥). This line he can locate between +2 and +-3, and also between —2 and —3, since |/5 is both + and — . + Transcendental functions afford other forms of imaginary expressions ; for example, sin-^ 2, sec^ >^, log (—120), log (— w), etc. But our limits forbid the con- sideration of the interpretation of imaginaries, except in the most restricted sense, as indicating incompatibility with the arithmetical sense of the problem. IKTEEPRETATIOI^ OF EQUATI0:N'S. 427 unequal ? When h =^b' and a =^ a' ? When a=z a' and h and b' are unequal ? What are the signs of x and y when by h' and a > «', the essential signs of a, a' , b, and b' being + ? When b> b' and a 4a; + 5 ; and for x nega- * It is to be observed that the relation a — — -, requires that a and a' have dif- ferent essential signs ; while the relation a' — a requires that they have the same essential signs. IKTERPKETATIOK OF EQUATIOiq-S. 429 tive, we have x'^ + 4x—5, which is positive when x- + Ax>5. The former inequality gives x^ — ix + 4 > 9, or x>5; and the latter gives x^ + 4x + 4:>9, or a; > 1. Hence for positive values of x greater than 5, y is real, and for negative values of x numerically greater than 1, y is real. The 4th inquiry is answered by this : y is imaginary for all values of a; between — 1 and +5. 5th. To ascertain what + values of X render y + , and what — , we observe that — {2x — 4) ± -y/a;^ — ix — 5 can only be + when the + sign of the radical part is taken and when y^x^ — 4a?— 5 > 2a; — 4. This gives x<2 ± ^/—B, i.e., an imaginary quantity. Hence y is never + for a; + . Taking the negative sign of the radical, we see that both parts of the value of y are — , and conse- quently y is real and negative for all + values of x which render y real, i. e., for values greater than 5. Finally, for x — we have ^ = 2a;4-4 ± 's/x^^^ — 5. Now when we take the + sign of the radi- cal, both parts are + ; hence this value of y is always plus. When we take the — sign of the radical, y is negative if 2a; + 4 < ^/x'^ + 4x — 5. But this gives a;< — 2 ± /y/ — 3. Hence y is never negative for any negative value of x. Therefore both values of y are positive and real for all negative values of x numerically greater than 1. 5 to 22. Discuss as above the values of y in the follow- ing ; i. e., 1st. Show how many values y has i?i gerieral, and whether they are equal or unequal; 2d. For what particular value or values of x, y has but one value ; 3d. For what values of x, y is real, and for what imaginary; 4th. For what values o^ x, y is +, and for what — ; 5th. Also determine what values of x render y infinite : 5. y^^2xy—2x^—4.y—x-^10 = 0;* 6. y^—2xy-{-2x^—2y + 2x = ; 7. y^-]-2xy + x^—Gy-^d = 0; 8. y^-{-2xy-\-Sx^—4:X = 0; 9. y'^—2xy-{-3x^-i-2y—4:X—3 = 0; 10. y^-\-2xy—3x^—4:X = 0; 11. y^—2xy-\-x^-{-x = 0; * In all cases solve the equation for y in the first place. In this example y = — x + 2 ± V3a;*— 3a;-^. 430 IKTERPRETATIOK OF EQUATIONS. 12. y'^—2xy + x^—4.y + x + 4. = {)', 13. y^—2xy + x^ + 2y + l = 0; 14. 2/2— 2ic2— 2^ + 6a;— 3 = ; 15. y'^—%xy—dx^-2y + 1x—l = 0; 16. y^—2xy—2 = 0; 17. y^—2xy-]-2y-{-^x—S = ; 18. 4?/2+4^2_|_2t/_3a; + 12 = ; 19. 3^2_8a;2 = 12 ; 20. 12?/2 + 4x2 = 20; 21. x^ + y^ = 16; 22. x^—y^ =z 20. 198, Arithmetical Interpretations of Nega- tive and Imaginary Solutions, 1. A is 20 years old, and B 16. When will A be twice as old as B ? SuG's.— We have 20 + a? = 2 (16 + a*) ; whence a; = — 12. The arith- metical interpretation of this result is that A will never be twice as old as B, but that he was twice as old 12 years ago, i. e,, when he was 8 and B 4. 2. A is « years old, and B b. When will A be w times as old as B? For ^>1 what are the possible relative values of a and & consistently with the arithmetical sense of the problem ? Interpret for a'>nb, a =z nh, a<^nb when w>l. Also for n=^l, aynb, ah, (2) when ab, x positive, which shows that the point at which they are together is at the right of B, i. e., in the direction which they are traveling. The time, tIoi" )> is positive, which shows that they are together after passing A and B, For ac. 432 Il!TTERPKETATION OF EQUATIONS. inTi T ho he J ^ ac ac When a = T), x = — - = - = oo , and. c + x = — -=■ == - = go a—h a—h wliicli indicates tliat they are never together. When c = 0. In this case x = = 0, and c + x = — - = 0, for a and 6 unequal, a—b a—o indicating that they are together when they are at A and B. This is evidently correct, since A and B coincide in this case. When a = b, X = — ^ = K > aiid c + x = -p:, which shows that they are always to- a—o gether, - being a symbol of indetermination which in this instance may have any value whatever, as we see from the nature of the problem. 194, ScH. — The student should not understand that the symbol - always indicates that the quantity which takes this form has an indefinite number of values. It is frequently so, but not necessarily. The indetermination may be only apparent, and what the value of the expression is must be determined from other considerations. The Calculus affords the most elegant general methods of evaluating such expressions. But the simple processes of Algebra will often suffice. Thus for x = l, T^^ =^, But zr^ = 1+x + x^, which, for x = l, 1—x 1—x 1—x^ is 3. Hence — '— = 3, for x = 1. Here the apparent indeterminar 1—x tion arises from the fact that the particular assumption (that a; = 1) causes the two quantities between which we wish the ratio, viz., the numerator and denominator, to disappear. Let the student find that ^-I^ =31 for x = l. (See also 190, 3d part of demon- 1—x+x^ — x^ stration.) 4. Two couriers starting at the same time from the two points A and B, c miles apart, travel foivard eacli other at the rates a and b respectively. Discuss the problem with reference to the place and time of meeting. (Consider when ayb, a 1, ^i < 1, ?z == 1, w = 0. 6. Divide 10 into two parts whose product shall be 40. Solution and Discussion. — Let x and y be the parts, then x -^ y = 10, xy = 40, and x = 5 ± ^ — 15, y = 5 t ^ — 15. These re- sults we find to be imaginary. This signifies that the problem in its arithmetical signification is impossible : this indeed is evident on the face of it. But, although impossible in the arithmetical sense, the values thus found do satisfy "Chq formal, or algebraic sense. Thus the 8um of 5 + /y/— 15 and 5 — /\/— 15 is 10, and the product 40. 7. The sum of two numbers is required to be a, and the product h : what is the maximum value of 5 which will ren- der the problem possible in the arithmetical sense ? What are the parts for this value of 5 ? 8. Divide a into two parts, such that the sum of their squares shall be a minimum, SuG's.— Let X and a — « be the parts, and m the minimum sum. Then a;2 + (a? x is imaginary. Hence the least value which we can have is 2m = a?, or m = \a\ 9. Divide a into two parts, such that the sum of the square roots shall be a maximum. 10. Let d be the diiference between two numbers: re- quired that the square of the greater divided by the less shall be a minimum. 434 Il^TERPRETATION OF EQUATIONS. 11. Let a and l be two numbers of which a is the greater, to find a number such that if a be added to this number, and h be subtracted from it, the product of this sum and this difference, divided by the square of the number, shall be a maximum. Sug's. — Let n be the number, and m the required maximum quo- x- i mi , XT j-^- n^ ■¥ {a — h)n — ah , tient. Then by the conditions ~ • = m, whence we find a — h ^a? + 2a& + &^ — ^ahm 2(1 - m) ^ 3(1 - m) From this we see that the greatest value which m can have and ren- -j-^ . This gives n = — ^r—. x = . iab ^ 2{1 — m) a — b , , . (a + bf mi . . a — b 2ab der n real is m = ^ . / . This gives n = — — ^ = 12. To find the point on a line passing through two lights at which the illumination will be the same from each light. Solution. — Let A and B be the two lights, and XY the line passing -e- D A B D through them. Let a be the intensity of the light A at a unit's dis- tance from it, 6 the intensity of B at a unit's distance from it, c the distance between the two lights, as AB, and x the distance of the point of equal illumination from the light A, as AD (or AD'). Then, as we learn from Physics that the illuminating effect of a light varies inversely as the square of the distance from it, we have for the illumination of the point D by light A — , and for the illumination of the same point by light B, ^ ^ - _ g . But by the conditions of the problem these efiects (c - xf are equal ; hence we have the equation to be discussed ; viz., ft _ b IKTERPRETATIOK OF EQUATIONS. 435 or . 1 = — 2:-_ ; or - = -^ ^^ — » or, finally, a; = c — ~- — , and x — - which are the values of x to be discussed. Discussion. — I. Let c he finite and > 0. 1. When a >'b,x = c —~ =- > t c, since — =.-^- a >b. This is as it should be, since for a > h the point of equal illumination will evidently be nearer to B than to A. Again, the other value of X gives x = c — = > c, since — = —is + and > 1, y^a — ^b ysja — y& when a >b. Hence we learn that there is a point beyond B, as at D', where the illumination is the same from each light. If we assume ^/a = 3y^&, AD = f c, and AD' = 2c. 2. It is evidently unnecessary to consider the case when a-s 2'ext-2^ooks, OOLTON'S NEW GEOGRAPHIES. The whole subject in Tivo Books, These hooks are the most simple, the most practical, and ifest adapted to tJie wants of the school-room of any yet published. I, Colton^s Keiv Introductory Gcof/rajyli^/, With entirely new Maps made especially for this book, on tlie most improved plan ; and elegantly Illustrated. IT, Coif on' s Common School Geography, With Thirty - six new Maps, made especially for this book, and drawn on a uniform system of scales. Elegantly Illustrated. This book is the best adapted to teaching the subject of Geog- raphy of any yet published. It is simple and comprehensive, and embraces just what the child should be taught, and nothing more. It also embraces the general principles of Physical Geog- raphy so far as they can bo taught to advantage in Common Schools. For those desiring to pursue the study of Physical Geography, re have prepared CoUon-s Physical Geography, One Vol. 2to. A very valuable book and fully illustrated. The Maps are compiled with the greatest care by Geo. W. Colton, and repre- sent the most remarkable and interesting features of Physical Geography clearly to the eye. The plan of ColtorCs GeograpJiy is the best I have ever pcen. It meets the exact wants of our Grammar Schools. The Bevieiv is unsurpassed in its tendency to make thorough and reliable scholars, I have learned more Geog- raphy that is practical and available during the short time we have used this work, than in all my life before, including ten years teaching by Mitchell's plan.— A. B. Hetwood, Pnn. Franklin Gram. School, Lmvell, Mass. So well satisfied have I been with these Geographies that I adopted them, and have procured their introduction into most of the schools in this county. James W. Thompson, A.M., Prin. of Centreville Academy, Maryland. Any of the above sent by mail, post-paid, on receipt of price. Sheldon d- Co???pa??J^'s Text-liooks HISTOEIES OF THE UNITED STATES. By Benson J. Lossing, author of " Field-Book of the Revolu- tion," " Illustrated Family History of the United States," &c. Lossiuf/'s Primary History, For Beginners. A charm- ing little book. Elegantly illustrated. 238 pages. Losshif/'s Oiiflhie History of the United States. One volume, 12mo. We invite the careful attention of teachers to some of its leading points. In elegance of apjiear- anee and copious illustrations, both by pictures and maps, we think it surpasses any book of the kind yet published. 1. The work is markedby tmcommon clearness of stntemeut. 2. The narrative is divided into SIX mSTINLT ruiiions, namely : Discoveries, Settlements, Colonies, The Hcvolution, Ihe Aation, and The Civil War and its consequences. 3. The work is arranged in short sentences, so that the substance of each may be easily comprehended. 4. The most imi^ortant events are indicated in the test by heavy- faced letter. 5. Full Qtiestions are framed for every verse. G. A J^ronouncing Vocabulary is furnished in foot-notes wherever required. 7. A Brief Synojisis of topics is given at the close of each section. S. An Outline History of IMPORTANT EVENTS is given at the close of every chajtter. 9. The work is x>^ofusely illustrated by Maps, Cliarts and Plans ex- planatory of the text, and by carefully-drawn pictures of objects and events. Lossing^ s School History, 383 pages. Containing the National Constitution, Declaration of Independence, Biographies of the Presidents, and Questions. This work is arranged in six chapters, each containing the record of an im- portant period. The First exhibits a sreneral view of the Aboriginal race who occupied the continent %\hen the Europeans came. The Second is a record of all the Discoveries and preparations lor settlement made by indi- viduals and governments. The Third delineates the progress of all the Setile- «