I LIBRARY OF THE University of California. GIFT OF Class # i /J a^-.^^^J^^M'Cz^^ \. Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/completealgebratOOsmitrich ECLECTIC EDUCATIONAL SERIES COMPLETE ALGEBRA TO ACCOMPANY KAY'S SEKIES OF MATHEMATICS BY * GEORGE W. SMITH Woodward High School, Cincinnati, Ohio Zbc lEclectic press VAN ANTWERP, BRAGG, AND COMPANY CINCINNATI AND NEW YORK 'Q; fr. Copyright, 1890, by VAN ANTWERP, BRAGG, AND COMPANY PREFACE. Thi8 work was commenced sixteen years ago at the earnest solici- tation of numerous teachers, who were dissatisfied with the text- books then in use. That they were not alone in their opinion is evidenced by the number of new treatises, or revisions of old ones, printed since that time, and now used in the schools of this country. The crudeness of even the best Algebras of a quarter-century ago was mainly owing to the fact that, as a rule, mathematicians neg- lected the elementary branches for the more attractive fields of Higher and Applied Mathematics; hence blunders and inconsistencies were allowed which otherwise would not have been tolerated. The wonderful progress made in the Natural Sciences, and the extended use of Algebra in the treatment of Geometrical Magnitudes, have finally called the attention of educators to the necessity of improving the elementary treatises, and more rigidly limiting the meaning of the signs. That this agitation comes none too soon is evident to every thoughtful teacher, and can be readily seen by auy one who compares the various text-books used in our schools. Note the fol- lowing inconsistencies: In some text-books now before me, 6 : 7 equals f; in others, 6 : 7 equals |. In some, 6 -f 4 X 2 = 20; in others, 6 -|- 4 X 2 = 14. Of course, the meaning and use of a sign depend upon agi'eement, but it is of extreme importance that we do agree in such matters. In the same work, too, statements incompatible with each other are made; thus, a -i-bc and a -i-b X c are said to have different values, and yet be and b Xc are, in all woi'ks, said to have one and the same meaning. Since a -h be and a -i- b X e differ only in She use of b Xc for be, it is plainly necessary that one or the other of these two statements be changed. One of the objects in writing this book is to urge the adoption of the following law for Numeri- cal Values; viz.,(l) Find the value of each term separately; thus, 6-f-4 X 2 = 6 -f- 8 = 14. (2) In finding the m,lue of a term, begin at the RIGHT and use the signs in their oi'der; thus, 6-f-4x2 = 6-r-8 = f. In other words, the jm'tion of the term to the left of the division sign is the DIVIDEND, and the part to the right is the divisor. This law (iii) 188989 PKEFACE. is at least as easy of application as the one in vogue in many American works, and has the merit of allowing no inconsistencies, besides per- mitting the use of the very important law that, under all circumstances, a6 and a X 6 have the same meaning. Owing to a complication of causes, the preparation of this work has been so long delayed that many of the improvements contem- plated when it was planned have been made since by others. The author hopes, however, that his work is of suflBcient value to warrant its being received favorably by the American public. The task of preparing an elementary treatise is far from being an easy one, and this difficulty is further enhanced by the immaturity of many beginners. For this reason most authors have prepared • their works in two or even three divisions : Part I. for beginners who are immature or badly prepared; Part II. for more advanced stu- dents. The opinion is rapidly gaining ground, however, that this method of studying Algebra consumes too much time and does not tend to produce the best results. In conformity with this opinion the present work is intended to combine in one volume all the prin- ciples a pupil needs from the time he begins the study of Algebra until he enters college, together with additional chapters for the benefit of such as do not have the advantages of a university educa- tion. Special care has been taken to adapt this treatise to the needs of those students who are without the aid of a teacher; hence every principle of importance is carefully explained and profusely illus- trated by examples. As in other text-books, the first chapter is devoted to definitions. This is partly done for convenience of reference, and partly because many teachers prefer that their pupils study the meanings of the various terms employed before proceeding to subsequent chapters. Some of the best teachers, however, believe that the definition should accompany the word when actually used; that the pupil should not be required to study the meaning of a term until he has some idea of the purpose of such study. For the benefit of this class, whenever a word defined in Chapter I. is employed thereafter, reference is made to the article in which the definition is found. This plan is pursued until it is assumed that the student has learned the meaning of each term used. The propriety of introducing at such an early stage the study of (iv) PREFACE. negative and fractional exponents may be questioned by many able teachers; and with weak pupils it is perhaps best to postpone the consideration of such topics to a later period. The sequence of subjects as herein contained has been carefully tested, with results which are eminently satisfactory; but some teach- ers may prefer to change the order of the topics. Thus, as mentioned above. Fractional and Negative Exponents may be treated many chapters later; Powers and Roots may also be considered later, whilst many of the principles of Factoring may be introduced immediately after Division, etc. The arrangement of the topics being to a con- siderable extent arbitrary, an effort has been made to render each chapter as nearly independent as possible, so that the order in which certain subjects are taken may be varied at the option of the in- structor. Throughout this work it is assumed that the teacher would rather omit than mpply, and would rather postpone topics for future consideration than be compelled to introduce them or bring them forward. For reasons specified previously, the subject of Numerical Values has received unusual care and attention, and the same may be said of Factoring and Equations. The introduction of Equations so early and so often is due to the growing desire among teachers to cultivate the reason of the pupil at the same time that he is studying the mere mechanical parts of the Algebra. The subject of Detached Co- efficients, also, is treated more fully than in other elementary works, and the use of this method of work is shown in the treatment of subjects involving division (the Greatest Common Measure, for in- stance). The time seems to be rapidly approaching when this plan will supplant the longer and more cumbrous w^ays now in vogue. In the preparation of this work it was thought best not to dis- pense with rules, but the student is recommended not to commit them to memory. They are intended merely as guides to the pupil, and to aid him in stating the processes employed in solving the various problems. In writing these rules, no attempt has been made to secure conciseness or elegance at the expense of clearness, the object being to state the successive steps in the solution. Many teachers prefer that the examples submitted for ordinary class work shall be accompanied by the answers; others prefer problems unaccompanied by answers; a third glass adopts a middle (V) PREFACE. course, by furnishing the results until the principle considered is understood, and then leaving the pupil unaided, to apply the principle. In this work the more difficult problems are worked out, then similar problems with their answers are given, and finally at the end of each chapter an " exercise" is appended, including examples unaccom- panied by answers. It is hoped that a sufficient number of examples is submitted to satisfy each and every requirement. Only the strongest and best pupils are expected to solve all the problems, and it is well for the student to remember that quality, not quantity, is the test of excellence, and that one solution thoroughly understood is better than a dozen learned by rote. An examination of this work will show that all complicated problems have been carefully excluded, and that none are admitted which are too difficult for studious pupils of medium ability. As far as practicable, Ihe examples are classified, the simplest of each class, as well as the simplest classes, preceding the more difficult. The matter contained in this work is to some extent original, but the best English and American works have been freely consulted, the chief sources of supply being Colenso's Algebra of 1849 and Todhunter's Algebra of 1863. Care has been taken not to trespass upon the works of recent American authors; but whenever a propo- sition occurs in three or more books, it has been considered common property, and treated accordingly: certain propositions not coming under this rule are also included, because they were in manuscript, and in the hands of the pupils of Woodward, previous to their publi- cation elsewhere. If the student has a fair knowledge of Arithm^etic before beginning the study of Algebra, he can easily accomplish the wJiole woi^k of this text-book in three hundred recitations of one hour each. If but one school year be devoted to the subject, it is advisable to omit portions of these subjects; Numerical Values, Negative Numbers, Indeter- minate Equations, Maxima, and Minima, Factoring, Variation, Probabilities, Binomial Theorem, and Annuities. It might be well, also, to learn but one method of Multiplication, Division, and solving Quadratics, and to exclude Equations above the second degree. Any amendments or corrections will be thankfully received. G. W. Smith. Woodward High School, January, 1890. (vi) CONTENTS. CHAPTER PAGE I. Definitions, 11 Quantity and Number 11 Symbols of Numbers, 13 Principal Signs, 13 Factors, Powers, and Roots, 15 Symbols of Aggregation— Miuor>Symbols, ... 18 Algebraic Expressions, 19 Axioms, 22 Exercise One — Problems for Solution 30 II. Negative Numbers, 31 Exercise Two— Problems for Solution, .... 37 III. Numerical Values 39 Exercise Three— Problems for Solution, ... 42 Specimen Paper, 43 IV. Reduction of Terms, 44 Exercise Four— Problems for Solution, .... 46 V. Addition; Subtraction; Brackets, 47 Addition, 47 Subtraction 47 Brackets, 49 Exercise Five — Problems for Solution 51 VI. Multiplication 52 By Detached Co-efficients, 58 Exercise Six— Problems for Solution 61 VII. Involution 62 Monomials, 62 Binomials, 63 Exercise Seven— Problems for Solution, ... 66 (vii) Vlll CONTENTS. CHAPTRR PAGE VIII. Division 67 A Monomial by a Monomial 68 A Polynomial by a Monomial, 68 A Polynomial by a Polynomial, ....... 69 Synthetic Division, 73 Exercise Eight— Problems for Solution, . = . . 76 IX. Evolution, 77 Monomials, 77 Square Roots of Polynomials . 78 Square Roots of Arithmetical Numbers, ... 82 Cube Roots of Polynomials 85 Cube Roots of Arithmetical Numbers 87 Higher Roots, 90 Exercise Nine — Problems for Solution, .... 91 X. Simple Equations— One Unknown Quantity, . 92 Exercise Ten — Problems for Solution, .... 99 XI. Simultaneous Equations of the First Degree, 102 Elimination by Addition or Subtraction, . . . 103 Elimination by Comparison, 103 Elimination by Substitution, 104 Three or more Unknown Quantities, 106 Exercise Eleven — Problems for Solution, . . . 108 XII. Elementary Quadratic Equations, 110 Affected Quadratic Equations, 112 Exercise Twelve — Problems for Solution, . . . 113 Specimen Paper, 114 XIII. Factoring, 115 General Remarks on Factoring 132 Exercise Thirteen — Problems for Solution, . . . 134 XIV. Common Factors and Multiples, 135 Highest Common Factor, .135 Lowest Common Multiple, 143 Exercise Fourteen — Problems for Solution, . . 146 XV. Fractions, 147 Reduction of Fractions 148 Addition and Subtraction of Fractions 155 Multiplication of Fractious, 157 CONTEXTS. IX PAGE Division of Fractions 158 Miscellaneous Propositions, 161 Exercise Fifteen — Problems for Solution, . . . 163 XVI. Simple Equations — Continued from Gliap. X, . . 165 Exercise Sixteen— Protlems for Solution, . . . 168 XVII. Problems Involving Simple Equations, . . . 170 Two or more Unknown Quantities, 174 Exercise Seventeen— Problems for Solution, . . 176 XVIII. Indeterminate Simple Equations, 180 Exercise Eighteen— Problems for Solution, . . 183 XIX. Radical Expressions, 185 Imaginary Expressions, 194 Square Root of a Binomial Surd, 195 Equations Involving Radicals 199 Exercise Nineteen— Problems for Solution, . . 202 XX. Quadratic Equations— C(97i or <. Thus, 7 > 4 is read, seven is greater than four; and 4 < 7 is read, four is less than seven, 19. The Symbols of Operation are the sign of addition, + ; the sign of subtraction, — ; the sign of multiplication, X ; the sign of division, -?- ; the signs of involution, ^, ^, *, etc.; the signs of evolution, |/, |/, |/, \/, etc. These signs have the same meanings in Algebra as in Arithmetic, but -|- and -- are also used as signs of opposition. See Chapter on Negative Numlers, Art. 78. 20. The Sign of Addition, + (read plus, meaning more), signifies that the number to which it is prefixed is to be added. 21. The Sign of Subtraction, — (read minus, meaning less), signifies that the number to which it is prefixed is to be subtracted. 22. The Double Sign, ± or =f , signifies that the number to which it is prefixed is to be both added and subtracted. Thus, 5 ± 3 is equivalent to the two statements 5 -|- 3 = 8 and 5 — 3 = 2; also, 6 =F 5 is equivalent to 6 — 5 = 1 and 6 + 5 = 11. The symbol ± is read plus or minus, and the symbol qp is read minus or plus. The upper sign shows which operation is to be performed first. 23. The Sign of Multiplication, X (read times, into, or multiplied by), signifies that the numbers between which it is placed are to be multiplied together. These numbers are caWq^l factors, and the result is called the product. Multiplication may also be expressed by placing a period (. ) between the factors. Thus, 2.3.4 = 2x3x4;6.8.a.6 = 6x8Xfl^X^, etc. The period is not used as a sign of multiplication whenever it can.be mistaken for a decimal point. For example, 6.4 means 6^, and not 6x4. DEFINITIONS. 15 When not more than one of the factors is expressed arithmetically, multiplication is commonly denoted by writ- ing the factors in close succession. Thus, a X b X c = abc; 4oabc = 4:6xaXbXc;8x6XaXbxc = 8x Qabc; etc. 24. The Sign of Division, -^ (read divided by, or simply by), signifies that the number which precedes it, called the dividefid, is to be divided by the number which follows it, called the divisor. _ The result is called the quotient. Division may be expressed by writing the dividend above a horizontal line, and the divisor below the line; thus, 24 -^ 6 = 2^*. The sign of a ratio, : (read is to), is also used as a sign of division; thus, 18 -^ 6, Y-, 1^ • ^j have the same meaning, namely, that 18 is to be divided by 6. FACTORS, POWERS, AND ROOTS. 25. The Factors of a number are those numbers whose product is that number. Thus, 24 may be separated into the following sets of factors: 24xl;12x2;8x3;6x4; 6X2X2; 4X3X2; 2X2X2X3. Similarly, 6abc = 2 .3 . a.b. c = 2 . 3abc = 2a . Zbc, etc. 26. Factors expressed by letters are called literal factors. Factors expressed by figures are called numerical factors. 27. The Co-efficient.— When two factors form a product, either of the two is the co-efficient or co-factor of the other. In the product Qabc, 2a is the co-efficient of ?>bc; 3b is the co-efficient of 2ac; 6 is the co-efficient of abc; 1 is the co- efficient of 6abc; 6c is the co-efficient of abj etc. 28. The word co-efficient is frequently used in the sense of numerical co-efficient. Thus, in Gabc the numerical co- efficient is 6. When no numerical co-efficient is expressed, 1 is understood. Thus, abc has the same meaning as labc. 16 ALGEBKA.. 29. When the co-effioient of a factor is an integer, that factor is called a divisor or 7neasnre of the product, and the product is called a multiple of the factor. For instance, 6 X i = 3. Here ^ is a measure of 3, and 3 is a multiple of -J, but 6 is 'iiot a divisor of 3. When both factors are integers, each is a measure of the product, and the product is a multiple of either. 30. When the product of two factors is unity, each factor is called the reciprocal of the other. Thus, | and | are reciprocals. Hence, the reciprocal of a number is unity divided by that number. The reciprocal of « is 1 -r- a; of 3 is i; of 1 is 1; of | is f ; of i is 2; of 2i is |; etc. 31. When all the factors of a product are equal, the product is called a power of one of those factors, and each factor is called a root of the product. Thus, 5 x 5 is the second power of 5, and 5 is the second root of 25 ; 5x5x5 is the third power of 5, and 5 is the third root of 5 X 5 x 5. Similarly, 2 is the fourth root of 16, and 16 is the fourth power of 2. In like manner, aa is the second power of a, aaa is the third power of a, aaaa is the fourth power of a, etc. ; a is the second root of aa, the third root of aaa, the fourth root of aaaa, etc. Every number is the first power of itself, also the first root of itself. It is evident that the third power of the third root of any number is that number; that the fourth power of the fourth root of any number is that number, etc. 32. The Exponent is a small figure or letter placed above a number and a little to its right, to show how many times the number is taken as a factor. Thus, a^ = a; a^ = aa; a^ z= aaa; a^b = aab; ab^ = abb; c^W = aaabb; etc. a^ is read, the second power of a, or the square of a, or a squared; a^ is read, the third power of a, or the cube of a, or a cubed; a^ is read, the fourth power of a, or a to the fourth power, or a to the fourth; etc. DEFINITIONS. 17 33. The meaning of co-efficient must be carefully distin- guished from that of exponent. Thus, 4:a = a -\- a -\- a -{- a; here 4 is the co-efficient. a^ = aXaXaXa; here 4 is the exponent. If « = 3, 4« = 4 X 3 = 12; whilst a* = 3 X 3x 3 X 3 = 81. 34. The exponent affects only the factor above which it is placed. Thus, ab^ = a . b . b . b. If « = 3 and 6 = 3, 6?^>3 = 3 X 2* = 3 X 8 = 24. 35. To extract the square root of a number means to find a factor whose square or second power is equal to that number. To extract the cube root of a number means to find a factor whose cube or third power is equal to that number. To extract the fourth root of a number means to find a factor whose fourth power is equal to that number. To extract the nth. root of any nuniber means to find a factor whose nth power is equal to that number. 36. The Radical Sign, 4/, when placed before a number, denotes that a root of that number is to be extracted. The index is a small figure or letter placed over the radical sign to denote what root is to be extracted. When no index is written, 2 is understood. Thus, the square root of a is denoted by Va or by ^i^j the cube root of a is denoted by Wa; the fourth root of a is denoted by Va; etc. If « = 64, Va = 8, because 8* = 64; Va = 4, because 4^ = 64; Va = 2, because 2" = 64; etc. 37. The root of a number is also denoted by the frac- tional exponeiit. Thus, a^ denotes the square root of a; a^ denotes the cube root of a; a^ denotes the cube root of a^; a^ denotes the fourth root of a^; etc. Hence, the numera- tor of the fractional exponent denotes the power, and the denominator denotes the root. For example, 8* = 4, be- cause the cube root of 8 is 2, and the square of 2 is 4. Alg.-2. 18 ALGEBRA. Similarly, 64-^" = 16, because the cube root of 64 is 4, and 4^ = 16, etc. SYMBOLS OF AGGREGATION^— MINOR SYMBOLS. 38. The Symbols of Aggregation, the hracJcets [], the parenthesis (), the braces { [, the vinculum , the bar \ , denote that all the numbers inclosed are to be treated as a single number. Thus, [6 + 4 - 7] X 5 - 8, (6 + 4 - 7) X 5 - 8, {6 + 4 - 7} X 5 - 8, 6 + 4-7 X 5 - 8, 6 X 5 — 8, have the same meaning; namely, 6-1-4 — 7 4-4 is to be taken as a single number, this result — 7! is to be multiplied by 5, and from the product 8 is to be taken. Solution: (1) 6-^4-7 = 3; (2) 3x5 = 15; (3) 15-8^=7. A^is. 7. Similarly, 6x54-2 x3 = 30 4-6 = 36. (6 X 5 4- 2) X 3 = (30 4- 2) X 3 = 32 X 3 = 96. 6X(5 4-2)X3 = 6X7X 3 = 126. 6 X (5 4- 2 X 3) = 6 X (5 4- 6) = 6 X 11 = 66. ^ ... 4-^2x3 4 4-6 10 o T+ -iiT In like manner, — — = — - — = --- = 2. It will be /i —\- O observed that the line separating the numerator from the denominator is a vinculum, used in a special sense. 39. The Symbols of Continuation are dots, . . ., or dashes, , and are read, aiid so on, 40. The Symbol of Deduction is .*. (read hence, therefore, or consequently). 41. The Symbol of Reason, •.-, is read since or because. 42. The Symbol ~ is sometimes used to denote the dif- ference between two numbers. Thus, a^Z* is equal to a — b or to 6 — a, according as a is greater than b or less than b. DEFINITION'S. 19 EXAMPLES. 1. Give the simplest factors of ISa^Wc, Ans. 2, 3, 3, a, a, h, h, l, c, 2. In 18a^% what is the co-efficient of 9b ? Ans. 2a^b^c. 3. Write the reciprocal of aj a — b; 1^. Ans. -; t; ^. a a — b 5 4. Give the second, third, and fifth powers of 3. Ans. 9; 27; 243. 5. Give tlie second, third, and sixth roots of 64. Ans. 8; 4; 2. 6. Give 2 the exponent 3; and 3 the exponent 2. Ans. 2^ = 8; 3^= 9. 7. From 3 with an exponent 5 take 3 with a co-efficient 5. Ans. S'^ - 5 X 3 = 243 - 15 = 228. 8. [2 xG-3]x(5-2)=:? Ans. [12 - 3]x (3) = 27. 9. 55 X 4-2 X 3}X2 x5-4=? Ans. j 20 - 6 } X 10 - 4 = 14 X 6 = 84. 10. Find the value of 16*; 8^; 9i Ans, 8; 16; 3. 11. Find the value of ba^ when a = 16. Ans. 20. 12. Simplify a^^b, when a = 3* and b = 4^ Ans. 81-64 = 17. ALGEBRAIC EXPRESSIONS. 43. An Algebraic Expression is a collection of letters, figures, and signs used to denote a number. Thus, Ga is the algebraic expression for six times the number denoted by a. 20 ALGEBRA. 44. A Term is an algebraic expression not connected wifh any other by the sign + or — , or it is any one of the parts of an expression which are so connected. Thus, 3a -^ be is a term; a^ and + be are the terms of a^ -\- be. 45. A Positive Term is a term which has the sign + pre- fixed to it. Thus, + a^bc is a positive term. When the first term of an expression is positive, its sign need not be expressed. Thus, a^ + ^^ ^^^ the same mean- ing as -|- ^^ + ^^• 46. A Negative Term is a term which has the sign -- prefixed to it. Thus, in ab — 3e, — 36' is a negative term. 47. A Simple Term is a single expression not having parts separated by + or — . Example: 3ab -^ c. 48. A Compound Term is a collection of terms united by one or more of the signs of aggregation. For example, each of the following is a compound term: a(b -\- c); (ab + c); {aJr[b-c]\; {a -}-[b -^ {a - d)]}. 49. Similar or Like Terms are terms which have the same letters and the corresponding letters aifected by the same exponents. Thus, da^c^ and — 5a^c^ are like terms; but 3a^bc^ and 5aW are unlike terms. 60. The Dimensions of a term are its literal factors. 61. The Degree of a term is equal to the number of its dimensions, and is determined by taking the sum of the exponents of its literal factors. Thus, ba^c^ is of the sixth degree, because the exponent of « is 2, of ^ is 1, and of c is 3, and the sum of these exponents is 6. 62. A Monomial is an expression which contains a single term, simple or compound; as, ba^b -^ c or 7 [a -\- b]. 63. A Polynomial or Compound Expression is an expres- sion which contains two or more terms. 64. A Binomial is a polynomial of two terms. . DEFlNITIOIfS. 21 65. A Trinomial is a polynomial of three terms. 56. The terms of a polynomial may be written in any order if the sign of each term be retained. Thus, a-\-b — c = b-\-a — c=b— c-{-a = a — c-{-b, etc. 57. A Homogeneous Polynomial is a polynomial whose terms are all of the same degree. Thus, Qx^ — 3a^y + 4a;y is homogeneous, for each term is of the fourth degree. 58. A polynomial is arranged according to the descend- ing powers of some letter when the value of the exponent of that letter in each term is less than that of the preceding term. A polynomial is arranged according to the ascending powers of some letter when the value of the exponent of tliat letter in each term is greater than that of the pre- ceding term. Thus, x*" — ax^ -\- b7? — c is arranged accord- ing to the descending powers of rr, and a — bx -\- c:)^ — ^ is arranged according to the ascending powers of x, 59. The Numerical Value of an algebraic expression is the number obtained by giving a particular value to each letter, and then performing the operations indicated. Thus, if « = 5, * = 4, and c = 2, a'c — Z'c + «^> = 5 x 2 -4X2+5X4 = 10 -8 + 20 = 22. 60. A Formula is an expression of a mathematical truth by means of symbols. 61. A Rule is a concise statement of the method of find- ing one side of a formula when the other side is known. EXAMPLES. 1. How many terms in 2« + 33 X c — d^f Arts. ^. 2. What is the degree of 2^^a^b'^c9 Ans. Seventh. 3. What is the value of aP when a = 3, b = 2? A7is. 24. 4. 6'*+4 X 2 = ? " A?is. 44. 22 ALGEBRA. 6. 9^ X 8t X 16^ = ? ^^s. 3 X 4 X 8 = 96. Q. In how many ways may 2^^ — 3^ + ^ ^^ written ? Ans. 6. 1. Write twice the square of the cube root of a. __ A ns. 2a^ or 2 Va^. 8. From the cube of l take five times the square of a. Alls. ¥ — ba\ 9. Indicate the {a + b) power of the fifth root of (a — b). a + b a + b Ans. {a— I) 5 or [Va — b] 10. Indicate the a root of the cube of ^. 3 ^ Ans. (iYov V(^)3. 62. The principal Axioms employed in algebra are the following: I. Numbers which are equal to the same number are equal to each other. II. If equal numbers be added to equal numbers, the sums will be equal. III. If equal numbers be subtracted from equal numbers, the remainders will be equal. IV. If equal numbers be multiplied by equal numbers, the products will be equal. V. If equal numbers be divided by equal numbers, the quotients will be equal. VI. If the same number be both added to and subtracted from another, the value of the latter will not be altered. VII. If a number be both multiplied and divided by another, the value of the former will not be altered. VIII. If equal numbers be added to unequal numbers, the sums will be unequal. DEFIN^ITIONS. 23 IX. If equal numbers be subtracted from unequal num- bers, the remainders will be unequal. X. If unequal numbers be multiplied by equal numbers, the products will be unequal. XL If unequal numbers be divided by equal numbers, the quotients will be unequal. XII. If unequal numbers be added to unequal numbers, the greater to the greater and the less to the less, the sums will be unequal. XIII. If a be greater than ^, and h greater than c, then a will be greater than c. XIV. Like powers of equal numbers are equal. XV. Like roots of equal numbers are equal. XVI. A number may be substituted for an equal number in any expression. 63. The following principles, derived directly from the arithmetical definitions of Addition, Subtraction, Multipli- cation, and Division, though not strictly self-evident, mere- ly require illustration to be perfectly understood: I. Adding two or more numbers is equivalent to adding their sum. 4 + 6 + 4 + 5 = 4+ (6 + 4 + 5) =4 + 15 = 19. II. Subtracting two or more numbers is equivalent to subtracting their sum. 17 — 6 — 4 — 3 = 17— (6 + 4 + 3) = 17 - 13 = 4. III. If any number be diminished by an equal number, the remainder is zero (0). 6 — 6 = 0. IV. Adding any number and then subtracting a less number is equivalent to adding their difference. 7 + 8 — 5 = 7 + 3 = 10. ^ 24 ALGEBRA. V. Adding any number and then subtracting a greater number is equivalent to subtracting their difference. 21+3-7 = 21 -4 = 17. VI. Multiplying two or more numbers by the same num- ber and adding the products, is equivalent to multiplying the sum of the numbers by that number. Thus, 5x6 4-3x6 + 10 X6 = 30 4- 18 + 60 = 108, and (5 + 3 + 10) X 6 = 18 X 6 = 108. Similarly, it may be shown that for any values of a, l, c, and d, ad -\- bd -\- cd = (a -{- b -\- c)d. VII. Multiplying two numbers by a number and taking the difference of the products, is equivalent to multiplying the difference of the numbers by that number. Thus, 5 X 6 - 3 X 6 = 30 - 18 = 12, and (5 - 3) X 6 = 2 X 6 = 12. In like manner it may be shown that for all values of a, b, and c, ac '^ be = (a ~ b)c. VIII. If any number be divided by an equal number, the quotient is unity (1). That is, — = 1. IX. Dividing two or more numbers by the same number and adding the quotients, is equivalent to dividing the sum of the numbers by that number. Thus, - — I 1 •^ ' 4 ^ 4 ^ 4 ^ 6 + 2 + 4 = 12, and -^ + 8 + 16 =,i^_ = 12. Similar- ly, for all values of a, b, c, and d, ^ 4- -7 + -^ = ., . •^ d ^ d ^ d d X. Dividing two numbers by the same number and taking the difference of the quotients, is equivalent to dividing the difference of the numbers by that number. Thus, -J - = 9 — 6 = 3, and = — r= 3. In 4 4 4 4 like manner, for all values of a, b, and c, it may be shown ,, , a b a '^ b that — ^ — — . ceo DEFINITIONS. aJO EXAMPLES. What number must x represent in each of the followmg? 1. a: is 5 moi-e than a. Aiis, x = a -}- 5. 2. a; _ 5 = 11. A71S. x = 16. 3. X = y and y = 7, Ans. x = 1. 4. ^x = 60. A ns. X = 12. 5. a; -^ 3 = 8. Ans. x = 24. 6. The sum of a; and 12 is 20. A7is. x = 8. 7. ic is « less than d, Ans, x — I — a. 8. 3a; + 4 = 16. Ans. x = 4. 9. 5rc — 2 = 48. Ans. x = 10. 10. 3a: + 4a; = 42. A7is. x = G. 11. fa; = 12. A71S. x = 18. 12. a: is 6 less than b -\- c, Ans. x = b -\- c — Q. 13. i/x = 4. ■ Ans. X = 16. 14. x^ = 36. Ans. x = 6. 15. V^= 3. Ans. x — 27. 16. a;t = 4. -4ws. a; = 8. 17. V^+ 5 = 21 — 14. J«5. a; = 16. 18. 3^ = 81. Ans. a; = 4. 19. 2^ + 1= 64. Ans. a; = 5. 20. 2a;^ = i^. ^W5. a; = 2. 21. 11 — a: = 2. Ans. a; = 9. 22. 18 — 5a; = 3. ^?i5. a; = 3. 1 23. 16* = 2. Ans. x = 4. Ala.— 3. 26 ALGEBRA. X 24. 362 = 6. Ans. x = l. 2b. 12 — X = X — A:. Ans. X = 8. 26. The sum of x and 10 is «. Ans. x = a — 10. 27. a;^ = 8^. Ans. x = 4:. 28. a;^ = lei ^^s. a; = 512. 64. The preceding examples show how algebraic symbols are used for asking questions about number. The pupil should state what question is asked in each case. Thus: Ex. 2. "What number is that from which if 5 be taken, the remainder will be 11? Ex. 5. What is the dividend if the divisor be 3 and the quotient 8 ? Ex. 8. If 4 be added to three times a certain number, the sum is 16. What is the number? Ex. 13. What number is that whose square root is 4? Ex. 14. The square of what number is equal to 36? Ex. 16. W^hat number is that of which the square of the cube root is 4? Ex. 18. To what power must 3 be raised to produce 81 ? Ex. 19. The 19 th is a combination of two questions. First, What power of 2 is 64? Ans. 6th power. Second, What number added to 1 will produce 6 ? Ex. 23. 2 is what root of 16? Ex. 24. To what power must the square root of 36 be raised to produce 6 ? Ex. 25. 12 exceeds a certain number as much as that number exceeds 4. What is the number ? Ex. 27. What number is that whose cube is equal to the square of 8 ? 65. After translating the algebraic expression into or- dinary language, it is well to reverse the process and learn how to express problems in algebraic language. DEFIXITIONS. 27 EXAMPLES. 66. The following examples show how Algebra is em- ployed in the solution of problems: 1. John bought an apple and an orange for 12 cents, paying three times as much for the orange as for the apple. What was the price of each ? Let X represent the number of cents the apple cost. Then 3a: — the number of cents the orange cost. Both cost the sum of x cents and Zx cents = 4a; cents. But both cost 12 cents. .-. 4a; = 12, (Ax. I.) whence a; = 3, (Ax. V.) and 3a; = 9. (Ax. IV.) Therefore an apple cost 3 cents and an orange cost 9 cents. 2. The sum of two numbers is 36, and one of them is five times tlie other. What are the numbers ? Let x represent the smaller number. Then bx = the larger number. Their sum ^ x -\- 5x = Qx, But their sum is 36. . •. 6a; = 36, (Ax. L) whence x = G, (Ax. V. ) and 5x = 30. (Ax. IV.) Therefore the numbers are 30 and 6. 3. The sum of three numbers is 90. The second is twice the first, and the third equals the sum of the first and second. What are the numbers ? Let X denote the first number. Then 2a; = the second number. And X -\-2x = 3x denotes the third number. The sum of the three numbers is a; + 2a; + 3a; = 6x. But their sum is 90. . •. 6a; = 90, (Ax. L) whence x = 15, (Ax. V.) and 2a; =30, (Ax. IV.) a; + 2a; = 45. (Ax. IL) Therefore the numbei's are 15, 30, 45. 28 ALGEBRA. 4. The sum of three numbers is 1888. The second is four times the first, and the third is 18 more than the sum of the other two. What are the numbers? Let X denote the first number. Then 4:X = the second number. And X -^ 4:X = bx = the sum of the first and second numbers. .' . bx -\- IQ = the third number. The sum of the three numbers is a; -f- 4a; -f 5a: -|- 18 = IQx + 18. But the sum of the three numbers is 1888. . •. (Axiom I.) 10.^ + 18 = 1888. Subtract 18 from each side. r,lOx = 1870, (Ax. III.) whence x = 187, (Ax. V.) and 4.x = 748. (Ax. IV.) Also, S.-c + 18 = 5 X 187 -|- 18 = 953. Therefore the numbers are 187, 748, 953. 5. The sum of two numbers is 50, and one of them is two thirds of the other. What are the numbers? Let Zx represent the larger number. Then 2x = the smaller number. Their sum z=3x -\-2x = 6x. But their sum is 50. .*. 5a; = 50, (Ax. I.) whence a: = 10, (Ax. V.) 3a' = 3 X 10 = 30, and 2a; = 2 X 10 = 20. Hence the numbers are 20 and 30. 6. The sum of two numbers is GO, and their difference is 10. Wliat are the numbers ? Let X denote the smaller number; then the larger must be 10 more than x, that is, x -\- 10. The sum of the numbers is a; + a; -f- 10 = 2a: -|- 10. But their sum is 60. . • . 2a; -f 10 = 60, (Ax. I. ) DEFIXITIONS. 29 Subtract 10 from each, 2x = 50, (Ax. III. ) whence x = 25, (Ax. V.) and a; +10 = 35. (Ax. 11.) Therefore the numbers are 35 and 25. 7. A has as many quarters as B has dimes: together they have $2. 10. How much money has each ? Let X represent the number of quarters A has. Then x also denotes the number of dimes B has. Since one quarter is worth 25 cents, .'. x quarters are worth 26x cents; therefore the value of A^s money is 25a; cents. Since one dime is worth 10 cents, .'. x dimes are worth 10a; cents; therefore the value of B's money is 10.z; cents. A and B together have 25a; -f 10a; cents. But they together have 210 cents. .-. 25:i; + 10a; = 210, (Ax. I.) whence 35a; = 210, and x = Q. (Ax. V.) Therefore A has 6 quarters, worth $1.50, and B has 6 dimes, worth 60 cents. It must be carefully noted that x denotes the number of units; thus, x represents the number of dollars, feet, per- sons, days, etc., as the case may be. Therefore, x must never be put for money, length, weight, etc., but only for the number of units of length, weight, etc. 8. Divide a line 42 inches long into two parts, so that one may be three fourths of the other. A7is. 18 in.; 24 in. 9. The property of two persons amounts to $4800, and one of them is seven times as rich as the other: what is the property of each ? Ans. $4200; $600. 10. Divide $42 between A and B, so that for each dime A receives, B may receive a quarter. Ans. A, $12; B, $30. 11. The sum of two consecutive numbers is 21. What are they ? A7is. 10 and 11. 30 ALGEBKA. Exercise I. 1. Divide 1760 among A, B, and C, so that B shall have $50 more than A, and $135 more than B. 2. The difference of two numbers is 12, and one of them is 4 times the other. What are the numbers ? 3. The difference of two numbers is 84, and one of them is three fourths of the other. What are the numbers ? 4. In a company of 133 persons, there are four times as many women as children, and twice as many men as chil- dren. How many men are there ? 5. Find three consecutive numbers whose sum is 45. 6. A has $40 more than B, and three times the number of dollars A has equals five times the number B has. How many dollars has A ? 7. A bought X horses at $100 each, 1x cows at $30 each, and ^x sheep at 112 each. He paid $1100 for all. Find the value of x. 8. Seven times a certain number exceeds four times the number by 84. Find the number. 9. A is twice as old as B, and three times as old as 0. The sum of their ages is 66 years. How old is B? 10. If 6 times a certain number be diminished by 10, the remainder will be 10 more than twice the number. Find the number. 11. Find the number which exceeds the sum of its half, fourth, and sixth by 5. 12. A gave an equal number of nickels, dimes, and dollars, in payment of a bill of $23. How many of each did he give? CHAPTER II. NEGATIVE NUMBERS. 67. There are certain quantities which are so opposed to each other in character that any number of units of the one taken together with the same number of units of the other, would neutralize each other. Thus, if a person's income were 1600 and his outlay were $500, his capital would be increased by 1100, because $500 of the income would be neutralized by the $500 outlay. If his .income were 1600 and his outlay $800, his capital would be diminished by $200, because his income would be neutralized by $600 of his outlay; hence the rest of his outlay must be taken from his capital. If A gains $10 and loses $6, his net gain is $4, because $6 of his gains are neutralized by the $6 lost. If he g?iins $10 and loses $12, his net loss is $2. If a person go 100 yards in any direction and then retrace his steps for 60 yards, he will be 40 yards from his starting-point, because the 60 yards forwards are neutralized by the 60 yards backwards. If he go forwards 100 yards and then back- wards 120 yards, he will be 20 yards from his starting- point, but in a contrary direction from his original motion. If the mercury in a thermometer stand at 10° above zero, a fall of 15° will cause it to stand at 5° below zero. If a vessel in 5° north latitude sail south through 7°, she will then be in 2° south latitude. Similar relations exist between east longitude and west longitude; motion to the right and to the left; time before and after a fixed date, etc. (31) 32 ALGEBRA. 68. The discussion of the following problem will tend to explain the principles considered in this chapter: " How much will A's capital be increased if his income be ^a and his outlay '^cf Since A's income tends to increase his capital, $a must be added to his capital. Since his outlay tends to diminish his capital, $c must be subtracted: hence A's capital will be increased by as many dollars as a exceeds c. We say, therefore, that A's capital is increased by {a — c) dollars. If a > c, no difficulty will arise in considering this answer, for, whatever values be given to a and c, to find a — c merely consists in subtracting a less number from a greater. It a = c, a— c = 0, and the answer is still true in an arithmetical sense. If a < c, in order to find a — c, we are required to subtract a greater number from a less, which, in an arithmetical sense, can not be done. In arithmetic we deal only with the natural series of num- bers; viz., 0, 1, 2, 3, 4, 5, etc.: these numbers increase to the right, each succeeding integer being obtained by adding one to the preceding integer. Since addition con- sists in counting to the right or forwards, subtraction, being the contrary of addition, must consist in counting to the left or backwards. To subtract 10 from 7, we count backwards from 7 to 0, but can go no farther, for there the natural series ends. Thus, we can subtract only 7, and there remain 3 units to be subtracted. 69. It is important that the answer to the problem in the preceding article, and to similar problems, shall in all cases be {a — c), and that for all values of a and c, a — c shall denote that a is to be added and c is to be subtracted. To enable us to subtract a greater number from a less, it is necessary to assume a new series of numbers, beginning at zero and extending to the left. To each of these numbers the sign — is prefixed, and to each of the natural series the sign -|- is prefixed. Numbers preceded by — are called Of THE \ . UNIVEPSITY j V p ^f" /negative k umbers. 33 negative numbers, and numbers preceded by + are called positive numbers. The algebraic series of numbers is written thus: .... - 5, - 4, - 3, - 2, - 1, ± 0, + 1, + 2, + 3, + 4, ... . or thus: 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, The sign — is never omitted. is either -j- or — , since it is the starting-point of both series. 70. A familiar example of the use of the algebraic series of numbers is furnished by the scale of an ordinary ther- mometer. A certain point is fixed upon as zero. Degrees above zero are + 1> + 2, -f 3, -|- 4, . . . . ; degrees below zero are — 1, — 2, — 3, — 4, .... , If the mercury descend 20° from a point 12° above zero, the result is found by subtracting 20° from 12°; that is, by counting downwards 20° from 12°: by so doing we arrive at 8° below zero, and the tempemture is recorded as — 8°. If the mercury descend 5° more, we subtract 5° from — 8°; that is, we begin at — 8° and count 5° downwards, and arrive at — 13°. If now the mercury rise 10°, we add 10° to — 13°; that is, we begin at 13° below zero and count 10° upwards, and arrive at — 3°. A rise of 7° additional will give + 4°, etc. 71. (1) To add 6 to 5, we begin at + 5 and count 6 units in the positive direction; that is, forwards or to the right, and arrive at -f 11; that is, 5 + 6 = H- (2) To add 5 to — 3, we begin at — 3 in the negative series and count 5 units forwards and arrive at -|- 2; that is, - 3 + 5 = + 2. (3) To add 4 to — 10, we begin at — 10 and count 4 units forwards and arrive at — 6; that is, — 10 + 4 = — 6. (4) To subtract 10 from 7, we begin at 7 and count 10 34 ALGEBRA. units in the negative direction ; that is, backwards or to the left in the algebraic scale of numbers, and arrive at — 3; that is, 7 - 10 = - 3. (5) To subtract 4 from — 3, we begin at — 3 and count 4 units backwards and arrive at — 7; that is, — 3 — 4 == — 7. In other words, (1) To add a positive number means to count so many units forwards or to the right, in the alge- braic series of numbers. (2) To subtract a positive number means to count so many units backwards or to the left in the algebraic series of numbers. 72. The Absolute Value of a number is its arithmetical value, taken independently of the signs + and — . Thus, + 4 and — 4 have the same absolute value; that is, four units. 73. The Algebraic Value of a number is its absolute or arithmetical value taken in connection with a sign -|- or — . Thus, -j- 4 and — 4 have different algebraic values. 74. Two numbers which are both positive or both nega- tive are said to have like signs. If one be positive and the other be negative, they are said to have U7iUke sig7is, 76. (1) To add - 12 to 7. In (71), it was shown that in adding a positive number we follow the arithmetical meaning of addition, and count so many units to the right; we shall therefore agree that to add a negative 7iumber means to count so many units to the left in the algebraic series of numlers. Beginning at 7 and counting 12 units to the left, we arrive at — 5 ; that is, 7 + (— 12) = — 5. . • . 7 + (— 12) is equivalent to 7 - 12. (2) To add — 5 to — 4, we begin at — 4 and count 5 units to the left and arrive at — 9. That is, — 4 -|- ( — 5) = -4-5 = -9. 76. Since the subtraction of a positive number means to count so many units to the left (71), therefore we shall NEGATIVE NUMBERS. 35 agree that the sitUr action of a negative numler is ^jer- forrned hy counting so many units to the right, (1) To subtract — 6 from 4, we begin at + 4, count 6 units to the right, and arrive at + 10. That is, 4 — (— 6) = 4 + 6 = 10. (2) To subtract — 7 from — 3, we begin at — 3, count 7 units to the right, and arrive at + 4. That is, — 3 — (— 7) = -3 + 7=+4. 77. From the illustrations given in Articles 71, 75, and 76, we see: (1) Adding a positive number and subtracting a nega- tive number both mean counting to the right. Hence, the addition of a positive number prodiices the same result as the subtraction of a 7iegative munber having the same absolute value. Example: 34-(+4)=^7. 3— (— 4) = 3 + 4 = 7. (2) Subtracting a positive number and adding a nega- tive number both mean counting to the left. Hence, tlte subtractio7i of a positive number produces the same re- sult as the addition of a negative 7iumber having the same absolute value. Example: 6 — (+ 2) = 6 — 2 = 4. 6 + (- 2) = 6 - 2 = 4. 78. It should be carefully noted that the signs -f- and — are used for two distinct purposes: (1) Arithmetically, as signs of operation, to connect numbers with each other by addition or subtraction. (2) Algebraically, as signs of opposition, to indicate in which series — the positive or the negative — a given number belongs. It should be further noted that, (3) The sign + placed before a term does not change its algebraic sign. Thus, -f (-f rt) = + a, and -f- (— ^,) = —a. (4) The sign — placed before a term changes the algebraic sign of that term. Thus, — [-\- a) = — a, and — (— a) = -\- a. 36 ALGEBRA. ' 79. The principles considered in the preceding article may be still further extended. (1) Since prefixing a sign + has no effect upon a term, therefore any number of -\- signs may be prefixed without affecting the term. Thus, + ! + [ + ( + ^0]}=+ «. and + {+[+(-«)]!=-«. (2) Since prefixing a sign — changes the sign of a term, then prefixing a second minus changes the sign again, and thus gives the original sign; prefixing a third — changes the sign, etc. Examples: —[—{-\-a)]=-\- a; —[—(—«)] = — a; —{ — [—{— a)]] = -{- aj etc. Hence: 7/ « ter?7i he affected ly an odd numler of minus signs, its essential sign ivill he minus. If a term he affected hy an even num- her of mimis signs, its essential sign toill he plus, 80. By Essential Sign is meant the sign properly belong- ing to a term when affected by but one sign, -)- or — . 81. The use of + and — as signs of opposition is carried into the treatment of exponents (32). Thus: „ aa aaa aaaa aaaaa , i k i. nc^ k tttt \ a^ =—- = = = , etc. (Art. 62, Ax. VII.) 1 a aa aaa ^ ' g _ aaa __ aaaa _ aaaaa _ aaaaaa 1 a ~' aa aaa ' ^ aaaa aaaaa aaaaaa ~ 1 ~ a ~~ aa ' - aaaaa aaaaaaaa ^ a^ = — z — = , etc. 1 aaa In the various forms of a\ there are two a's in the divi- dend more than in the divisor; of a^, there are three more; of «*, four more; of a^, five more, etc. That is, a posi- tive exponent denotes that the number next to which it is placed occurs so many more times as a factor in the divi- dend than in the divisor. NEGATIVE NUMBERS. 37 82. If any number occur as a factor an equal number of times in both dividend and divisor^ the exponent of that number is zero (0). Tims: «^"_„o. «*-,,o. l-y^. etc __«, -_«, ^3-6, etc. Since ^ = l(Art. G3, VIIL), ft" and since —=«*>,.•. a^ =1 (Art. 62, Ax. I.) Therefore, any number ivith an exponent of zero is equal to unity. 83. Since we use the positive exponent to denote that the number affected by the exponent occurs as a factor so many more times in the dividend than in the divisor (81), and since we use the zero exponent to denote that the number affected by it occurs an equal number of times in both dividend and divisor (82), therefore, the negative exponent denotes that the number affected by it occurs as a factor in the dividend a fewer number of times than in the divisor. Thus : 1 = 2-«; | = 3ft-*; A = | = 5 X S'^; 2^ 'ft* ' 27 3^ -— ■ = ab~^c~^; etc. Exercise II. Simplify the first five problems : 1. 6 + (- 5); _ 6 - 5; 4 - (- 7); 5 + (- 3). 2. -[-(-5)]; -[-[-(-4)]}; -{_[-(-_ 8)]}. 3. -2-[-4]; -5-{-[-(-9)]}; -(_6)-[-(-2)]. 4. 6'; - (4«); 4 X Q'; 7" X 3"; - 2 X 9°. 5. 3-^; 4-2; 3 X 5-^; 4"^ X 2"^; 3« X 6"^ 38 ALGEBRA. 6. A was boru 10 B.C. and lived 67 years. When did he die? 7. D died 17 a.d. at the age of 49. When was he born? C. E was born 205 B.C. and lived 80 years. When did he die? 9. F was born 43 B.C. and died 12 a.d. How many years did he live ? 10. The zero of the Mohammedan calendar is the date of the Hegira, 622 a.d. Time since 622 a.d. being -|-, and before 622 a.d. being — , what are the following dates according to this calendar: 1888 a.d.? 300 a.d.? IB.C. ? 753 B.C.? 218 B.C.? 11. A has $4750 and owes $927. How much is he worth ? 12. B has 11800 and owes $3000. How much is he worth ? 13. went 125 steps forwards, then 49 steps backwards, then 20 steps forwards, then 74 steps backwards. How many steps did he take ? How far is he from his starting- point ? 14. A goes 75 steps in a minute, and B 60 steps. How far apart will they be at the end of five minutes, if they walk in the same direction from the same starting-point ? In opposite directions? CHAPTER III. NUMERICAL VALUES. 84. In" Algebra, a letter may stand for any number which we wish it to represent. Thus, a may represent 2, 5, 10, 125, i, f, 0, — 2, — 7, — f, — 2^, or any other number, positive or negative, fractional or integral. It must not be understood from this, however, that the letter has no determinate value. Its value is fixed for the time being ; it can not represent two different numbers in the same problem, but on a different occasiori the same letter may be put for any other number. 86. The numerical value of a simple term (44, 47), con- taining no sign of division, may be found from the preced- ing definitions and principles. To illustrate the various cases, let a = 8 in each of the following examples : 1. rti =z ? This denotes that the cube root of 8 is to be extracted, and the result raised to the second power. The cube root of 8 is 2, and the second power of 2 is 4. . • . when a = 8, at = 4. (Arts. 31, 35, 37.) 2. a-t = ? Since ftt = 4^ then a''^ — \. (Art. 83.) 3. «-^ = ^=^. (Art. 83.) 4. V'rt^zz: at = 8^= (2)*= 16. In the following examples, let a = 9, and J = 16 : 5. aU-t = 9i X 16-* = 3 X i = |. a _ 9 ^' :s - 4- 7. 4a^J-i = 4 X 3 X J = 6. (39) 40 ALGEBRA. 86. To find tlie numerical value of a simple term con- taining but one sign of division: Find the value of the expression (43) 'preceding the division sign (85), tlien of that following the division sign, and divide the former hy the latter. Examples. — i. ah'^ -^ c^ d^ = ? when a = S, b = 2, c=lQ, d = 8. ab-^ = 8 X i = 4. c^d^ =: 4 X 4 = 16. 4-^16 = i Atis, 2. ia -^bhid' =? when a = 12, b = 9, c = 8, d = 6. 4a = 4 X 12 = 48. b^cid^ = 3x2x1 = 6. 48 ^ 6 = 8 A71S. After indicating the work, the dividend may be placed above a horizontal line, and the divisor below it, and can- cellation may be employed. 87. To find the value of a simple term containing two or more signs of division: (1) Fi7id the values of the expressions separated by the division signs, (43.) (2) Commence at the right of the term, and use the signs in their order. Examples.— 1. 16 -^ 8 -^ 2 = 16 ^ 4 = 4. 2. 18 X 8 -^ 12 X 9 ^ 8 = 144 -^ 108 ^ 8 = lOf. 88. To find the numerical value of a polynomial (53) consisting of simple terms: (1) Fi7id the simplest value of each term (85, 86, 87). (2) Add together all the positive term,s (45). (3) Add together all the negative terms (46). (4) Find the difference of the absolute values of these su7ns (72). (5) Prefix to the remainder the sign of the sum whose absolute vahie is the greater. I^-UMEIIICAL VALUES. 41 Example. — Find the value of at -f jic^ — ^>i -f- as — « ~^^^6'~S when a = 8, b = IQ, c =^2. 1st term = at = st = 2' = + 4. 2d term = + ^^^^ ==+2x4:=+ 8. 3d term =:-^i-^rtt=-4-^-2 = -2. 4th termor —a-^b^c-^= — ix4xi = — 1. 4_^8-2-l = 12-3 = 9. 89. To find the numerical value of a compound term (48): (1) Find the value of the expression inclosed by the in- nermost sign of aggregation (43, 38). (2) Remove this sign of aggregation (78, 79). (3) Proceed i?i like manner with the remaining signs of aggregation {removing one at a time); until all are removed. Example. — If a = 4, ft = 3, c= — b', d = 0, find the value ot - {a - [b + 2c - (a + d)]], /? + t? = 4 + = 4. . • . - (a + fZ) = - ( + 4) = - 4. J + 2c-4 = 3 + 2(-5)-4 = 3-10-4 = 3-14r= -11. .-. - [^ + 2c-4] = - [-11] = + 11. (Art. 79.) a + 11 = 4 + 11 = 15. .-. - la + 11 15. 90. From (88) and (89) the numerical value of any poly- nomial may be found. EXAMPLES. When a = 4, J = 8, c = 1, d = 0, find the value of: 7. a^—b-j--. Ans. — 6. Q. c -^ a~^ b -\- d. A ns. ^. 9. -abc''-d\ Ans. -32. 10. (at + ^ — cd)^. A)is. 4. 11. dac — bk A us. — 2. 12. ab — ac — bd. Ans. 28. 1. a^X b^ X c'. A71S. 8. 2. a^b~^c. A ns. 4. 3. a'bh\ Ans. 2. 4. ab^ -^ c. A ns. 8. 5. a^ b^ -^ c. A ns. 4. 6, c^ ab-^' A71S. 1. Alg.-4. 42 ALGEBRA. Exercise III. Given a = 2, b = 3, c = i, d = 6, m =1, r = 0, find the value of each of the following expressions : 1. ^7^ + i/2^2 _ ^^^^ 5 2. 3 V2c + 5 |/9^ - t'Sam. 3. he Vba -j- d -\- m — (d — a) V^o + r. 4. {(a-^c^Y-^{a-c^Y\\ 5. |/(« + ^ + c)(2^-w). 6. [{a + ^)M^ + l)]H(^ + ^)^. 7. (c — ft)(c -|- a) 4- rt??i^ — 2 (a + c + ^) — fl!6?r. 8. (c + %dY H- (^ + ^)^ - [be {2d + «)]i 9. {a-{-be -\- 2m + ^')^ — abmr. 10. «^ 4- Jr 4- c^ m — (^ 4- m 4- r). 11. 'Va -{-[2b + e(d + mr)] inf 12. V3« 4- 2 [3^ - 6- (^ - 2m) 4- 2«] 4- 2^c. Show that the numerical values of the following are equal: 13. (d^b){d-b) = d^-b\ 14. {d^- a^)-^ {d-a) =d^^ad^a\ 15. [d^ 4- a^) -^ {d-\- a) =^d^~ ad 4- a\ 16. (6^ 4- «)' 4- ((? - «)2 = 2 (£?2 4- «')• 17. {d 4- «)' - (^ - ay = Aad. 18. (d 4- «)^ 4- (^^ - «)' = 2^Z (^^ 4- 3a^). 19. (5 4- af - {b - af = 2a {3b' -{- a'). 20. {a' 4- b') (6'2 4- 6^2) = (ac 4- ^^)' + {be - adf. 21. a' 4- ^'^'^ + ^* = («' + ^^^ + ^') (^^' - fif^ 4- ^'). NUMERICAL VALUES. 43 SPECIMEN" PAPER. 1. Find the value of c^c — Sab^ — 5 {a — b -{- c), when a^ -2, b = '3, c= -4:. + 1st term = a^c = (- 2y (- 4) = (+ 4) (- 4) = - 16. 16 2d term = - Sub^ = - 3(- 2) (S)^ = - 3(- 2X9) = + 54. 54 3d term = - 5(a - 6+c) = - 5(- 2 - 3 - 4)=- 5(- 9)= +45. 45 + 99 -16 = + 83. 99 16 Am. 83. 2. Find the value of a — {c — b [2a — b — c (a^ -{- b)]}, when « = — 3, Z> = 4, c= — 2. (a2 4.j)=:(+9 + 4) = (+13). [2a-b-c (13)] = [- 6 - 4 - (- 2) (13)] = [-6 - 4 + 26] = [+ 16]. {c_J[+16]} = i-2-4[l6]i =[_2-64] = [-66]. a-{— 66( =-3 + 66 = + 63. Am 63. 3. Find the vahie of «« — Sbc^ + 2b^ c — 5, when a 3:3, b = - S, c = d. 1st term = J = 32* = 2^ = 16. -f 16 — 2d term = - 3^6* = - 3 (- 8)(9)i = - 3 (- 8)(3) = + 72. 3d term = + 2 J^r = + 2(- 8)^(9) = + 2 (- 2)"^ (9) = + 2(+4)(9) = 72. 4th term = - 5. 72 72 5 + 160-5 = + 155. 160 5 Am. 155. 4. {a -by -2 {a + Z»)« + 5 (abf -\- (b - af = ?, when a = 2,b=:-3. + 1st term = {a - by = (2 + 3)« = (5)-^ = + 25. 25 2d term=-2(a+&)3=-2(2-3)«=-2(-l)3=-2(-l)=+2. 2 8d term = + 5 (abf = + 5 (- 6)'^ = + 5 (36) = 180. 180 4th term = -\-{b - a)^ = ^{- 3 — 2f = (- 5)^ = - 125. + 207 -125 = +82. 125 207125 Am. 82. CHAPTER IV. REDUCTION OF TERMS. 91. Reduction is the process of changing the foim of an expression without altering its value. 92. Reduction of Terms is the process of uniting terms and finding the simplest expression for the result. 93. (1) To simplify Za + 5«. If a stand for 12, it is required to find the sum of 3 dozen and 5 dozen, which is evidently (3 + 5) dozen = 8 dozen, li a =^ 20, the sum of 3 score and 5 score is evidently (3 + 5) score = 8 score. In like manner, for all values of ^, 3a^ -|- 5^ = (3 + 5) ^ = Sa, Similarly, 4:ah -f lah = (4 + 7) aJ = llah, and Qair- + ^a¥ = (6 + 8) a¥ = 14:ah\ Since the method of work is plainly the same for all other numbers, therefore, for every value of a, l, and c, ac ^hc = {a-\- h)c, (Art. 63, VI.) (2) To simplify — 3« — la. Subtracting two numbers is equivalent to subtracting their sum. (63, II.) . • . - 3a - 7a = - (3 4- 7) « = - 10a. Therefore — ca — ia = — {c -\- h) a. (3) To simplify la — 4a. Reasoning as in the first problem, since 7 dozen — 4 dozen = (7 — 4) dozen, and since 7 score — 4 score = (7 — 4) score, and similarly for all values of a, therefore 7a — 4a == (7 — 4) a = 3a, If, instead of 7 and 4, any other numbers be substituted, the method of work is the same. ,'. ba — ca ^^ (b -~ c) a, (44) REDUCTION OF TERMS. 45 94. (1) To simplify 4 (^ - c) + 8 (Z> - c). Ub = 12 and c = 3, then b— c = 9, and the sum of 4 nines and 8 nines is (4 + 8) nines = 12 nines. If ^ = 5 and c = 2, then b — c==d, and the sum of 4 threes and 8 threes is (4 + 8) threes = 12 X 3. In like manner, for all values of b and c, 4:{b-c)-\-8(b-c) = (4.-{- 8) ib-c)= 12 {b - c). Since the solution is evidently similar if any other par- ticular numbers be substituted for 4 and 8,.'. m (b — c) + a(b — c) = {m + c(){b — c). (2) To simplify 9 {x — y) — 4(.7: — y). Reasoning as before, we find the result to be {9 — 4){x — y) = b {x — y). Therefore c(x — y) — b (x — y) = {c — b)(x — y). (3) To simplify a(x — y) -}- bx — by. Since bx — by = b (x — y)f therefore the given expression is equal to «(^-2/) + ^(^-y) = (^ + *)(^-'y)- . 96. (1) To simplify a* — a. a^ — a = aaaa — a = (aaa — 1) « = («* — 1) a. Similarly, a^ — a^ = aaaaa — aaa = (aa — 1) aaa = {a^ - l)a\ (2) To simplify 6«'' — Sa. Since each term is divisible by 2«, therefore 6a' -8a = 2a (3r?.« - 4). 96. (1) a' - a -{-a'- a' = a(a' - 1) + «' («' - 1) = (a 4- a') (a^ - I) = a(l + ^0 («' - !)• (2) 3a{x + y)-2b(x + y) + {x-2y)(3a-2b) = (3a - 2b)(x + y) -\- {x - 2y){3a - 2b) = ■ (3a - 2b){x -]-y + x- 2y) = (3a - 2b){2x-y). (3) a{x-y) - b{x-y) - c{x-y) = {a-b-c)(x-y), (4) (a - b){3x-y) - 2b [Zx - y) -\- (x-y){a-U) = (33: - y){a - b - 2b) -\- {x - y){a - 3b) = {3x - 'y){a - db) -^{x- y)(a - 3b) = (a - 3b){3x -y-{.x-y) = (a-3b){ix- 2y) = 2{a-db){2x-7/). 46 ALGEBRA. 97. From the illustrations given in (93-96), inclusive, we derive the following Eule for the Reduction of Terms having a Common Factor : (1) Remove from each term tlie common factor, (2) Inclose the resulting terms in parenthesis, (3) Write the common factor , either lefore or after the parenthesis, with the sign X , expressed or understood, be- tween them. (4) Write the sign -\- hefore the whole. (5) In lihe manner combine the resulting term§, and so continue until the expression is in its simplest form. Exercise IV. Simplify each of the following expressions: 1. 5(2 4- 9a^ - 2a — 6«. 9. a^ + tt^ + a + 1, 2. 3^5 — ac. 10. x^ — x^ -\- X — 1, 3. a" - 3f?. 11. «r6 _ ^4_|- fl^3 _|_ ^2_ fj,-\-\, 4. b{b-c)-2{b- c). 12. {a-\-b)x-^{a-b) X. 5. —{a — b)-^l{a — b). 13. {a — b)y-\-{a — b) z, 6. ab -\- 3^. 14. {a — c)bd— {a~ c) ab, 7. a'' - a. 15. (a -^ b) ^ -^ {a - b) Vx. 8. ab-^-a^bc-^c. i6. a^ - a% + ab^ — b\ 17. (a-b^ c)x^ + [U + 3^> - c)x^. 18. (3a - 2b){x -y)J^{a- 2b){x - y) -j- 4(a - b){x + y). 19. 6a — Qb -{- ca — cb ~ 2a -i- 2b. 20. {2a -b)(c- 2d) + (2a - b){2c +5^/). CHAPTER V. ADDITION; SUBTRACTION; BRACKETS. 98. Addition in Algebra is the process of uniting expres- sions with their proper signs, and reducing the result to its simplest form. This definition agrees with the principle advanced in (78), that addition does not change the algebraic sign of a term. Thus: -{-{-}- a) = -^ a, and -{■ { — a) = — a. 99. The arithmetical notion of addition applies only to the addition of a positive number to a positive number; but the use of negative numbers requires us to extend this defi- nition. The addition of + ^ to + ^ gives a -\-h; of — i to -f a gives a — b; ot -\- b to -^ a gives — a -\- b; of. — b to — a gives — a — b, 100. The result obtained by addition is called the algebraic sum. When the numbers to be added have like signs (74), the algebraic sum has the same absolute value as the arithmetical sum. When the numbers have unlike signs, the algebraic sum has the same absolute value as the arithmetical difference. 101. Subtraction in Algebra is the process of finding a number which, added to a given number, will produce a given sum. With reference to this operation, the given sum is called the minuend, the given number is called the subtrahend, and the required number is called the re7nain- der or difference, (47) 48 ALGEBRA. 102. The algebraic definition of subtraction agrees with the arithmetical definition; but, as the algebraic sum has a more extended meaning than the arithmetical sum, so the operation of subtraction has a correspondingly extended meaning. In both Arithmetic and Algebra, subtraction is understood to be the contrary of addition; if the addition of a term has a certain effect, then the subtraction of that term has a contrary effect. Since -\- (-\- a) = -^ a, then — {-\- a) = — a; also, since -{-( — «) = —«, then — (— a) = -\- a. These results agree with the principles advanced in Arts. 77 and 79. 103. Rule for Subtraction : (1) Write for the remainder all the terms of both minu- end and subtrahend, giving to each term of the minuend its oicn sign, and to each term of the subtrahend a sign contrary to its oiun. (2) Reduce the result to its si?nplest form (Chapter IV.). Verification: Add the remainder to the subtrahend; the sum must be the minuend. KoTES. In subtraction, and also in addition, it is convenient to write like terms under each other. After the pupil has become expert in the use of the rule given above, it is recommended that the operations of changing the signs of the subtrahend, and of reducing like terms, be performed men- tally. 104. The truth of the principle that the subtraction of ^ — c is equivalent to the addition of — ^ + c, may also be shown as follows : First Method. To subtract b — c from a means that c is to be taken from b, and this remainder is to be taken from a. If b be taken from a, the remainder is ^^ — b; but this remainder is c units too small, for c units too many have been taken from a. .*. the true remainder is found by adding c to a — b. That is, a — (b — c) = a — b -\- c. BRACKETS. 49 105. Second Method. a = a-]-b — b-{-c — c (Art. 62, Ax. VI.). Since adding (b — c) to any number means to unite b — c with that number (98), therefore subtracting b — c means to remove b — c from the sum. It b — c be removed from a-{-b — b-^-c — c, the remaining terms will he a — b-\- c, . • . a — (b — c) = a — b -\- c. 106. Brackets. — The following laws for the use of brack- ets (38) depend upon the preceding principles : (1) When an expression ivWiin brachets is preceded by -{-, the brackets may be removed (79). (2) Any number of terms in an expression may be in- closed by brackets, and the sign + placed before the tvhole. (3) When an expression within brackets is preceded by the sign — , the brackets may be omitted if the sign of eveyy term within the brackets be changed (79). (4) Any number of terms in an expression may be in- closed by brackets and the sign — placed before the whole, provided the sign of every term tvithin the brackets be chaiiged, 107. Expressions may occur with more than one pair of brackets; these may be removed in succession. Thus, a-\b-c-{-d+e-f^g-]\=a-{b-c-'{-d-^e-f-Yg^\ = a— \b — c-\- d — e -\-f — g\ = a ~b -\- c — d -^-e —f-{-g J or, proceeding in a different order, a-{b- c - [-d -{-e-f-g]]=a-b-\-c-{-[-d-{-e-f^] = a — b-\-c — d-\-e —f—g = a — b-\-c — d-\-e— f-\-g. Reversing the above, we may introduce more than one pair of brackets. 108. The rules for removing more than one pair of brackets may be stated as follows : First Method. Remove the innermost pair of brackets; next, the innermost of all that remain, and so on, Alg.— 5. 60 ALGEBRA. SEC02frD Method. Re7nove the outermost pair of brack- ets; next, the outermost of all that remain, and so on. Make no change in any term until the brackets inclosing that term are removed. 109. Brackets may all be removed at once by observing the rules mentioned in (79). Every term inclosed in one pair of brackets is affected by two signs : (1) the sign prefixed to the term; (2) the sign prefixed to the brackets. If this pair of brackets be inclosed in another pair, each term is affected by three signs, and so on. If an odd number of these signs be negative, the essential sign (80) of the term will be minus. If an even number be negative, the essential sign of the term will be ylus. Thus, in a - \b - c - I- d -\- e -\- {- d - e-f-g)]}, a is +, being affected by no minus ] b — c, being affected by the sign prefixed to the brace, becomes — b -\- c. The terms inclosed by the brackets keep their own signs, because the — before the brace and the — before the brackets are to- gether equivalent to +. . * . those terms become — d-\- e — d-e-f-{-g. 110. Since -{- {a — b — c) = — {— a -\- b -\- c), therefore: If the exponent of an expression i^iclosed in brackets be unity, the sign of each inclosed term may be changed, 'provided the sign prefixed to the brackets be changed. EXAMPLES. 1. Add a — w; a — Zx; 2a — 2; 3 — 2x; x — 4. Ans. 4a — 5x — 3. 2. Add {a 4- ^); 2 (« + x); 7 (« + x). Ahs. 10 (a + x). 3. Add S{a - b)j -5{a- b); c(a - b). Ans. (c — 2) {a — b), 4. Add a{x- y); -b(x — y); c(x — y). Ans. (a — 15 -f- c) (a; — y). BRACKETS. 51 5. From 2x — y — dz take x — Ai/ — z, Ans, X -{-Zy — 2z. 6. From ax take bx, Ans. (a — b)x. 7. From 2a^ - P take a^ - a% + aV. Ans, (a-^V)(a''-¥), 8. From a{x — y) take — b {x — y). A71S. (a -{-b){x — y), 9. From {a — b -{- c) d take (2« — b— c) d. Ans. (— a + 2c) f^. 10. From ax — by -\- ex take (1 -\- a)x -\- by. Ans, (c — \)x — 2by. Exercise' V. 1. Add {a + b) x; (b — a) xj ex. 2. Add (3& - 2c + 1) x; (- b -\- Sc) x, 3. From 1 {a — b) take 3 (5 — a). 4. From (4a + 2b) {2x — y) take (2a; - y) (3a - b). Simplify each of the following expressions: b. a—[U— \a — {2c — c — b) — U\ - a-{-c\. 6. 1- \1 - a -[{- a" -\- a) -I'l-l- a-\-l -cF^^I]' 7. 1- {i-a-[a-{\-a')-\-{l-'a-~^^^a)']}. 8. 2a - !2a - (4a - 8 - "4a -43)!. Collect the co-efficients of the like powers of x. 9. x^ -\- ao(^ — bx^ -\- ax — bx -\- a — 1. 10. a? -\- y^ -^ z^ -^ xy -\- xz -\- yz. 11. ^ + ^y — ^z-\-oi?-\-xy. 12. aa:' — aic* — arc + a — Jar^ + Ja:^ — Ja; — 5. CHAPTER VI. MULTIPLICATION. 111. Multiplication, as defined in Arithmetic, is a short method of adding equal numbers together. With reference to this operation, one of the equal numbers is called the multiplicand, the number of those equal numbers is called the multiplier, and the sum is called the product. The multiplicand and multiplier are i\iQ factors of the product. These words are used similarly in Algebra, but their mean- ings must be extended to include negative numbers. 112. Multiplication in Algebra is a process of adding as many numbers, each equal to the multiplicand, as there are units in the positive multiplier; and it is a process of subtracting as many numbers, each equal to the multipli- cand, as there are units in the negative multiplier. 113. As adding a term does not change its sign, and as the sum of any number of positive terms is positive, and of any number of negative terms is negative; therefore, if the multiplier be positive, each term in the product and the corresponding term in the multiplicand will have like signs. Thus, + 5 X (+ 4) = + 20, and + 5 X (- 4) = - 20; also, -\- a X {-\- h) = -\- ah, and -{- a X {— h) = — ai. Since subtracting any number of terms is performed by changing their signs, therefore, if the multiplier be nega- tive, the corresponding terms in the multiplicand and product will have unlike signs. Thus, — aX {-\-h)= — ah, and — a{ — h) = -^ah. (52) MULTIPLICATION. 53 114. Since -f «(-[- Z>) = -f ab, and — a(— J) = 4- ab; therefore. If two factors have like signs, the sign of their product is plus. Since — a X (-}- h) = — ah, and -\- a X {— h) = — ah; therefore. If two factors have unlike signs, the sign of their product is mitius. 115. The product of any number of factors will be negative if the number of negative factors be odd (1, 3, 5, 7, 9, 11, etc.). (Art. 79.) The product of any number of factors will be positive if the number of negative factors be even (0, 2, 4, 6, 8, 10, etc.). (Art. 79.) Thus, 2(-3)(-4)(-5) = -120. -2(-3)(-4)(-5) = + 120. 116. From (115) it follows that the signs of an even number of factors may be changed without affecting the sign of the product; but if the signs of an odd number of factors be changed, the sign of the product will be changed. 117. (1) a^ X rt* = aaa X aaaa = d' = a^ + K (2) a^ X a^ Xa^ = aax aaa X aaaa = a* = a^ + s + i^ (3) a-'xa' = - xaaa = ^^ = a' = a-' + \ ^ ' aa aa , .S A 1 « 1 1 , ... (4) a-^Xa = Xa = = = -3 = rt-' = a-* + ^ ^ ' aaaa aaaa aaa a. (5) rt-2 X a-« = — X — = -^ = -5== «-' = a-^ + <-'>. ^ ' aa aaa aaaaa a° In like manner, for all values of m and n, a^ X a"" = ^m + n^ That is. The powers of a number are multiplied by adding the exponents. 118. The product of the same powers of different letters may be indicated by writing the letters within brackets, and placing the exponent over the whole. Thus, a^ X b^ = 64 ALGEBRA. aahh = ab X ah = (ahf. Similarly, a^ x h^ X (^ = (ahcy. That is, O'Xl'^^ {obY; d^xV^xe^ (alcY ; a^XlfX c^ Xd^ = {al}cdY; and so on for any number of factors. 119. The principles explained in the preceding articles, being true for all numbers, must necessarily be true for particular numbers, expressions in brackets, etc. Thus, 2^X 2* = 2^; 3i X 3t = 3^; {a - hf X (a ~ hf = {a - bf; 5 - i X 5t = 5^ ; {abf X (abf = {abf ; {a + bf X {a-{-b)-^= (a-{-b)'-' = (a-i-by=l; etc. 120. Erom the principles explained in Arts. 114 to 119 inclusive, we derive the following Kule for the Multiplica- tion of two or more Monomials : (1) Sign". If the number of negative factors be odd, the sign of the 2Ji'oduct is minus. Otherwise, the sign of the product is plus. (2) Numerical Co-efficient. Fi7id the product of the numerical co-efficients as in Arithmetic. (3) Literal Factors. Write in succession all the let- ters occurring in the factors, using each letter but once. (4) Exponents. To find the exponent of any letter in the product, add the exponents of that letter in the factors. 121. To multiply a polynomial by a monomial: Multiply each term of the multiplicand by the multiplier, and add the products (63, YI. and VII.). EXAMPLES. 1. a'^bc X a^b^c' = ? Ans. a'b'c\ 2. x^y^z X xyz~^ = ? Ans. xiyh~^. 3. — abcr^ X a^bc = ? Ans. — a^b^c\ 4. - a^^c x{-abc-^)=? Ans. + a'b'c-\ 5. - 2a^^c X (- 5«Z»2c) = ? Ans. + lOa^^e". 6. aH^c~^ X (—iab-^c) =? Ans. — 4:ah-^c^, MULTIPLICATION. 55 7. G^X 6-^ = ? A71S. 6-^ = 1 8. (x -\- y — z)x = ? Ans. x^ -\- xy — xz 9. (x — x^ — l)x^ = ? Alls, x^ — of — x^ 10. (a — b — c) (— 5a) = ? Ans. — bac + 6bc + 6c^, 11. iix" - fx - 3) (- 6x) = ? Ans, - ^x^ + 4a;2 + ISx 12. [(a - hf - (a - h)'\{a - b) =? Ans. {a - by-(a - b) 13. (6* — 6«'— 6^)x6*-«' = ? Ans. 6^-^—6^ — 6'^-» + ^, 14. [a — (b — c)] (— rtZ>) = ? Ans. — a^b -\- a¥ — abc, 122. To multiply one polynomial by another: M^dtiply each term of one factor by each term of the other, a7id add the partial products. This rule is a result of the preceding principles. Thus, multiplying by a-\-b is equivalent to multiplying by a and by b separately, and adding the products; multiplying by a — & is equivalent to multiplying by a and by b sepa- rately, and subtracting the second product from the first; but subtracting b times a number is equivalent to multiply- ing it by — Z> and adding the result. 123. For convenience in collecting like terms, the fol- lowing additional steps are recommended : (1) Arrange both expressions according to the descend- ing powers of a letter common to both, or according to the ascending powers of that letter (58). (2) Supply ciphers in place of the missing powers of the letter of arrangement, in order that the exponents of that Utter may have a common difference in the successive terms. (3) Write the multiplier under the multiplicand, the first term under the firsts the second under the second, etc. (4) Multiply each term of the multiplicand by each term of the multiplier, placing the first term of each partial product under the term of the multiplier which produces it, (5) Add the colum?is. b(j ALGEBRA. 1. Multiply a^ - 2al + 35= by a^ + dab - Vj\ a" - 2ab + db^ a^ + dab - W a* - 2a^5 + daW + da^b - Mb^ + ^aW - Wb^ -f 4:ab^ - GZ>* a* 4- rt'5 - 5ft2^>2 + 13«/^» - G^>* 2. Multiply 1 + 2« + rt* - 3^2 by a^ - 1 - 2a, «* + - 3«2 + 2« +1 «3 _|_ _ 2« - 1 -2a' - + 6«^ - 4^2 _ 2a -a' - + 3^2 - 2« - 1 a^ + - 5ft^ + a* + 7a^ - a^ - 4a - 1 3. Find the product of (x + a)(a; + b){x + 6-). a; -\-a X -{-b a^ + a\x ^^ + (^ + ^)^ + (^b X -\-c a? -\-a -Vb a^ + aZ* + ac -^bc + a^c 4. Multiply X -\- x^ -{- 1 hj X — x^ -{- 1, X -\-x^-\-l X —2:^+1 x^ -\- X'^ -\-x — X^ — X — x^ + a; + a;i + 1 aj^ + O +a; + +1 MULTIPLICATION-. 57 5. Multiply X — 1 -\- x-^ hj X -{- 1 -\- x'K X - l-\-x-^ X 4-1 + a:-^ x?-x-\-l -\-x-l-\-x-'^ + 1 - a;-i + a:-2 2:^ + + 1 + +~r^ 6. Multiply 9a2 + Z^^ + 3a^>-6« + 4 + 2^ by 3a + 2-^>. M + 3a (^> - 2) + (Z/^ + 2Z> + 4) 3rt - (^-2) 21a^ + 9«2 (^ - 2) + 3a (b^ + 2<5> + 4) - 9a^ (^ - 2) - 3a {b^ - 4^> + 4) - {b^ - 8) 27^+0 H- 3a (6^) - (^»» - 8) = 27a« + ISab - (P - 8). It will be observed that {b -'^) X {b -2) = b^ - ^b-\- 4, and (^2 + 2b + 4) (^ - 2) = ^'^ - 8. 7. Multiply o(^ -{- y"^ — xy -\- x -^ y hy X -\- y — \. x^-x(y-\)-{- (/ + y) ^ + (y - 1) 01? - x'(y - I) -^ x(y^ + y) J^x^(y-\)-x(f-2y-^\) + {f-y) a:« + +a;(3y-l) -^ (f - y) 8. Multiply a^ — a; (a — i) -|- Z>^ by a -|- ^ + ^' - :r (a - ^) + (a^ + ^2) ^ + (^ + M -x'(a-b)^x{a'^b'') -x(a?-l^)-^ (a^ + a^J + ab^ + Z>«) 58 ALGEBllA. EXAMPLES. By multiplication show that : 1. {a^hY = a'-{-2al)-{-hK 2. {a - hf =0"- "lab + l\ 3. {a^V){a-l) = a^-l\ 4. («^ -ab^ b')(a -{- b) = a^ -\- b\ 5. {a^ -\-ab^ b'){a - b) = a' - b\ 6. {a" - 2a + l)(a2 + 2a H- 1) = a* - 2a« -f 1. 7. (a' + 2^2 + 4a + 8)(a - 2) = a* - 16. 8. (a' - ^ab + 2^2) (a^ + Ub + 2Z>^) = a* + 4&*. 9. (a*-a^+l)(a^ + l) = a« + l. 10. (a* + a« + l)K-«' + l) = ^'+«*-f 1. 11. {a + ^)^ (a - by = n'- 2a^^ + Z>*. 12. (a; - 3)2 (a; + 3)' = x^- 18^2 + 81. 13. {a -{- x){a — x)(a^ + x^) = a* — a:*. 14. (a^ - 2)(a'^ - 3) = a^ - 5a^ + 6. 15. (3^ - 1)2 = 9^ - 2 (S"') + 1. 16. (a^ - byy =-- a^^ - 2a^^>^ + ^>2^. MULTIPLTCATIOJ^ BY DETACHED CO-EFFICIENTS. 124. An examination of the solutions in (123) will show that if the terms of the two factors be arranged and graded (by supplying ciphers in place of the missing powers of the letter of arrangement, and collecting like powers in brack- ets): (1) The like powers of the letter of arrangement will be in the same column. (2) The powers of that letter will descend or ascend in the product as in the two factors. (3) If both factors be arranged according to the powers of two or more letters, the product will also be so arranged. MULTIPLICATION^. 59 By observing these laws, the letter or letters of arrange- ment may be omitted in the solution and placed in the result. Compare the following solutions with those in the preceding article: 1. (a' - 2ab + 3b') X (a' + dab - 2b') = ? 1-2 + 3 1 + 3-2 1-2 + 3 +3-6+9 -2+4-6 1 + 1-5 + 13-6 A71S. rt* + a^ - 5a^b' + 13^^'' - 6^**. 2. (a* - oa' + 2« + 1) X (a' - 2a - 1) = ? 1+0-3+2+1 1+0-2-1 1+0-3+2+1 -2-0+6-4-2 -1-0+3-2-1 1+0-5+1+7-1-4-1 Ans. flr7 -^ - 5a^ + a* + 7a^ - a^ - 4a - 1. 3. {^ + a) (x + b) {x 1 + a 1 + 5 1 + a + 5 + a5 + c)-- = ? l-}-(a + b)i-ab l-\-c 1 + (a + J) + ad + c + (aiJ + be) + abc 1 + (a + & + c) + (a& + ac + 5c) + abc Ans. a^ -{- (a -{- b -{- c) x' -i- (ab -{- ac -j- be) X -\- abc. 60 ALGEBilA. 4. (9«2 + Z»^ 4- 3^5 - 6« + 4 + 21) X (3« + 2 - Z*) = ?. Arrange according to the descending powers of a, and omit a. 9 + 3 (^ - 2) + (^2 + 21) + 4) 3- {b-2) 27 + 9 (5 - 2) + 3 (^2 + 2^> + 4) _ 9 (^, _ 2) - 3 (^*« - 4^> + 4) - (Z.3 - 8) 27 + +3 (6^) - {h^ - 8) Ans. 27«^ + 18a^> - (b' - 8). 125. The principles of the preceding article may be employed for the purpose of verifying results obtained by multiplication. Thus: Omit the letter or letters of arrange- ment, and simplify the results. The sum of the co- efficients in one factor, multijMed hy the sum of the co- efficients in the other, must be equal to the sum of the co-efficients iii the product. For instance, in Ex. 1 (124), the multiplicand is 1— 2 + 3=2; the multiplier is 1 + 3-2=2; and the product is 1 + 1-5+ 13 -6 = 4. EXAMPLES. By detached co-efficients show that: 1. (a;* - 2a;3 + 3.^^ - 2^; + 1) (:^-* + 2a;3 _|_ 3^ ^ 2a; + 1) = Q^ + 2x^ + 3a;* + 2r^ + 1. 2. {7? - bx^ -\-12^a? - a? - X ■\- 2) (a;2 _ 2a; - 2) = x' - W + 21^;^ - 17a;* - 250.'^ + %j? - 2a; - 4. 3. {x - l)(a; - 3)(a; + 3)(a; + 1) = a;* - lOa;^ _|_ 9^ 4. [x'-x-^- l){x' + a; + l)(a;* - a;2 + 1) = a;« + ic* + 1. 5. {x + af = a;5 + 5a;*a + 10a;V + lOa;^^^ + 5a;a* + a\ 6. {x - ly = a.-^ - 6a;^ + 15a;* - 20a;3 + 15a;2 - 6a; + 1. 7. {x - 2) (a; - 3) {x + 3) (a; + 2) = a;* - 13a;2 + 36. 8. (a; — a){x -{- b) (x — c) = a^ — (a — b -{- c) x^ -{- {ac — ab — be) x + abc. 9. (- y -iy= -y^- 5f - lOy^ — 10/ - by - 1, MULTIPLICATION. 6^ Exercise VI. Perform the multiplication indicated: 1. (2a^ + ix' + Sx + 16){3x - 6). 2. {x^ + 4:X^ + 5a; - 24)(r2 - 4a; + 5). 3. {a^ -4:0^+ 11a; - 24)(a;2 _^ 4^ _|_ 5). 4. (a;^ - 2ar' + 3a; - 4)(4a;' + Sa;^ + 2a; + 1). 5. (^-xy + y^-i-x-{-y-{- l)(x-{-y- 1). 6. (a;2 - 3a; + 2)1 7. (a; — a){x + ^)(a; + c). - 8. (x' + 2a; + 1)1 9. (ar» - 3ar» + 3a; - l)(a;2 _ 2a; + l)(a; - 1). 10. (x'-x- l)(2a:« + 3)(a.'^ + a; - l)(a; - 4). 11. («' - 2 + a-^)(a^ + 2 + a-'). 12. (a^ -\-a^ — a~^){a^ — a^ -\- a~^). 13. {a^ + b^ + a-^b){ab-^ -a^-{-i^), 14. (at - aU^ + aib^ - bi)(d + a^^i + ai^i + b^). Simplify the following expressions: 15. (a; + yY - {x - y)\ 16. (a;2 + 2a; + 1)^ - {x^ - 2a; + 1)^ 17. {x + 2)* -(x- 2)*. 18. {a + J)(^ + c) - (c + ^)(^ 4- a) _ (« _j- c)(^ - ^). 19. {a-\-b-\-cf-Z(a^ b)(a + 6-)(J + c). 20. (a + 5 + c)^— a (^ + c—a) — b {a -\- c—b)— c(a -\- b—c), 21. (ar^ + a;y + t/'^)' - (^' - '^^V + y'Y- 22. (a;^ + a;y — ?/^)^ — (a;* — xy — y^)\ 23. (a; + 2/)^ - 4a;i/ (a; + yf + 2a;V«. 24. [(x + yY+{x-yYY, 25. Substitute y + 2 for a; in a;* — 2a;* — a;^ -|- 4. CHAPTER VII. INVOLUTION. 126. Involution is the operation of raising an expression to any required power. Every case of involution is merely an example of multiplication in which the factors are equal. Thus : (- 2a'y = (- 2a') (- 2a') (- 2a') = - 8a\ MON^OMIALS. 127. To raise a monomial to any power : Multiply the exponent of each factor of the monomial hy the exponent of the power. If the given expression he negative, and the exponent of the power be an odd numher, the product will he negative ; otherwise the product loill he positive (115). (1) (- 2aWy = - 2'a'h'' = - ^2aW^ (2) (- 3a'hc')'' = + d'a'b'c^' = + Sla'hYK (3) {2a^l^cY = 2htW(^ = 64:aW(f^. (4) (3V^.)* = 3 %«Z>*. (5) (- 3abic-')' = + d'a^h'c-'^ = Sla'h'c-'. (6) _ (_ 2aHh-^)' = - (4- 2'a'b'c-') = - Q4:a^¥c-\ (7) {-2ah')-'' = -2-'' a-^h-^^ = - -i^a-H-^\ (8) (— 2a^^)" = + 2"a"Z'^" if n be any even integer, but (— 2a^^)" = — 2"a"J^" if 7i be any odd integer. 128. Since every even power is positive, therefore : If any expression in brackets he affected hy an even exponent, the sign of each inclosed term may he changed without altering the value. Thus: (63) INVOLUTION^. 63 (a - by ={b- aY; {a - hf = (b - a)'; (a-b-cy=(-a-\-b-ir cY; etc. If any expression in brackets be affected by an odd ex- ponent, the sign of each inclosed term may be chayiged, provided the sign preceding the brackets be changed. Thus: (a^bY=-{b-aY\ (a-b^cY=-{-a-Yb-cY', etc. BINOMIALS. 129. By multiplying {a + b) by {a-\-b)y and that product by {a + b)y etc. (123), the following formulas are obtained: (a + by = «* + '^ab + b\ \a + by = a» + ^a^b + ^ab^ + b\ (a + by = «* + ^a^b + Qa^b^ + 4« J' + bK ' (a + by = tt*^ + ba^b + IQa^b'' + lOd'b^ + 5«J* + b\ {a + ^')« = «* + ea'^^^ 4- 15a*Z''' + 'ZWb^ + IS^^Z** + 6ft J^ + b\ By multiplying (« — b) by (« — J), and that product by (a — b)f etc., the following formulas are obtained: (a - by = a^ - 2ab + b\ (a - by = a^- Zn^ + 3«J' - b\ (a - by = a' - 4:a^b + 6a^b^ - 4:ab^ + bK (a - by = a^ - ba'b + lOa^^ - lOaW + bab* - b\ (a - by = a' - Mb + 15a**« - 20^'^^ + Iba^b*' - Qab^ + b\ By means of the Binomial Theorem, or Newton's The- orem, as it is sometimes called, any power of a binomial may be found without performing the successive multiplications. This theorem may be briefly stated thus : In the expansion of (a -\- by, n being any positive integer: (1) The number of terms is greater by one than the ex- ponent of the power to which the binomial is raised. Thus, if w = 8, the number of terms isw-}-l = 84-l = 9- (2) The exponent of a in the first term is the same as the exponent of the power to which the binomial is raised, and it decreases by one in each succeeding term. 64 ' ALGE1311A. (3) 1) appears in the second term with an exponent of 1, and its exponent increases by one in each succeeding term. (4) The co-efficient of the first term is 1. (5) The co-efficient of the second term is the same as the exponent of the power to which the binomial is raised. (6) The co-efficient of each succeeding term is found by multiplying the co-efficient of the preceding term by the exponent of a in that term, and dividing the product by a number greater by one than the exponent of h in that term. (7) Terms which are equidistant from the middle of the series, have equal co-efficients. If the number of terms be odd, there is one middle term, and its co-efficient is greater than that of any other term. If the number of terms be even, there are two middle terms whose co-efficients are equal, and greater than those of the other terms. (8) If a be positive and h be negative, the terms in which odd powers of h occur, are negative. 130. Expanding an Expression is performing the multi- plication indicated by multiplication-signs or exponents. 131. Given powers of any binomial may be expanded by means of Newton's Theorem. For example: Expand (2x — 3yY. For 2x put a, and for 3^ put h. Then (2x - ^Y = {a- by. By Newton's Theorem, (a - if =■ a^ — 6a'b + lOa^^ — lOa^b^ -i- 6ab^ - b\ . •. (2x - 3yf = (2xY - 5 (2^:)* (3?/) + 10 (2xf (dijf - 10 {2xY (3yY + 5 (2x) {ZyY - (3//)^ = Z2x^ - 2^0x^y + TZOxY - ioSO^Y + 810:ri/* - 243?/^ In like manner, powers of other polynomials may be ex- panded. Thus {a-\- x — yY may be expanded by inclos- ing X — y m brackets, and applying Newton's Theorem to the resulting binomial, as follows ; [« + {x-y)Y = «« + Za' (x-y) + 3« {x-yY + {x-yY = a^ + 3A-3a2|/ + 3ax^-6axy-}- Zay'' -j- 7?-ZQ^y-\- Zxy^-yK INVOLUTION. 65 132. It may be shown by actual multiplication that : (« + Z> + c)2 = a^ + Z'^ + c' + 2a {b-^c)-\- 2bc. (a-h + cy = a^-{-b^ + (^^2a(-b-i-c)- 2lc. The following rule may be observed to hold good in the above and similar examples: The square of any polynomial consists of the square of each term, together with twice the product of each term hy the sum of all the terms which folloio it, 133. Since {a -\-h ^ cf = a^ -\- U" -\- (^ + Sa^ {p j^ c) -\- 3^2 (rt + c) + 36-^ (a^h) -{- 6abc, and (f^ - ^ + cY = a^ - b\-\- c^ + 3^^ (- J + c) + W (« + c) + 36-2((? -b)- Qabc, therefore: The cube of any polynomial consists of the cube of each term, plus three times the square of each term into the sum of the other terms, plus six times the product of the terms taken three at a time, 134. By detaching the co-efficients, the results obtained in accordance with the preceding formulas may be verified, as may also the following : (1) In the expansion of {a -\- by the sum of the co-effi- cients is (1 + 1)" = (2)^ (2) In the expansion of {a — by the sum of the co-effi- cients is (1 — 1)" = (0)" = 0. Since the sum of all the co- efficients is zero, the sum of the co-efficients of the positive terms must be numerically equal to that of the negative terms. But all the odd terras are positive, and all the even terms are negative; therefore: In the expansion of (a ± by the sum of the co-efficients of the odd terms equals the stim of the co-efficients of the even terms. (3) In the expansion of {a-\-b -\- cY the sum of the co- efficients is (1 4- 1 -f- 1)' = (3)2 = 9. In (a + Z> 4- cY the sum of the co-efficients is (1 + 1 + l)^ = (3)^ = 27. (4) In the expansion of {a-\-2b — 3<;)" the sum of the co-efficients is (1 + 2 - 3)" = (0)" = 0; etc. Alg.-6. 66 ALGEBRA. EXAMPLES. \ Expand the following expressions : 1. {x - y)\ Ans, x' - Ix^y + 21^y - 35^;*// + %^7?y^ — 2ixY + "^^y^ - / 2. {x-\-yy. Ans. a;* + 8^;^?/ + 28.ry -f 56^y + 70.cy ^ 56:cy + 28:cy + ^xy' + /, 3. {x — y)~ *. ^7^5. (re* — 4:0^y -\- ^x?y^ — Axy^ + 2^) ~ ^' 4. (2^ - 3^')^ Ans. Sa^ - 36«2^> + 54aZ»2 - 27^>^ 5. (1 - 2:c + .^2)1 ^^^5. 1 - 4a; + 6a:2 - 4a;3 + 2:*. 6. (1 - 22; + 32;«)2. ^^5. 1 - 4.'z; + lOx^ - 12cc^ + 9:?;*. 7. (rt + Z>-c)^ ^W5. «^ + Z*^-6'^+3ri2^»-3a^c+3aJ«- 3^2^ 4- 3«6-^ + 3^6'^ - Qahc. 8. (1 - 3.^ + 3.^'2 _ .^3)2^ ^^^^^ 1 _ 6a; + ISa;^ - 20.^-3 -f 152;* - 60;^ + :r<^ Exercise VII. Expand and simplify the following expressions: 1. (X - yf. 13. («i-3)*. 2. {x-y)-K 14. (-2a'* + «-3«-y. 3. {a - 2by. 15. i-x-yy. 4. {3x-2y)-K 16. {x-2x^-\-x-^y. 5. {x + ^)* - (^ - !/Y^ 17. (bx ± ayy. 6. (x- 1 + ^-1)2. 18. (1 ± 2xy. 7. {x^-2-\-x-^y. 19. {a + Ix + c:?;^)^. 8. (a^ - 2a + 1)^ 20. (« - 2Z> + c)^ 9. {x-^y- ly. 21. {ax^x^)-\ 10. (a— & — c4- dy. 22. (l+^4-^y-(l-^+2^^) 11. [(a+J)«+(«,-J)2p. 23. {^-2x^x'y-{2-xy. 12. [(a 4-1)3 + (« -1)3]- K 24. {l^2x-Zx^^^xy.^ CHAPTER VIIL DIVISION. 135. Division is the process of finding a factor which, being multiplied by a given factor, will produce a given product. 136. "With reference to this operation, the product is called the dividend, the given factor is called the divisor, and the required factor is called the qiwtient. 137. Since the dividend is the product of the divisor and the quotient, it follows from the law of the signs in mul- tiplication that if the dividend be positive, the divisor and quotient have like signs; if the dividend be negative, the divisor and quotient have unlike signs. That is: iX) -\-ah-^{-\-a) = -^h; (3) - ah -^ {-^ a) = -h; (2) + fl^> -^ (- a) = - b; (4) - ab -^ (- a) = -\-l). From the first and fourth of these formulas we derive the rule: If the dividend and the divisor have like signs, the sig7i of the quotient is plus. From the second and third of the above formulas we derive the rule: If the dividend and the divisor have tm- like signs, the sign of the quotient is minus, 138. Since the numerical co-efficient of the dividend is equal to the product of the co-efficients of its two factors, therefore : The numerical co-efficiefit of the quotieiit is found by dividing the numerical co-efficient of the dividend by that of the divisor. Thus, since 5a X 46 = 20ab, there- fore 20ad -J- 46 = ba, (67) 68 ALGEBRA. 139. Since the exponent of any letter in the dividend is equal to the sum of the exponents of that letter in its two factors (117), therefore: The exponent of any letter in the quotient is found hy subtracting the exponent of that letter in the divixor from its exponent in the dividend. A letter which has the same exponent in both dividend and divisor, is not written in the quotient. (82.) Examples. — i. Since a^ x a^ = a^, ,-. a^ ^ a^ = a^, and a^ -^ a^ = a^. 2. Since a^Xa~^=a^, .-. a^-^a^=a~^, and a^4- rt~^=rt^ 3. Since a-"" h X a'b^ ■= h\ ,- . h"^ -^ a-^ b = a%\ and b'-^a^' = a-H, A MONOMIAL BY A MONOMIAL. 140. By means of (137, 138, 139) we may divide one monomial by another. EXAMPLES. 1. a^ -^ ab = a\ 4. rt^ + ^ -^ a^-' = a^. 2. a*b^ -^ a^b^ = a^, 5. a'b^c ^ a^bc^ = b^c-\ 3. a"^ -r- a"^-' = a\ 6. {a - bf-^{a-bf = {a - bf. 8. {a - cY^'' -^ {a - cY-'' = (a - cf. 9. a^b-^c^-^a'bc''=a-H-*c''-\ 10. 2* + J' -^2^ + 2^2^-1 A POLYNOMIAL BY A MONOMIAL. 141. Since {a — b) c = ac — be, . • . (ac — be) -i- c = a — b. Since [a — b){—c) = — ac -{-be, .*. (— ac -\- bc)-^{— c) = a — b. Hence, to divide a polynomial by a monomial. Divide each term of the dividend by the divisor. DIVISION-. 69 EXAMPLES. 1. {a'h - aW - aP) ^ ah = a - a^h - l\ 2. (ac — a(?-\- ac") -^ ac = 1 — c -{- c^, 3. {ah^ + a'b^ - a^') -^ a^l? = a-^-\-l)- ah\ 4. (_«"* + a "* + *-« "^ + *) -|-rt"»-2= _fl2 -I- «*-««. 5. (a^h^c-ah(^^alc)-^{-aljh^) = -ahc-^^b-^-h-''c-K A POLYNOMIAL BY A POLYNOMIAL. 142. If the divisor (given factor) = a -\-i -\- c, and the quotient (required factor) = x -\- y -\- z, ( ax 4- bx + ex then the dividend (product) — \ -\- (f^V -\- by -\- cy \ -\- az -{- bz -\- cz. The first term of the dividend, ax, is the product of the first term of the divisor, «, by the first term of the quo- tient, X ; hence, to find the first term of the quotient. Divide the first term of the dividend by the first term of the divisor. If the partial product formed by multiplying the entire divisor by x be subtracted from the dividend, the first term of the remainder, ay, is the product of a, the first term of the divisor, by y, the second term of the quotient. Hence, to find the second term of the quotient. Divide the first term of the first remainder by the first term of the divisor. In like manner, if the partial product formed by multi- plying the entire divisor by y be subtracted from the first remainder, the first term of the second remainder, az, is the product of ^ + 6-V a-b-{-c. Ans. ce - 2ab -^ b^ - ac ^ be -\- c\ 18. a" (5 + c) + W (a - c) + 6"2 (^ - ^) H- rt<^c; a -\- b -\- c. Ans. a{b -\- c) — be. 19. {a — by — 2 {a — b) e -{- c^; a — b — c. Ans. a — b — c. 20. (a-bY-^^a-bYc+3(a-b)€^-\-(f; (a-bY^2(a-b)e+c'. Ans. a — b -\- c. 21. W - 24a: + 58a;-i - 21a;- V To; - 3a;- ^ ^7^5. a;2- 3 ^lx-\ 22. (.r^ + ^^) {x' - f); (a;^ - xy -f /) (2; - y). Ans. x^ -\- 2xy -\- y^. divisions', 73 SYNTHETIC DIVISION". 144. An examination of the solutions in (143) will show that they may be abbreviated in several respects : (1) The products of the first term of the divisor by the various terms of the quotient are unnecessary, as they are the same as the first terms of the corresponding dividends. (2) Since only the first term of each remainder is used for purposes of division, the other terms need not be found. (3) By changing the signs of the terms of the divisor in the first instance, the subtraction of the various partial products is avoided. (4) Since the like powers of the 'letter of arrangement occur in the same column and follow a uniform law, such letter may be omitted in the solution and supplied in the answer. These suggestions are embodied in the following solutions of four problems considered in the preceding article. For the sake of conciseness, the divisor is written at the left in a vertical column. The first term of the divisor is separated from the others, because it is used as a divisor and not as a multiplicand. The signs of all the other terms of the divisor are changed. The quotient is written below the dividend and partial products. -6 ft*- 4a'- 19^2 4- 5^2 + 106ft-120 — 6a -lOOff -1-120 Without detaching co-efficients. a»+ a -20 I -f +0 Quo. a^ -}- a - 20. -f- 106 — 120 By detached co-efficients. +5 -6 Alz. 1-4-19 -f5- 6 + 5 -6 -100-fl20 l--fl-20|+ + -7. Quotient, a^ -\- a — 20. 74 ALGEBRA. Explanation". — After the terms are arranged and graded, divide the first term of the dividend, rt*, by the first term of the divisor, c^, and write the first term of the quotient, c^, under the dividend, as shown above. Multi- ply 5a — 6 (the terms of the. divisor, except the first, with contrary signs) by c^, and write the product under the dividend in the second, third, etc., columns. Add the second column and divide the sum, d^, by a^, the first term of the divisor. Write the quotient, -f a, under the column added. Multiply 5a — 6 by a, and write the product under the dividend, beginning in the third column. Add the third column, and divide the sum, — 20^^, by c^. Write the quotient, — 20, under the third column. Mul- tiply 5« — 6 by — 20, and write the product as before, beginning in the fourth column. The sum of the other columns being 0, the division is exact, and the quotient is 0? -^ a — 20. It is well to draw a second vertical line, to denote where the division ends. The number of columns to the right of this line is one less than the number of terms in the divisor. If the letter of arrangement be omitted, the work is still shorter, as is shown in the second solutions. x^ ^6 _ 4^ _^ 3^* _^ ^^ + 4a;2 + - - 15 + 3 -0 + 52;^ + + 15 -fl + + 3 a;» - 3:c« + +5 11-4+3+2 + 1 + 3 3-0 3 + + 5 1+0 +0+0 4 + 0-15 By detached co-efficients. -9 + 5 + + 15 +0+0+ la;^ - 30.-2 + 0.Z; + 5 = a;2 - 3^_|_ 5. DIVISION". 75 3. Divide x^ + x^y + x^z — xyz y^z yz^ \i^ X? — yz. + x^J^x^y-^z) + -\-xyz + -^ {ijh -\- yz^) a: + (y^z) + + + '\ry^ 1 + (2/ + 2;) - 2/^ - (^'^ + 3/^') + + ?/^ + + {fz + y^^) l + (y + ^). . 4. Divide a:^ + ^^ 1^7 ^ ~" ^' 1 Quotient, x -^ {y -\- z). 7^-^ ax + a'' -|-2a» 1 + + + 1 + 1 1 + 1 + 1 + 1 +_1 + 2 §wo., a:^+ ax + rt^+ 2a» ic— a EXAMPLES. Divide the first expression by the second: 1. - 10a;* + Ibx" - bx^; - bx". Ans. + 2x^ + a; - 3. 2. x^ — 40x - 63; X - 7. Ans. x^ + 7a; + 9. 3. 2*+ x^- 9a;2- 16a; - 4; a^ -\- 4x -{- 4. Ans. x^- 3a; - 1. 4. x'-a:i^-{-(b- - 1) a;^ + «a; - o; x^ - 1. A ns. a;* — ax + J. 5. l-\-x'y-^-x- -'y-yh a;t - ip. Ans. ^•^^-^ + 1 + x-^-\-x -y. 6. o;* — ?/*; a;~^ + «/~\ ^?is. a;*«/ — x^y^ + a;y — xy*, 7. a;^ — ?/^; x~^ — y~^' Ans. x^ y^ + ^^^^ + ^^^ + x^y^' 76 ALGEBEA. Exercise VIII. • Divide the first expression by the second : 1. - Sa'dc^; 4.abc. 2. 6a'b - 12aP + 18a'b^- - 3ab, 3. {771 — ny {m -\- nY; — {m — n)(m -}- ny, 4. a' -6a-\- 5; cf -2^ + 1. 5. a;* + 42; + 3; x"" -\- 2x -\- 1. 6. a' - 4.a'b' - 8aW - llab' - 12b'; a^ - 2ab - ZbK 7. x^ — {a -{- b) x^ -\- (c -\- ab) X — ac; x — a, 8. x' — ax^ -\- b:it? — b:ii? -\- ax — \\ x — \, 9. a^ — b^ — 0^ — dabc; a — b — c. 10. a^ — 8f + 6xy -{- 1; 1 — 2y -\- x. 11. a' - b'j {a -b)(a-{- b). 12. {a + by + 86^- a^b-\-2c. 13. 2a-^- 3a-2+«-*- «-^• -a-\ 14. (a -^iy-2{a-\-iy-^a-\-l', a-\-l, 15. {a — b)^ — (a — b) cd -\- a — b; {a — b)^. 16. 3^ {a-b)- 3^ {a - by - 3* (« - bf; 3"^ (^ - b)-\ 17. {a-{-b-cY-{a-\-b-o)''-y -{a^b-cY; («+Z'-c)^+»'. 18. a;* + a;2 + l; 2^-0:^+1. 19. (a' - 2bcy - m&; a" - 4:bc. 20. (a^-b'){a'-b'); {a - by. 21. a^ - 2a^ + 2a- ^ - a'^j a^ + 1- a'k 22. 2a^ + 2b^ + 2c'- 6aJc; (a - by + (^> - c)^ + (c - «)l 23. {a^ - by; {a^ -^ ab -{■ by. 24. a-^-{b-c)-^; a-^- (b - c)-\ CHAPTER IX. EVOLUTION. 145. Evolution is the process of finding any required root of an expression; that is, evolution is the process of finding a number which, being raised to -a proposed power, will produce a given number. MONOMIALS. 146. Since evolution is the opposite of involution, there- fore the rules for the extraction of roots are formed by reversing those given in Chapter VII. Since ((fY = a'"" (Art. 127), therefore l'o^= a\ Since (- 2^^^)=' = - 2V = - 8a*, . • . v^^Ts^ = - 2a\ Since (± a^y = + a'\ . • . i/^=_± a\ Since (±rt-y = +«-«, .-. Vii^^ = ± a-\ Since (-J = -3, therefore the cube root of -7-3 is j-. Since no even power is negative, therefore an even root of a negative number is impossible. Rule for Extracting any i-equired Root of a Monomial: (1) Sign. If the given monomial be positive, any odd root will he positive, and any even root will have the double sign (±). If the given monomial be negative, any odd root will be negative, and any even root will be impossible. (2) Numerical Co-ej^ficient. Separate the given nu- merical co-efficient into as many equal factors as there are units in the index of the root. One of these factors will be the numerical co-efficient of the required root. (77) 78 ALGEBRA. (3) ExpoiTEN'T. Divide the exponent of each factor ly the index of the root; the quotient will he the exponent of that factor in the required root, 147. If the root of a number expressed in figures is not readily detected, it may be found by resolving the number into its prime factors. For this purpose the following prin- ciples may be advantageously employed: (1) Every number is divisible by 2 when its right-hand digit is 0, 2, 4, 6, or 8. (2) Every number is divisible by 2^ = 4 when the number denoted by its two right-hand digits is divisible by 4. (3) Every number is divisible by 2^ = 8 when the number denoted by its three right-hand digits is divisible by 8. (4) Every number is divisible by 3 or 9 when the sum of its digits is divisible by 3 or 9. (5) Every number is divisible by 5 when its right-hand digit is or 5. (6) Every number is divisible by 11 when the sum of the digits in the odd places and the sum of the digits in the even places differ by zero, or by a multiple of 11. (7) Every number is divisible by a composite number when it is divisible by all the factors of that number. 148. An integer is prime when it is divisible by no integer except itself and unity. Examples: 2, 5, 7, etc. An integer is composite when it is divisible by some other integer besides itself and unity. Examples: 4, 6^ 9, etc. SQUARE ROOTS OF POLYNOMIALS. 149. Since the square of a -\- i \^ a^ -\- 2ab -{- i^, .'. the square root of a^ -f~ ^^^ -\- b^ is a -\- b. Since the square of a -{- b -{- c is {a -\- by -{- 2 (a ■-{- b) c -{- c^y . • . the square root of (a + ^)2 + 2 (a -^b) c-\-c^ is (a -i-b)-\- c. Similarly, the square root of (a -\- b -\- cy -\- 2 (a -\- b -^ c) d -^ d^ is (a -\- b -{- c) -{- dj etc. All of these squares are of the EVOLUTION. 79 general form, a^ + 2^*^^ + ^^ i^ which a stands in succes- sion for one term, tAVO terms, three terms, etc. The rule for the extraction of the square roots of pol3'noraials is formed, therefore, by finding in what manner a + J is derived from a^ + 'iah + h^, or its equal, a^ -\- (2« + h) h. 2a-]-I) 2ab + P 2ab + b^ The first term, a, of the root is evidently the square root of the first term, a^, of the given expression. If a^ be sub- tracted from a^ + 2ab -\- b^, the remainder is 2ab + b^- Therefore b, the second term of the root, is obtained by dividing 2ab, the first term of this remainder, by 2a; that is, by twice the part of the root already found. Multiply 2a -{- b by b, and subtract tlie product, 2nb -f ^^ from the remainder. If the root contain more terms, we pro- ceed with a -{- b iis we formerly did with a; its square, a^ -\- 2ab -\- b^, has already been subtracted from the given expression, and the first term of the remainder is 2 (« -[~ ^) ^j so we divide this term by 2 {« + ^)^ ^^id the quotient is c, the third term of the root. Multiply 2 (« -|- J) + ^ by c, and subtract the product, 2 (« + ^) c + ^^i from the second remainder. This process must be continued until the required root is found; the trial divisor in each case being double the part of the root already found, and the complete divisor being the sum of the trial divisor and the new term of the root. Examples.— 1. '/SG:?.'^ - 60x?/ + 25/ = what? 36:^2 - mxy + 2'o\f \^x-hij 36r?;2 12a; — hy - QOxy + 25//2 — QOxy + 25i/2 Ans, ± {6x ~ 5y). - 2x' - x^ 2x^ + 22; + 1 2x^ + 2a; + 1 80 ALGEBRA. The square root of the iirst term is Qx, and Gx is placed at the right of the given expression for the first term of the root. The trial divisor is 12a;, and the second term of the root, — 5?/, is obtained by dividing — QOxy by 12a:. The complete divisor is 12a; — 61/. 2. Find the square root of x*" — 2x^ — x^ -\- 2x -^ 1. x^ -2x^- a;2 + 2a: + 1 1 a:^ - a; - 1 2a;^-a; 2a;2 — 2a; - A71S. ± (a^ — X — 1). The square root of a;* is a;^, . * . x^ is the first term of the root. The first term of the remainder is — 2a;^ and the first trial divisor is 2x^; . • . — 2a;^ -^ 2a;^ = — ,'c is the second term of the root. The complete divisor is 2x^ — x, and the product of 2a;^ — x into the second term of the root is — 2a:* + x^. The second trial divisor is 2a;^ — 2a*, and the first term of the second remainder is — 2a;l Since — 2a;^ -^ 2a;^ = — 1, . • . — 1 is the third term of the root. The second complete divisor is 2a;^ — 2a; ~ 1; etc. 150. The fourth root of an expression may be found by extracting the square root of the square root ; the eighth root, by extracting the square root of the fourth root; the sixteenth root, by extracting the square root of the eighth root; etc. 151. Since (a ± hf = a" ± 2ab + h\ therefore: (1) The square root of a trinomial, which is a perfect square, may he found by extracting the square roots of the ter^ns luhich are squares, and placing the sign of the re- maining term between them. Thus, the square root of 9a;^ — 6a; + 1 is ± (3a; — 1), because the square root of Gar* is 3a;, of 1 is 1, and the sign of the remaining term is — . EVOLUTIOiT. 81 (2) If the first and tliird terms of a square he givetiy the second term is found by taking twice the product of the square roofs of the given terms. Example : The first and third terms of a square being 9a:* and 4?/^ tlie second term is ± 2 V^ X Vltf = ± 2 X 3a;2 X 2/ = ± 12a:y. (3) If the first and second terms of a square he given, the third term is found hy extracting the square root of the first term, dividing the given second term hy twice this root, and squaring the result. Example : The first and second terms of a square are IGa;^ and ± 2^xy. To complete the square proceed as follows : (1) 1^16^ = 4^:. (2) 2 (4a:) = 8a:. (3) 24a:?y -^ 8a: = 3^. (4) (3//)'^ = ,9/. . • . the third term is -|- ^y^, and the complete square is IQj?. ± 24a;// -f 9y^. (4) A binomial can not be a perfect square, for the square of a monomial is a monomial, and the square of a binomial is a trinomial. EXAMPLES. Extract the square root of each of the following : 1. 4rt* - 12^'* + 25ri« - 24rt + 16. Ans, 2^^ - 3« + 4. 2. 16a:« - 24a:5 + 25a:* - 20:^ + lOa:^ - 4a: + 1. Ans. 4a:' - 3a:2 + 2a: - ^^ 3. 9^2 + 12«^ 4- W' + Grt6' + 4Z>c + t\ Ans. 3^ + 2^ + c. 4. 1 - G.f + 15a;2 - 20ar* + 15a;* - Ga.-^ -f x\ Ans. 1 - 3a: + Sa:^ - t?. 5. 4a« + 5rt« - 11a* - 4^5 + 14r?' - 12^ + 4. Ans. 2«' — a"^ — 3^ + 2. 6. 25a:* - 30«a:' + ^Mx^ - 24:a^x + 16aK Ans. 6a^ — Sax -\- 4:a\ Extract the fourth root of : 7. a* - 4.a^ + Qa^^ - 4ab^ + hK Ans. a - h. 8. a* - M + 24^2 - 32rt + IG. Ans. a - 2. 9. IGrt* - 9Ga'^ 4- 2l6a''h^ - 216ah^ + 8U*. Ans. 2a - 3h, 82 ALGEBRA. Extract the eighth root of : 10. a^-8x^-{- 28x' - 562;^ + 70a;* - 56.r' + 28^:^ -8x-^l. Ans, X — 1. 11. [(x^ + x-'Y - 4 (a; + x-y + 12]l A7is, x - x-\ Complete the square in the following expressions : 12. fl^* — 6^2 _|_ ( y ^^^g^ ^i _ Q^2 _p 9^ 13. 16a' - 2ia'b' + ( )• A7is. 4V - 24:a'b^ + 9Z»*. 14. 9a^ - ( ) + 4Z>*^ .-l^is. 9a^ - 12a^b'' + 4^*. 15. (a + Z,)* - 4 (a + /^)n« - ^0 + ( ). Ans. (a + ^)* - 4 (^ 4- bf (a - b) + 4: {a - b)\ 16. {a-bf^{ ) + 81c^ ^jz."?. (6f - Z')2 -f 18 (a - Z>) c + 8l6'2. SQUARE ROOTS OF ARITHMETICAL NUMBERS. 152. The method of finding the square roots of arith- metical expressions is similar to the preceding, but in the first place it is convenient to mark off the figures in periods. Since 1 = 1\ 100 = 10^ 10000 = 100^ 1000000 = 1000^ etc. ; therefore the square root of any number between 1 and 100 lies between 1 and 10; the square root of any number between 100 and 10000 lies between 10 and 100; the square root of any number between 10000 and 1000000 lies between 100 and 1000; etc. In other words, the square root of any number expressed by one or two figures is a number of one figure; the square root of any number expressed by three or four figures is a number expressed by two figures; the square root of any number expressed by five or six figures is a number expressed by three figures; etc. If, therefore, a dot be placed over the units' figure of a number, and also over every alternate figure, the number of dots will be equal to the number of figures in its square root. EVOLUTION-. 83 To extract the square root of 5329. 5329 4900 140 + 3 429 429 I 70 _i_ 3 Since the square of 70 is 4900, and the square of 80 is 6400, there- fore the square root of 5329 must be greater than 70 and less than 80. Taking 70 as a in the general for- mula a^ + 2ah -\-W- 5329, and subtracting a^ = 4900 from a^ + 2ab -}- b^ = 5329, we find 2ab -j- b"" = 429. Since 2a = 140, then b can not be greater than 3. Taking Z» = 3, 2a -{-b = 143, and bi2a -{- b) = 429. There being no re- mainder, 73 is the required square root. In practice the ciphers are usually omitted; thus, 5329 I 73 49 143 429 429 L53. The same method will apply to numbers of more than two periods by considering a in the formula a^ + 2ab -{- b^ = (a -{• by at each step to represent the part of the root found; that is, a represents so many tens with respect to the next figure of the root. 1. Find the square root of 11607649. 11607649 I 3407 9 2a = 60 2ai = 680 2ao 6800 64 260 6807 256 47649 47649 154. From the arithmetical rule for the multiplication of decimals, it is plain that if a number have any number of decimal places, its square will have twice as many ; therefore the number of decimal places in every square decimal will necessarily be even, and the number of decimal 84 A.LGEBRA. places in the root will be half that number. Hence, if the given square number be a decimal, and therefore one of an even number of places, place a dot over the units' figure, and then over every alternate figure on both sides of it. The number of dots to the left of the decimal point will be the number of figures in the integral part of the root, and the number of dots to the right the number of decimal places. The square root of 11607649 being 3407, the square root of 116076.49 is 340.7; of 1160.7649 is 34.07; of 11.607649 is 3.407; etc. 155. If a number have an odd number of decimal places, its exact square root can not be determined. Thus, . 144 is not a perfect square. In every case, if there be a remainder after finding as many figures in the root as there are periods in the proposed number, the exact root can not be deter- mined. The approximate root may be found to any required degree of accuracy by annexing ciphers and continuing the operation. 2 and 3. To find the square roots of 12 and of .144 to five places: 12.0000000000| 3.46410 9 64 300 256 686 4400 4116 6924 28400 27696 69281 70400 69281 92820 111900 1^12 is between 3.46410 and 3.46411. ,1440000000 9 .37947 67 540 469 749 7100 6741 7584 35900 30336 75887 556400 531209 25191 4/.144 is between .37947 and .37948. EVOLUTION. 86 EXAMPLES. Extract the square root of : 4. 33124. Ans. 182. 8. 6.5536. Ans. 2.56. 5. 47961. A71S. 219. 9. .194481. Ans. .441. 6. 4643.0596. Ans. 68.14. lo. 3.515625. Ans. 1.875. 7. 285.61. Ans. 16.9. ii. 3. Ans. 1.7320 +. Extract the fourth root of : 12. 6.5536. Ans. 1.6. 14. .00028561. A7is. .13. 13. 19.4481. Ans. 2.1. 15. 430'4.6731. Ans. 8.1. CUBE ROOTS OF POLYNOMIALS. 156. Since the cube of {a + b) is «' + 3^2^ + 3ah^ + ^.3, . •. tlie cube root of a^ + 30,^ + 3a^2 + Z/Ms a + ^>. The cube being given, a geneml rule for the extraction of the cube root may be deduced by observing in wliat manner a and b may be derived from a^ -\- 'da^b + ^ab^ + b^ or its equal, a^ + {da^ + dab + b^) b. a^ + '^a^b ^ 'dab'' -{- V" \a -\- b 3^2 + Ub + Z>* The first term, a, of the root is evidently the cube root of the first term, a', of the given expression. If the cube of a be subtracted, the remainder is da^b + Zab^ -\- b^; therefore the second term, b, of the root is found by dividing dc^b by 3a^ ; that is, by three times the square of the first term of the root. To this trial divisor add dab + W for the complete divisor. Multiply the complete divisor by b, and subtract the product, dc^b -f dab^ + ^^ 86 ALGEBRA. from the remainder. If there be but two terms in the root, this finishes the operation. If the root contain merer terms, the same method may be employed, by considering a to represent the part of the root already fonnd, and b to represent the new term of the root. In every -case, Tlie trial divisor is three times the square of the part of the root already found, and The complete divisor is formed by addiiuj to the trial divisor three times the product of the part of the root already found times the neiu term of the root, and the square of the neiv term of the root. If the root hQ a -{- b -{- c -\- d, (1) To find b, the trial divisor is 3fl^, and the complete, divisor is 3^^ -j- '^ab + W. (2) To find c, the trial divisor is 3 (fl^ -[- ^)^ ^^^ ^^e com- plete divisor is 3 (ri + by -\- ?> {a -\- b)c -{■ c^. (3) To find c?, the trial divisor is 3 (^5 -j- ^ + ^)^ ^^^ the complete divisor is 3 (« -f ^ + <^Y -\- ^^ {a -\- b -\- c) d -\- d^. Ex. 1. Find the cube root of 21x^ + b4:x'y -f ?>Qxy^ + ^K %W+ 542!^+ 36a;y'+ Sy'\^^ 2y 27x^ 3«2 = 3(8^^)2 =27i»* I 54x''y-{-dQxf-\-Sy^ 'Sab -{-b'= -I- I8a^ + 4y2 54x'-y-\- mxy'-{- 8y^ 3^2 _|_ 3^5 _|- b-2 = 27«^+ l&zjy -j- 4y^ Cube root, 3aj + The expression being arranged according to the descend- ing powers of x, the first term, 272;^, corresponds to a^ in the general formula; therefore the cube root of 2'7x^, which is 3x, is the first term of the root. The trial divisor is three times the square of 3x, that is, 27x^. Since 54a;^?/ -^ 27^;^ = 2y, therefore 2y is the second term of the root. To 27x^ add three times the product of the two terms of the root, 18xy, also the square of the second term, 4y^, and the sum is the complete divisor, 27:?:^ + I82;?/ + 4_z/^ Multiply this complete divisor by 2y, and subtract the product from the remainder. EVOLUTION. 87 Ex. 2. Find the cube root of x^ — Qx^ -\- 9x* + 4.i;^ — 9;i.2 _ (5^ _ 1, jins. x^ — 2x — 1. ~-6x'-{- ^x' + 43^-9x'-6x-l 3a« =3aj4 3(a + 5)2 = ar-* -12^^ + 12a;2 8(a-j-6)c= - dx'-\-Qx + c' = ' +1 3a^-l2x^-\- 9a;''' 4- 6a; + - 3ar* + 12a;=* - 9a;« - 6a; — 1 - 3ar* + 12a;3 - 9a;2 - 6a; - 1 The expression being arranged according to the descend- ing powers of x, extract the cube root of x^, and the result, x^f is the first term of the root. Subtract x^. The trial divisor being 33^, and the first term of the remainder being — 6x^, . • . — 6x^ -^ 3^* = — 2x is the second term of the root. To 3:?:* add 3 (.^•2)(- 2x) = - Gr'; also add (- 2xy = 4:X^, and multiply the sum, 33^ — Qx^ -\- 4:X^, by — 2x. Subtract the product, — G^* + 12a;* — 8^^ from the first remainder. The second trial divisor is 3 (:c* — 2xY = 3a;* — 122r* + 12a;*. The quotient of the first term of the re- mainder, — 3a"*, by the first term of the trial divisor is — 1; . •. — 1 is the third term of the root. To the trial divisor add 3 {x^ - 2x)(- 1) + (- 1)'' = - 3a;2 + 6a; + 1, and multiply the complete divisor by — 1. Subtract the product from the second remainder. CUBE ROOTS OF ARITHMETICAL NUMBERS. 167. Since 1 = l^ 1000 = 10^ 1000000 = 100^ etc., it follows that the cube root of any number between 1 and 1000 is between 1 and 10 ; the cube root of any number between 1000 and 1000000 is between 10 and 100; etc. In other words, the cube root of any number which has one, two, or three figures, is expressed by one figure; the cube 88 ALGEBRA. root of any number which has four, five, or six figures, is a number of two figures; the cube root of any number which has seven, eight, or nine figures, is a number of three figures; etc. Hence, if a dot be placed over every third figure of a cube, beginning with units' figure, the number of dots will be equal to the number of figures in its cube root. 158. If a number contain any decimal figures, its cube will contain three times as many. Hence, there must be one decimal place in the cube root for every three decimal places in the cube. If a given cube have decimal places, and a dot be placed over the units' figure, and over every third figure on both sides of it, the number of dots to the left of the decimal point will be the number of figures in the integral part of the root, and the number of dots to the right will be the number of figures in the decimal part of the root. 159. From the preceding article it follows that if the number of decimal places in a given number be not exactly divisible by three, the number is not an exact cube. If there be a remainder left after finding as many figures in the root as there are periods, then the given number is not a perfect cube. The approximate cube root of an imperfect cube may be found to any required degree of accuracy by annexing ciphers and continuing the work. 160. After pointing off the number into periods of three figures each as explained in the preceding articles, the work is continued, as in the extraction of the cube root of alge- braic expressions, by applying the formula (a -\- by = a^ + (da^ -\- dai + b^) b ; a representing the first figure of the root, then the first two, next the first three, and so on. In every case a represents so many tens, and b represents so many units. EVOLUTIOX. 89 1. Find the cube root of 820025.856. 820025.856 |93.6 a^z=z 9^ = 729 3«2 = 3 (90)2 ^ 3flZ> = 3(90)(3) = b^= 3' = 24300 810 9 91025 da^ -}- 3ab -i- b^ = 25119 2594700 16740 36 75357 3a'^ = 3 (930)2 ^ 3a'y = 3(930)(6) = b'' = 62 = 15668856 - 3a'2 + 3^'^'+ b'' = 2611476 15668856 L .- 161. If the successive terms of the root be represented by a, b, Cf etc., it will be observed that the first complete divisor = 3^2 -|- ^ab + ^^ and the second trial divisor = 3 (« + Z')2 = 3rt2 -f Gab + ^b\ Hence : To the first complete divisor add its second term and twice its third term, and the sum ivill be the second trial divisor. In like manner, each trial divisor may be found from the preceding complete divisor. EXAMJ>LES. Extract the cube root of : 1. 27a;* — b^x^y -j- ZQxy* — Sy\ Ans. 3x — 2y. 2. 82;«-36.r;^+66a;*-63a;*+33^-9^'+l. Ans. 22;2-3:z:+l. 3. x'y-^ - Qx*^ + 12:t^f - 8y\ Ans. x\i/-^ - 2y~. 4. rt»-9a»+36a^-84a«+126a^-126ri*+84«*-36a2-f9^-l. Ans. a* — 3^2 4- 3r^. - 1. 5. x'-\- QxY + Ua^y"- + 8?/«- 3 (.^2 + 2yy + 3.^2 + 6^2 _ j^ J^5. a;2+ 2^/2-1. 6. 1 + 12a; + 60a;2 + 160^;^ + 240a;* + 192a;^ + 64a,«. Ans. 1 + 4a; 4- 4a;'. Alg.-8. 90 ALGEBRA. 7. 34012.224. Ans. 32.4. lO. 6321363049. Ans. 1849. 8. 148:035889. Ans, 5.29. n. 1.073741824. A7is. 1.024. 9. .244140625. Ans. .625. 12. 3 (3 places). Ans. 1.442. HIGHER ROOTS. 162. Since the sixth power of a number is the square of the cube, or the cube of the square, therefore the sixth root of a number is the square root of tlie cube root, or the cube root of the square root. Since the ninth power of a number is the cube of the cube, therefore the ninth root of a number is the cube root of the cube root. In like manner, the tenth root is the fifth root of the square root, the twelfth root'is the cube root of the fourth root, etc. That is, when the index of the root is a composite number, the root may be found by extracting the root indicated by one factor of the given index, then the root of the result as indicated by one of the remaining factors of the given index, and so on until all the factors have been used. The final result is the root required. Thus : 4/64 = ^ VU= \/'±S = ±2, |/256 = \/ V~vWq = V VlQ= VI = ± 2. 163. The root of any expression may be found by apply- ing the formula for the corresponding power. For example, since (a + b)' = a* + 4:a'b + 6aW + 4:aP -f b' = a' -\- i (4<^3 _|_ g^2j _|_ 4^j2 _|. ^3^^ therefore the first term of the root is the fourth root of the first term of the given power. The trial divisor is Aa^, and the second term of the root, b, is found by dividing 4fl^J (the first term of the remainder left after subtracting «* from the dividend) by the trial divisor. The complete divisor is {ia^ -\- 6a^ + 4(7 Z>^ + ^^)- Similarly, the fifth root is found by extracting the fifth root of the first term of the power, dividing the first term of the remainder by 5a*, etc. ; the sixth root is found by EYOLUTIOX. 91 extracting the sixth root of the first term of the power, dividing the first term of the remainder by 6a^, etc. The trial divisor for finding the 7iih root is na""'^. That is: I?i t//e extraction of any root, the trial divisor is formed by raising the root found to a power whose exponent is less by unity than the index of the root, and muUi2)lyi7ig this power by a number equal to that index. Exercise IX. Extract the square root of : 1. ^xY; 26a'b''j Sla' (b - cf; 36a^* {x - y)\ 2. {a^ - 2ab "+ b'Y; (4:x' - 4:X -{- If; (a« - 6a + 9)^ 3. (a-b-{- cy\- a'^-^b- c)'""*; ir*'*" (a - 2by\ 4. {x^-2xy-]-yy; (9a^-6a-{~lY''; {(a-bY-2{a-b)-\-'i]\ 5. 9a^ - Gx" - 53^ -{- 2x -{- 1. 6. lesf^ - 40rf^ + a:* + 46r» - ll:^-^ - 12a; + 4. 7. 177241; 5431G9; 14356521; 17.338896. 8. To four decimal places: 2.5; 14.4; .169; .3; .03. Extract the cube root of : 9. 8:c« + 48^;^ + 60^* - 80^* - 90x^ + IO82: - 27. 10. x""- 3a;»+ Gx'- 10:k«+ 12a:'*- 12a;*4- 10a;'-6a^+ 3a; - 1. 11. a«+&'+6'^+3«'Z>+3rt2c+ 3r7 J«+ 3b^c + 3«c«+ 3^^^+ 6abc, 12. 74088; 110592; 103.823; 884.736. 13. 12.812904; 33076.161; 102.503232. 14. To two decimal places: 1.25; .08; 16; 12.8. Extract the sixth root of : 15. a-* - 6a-« + 15a-* -- 20^-' + I5a-^ - 6^-^ + 1. 16. 729; .004096; 15626a''b-'\ CHAPTEE X. SIMPLE EQUATIONS INVOLVING ONE UNKNOWN QUANTITY. 164. An" Equation is a statement that two expressions are equal. Thus, 3x — S = 10. The expression to the left of the sign of equality is called the first memher or first side, and the expression to the right, the second member or second side. 165. An Identical Equation is one in which the two sides are equal for all values of every letter employed. Thus, (a -\- b)(a — b) = a^ — b^ is an identical equation. An identical equation is also called an identity. 166. An Equation of Condition is one which is true only for particular values of one or more of the letters employed. Thus, it" + 4 = 7 is true only when a; = 3. When the word equation is used in Algebra, it is under- stood that equation of condition is meant. 167. A letter to which a particular value must be given in order that the statement contained in an equation may be true, is called an unhnown quantity. Such particular value of the unknown quantity is said to satisfy the equa- tion, and is called a root of the equation. 168. To Solve an Equation is to find the value of the un- known quantity. (92) ECiUATIONS INVOLVING ONE UNKNOWN QUANTITY. 93 169. A Simple Equation, also called an equation of the first degree, is one which contains only the first power of the unknown quantity. 170. To solve an equation, we separate the known from the unknown quantities by using the contrary operation to that indicated by the connecting sign. That is, if the known and the unknown quantities be connected by -|-, they may be separated by subtraction; if connected by x, they may be separated by division; etc. In performing these operations, the two sides remain equal throughout; if the value of one side be changed in any way, the same change must be made on the other side (Ax. II, III, IV, V, etc., Art. 62). To illustrate this principle, let it be required to find the value of x in the following examples : 1. 3a; — 1 = 14. Here 1 is connected with the unknown term, 3a:, by the sign — ; therefore we add 1 to each side. Whence Zx = 15. Now, since 3 and x are connected by the sign X, we divide each side by 3. .'. x = 5. 2. 5a; 4- 6 = 21. 6 being joined to 5x by -\-, we subtract 6 from each side. Whence 5a; = 21 — 6 = 15. x and 5 being connected by X, we divide each side by 5. . •. x = 3. 3. 4a; + 3 = 2a; 4" 11* By subtracting 2a; from each member, the unknown term disappears from the second side ; that is, 4a; — 2a; + 3 = 11. Now subtract 3 from each side, and 2a; = 8. . • . a; = 4. 4. 3a; -f 13 = 68 — 2a;. Here we add 2a; to each side to cause X to disappear from the second member, and we sub- tract 13 from each side to cause the known term to dis- appear from the first member. Whence 3a; -j- 2a; = 68 — 13; that is, 5a; = 55. . *. x = 11. 171. If terms containing the unknown quantity occur on both sides of an equation, it is customary, by means of Ax. II and III, to place them all on the first side, and. 94 ALGEBRA. in like manner, to put the known terms on the second side. This operation is called transposition. Thus, if we have X -\- a = h,\)j subtracting a from each side we find x = b — a; also, it x — a = b, hj adding a to each side we find X = b -{- a. In each case a is transposed from one side to the other, but its sign is changed ; hence. Any term may be transposed from one side of an equation to the other, pro- vided its sign be changed. 172. The signs of all the terms on both sides of an equa- tion may be changed, for this is equivalent to multiplying every term by (— 1). (Ax. IV.) In like manner, an equation may be cleared of fractions by multiplying every term of each side by a common mul- tiple of the denominators (29). Thus: X X 5. r- + 2 = 4; .•.-— = 4 — 2 = 2. Since x and 5 are connected by division, we multiply each term by 5. r , x=10. XXX 6. ^ + T ~ ^ = 5- Since x is connected by division with o 4 t) 3, 4, and 6, we multiply each term by 12. .' . 4:X -\- ^x — 2x =: 60. Whence 6x = 60, and x = 12. 7. 1 1 — = 7. Here multiply each term by x, X XX . • . 24 + 12 + 6 = 7x. Whence 42 = 7x, and a; = 6. 173. To solve an equation of the first degree, containing one unknown quantity. (1) If necessary, clear the equation of fractions. (172.) (2) Transpose all the terms containing the unknoion quantity to one side, and all other terms to the other side, remembering to change the sign of each transposed term. (171.) (3) Reduce each side to its simplest form. (4) Divide both sides by the co-efficient of the unknown quantity. EQUATIONS INVOLVING ONE UNKNOWN QUANTITY. 95 EXAMPLES. Find the value of x in each of the following equations : 1. 3a; + 7 = 9a; — 5. Ans, 2. 2. 3 + 2:c = 17 — ^x. Ans, 2. 3. 3 (a; - 2) = 4 (3 - x). Ans, 2. 4. 5 - 3 (4 - ic) = 4 (2a; - 3). Ajis. 1. 5. Ux - 21 (.'C - 3) = 10 - 21 (3 - x), Ans. 4. 6. 5 (2a; + 6) = 4 (3a; + 6). Ans. 3. 7. 4a; — 40 = 40 — a;. Ans. 16. 8. 4 (a; + 16) = 10 (a; + 1). Ajis. 9. 9. 2a; - 22 = 3 (a; - 22). ' Aiis. 44. 10. ax -{- b^ = hx + ci^' ^ns. a-\-b. 11. 5 (a; + 1) - 2 (6 - a;) - 2 = 3 (a; + 5). Ans. 6. 12. 3 (a; - 2) - 4 (3 - a;) = - 4 + 2 (a; 4- 1) - 3a;. Ans. 2. 13. I + I + I = 26 - 13 (13 - x). Ans. 12. 14. 2a; - 1 - 2 (3a; - 2) + 3 (4a; - 3) - 4(5a; - 4) = 0. A)is. f. 15. i(3a; - 1) - |(a; - 1) = J(a; - 3) - i(a; - 5) +-V^ A71S. 7. 16. J^-lf = 8f + 2(|2--l)-i(a; + 8). ^;?5. -8. X 4iX . oa; ^ w ■ a; a; , ^ . "•2-T + T=^' + 8-6- ^'^^•'^• 18. -^ + - (3a; - 4) + - = -^-. ^W5. 3. 5a;- 11 1 , ,, 11 1 .... 19. — ^ - - (a; - 1) ^ -- X - ~. Ans. 11. 2x — a2x + Zbx—a , 5a — 18b 20. ^ g— = -y-. ^««. ^ . cd — ah 21. (a -|- a;)(Z' -1- a;) — {c-\-x){d-\- x). Ans. a-\-h — c — cC 96 ALGEBRA. PROBLEMS. 174. One of the main uses of equations is to find num- bers which shall satisfy certain given conditions. In order to find such a number, two steps are necessary: (1) To express the conditions of the problem in algebraic language ; that is, to form the equation. (2) To solve the equation thus formed. It is impossible to give a precise rule by means of which every question may be readily stated in the form of an equa- tion. The first step is to understand fully the nature of the question; secondly, to denote the required quantity by one of the final letters of the alphabet; thirdly, to indicate, by means of signs, the same operations that it would be necessary to perform with the answer, to verify it. EXAMPLES. 1. Five times a certain number exceeds twice the number by 30; find the number. Let X represent the required number; then the problem becomes: Five times x exceeds twice x by 30; find x: or, bx exceeds 2x by 30; find x. The corresponding equation is 5x — 2x = 30. . • . X = 10. 2. Divide 140 into two parts such that one part shall be 2J times the other. Here it is convenient to represent the smaller part by 2x. Since the larger part is 2J^ times the smaller, then 5x is the larger. But the sum of the two parts must be 140, and their sum is 2x -\- 5x = 7x. We now have two ex- pressions for the same thing. . • . (Ax. I) 7x = 140. Whence X — 20, and the two numbers are 40 and 100. 3. Divide 27 cents between A and B so that I of A^s share shall equal J of B's. Let X represent the number of cents A shall receive. Since A and B together receive 27 cents, and since A alone receives x cents, therefore B must receive 27 — x cents. EQUATIONS INVOLVING ONE UNKNOWN QUANTITY. 97 But ^ of A's share must equal } of B's. . • . |a; = J (27 — a;). Now multiply both sides by 20. 4:x = 5 (27 — x) ; that is, 4.'c = 135 — 6x. . • . X = 15, A's share, and 27 — re = 12, B's share. 4. A and B together have |7; J of A's equals f of B's. How many dollars has each ? Let 3^; represent the number of dollars B has. Since ^ of A's = I of B's, . • . i of A's = 2x, and A's = 4:X. As A has 4:X dollars and B dx, both have Ix dollars. But together theyhave 7 dollars. . * . '7x = 7. Therefore a; = 1, 3^ = 3, B's share, and 4:X = 4, A's share. 5. Find two consecutive integers, the difference of whose squares is 25. Let X represent the smaller; then, since the integers are consecutive, a; + 1 is the greater. ... (^ _|_ 1)2 _ 2^ _ 25. Alls. 12 and 13. 6. A is three times as old §,s B, but in ten years he will be only twice as old ; how old is A ? Let X represent B's age. Since A's age is three times B's, .-. 3:c will represent A's age. In ten years B's age will be a; + 10, and A's 32: + 10. But then A's age will be twice B's. ,'.^x-^l0=2(x-\- 10). Whence x = 10, and 3x = 30. 7. A has 125 in half-dollars and dimes, the number of both being 170. How many half-dollars has he? Let X represent the number of half-dollars; then 170 — x must be the number of dimes. The X half-dollars must be worth 50x cents. The 170 — X dimes must be worth 10 (170 — x) cents. . • . 50a; + 10 (170 - x) = 2500. Whence x = 20. Ans, 8. Divide $33 between A and B, in such a manner that 4 times what A receives shall equal 7 times what B receives. Ans. A receives $21, and B $12. 9. How many dollars has A, if $5 more than ^ of his money is $20 less than | of it ? A ns. 150. Al£r.— 9. 98 ALGEBRA. 10. A and B together have $156; | of A's equals f of B's. How much has each? Ans. A, |48; B, 1108. 11. A and B together have $130; A and together have $60. B has $20 more than twice as much as 0. How many dollars has each? Ans. A has 10; B, 120; 0, 50. 12. A and B had equal sums of money; A gave B $30, and then had only half as much money as B. How much had each at first? Ans. $90. 13. A is five times as old as B; in 15 years he will be only twice as old. How old is A? Aiis. 25 years. 14. A is 50 years old, and B is 5. In how many years will B be :J as old as A? A7is. 10. 15. What o'clock is it, if the time past noon is ^ the time to midnight? Ans. 4 p.m. 16. A can do a piece of work in 10 days; A and B do the same work in 8 days. How long would it take B alone? Let X = the number of days B would take to do the work. Since A can do all the work in 10 days, he can do -f^ of it in one day, and y^ of it in 8 days. Since B can do all the work in x days, he can do 1 -^ a; of it in one day, and 8 -^ a; in 8 days. In 8 days both do (8 -^ 10) -{- (8 ^ x) (the work being rep- resented by unity). But in 8 days both do once the work. . • . (Ax. I) ^^ _!_ ^ :^ 1. jifis. 40 days. Verification: A can do -^^ in a day, and B ^j^; together they can do ^ig- -j- ^^q- = |. . • . they together can do the work in 8 days, which agrees with the given conditions. 17. A cistern can be filled by a pipe, A, in 8 hours, and emptied by a pipe, B, in 12 hours. How long would it take to fill f of the cistern if both pipes be left open ? Let a; represent the required number of hours. Since A can fill the cistern in 8 hours, it can fill | of the cistern in 1 hour, and a; -f- 8 in a; hours. EQUATIONS INVOLVING OKE UNKNOAVN" QUANTITY. 99 Since B can empty the cistern in 12 hours, it can empty ■fj of it in 1 hour, and a; -=- 12 in x hours. (x -^ 8) — (a; -^ 12) will remain in x hours. X X 3 But I remain in x hours. . *. (Ax. I) ^ — t;t = -r- o 12 4 Whence x = 18. Ans. 18 hours. 18. A can do a piece of work in 60 days. After working 20 days, B joins him, and together they complete the job in 16 days more. How long would it take B alone to do all the work? A7is. 40 days. 19. A can do a piece of work in ^ the time that B can; B in f the time that C can. The three can do the work in 18 days. How long would it take each of them alone? A71S. A, 33 days; B, 66 days; C, 99 days. 20. A works half as fast as B; together they can finish a piece of work in 12 days. How long would it take each alone? Ans. A, 36 days; B, 18 days. Exercise X. Find the value of x in the following equations: 1. 5x — 3 = S(x — 3). 6. ax — bx = ex — d. 2. {ix-2)(l-x) = i-(2x-l)\ 7.| + ^ = i 3. P + H^ + ^) = -^'' 8. f + f=l. 4. a-\-b a-2b^2a — b X 2x X 5. i+s+f,=-"- «■■<»-«=«-»■ -•-^+>-<-+')=^- 12. {x — a){x-^b) = {x- b) (x - c). X 2{r.~-^) 3>--4 X ^^'8 5 ~ 15 + 12* 100 14. ALGEBRA. X 6 ~ 8i = -»'-)-"t'+l- - 26|. 9a;- ■1 o: — 9 2a; — 3 x-^ 1 10 5 "^ 15 6 ~2* 15. 16. 5a; + 16 - 2.C + 3a = 5« + a: + 8. Solve the following problems: 17. Find two numbers whose difference is 2, and whose product exceeds the square of the less by 8. 18. The half of what number exceeds the third by 6 ? 19. A cistern is supplied by two pipes; the smaller alone can fill it in half an hour, and the larger in 20 minutes. In what time will they both fill it when running together? 20. It takes an inlet-pipe, A, one third as long to fill a cistern as it takes an outlet-pipe, B, to empty it. When both run, it requires 6 hours to fill the cistern. How long would it take B to empty it? 21. A can perform a piece of work in 4 days, B in 6 days, and C in 8 days. How long would it take all to do the work ? A and B ? A and ? B and C ? 22. Equal numbers of dollars, half-dollars, and quarters amount to $700. How many of each are there ? 23. Divide $440 among three persons, so that A may have f as much as B, and B f as much as C. 24. A and B engage in trade with the same capital. A gains $1600, and B loses $1900, and A's capital is now 8 times B's. With how much money does each begin ? 25. A gentleman divided a dollar among 12 children, giving to some 9 cents, and to the remainder 7 cents. How many children received 9 cents apiece ? 26. Divide 12 into two such parts that the difference of their squares shall be 48. 27. A is 54 years old, and B 9 5^ears. In how many years will A^s age be just four times B's? EQUATIONS INVOLVING ONE UNKNOWN QUANTITY. 101 28. A is 3 times as old as B, and in 3 years the sum of their ages will be 30 years. How old is each ? 29. Ten years ago A was twice as old as B; 20 years hence he will be f as old. How old is each? 30. A is 15 years older than B, and in 10 years he will be twice as old as B. How old is each ? 31. Two shepherds owning a flock of sheep agree to divide its value c^jually. A takes 20 sheep; B takes 30 sheep, and pays A $150. What is the value of a sheep? 32. One square field is 2 rods longer than another, and contains 1 acre more. How long is each ? 33. A lent $500, part at 4 per cent and the rest at 5 per cent: the whole annual interest received was $22. How much was lent at 5 per cent? 34. A lent $900, part at 4 per cent and the rest at 5 per cent, and he received equal sums as interest from the two parts. How much did he lend at 5 per cent? 35. In a mixture of wine and water, the wine composed 25 gallons more than half, and the water 5 gallons less than a third. How many gallons were there of each? 36. Divide 144 into two parts such that one part may be five sevenths of the other. 37. The time past midnight is one third of the time to noon. What o'clock is it? 38. The time past noon is one fifth of the time past mid- night. What o'clock is it? 39. If 20 men, 40 women, and 50 children receive $500 for a week's work, and 2 men receive as much as 3 women or 5 children, how much does each woman receive ? 40. A bought a number of apples, half at 2 for a dime, and half at 3 for a dime. By selling all at 4 cents each, he lost half a dollar. What was the number of apples ? CHAPTER XL SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 175. If one equation contains two unknown quantities, an unlimited number of pairs of values may be found that will satisfy the equation. Thus, in x -\- y = 16, if any value be given to y, a corresponding value for x may be found which will satisfy the equation. It y = 15, x = 1; if y = V^, x — ^', it y = 2J, x = 13J; etc. But if a second equation be given, independent of the first (that is, expressing different relations between the un- known quantities), only one pair of values of x and y can be found that will satisfy both equations. Thus, whilst X -{- y = IQ and 2x — y = 2, considered separately, may be satisfied by an indefinite number of pairs of values, there will be but one pair common to both; namely, x = 6, y = 10. These are therefore the roots of the pair of equa- tions X -}- y = IQ, and 2x — y = 2. 176. Equations which are to be satisfied by the same values of the unknown quantities, are called siinuUaneous equations. To determine the values of the unknown quantities; (1) The number of independent equations must be equal to the number of unknown quantities. (2) The given equations must be consistent; thus, 2x-\- y^ll and 2x-\-y = 15, can not both be satisfied by any pair of values. 177. Simultaneous equations are solved by combining the equations so as to obtain one equation containing one unknown quantity. This process is called eliminatio7i. (102) SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 103 Three methods of elimination are generally given: (1) By Addition or Subtraction. (2) By Comparison. (3) By Substitution. ELIMINATION BY ADDITION OR SUBTRACTION. 178.— 1. Solve: I. 3a; - 2?/ = 11. 11. 5a; -4?/ = 17. Multiply I by 2. III. 6a; - 4?/ = 22. Subtract II from iIII. IV. X 5. Substitute 5 for X in I. V. 15 - 1y = 11. .-. Verification: I. 3X 5 -2X2 = 15- - 4 = 11. II. 5X5-4X2= 25- - 8 = 17. 2. Solve: I. 5a; -1- 2^ = 30. II. 4a; -3//= 1. Multiply I by 3. III. 15a;+G// = 90. Multiply II by 2. IV. 8a;-6"y= 2. Add III and IV. V. 23a; * =92. .' . x = Substitute 4 for x in I. 20 + 2y = 10. • •• .V = 2/ = 2. 5. Hence, to eliminate an unknown quantity by addition or subtraction: Multiply the equations hy such numbers as will make the co-efficients of this ^inhnoion quantity equal iii the restilting equatio7is. If these equal quantities have unlike signs, add the re- sulting equations; if they have like sigtis, subtract one equation from the other. ELIMINATION BY COMPARISON. 179.— 1. Solve: I. 3a;-2y = ll. II. 5x- iy = 17. In I transpose 3x, III. -2y = ll- 3a;. Multiply III by 2. IV. - 4y = 22 - 6a;. In II transpose 5a;. V. _ 4^^ = 17 - 5a;. Expressions which are equal to — 4^/ must be equal to each 104 ALGEBRA. other. . • . 22 - 6a; = 17 - 5a:. Whence x = 5. Substi- tuting 5 for X in III, — 2// = 11 — 15. ,' . y = 2, , 2. Solve: I. 5:c + 2^ = 30. II. 4a; - 32/ = 1. In I transpose bx. III. 2// = 30 — bx. Multiply III by 3. IV. C^ = 90 - 15a;. In II transpose 4a;. V. — 3?/ = 1 — 4a;. Multiply V by — 2. VI. Q>y = -2 + Sx. Since the first sides of IV and VI are equal, the second sides must be equal (Ax. I.). . •. 90 — 15a; = — 2 + 8a;. AVhence a; = 4. In III substitute 4 for x. Whence 2?/ = 30 — 20. .-. y = f). Hence, to eliminate an unknown quantity by comparison, In each equation find the value of one of the unknown quantities (or of the same multiple of that quantity) in terms of the other, and place these values equal to each other. elimin"atio:n" by substitution". 180.— 1. Solve: I. 3a; -2y = 11. II. 5a; - 4// = 17. In I transpose 3a;. III. — 2y = 11 — 3a;. In II substitute for — 2y its value = 11 — 3a;. Whence, IV. 5a; -f 2* (11 - 3:c) = 17. . • . a; = 5. Substituting 5 for x in III. — 2?/ = 11 — 15. .• . y = 2. 2. Solve: I. 5a; + 2?/ = 30. II. 4a; - 3«/ = 1. In I transpose 5a;. III. 2y = 30 - 5a;. . •. IV. y = 15 — 5a; 2' In II substitute for y its value. .-. V. 4a; -3^15 -I")-- Multiply V by 2. 8a; - 90 + 15a; = 2. . • . X = 4. In IV substitute 4 for a;. . • . y = 16-10 1 = 5. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 105 Hence, to eliminate an unknown quantity by substitution: \ From one of the equations find the value of one of the unhnoiun quantities in terms of the other. Substitute this value instead of that unknoion quantity in the other equation. Note. Each equation must be simplified, if necessary, before the elimination is performed. EXAMPLES. Find tbe values of x and y in the following pairs of equa- tions: 1. Ix-by = 24; Ax - ?>y = 11. Ans. x = 17; y = 19. 2. ^x-^2y = 32; 20.?: — 3?/ = 1. Ans. x = 2; y = U. 3. 3x — 4:y = 18; 3x + 2y = 0. Ans. x = 2; y = —3. 4. 2x - iy = 11; 2x — 6y = 0. Ans. x = 6; y = 2, 6. 3a; = 2y; 6x — by = — ^. Ans. x = 2^; y = SJ. 1,15 11 1 7. - + -=--; = — — . Ans. X = Q; V = 4. X y 12 X y 12 ^ 8. 2x-' + 3y-' = /^; Sa;-^ + 2^"^ = t\- Ans. a: = 10; y = 12. 9.^-\-5y = 51; 5x -{- \y = 27. Afis. x = 6; y = 10. 7 , J be ac 10. «x = by; x 4- y = c. Ans. x = — -^; y = — —7. ^ , J ^ , « , ^ a^ — b^ a^—¥ 11. - + -=c; - + -=f/. Ans.x = 7-; y = -^ — r-- a: ?/ ^2/ ac — bd'^ad — bc 12. 5a; + 7y = 43; llo; + 9^ = 09. Ans. a; = 3; y = 4. 13. 3;/ — 7a; = 4; 5x^2y = 22. ^?i5. a; = 2; y = 6. 14. 11a; — 10;/ = 14; 5a; -|- 7y = 41. Ans. a; = 4; y = 3. 16. 13a; + lly = 4flj; 12a; — Gy = a. Ans. x = \a; y = ^a. 106 ALGEBRA. SIMULTA^^EOUS EQUATIONS CONTAINING THREE OR MORE UNKNOWN QUANTITIES. 181. If three simultaneous equations involving three unknown quantities be given, one of the three unknown quantities must be eliminated between two pairs of the equations; then a second between the resulting pair. Solve: I. 3a;— 2?/+ 2= 5. 11. 6x+ ^y-bz= 2. III. 2x- by-\-^z^ 7. IV = I X 5. Ibx - lOy -^5z = 25. V= 11+ IV. 20^'— 7y = 27. Contains only x and J/. VI = 1X4. 12^; - 8// + 4^ = 20. VII = VI — III. 10^ — ^y = 13. Contains only ic and 2/. VIII = VII X 2. 20.^; - 6ij = 26. IX = VIII - V. y =-1, In VII put - 1 for 2/. 10a; + 3 = 13. .•.x = l. In I put — 1 for y and 1 for a;. .• . z = 0. 182. If four simultaneous equations involving four un- known quantities be given, one of the unknown quantities must be eliminated between three pairs of the equations; there will result three equations containing three unknown quantities. Another unknown quantity is then eliminated between two pairs of these equations, the result being two equations containing two unknown quantities. A third unknown quantity is eliminated between this pair of equa- tions. Proceed in a similar manner if there be more than four unknown quantities. EXAMPLES. Solve the following simultaneous equations: 1. dx + 2y-4:Z=15', 5x-3y-\-2z=28; —x-{-3y-^4:Z = 2i. Ans. x = 7; y = ^i z = 4. 2. 4:X-dy + 2z = 9; 2x-{-5y-3z = 4; 5x + 6y-2z = 18. Ans. x = 2; «/ = 3; z = 6. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 107 3. X -\- y — z = 8x -\- 3y — 6z = — 4:X — y -{- oz = 1. Alls. X = 2; y = 3; z = 4. 1 1_^, 1 I L_l. 1 I 1-A ^ x~^ y ~ 12' x~^ z ~ 2' y'^ z ~ 12' Ans. a; = 3; ?/ = 4; z = Q, 6. 1 + 1 = 3. 3_3^2. i + ?i = i X y z' z y ' X z ' '6' Ans. x = i', y = -I; z = ^. PROBLEMS INVOLVING SIMULTANEOUS EQUATIONS. 183. A pound of tea and 3 pounds of sugar cost 96 cents, but if sugar rise 50 per cent and tea 25 per cent, they would cost $1.26. Find the price of tea and sugar. Let 4:X represent the number of cents 1 lb. of tea costs. Let 2y represent the number of cents 1 lb. of sugar costs. From the first condition, I. 4a; + 6y = 96. If sugar rise 50 per cent, then the original cost, 6y, would be increased by 3y. If tea rise 25 per cent, then the original cost, 4:X, would be increased by Ix. But the total increase in price is 30 cents. . * . II. x -\- 3y = 30. Solving these equations, we find that a pound of tea costs 72 cents, and a pound of sugar 8 cents. 184. What fraction becomes 1 when 3 is added to its numerator, and ^ when 2 is added to its denominator? Let X represent the numerator, and y the denominator. ic 4- 3 From the first condition, ^ =1. .' . I. x-{-d = y. X 1 From the second condition, — — - = — . . • . II. 2x=v-\-2, y -{-2 2 ^ ' .*. a; = 5, and y — 8. Ans. | is the fraction. 185. A purse holds 6 dimes and 19 nickels; W of it holds 15 dimes and 12 nickels; how many of each will it hold? Let X = the number of dimes that will fill the purse. 108 ALGEBRA. Let y — the number of nickels that will fill the purse. The size of the purse may be represented by unity (1). If X dimes or y nickels fill the purse, 1 dime will occupy - of it, and 1 nickel will occupy — of it. X y 6 19 Therefore 6 dimes occupy -, 19 nickels occupy — , and X y both occupy --| . X \j 6 19 But both just fill the purse. . • . I. — I = 1. In a similar manner, 11. = — -. X y %\. Multiply I by 5. III. — + — = 5. Multiply II by 3. jy. |^ + ^ = |J. Subtract IV from III. -- = ^^. . * . ^ = 21. ?/ 21 ^ Substitute 21 for ?/ in I. . • . a; = 63. Exercise XI. Solve: 1. 6a; - 5?/ = 1; \\x - 8y = 17. 2. 2x-\-y = 147; 2x — y = 77. 3. 7x + dy- 2z = 16; 2x -}- 5y -{- 3z = 39; 5x-y-{- 5z = 31. 4. 4:x-5y-^z = 6; 7x -lly-^2z = 9; 6x-12y-z = -3. 6. x-\-y-\-z-^v=14:; x-z=2y-2v-2; 2y-{-2z-]-3v=19-]-x; 5z + 4?/ + lOz + ^o-v = 80. 6. A, B, and can together perform a piece of work in 15 days; A and B can together perform it in 16 days; and B and C can together perform it in 60 days. Find the time in which each alone can perform the work. SIMULTANEOUS EQUATIONS OP THE FIRST DEGREE. 109 7. If B give A $50, they will have equal sums of money; if A give B $44, B's money will be double that of A. Find the money which each actually has. 8. Seven years ago A was three times as old as B, and seven years hence A will be twice as old as B will be. Find their present ages. 9. What fraction becomes equal to J when the numerator is increased by 1, and equal to i when the denominator is increased by 1 ? 10. A spends $3 in apples and pears, buying the apples at 4 for a dime and the pears at 5 for a dime; he sells half his apples and one third of his pears at prime cost for $1.30. How many apples and pears does he buy? 11. A owes $240, and B $510. Said A to B, ''Give me | of your money and I shall have enough to pay my debts," B answered, '' I can pay my debts if you give me J of your money. " How much has each ? 12. Divide $640 between A and B, so that 7 times A's share shall equal 9 times B's share. 13. What numbers are those whose difference is 20, and quotient 3 ? 14. Find four numbers, such that the sum of the first, second, and third shall be 13; the sum of the first, second, and fourth, ^5; the sum of the first, third, and fourth, 18; and the sum of the second, third, and fourth, 20. 15. Find three numbers which, being taken in pairs, shall produce sums of a, b, and c, respectively. CHAPTER XII. ELEMENTARY QUADRATIC EQUATIONS. 186. An" equation which contains the second power of the unknown quantity, but no higher power, is called a quadratic equation. Ex. x^ = 25; x^ -\- 4iX = 60. 187. If the equation contain the second power only, it is called Si pure quadratic; but if it contain both second and first powers, it is called an affected quadratic. Thus, 3.c^ = 36 — a;^ is a pure quadratic, and bx^ — 2:c = 3 is an affected quadratic. 188. To solve a pure quadratic: (1) Collect the unknown quantities on one side, and the hnown quantities on the other, (2) Divide loth sides by the co-efficient of the unhnown quantity. (3) Extract the square root of each side of the resulting equation. 1. Solve 3a;2 = 36 - x\ Transpose — x^. Ax^ = 36. % Divide by 4. x^ — 9. Extract square root. a; = ± 3. 2. If 3a;2 - 48 = 0, . • . 3:z;2 = 48, x" = 16, audi x ^ ± 4. 3. If 4:X^-9 = 19, then 4^^ = 28, a:^ ^ ^^ and x = ± ^J, Here x represents a number whose value can not be ex- actly ascertained, but an approximate value of it to any assigned degree of accuracy may be found. 189. It will be observed that there are two roots of equal value but of opposite signs. The square root of x^ = 9 (110) ELEMENTARY QUADRATIC EQUATIONS. Ill really admits of four forms: ±x = 4-3; that is, -f- ^ = + 3, -{- X = — 3, — X — -\- df — X = — 3. But the equation — X = -\-3 gives X = — 3, and the equation — x = — 3 gives X = -\- 3; hence the equation x = ± 3 expresses all the values, and there are only two roots. EXAMPLES. Solve the following equations: Answers. 4. lla:^ - 30 = 6x^ -f 24. x = ± 3. 5. ix^ -4: = ix^ -1. x= ± G. 6. {x + 3)2 = Gx 4-25. x= ±4:. 7. {2x - 3)2 = 18 - 12a;. a; = ± f . 8. IQx^ = 4(3 — a:)2. Extract sq. root. a; = 1 or — 3. x^~24 x^-37 ^ 9. ;; =8. X= ±7. 5 4 10. (a; — 4) (a; + 4) = 9. x = ± 5. 12. i|^ = i(^-9)'- a: = 10 or 8^. Multiply by 120, and extract sq. root, a; , 4 a; , 9 a; , 4 a; 14. j + -=^. a:=±4. 16. (3a; + i)' = 2a; + |. a; = ± |. 17. (a; - 3) (a; - 2) - (a;-4) (a; + 5) = 3(a; - 1)^ + 11. A71S. X = ± 2. 18. (a; + l)(a;-l)(ar»-f 2) = (a;2-l)2+9. Ans, x = ± 2. 19. I (3ar^ + 5) - i (a; - 1) (a; + 1) = J (4^^ + 9) - 13. ^7i5. a: = ± 3. 112 ALGEBRA. AFFECTED QUADRATIC EQUATIONS. 190. Every affected quadratic may be reduced to the form of a^x^ + lahx = c^. The first step in the solution of such an equation is to complete tlie square (151 (3)); that is, Add to each side the square of the quotient found from dividing the second term hy tic ice the square root of the first. The first side will then be a perfect square. The second step is to extract the square root of each side of the resulting equation. By placing ± before the second side, two simple equations will result. The third step is to find the value of x in each of these simple equations. 1. Solve 7^ — 4,x = b. (1) The square root of x^ is x. (2) Twice this root is 2x, (3) 4:r -T- 2a; = 2. (4) The square of this quotient is 4. . • . Add 4 to each side of the given equation. Whence x^ -^x-\-4: = 9. Extract the square root, and use the double sign on the second side only, for the reasons explained in Art. 189. Whence a: — 2 = ± 3. If a: — 2 = + 3, then x = b. If a; — 2 = — 3, then x = — 1, .*. re = 5 or — 1. 2. Solve a? + 13a; = - 12. Twice the square root of the first term = 2^;, and 13.7; -T-2x = J/; therefore we add (y)^ to each side. . • . a;2 + 13a; + (J/)' = - 12 + J-f ^ = -ifi. Extract the square root, x -\- J/- = ± y-. ,',x= - V- ± V- = - 1 or -12. 3. Solve 2x'^-{-x = 3. Since 2a;^ is not a perfect square, it is necessary to multi- ply or divide each term by such a number that the first term shall be a square. Multiply each side by 2. . • . Ax^ + 2a; = 6. Now proceed as before. . • , 4a;^ -|- 2a; -|- J = 6J. . • . a; = 1 or — #. ELEMENTARY QUADRATIC EQUATIONS. 113 4. Solve 3x^ -3-\-x = 3x — It?, Transpose and combine so that the first term shall con- tain all the ic^'s, the second term all the ic's, and the second side all the known terms. . • . ^o:*? — 2x = 3. The first term not being a square, multiply by 5. . • . 25a? - 10a; = 15, whence 252;^ - 10.T + 1 = 16. . • . 5.C — 1 = ±4, and re = 1 or — |. EXAMPLES. Solve: 6. x^-8x= - 12. X = ATiswers. G or 2. 6.a?-^5x-U = 0. X = 2 or — 7. 7. a? — 6x = 4x - IG. X = 8 or 2. S. a? + lOo; + 21 = 0. X = - 3 or — rv 9. x^ + X = 6. X ~ 3 or — 3. 10. (x -2) {x + 6) = 33. X = 5 or — 9. 11. {3x - 5)2 = 2O2:. X = 5 or 1- 12. 12:c2 _ 7rr + 1 = 0. X = ior i- 13. 2x'- 3 = H^ + 5) + 11. X = 3 or - ¥• 14. ^x - ia? = 5. X — 10 or 2. 15. ia;^ - ia; = 7. X = 4 or — 3i. 16. x = l-\- 110 -4- X. X = 11 or - 10. 17. (x -l)(x-2) = 12. X — 5 or — 2. 18. far^ + 3i = ix + 8. X = 3or - i- Exercise XII. Solve: 1. a? — bx= — 4. 5. 14a; — 3? = 46. 2. a?-4x+3 = 0. 6. {x - 2){x + 1) = 10. 3. 2a;^ - 7a; = - 3. 7. (x - 4)(a; - 5) = 0. 4. a;2 + 10a; + 9 = 0. 8. (a; - 14)(a; + 3) = 0. 9. (2a; + l)(a;-f 2) = 3a,-2-4. 10. (x - l)(x - 2) + (a: - 2)(a; - 4) = 15. Alg.— 10. 114 ALGEBRA. 11. {2x-3Y = 12. {Sx-2){x 13. {5x-3Y = 6^ + 1. - 1) = 14. 4(11:^ + 3). 14. x^-3=i{x- 3). 15. a;2 - 2ax + a^ = b^ 16. (x + af -¥ = 0. SPECIMEN PAPER. 1. Solve o(^ — Ux= - 63. No. Operation. Axiom. I x^-nx=:-Q^. II ! + (¥)' 2 a;-2_i6aj_|_ 8-2 ^04-63 = = 1. III VII 15 X -8= ± 1. IV III + 8 2 X = 8 ± 1 = 9 or 7. Ans. a; = 9 or 7. Verification: 1st, a; = 9. 1st side 2d side 2d, a; = 7. 1st side 2d side (9)'^ - 16(9) = 81 - 144 {ly - 16(7) = 49-112 -63. -63. -63. -63. 2. Solve 4:X^ — lU' - 2 = Ix^ - 20x - 32. No. Operation. Axiom. I 4a!-2 _ 11^ _ 2 = 7a?2 _ ^Qx - 32. II I transposed 2 and 3 4^2 _ iia; _ 7^2 _^ 2ac = + 2 - 32. III II simplified - 3a!2 + 9a! = - 30. IV III -^(-3) 5 aj2 _ 3a. = -f- 10. V IV + (f)^ 2 aj2 - 3a! + f = ^-. VI w 15 ^ - f = ± |. VII vi + i 2 a! = + |±| =5 or — 2. Ans. a! = 5 or Verification : 1st, a!= 5. 1st side: 4(5)2 - 11(5) - 2 = 100- 55- 2 = 43. 2d side: 7(5)^ - 20(5) - 32 = 175 - 100 - 32 = 43. 2d, a! = - 2. 1st side: 4(-2)'^-ll(- 2)- 2 = 16 + 22 - 2 = 36. 2d side: 7(-2)-2-20(- 2)-32 = 28 + 40 - 32 = 36. CHAPTER XIII. FACTORING. 191. Factoring is the process of finding the simplest numbers whose product is a given number. 192. — Case I. Monomials. The factors of a monomial consist of tlie factors of the numerical co-efficie7it, together with the various letters tahen as many times as there are units in their exponents. Thus : ^a^h& = ^ .^ . a . a .h . c . c . c ; \2a^ {b - cf = 2.2.3.a.a.(b-c).(b-c).{b-c). 193. — Case II. A polynomial whose terms have a com- mon factor is factored as in Chapter IV. Thus: 1. a" + "^ab = a{(i-\- 2b). 2. 2^^ - Ga^ = 2a^ (1 - 3«). 3. (X - yf -{X- yf = {x - yy\l - {x - ;/)}. 4. (a — b — cf— {a — b — c) = (a — b — c)[{a — b—c)—l]. 5. 2^ - 2* = 2* (2^ - 1) = 2* (7). 194. — Case III. A polynomial whose terms may be so combined that the resulting polynomial lias a factor com- mon to all its terms, is factored as in Chapter IV, Thus: 1. x^-^x^-\-x-{-i = x%x 4- 1) + (re + l) = (a; + 1) (r' + 1). 2. ab — ac — b^ -\-bc = a{b — c) ~b{b — c) = {b—c) (a—b). 3. a*- 2a-l-a''bi-2ab + b={a^-2a-l)-b{a^-2a-l) = (a^-2a-l){l-b). 4. [a-by- (b-aY={a-b)'+{a-b)'={a-b)'[(a-b-\-l)]. (115) 116 ALGEBRA. 5. {a -b-cy-{b-^c- af ={a-b- cy-{a ~b-cy = (a-b-cy[{a-b-c)-l], 195. After separating an expression into two factors, each of these factors must be separated, if possible, into factors, and so on until the various factors are prime. . 196. Before applying any of the following cases to the factoring of a polynomial, the factors which can be found by the preceding rules are removed from the expression. 197. — Case IV. A trinomial which is a jyerfect square is factored according to (149) or (151). Notice that a trinomial is a perfect square when two of its terms are squares and positive, and the remaining term is twice the product of their square roots (151). Formula: ft^ ± lab -\-y' = {a± b){a ± b), EXAMPLES. 1. a"" -4:a + 4: = (a-2){a- 2). 2. a^b' - Gab'c + 9c^ = (ab'' - 3c) {ab^ - 3c). 3. 4a^ - 12aU + 9b^ = (2r«t - Uy. 4. 2a^ - 8a^ + 8a = 2a {a - 2)^ 5. 18a^ - 12a + 2 = 2 (3a - 1) (3a - 1). 6. {a-by-2(a-b)-\-l = [(a - b) - If. 7. x-^ - Sx-'' + 16 = (x-'' - 4)2. 8. 2a» + I2a^ + 18a = 2a (a + 3)\ 9. a^ - 10a'' + 25 = (a' - 5)1 10. 3^-2 (3^) + 1 = (3^ - ly, 198. — Case V. A polynomial which is the square of three or more terms is factored by (149), or thus : Transform the given expression into a trinomial such that two terms shall be squares and positive, and the other term shall be tivice the product of their square roots. FACTORIXG. 117 1. To factor a^ -^ V -\- a" - 2aJ) - 2ac + 2bc. Here there are six terms, three being squares and three being products with even co-efficients. Arranged accord- ing to a, the expression becomes a^—2a{b-\-c)-{-{b^-{-2bc-{-(^). Now the first term is the square of a, and the third term is the square of {b + c) (Art. 197); also, the second term is twice the product of these square roots; . • . the given ex- pression is the square ot a — (b -\- c) = a — b — c. 2. a^ ^y - 4:a + 46 - 2ab -\- 4 = a^ - 2a {b -\- 2) -}- (b^ + 4:b + 4) = a'-2a(b-\-2)-i-{b-{-2y=(a-b- cf, 3. To factor «* + 2a^ + 3a^ + 2^ + 1. Here there are but five terms; . • . one of the terms must be separated into two. In sucli cases it is convenient to use Art. 149, but the example may also be factored thus: rt*+ 2«'+ ft^-{- 2a^-\- 2a-\-l = (a'-{- ay + 2 («« + «) + 1 = («'+« + 1)*. 4. a'+ b^ + 6-2 -\-d^- 2ab-2ac-\-2ad + 2bc-2bd -2cd = a" - 2a (b-^c- d)^(b^^(^-\-d^^ 2bc - 2bd - 2cd) = [a- {b-irc- d)Y' See Ex. 1. 199. — Case VI. A hmomial lohich is the difference of two squares. Formula: a^ — b^ = (a -{- b) (a — b). Rule. (1) Extract the square root of each square, (2) One factor will be the sum of these roots, and the other IV ill be their difference, EXAMPLES. 1. ^2 _ 4 = (r? + 2) {a - 2). 2. a'-a = a (a^ - 1) = a{a -}- 1) {a - 1). 3. a'-b'= {a^ + ^') (a' - b') = {a' + b') {a + b) {a - b). 4. (a - by- (c-dy= Ka-b) + {c-d)] [(a-b)-{c-d)^. 5. 4«» -^a = a (4^(2 - 9) = « (2« + 3) (2a - 3). 6. (a - by - (6'2 - 2rd + d') = (a - by - (c - dy = [(a - b) ^ (c - d)][(a - b) - [c - d)]. 7. 9^ - 4'' = (3' + 2^) (3-^ - 2"). 118 ALGEBRA. 8. a-'- 16 = («-i+4)(«-i-4). 9. J -26 = {a^ + 5) (ai - 5). 200. — Case VII. A polyno^nial luhich is the difference of two squares is factored iy collecting the terms of one square in one parenthesis, and the terms of the other squarr in a second parenthesis, and then using Case VI. 1. a' - 2ab ■^l^-&= («« _ 2ah + l^) - c^= (a-hf- &= \i^a-l)\c-\\{a-V)-c\ 2. ft'^ - 4rt - J2 H- 4 = (a^ - 4« + 4) - ^>2 = (a _ 2)2 -l^ = {a-2\b){a-%- b). 3. a^-i-b^- c' -d^ - 2ab - 2cd = (a^ - 2ab + b^) - {c" + 2cd + d^) = (a-bY-{c-{-dY=[(a-b)+(o+d)][(a-b)-{c+d)] = (a — b -{- c -\- d) {a — b — c — d). 4. a' - b' -ia - 6b - 5 = (a^ - 4:a-{- 4:) - {b^ -{- 6b + Q) = {a-2Y-{b+dY=[{a-2)^(b-\-^)][{a-2)-{b-]-3)] = (a-{-b-\-l)(a-b -6). 201.— Case VIII. Since (x+c)(x+d)=x2+(c+d)x+ cd, therefore a trinomial of the form x^ -|- ^^ + 1> can be resolved into two factors when a is the sum of two mwibers whose product is b. Thus: I. x'-\-6x-\- 5 = (a; + l)(^+5), since 5 + 1 = 6, and 5 X 1 == 5. 2.ay'-^x-{- 2=:(^-l)(.r-2), since -l+(-2)3=- 3, and -l(-2) = + 2. 3. x'-bz-U = {x - 7)(a;+2), since - 7 + 2 = ~ 5, and -7(2) =-14. 4. a;2-f4a;- 5 = (a; + 5)(.'r-l), since 5+{-l) = + 4, and 5(-l) = - 5. 202. A composite expression of the form x^ -\- ax -{-b is really the difference of two squares, and may be factored by Case VII. FACTOKING. 119 5. a;' - 4a; + 3 = (ar^ - 4a; + i) - 1 = (x - 2Y - 1 = (x-2-\-l)(x-2-l) = (x-l){x- 3). 6. x^-6x-7 = {x^ -6x-i- 9) - 16 = (a; - 3)^ - 16 = (a; - 3 + 4) (a; - 3 - 4) = (a; + 1) (x - 7). 7. a:^ _^ 5a; - 6 = (a;2 _|. 5^ _|. ^) _ ^ 3^ (3. _^ ^y - ^ = (^ + f + f)(^ + f-l) = (^ + 6)(.^-l). 8. x' - bx"-^ 4 = (a;* - 63^ + \'-) - ^ = (x' - If - | == (^- i + I) (^- 1 - 1) = (^^-1)(^'^- 4) - (a;+l)(a;-l)(a; + 2)(a;-2). 203. — EuLE. (1) Arrange the given trinomial, (2) Complete the square of the first two terms (151, (3) ). (3) Subtract the resulting third term from the given third term, and write the remainder as the fourth term, (If this fourth term is not minus a square, the given trinomial is prime. ) (4) Combine the first three terms in 07ie, and continue as in Case VII, EXAMPLES. 9. a;^ - 5a; + 6 = (a; - 2) (a; - 3). 10. ar' - 2a; - 3 = (a; + 1) (a; - 3). 11. a.-^ - 9a; + 20 = (a; - 4) (x - 5). 12. a;2 + 2a; - 3 = (a; + 3) (a; - 1). 13. x^ 4- 4.x + 3 = (a; + 3) (a; + 1). 14. 3a;2 - 3.r - 6 = 3 (a; + 1) (a; - 2). 15. x?-Q?- 20.r = X {x + 4) (a; - 5). 16. a;* + 5a;2 + 6 = (a;^ + 3) {x^ + 2). • 17. a;*-10.f2+9 = (x^ 1) (a; - 1) (a; + 3) (a; - 3). 18. 22^-2^ + ^-15 = (2^- 5)(2'-f 3). 204. In like manner, expressions of the form {cxY + a (ex) -f b may be factored. It is convenient in this case to let ?/ represent the square root of the first term. 120 ALGEBRA. EXAMPLES. 1. 4x' - 12a; + 5 = (2xy - 6 (2x) + 5 = / -62/ +5 = (y - l)(y - 5)- .-. 4.T^ - 12:c + 5 = (2:^: - l)(2:c - 5). 2 9.^2 - 15x ~U = (3xY - 5 (Sx) -U = y^-5i/-U = (2/+2)Cv-7). J^^5. (3.^■ + 2)(3x - 7). 3. 16:^*4- 82;^ - 3 = (4x-2)2 + 2 (4^;^) -3 = ?/2+2^-3== (y + 3)(y - 1) = (4a;« + 3)(4a^ - 1) = {4:X^-{-3){2x4-l){2x-l), 4. 9^' - 12.T - 5 = (3.?; 4- l)(3a; - 5). 5. 4:X^ - Ux -}- 6 = 2 {2x - l){x - 3), 6. 4:x'- 7.^- + 3= (.^^_l)(4a:-3). 7. 16.^*+ 4.?;2 - 2 = 2 (2.6-2+ i)(2j:+ 1)(2^- l). 8. 4.C* - nx^ + 4 =3 (:y + 2)(:i^ - 2)(2a; + 1)(2^ - 1). 205. — Case IX. To factor a trinomial of the form ax^+ bx -\- G, mMltiphj the expression hy a, factor the result ac- cording to Case VIII, and finally divide hy a. Thus: o o iO/ + 6)0/ - 8) = \{?>x + 6)(3:z; _ 8) = (a; + 2)(32: - 8). o '^r^ ^r OQ - (^■^')' - ^ (3^) - 60 _ y^ - 7y - 60 _ ^(^+5)(^-12) = M3^+ 5)(3a:-12) = (3.r + 5)(a; - 4). 3. Sx"" - 2x - S=: ^(16a;2 - 4a; - 6) = ^^^ _ ^ _ q^ ^ Uy - my +^)=i (4^ - 3)(4a; + 2) - (4x-3)(2:.+l). 206. Example 3 shows that it is not always necessary to multiply the expression by the co-efficient of x^. It is usually best to multiply by the least integer that will make the first term a perfect square. 207. If the multiplier be composite, it is sometimes necessary to divide one factor of the product by one factor FACTORIXG. 121 of the multiplier, and to divide the other factor of the product by the other factor of the multiplier. Thus: i[y-4.]y-9]=i(^x-^)(Gx-9) = (3x-2)(2x-3). EXAMPLES. 6. 2x^-7x^3 = (2x - l)(x - 3). 6. 6x^ -{-Ux + ^ = (2x + 3) (3a: + 2). 7. 3x^ + 10.2: + 3 = (3x + l)(rc + 3). 8. Sx^ - 20.C + 12 = (3x - 2) (a; - 6). 9. 3x^ + ICt - 57 = {x - 3)(3x + 19). 10. dx^ -2x- Q5 = (x- 5)(3.r + 13). 11. 3a:* - Ux^ + 4 = (3a;2 - l)(x + 2)(a; - 2). 12. ix' -j\x-^=i(x- i)(x + i). 208. — Case X. A trinomial, which ivould he a square if a square were added to its middle term, is factored like Case VI I. Thus: 1. a:* + 2a:2 + 9 = (a:* + 63;^ + 9) - 4a:* = (a:« + 3)^ - 4a:2 = {x" + 3 + 2a:) (a:^ + 3 - 2a). 2. 9a;*-13a:2+ 4 =(9a:*-12a:2 4- 4) -3^= {Zx^ - 2)^ - x^ = (3a:2-2+a:)(3a:2-2-a:) = (a:+l)(3a:-2)(a:-l)(3a:+2). 3. a:* - 10a.-3 + 9 =r (c* - B.c^ + 9) - 4a;2 = {x^- Zf - ^x^ = (a:2-3+2a')(a:2-3-2a:) = (a:-l)(a:+3)(a:H-l)(a:-3). Examples 2 and 3 may be factored by Case VIII. Thus: a:*-10a:« + 9 = (a:* - lOa.-* + 25)- 16 = (ar» - 5)* - 16 = {x'-l){x^-Q)= (x + 1) (a; - 1) (x + 3) (a; - 3). 209. — Rule. (1) Arrange the expression. (2) Write for the second term tivice the product of the square roots of the extreme tervis (151, (2) ). Alg.-ll. 122 ALGEBRA. (3) Subtract the second term from the given second terin, and write the remainder as the fourth term. (4) Combine the first three terms, and continue as in Case VII. 210. When the terms of a binomial are squares and positive, and twice the product of their square roots is a square, the binomial may be factored by the rule in (209). 4. a;* + 4 = (:c* + 4a;2 + 4) - ^x^ = (x^ + 2)^ - ^x"" = (a;2 4- 2 + 2x) (a;2 + 2 - 2x), EXAMPLES. 5. 4a;* + 1 = (2x^ + 3a; + l)(2a;2 - 2x + 1). 6. 64a;* +«/*= (8x^ + 4a;^ + y%Sx^ - Axy + if). 1, a^-{-x'-{-l = (x^-{-x-}- l)(x' - ^ + 1). 8. a;* - 7a;2 + 1 = {x^ + 3a; + l){x^ - 3a; + 1). 9. a;* - 23a;2 -|- 1 = (a;^ -f 5a; + l){x^ - 5x + 1). 10. a;* + 6x^y^ -\- 9?/* = (x^ + ^^ + ^f) (^^ — ^y -\ 11. a;* - a;2 + 16 = (x" + 3a; + 4){x^ - 3a; + 4). 3^^). 211. — Case XI. A binomial of the form a° — b", n bei7ig any integer > 1, can always be factored. (1) When n is an even number, «" — b^ is the difference of two squares, and can be factored as explained in Case VI. 1. «* - 4&2 = (rt^ + 2b){a' - 2b). 2. 4«* - 9^* = (2«2 + ?>b^){2a' - W). 3. ft* -¥= («2 + h^a" - b') = (a' + b'){a + Z»)(ft - b). (2) To factor «" — b'^ when ?i is an odd integer. (1) Extract that root of each term tvhose index is the least integer > I by ivhich n is divisible. (2) Divide the expressiofi by the difference of these roots. 4. a^-,b^={a- b)(a' + ab + b'). 6. a'-b'={a- b)(a' + a'^b + «'^' + ab' + Z^*)- FACTO RIXG. 123 6. a« - 8 = {a^ - 2)(rt* + 2«2 4- 4). 7. Sn^ -27 = (2a - 3)(4«2 + 6a + 9). 8. tj* - b' = (a^ - h'){a' + ft^Z'^ + i^«) = (« _ J)(rt2 + «5 + b^)(a' + a^Z'^ + Z/«). 9. a:® — a;?/^= a: (a; — y){x*' -\- a^y -\- x^y^ -\- xy^ + ^*)- 10. 8a^ - rt = rt (8 1. Thus: 1. a^ ^b^ =^(a-\- b){a^ - ab -\- b% 2. ft5 4- ^« = (a + b){a^ - a^b + a^l)' - ab^ + b% 3. a« + 8 = (a^ + 2)(ri* -2^^ + 4). 4. «« 4- Zi« = (a^ + ^')(«* - «'*' + ^0- 5. a* + b*' is prime, because 4 has no odd factor > 1. 6. «* + ^' is prime, because 8 has no odd factor > 1. 7. «!« + 1 = («2 + 1) (a* _ ^2 + 1) («i2 _ a« + 1). 213. — Rule. (1) Extract that root of each term whose index is the least odd number > 1 by which n is divisible, (2) Divide the given expression by the sum of these roots. That is, if the terms of the proposed binomial be cubes, the divisor will be the sum of their cube roots ; if fifth powers, the divisor will be the sum of their fifth roots, etc. EXAMPLES. 8. 27 a^ + S = {3a + 2)(9a'' - 6a + 4). 9. ft" + 64 = (a^ + 4)(«* - 4«2 + 16). 10. a^ + 27^2 = a^a-\- S){a^ - 3a + 9). 11. 3^ + 2^ = (3^ + 2^) (9^ - 6^ + 4^). 12. a'-\-l = {a-\- 1)(«2 - « + l){a' - «^ + 1). 13. a'^ + 1 = («* + 1)(^«' - «* + 1). 124 ALGEBRA. 214. The following examples involve Cases XI and XII. 14. a^ -64:= {a- 2){a^ + 2a + 4)(a + 2)(a'' - 2a -{- 4). 15. a^ -a^ = «2 {a - l)(a^ + « + 1)(« + !)(«' - a + 1). 16. a}' -h' = («2 - ^)(«* + a% + -b^){oJ' + ^)(a* - a^^* + //). 17. 2:c-« - 128 = 2(x-^ - 64) = 2{x-''- 2)(a;-2+ 2a;-i+4)(a;-i4- 2)(a;-2 - 2:r-i + 4). 18. a-^— 1 = 19. 3^^ - 2«^= (3^- 2^)(9^+ 6^+ 4^)(3'"+ 2^)(9^- 6^ + 4^). [Note.— Beginners may omit 215-220 inclusive.] 215. In Case XI we have assumed that a^ — If- is divisible \y^ a — h, if n be any integer; and in Case XII that «" -|- If is divisible \^ a -^h, if n be an odd integer. That these principles are true in any selected case may easily be shown by performing tlie division; but the general proof is some- what difficult for beginners, as it depends upon inductiun, a kind of reasoning not heretofore employed. 216. Mathematical Induction may be thus described: We prove that if a theorem is true in one case, whatever that case may be, it is true in another case, which may be called the next case. Now, by trial we prove that the theorem is true in a certain case, hence it is true in the next case, and hence in the next to that, and so on without limit ; therefore, the theorem must be true in every case after that with which we began. For example: 217. To prove that a" — Z*" is divisible by r? — ^. By division, ^^- = «»-i + Maf^^^^^ ^ .^^3 . a — a — b ^ ' The second term of the quotient is fractional in form, but, if rt""^ — ^"-^ be divisible by a — i, it will be in- FACTORIXO. 125 tegral in value; therefore, «" — Z/" will be divisible hy a — b if a*^-^ — ^z"-* be divisible hy a — b. That is. If the dif- ference of the same jyowei's of two quantities be divisible by the difference of the quantities themselves, then vjill the difference of the next Idglier powers of the same quantities be divisible by the difference of the qiiantities. Hence, since a^ — ¥ is divisible hy a — b, . • . «^ — ^ is divisible by a — b ; since a^ — V^ is divisible hy a — b, .' . a^ — b^ is divisible by a — b ; and so on without limit. Therefore, for every integral value of n, a^ — ^" is divisible hy a ~ h 218. ^-|" = rt"-^ + «"-2^' + a"-^Z.2+ . . . + By carrying out the division of «" — Z*" by « — b, it will be observed that the number of the term and the exponents of a and b have a constant relation to each other. This relation is expressed by the general term, -\- a*^~''b^~^. By substituting in this term 1, 2, 3, 4, etc., in succession, in place of ?•, all the terms are found. Thus, the 15tli term of r- is found by substi- a — b '' tuting 20 for w, and 15 for r. That is, «""*' b''~^ becomes ^80-15 ^15-1 _ ^5^u^ which is the required 15th term. 219. To prove that r/" + i" is divisible hy a -\- b when n is an odd integer, and is not so divisible when n is even. By division, [—- = a""-^ — a""-^ b -\ ^ -^ \ -^ a -{- b a-\-b Keasoning as in (217), we see that, If the stem of the same potoers of tioo quantities be divisible by the sum of the quantities themselves, then will the sum of the powers two degrees higher be divisible by the sum of the quantities. Hence, since a^ -\- b^ is divisible hy a -\- b, .' . a^ -\-b^ i^ divisible by ^5 + ^/ since a^ ~\- b^ is divisible hy a -{-b, . • . a^ -\-b^ h divisible by a -\- b ; and so on without limit. . • . if 71 be any odd integer, a" -}- Z*" is divisible hy a -\- b. 126 ALGEBRA. We see further that dl^ + ^" will not be divisible by « + J unless rt"~^ + Z>"~^is divisible by a -j- ^. Now since c^ -\- W is not divisible by a -{- h, therefore rt* -\- h^ is not divisible by a -[- h, therefore a^ -\- If is not divisible by a -\- 1), and so on to infinity. The quotient of («" + ^") ^{a^V) is like that in (218), excepting that the sign of each even term is — . 220. In like manner it can be shown : (1) a" — If' is divisible by «^ -f ^ when n is even, and is not so divisible when n is odd; (2) a]^ -\- If is never divisible by « — &. 221. — Case XIII. A polynomial luhich is a power, the difference of like poivers, or the nimi of like powers, may he factored hy using the Binornial Formula (129) in connec- Hon tvith the preceding cases. Thus: 1. a^ - %a^l) + '6al^ - h^ = {a - hf, 2. a* - 4^3^ + Mb^ - 4«&^ + Z>* = (« - hf. 3. 2x^ - 3a;2 _|_ 3,^ _ 1 ^ ^3 _|_ (^3 _ 3^2 _|_ 3^ _ i) _ x^-^{x- If = \x -]-{x - l)][x^ -x(x -!)-{- {x-iy] = (2x- l){x^-x^l). 4. x^- 4:^3+ 5^:2- ix + 1= {x^—4:X^-\- 6x^- 4^: + 1)- x^= (x - 1)* -x^ = [{x - If + x][{x - ly -x] = (x^ - X -{- l){x' - 3x -\- 1). 6. a^- 4:X^ -\-6x''-8x-3 = {x^-4:X^+6x^-4:X-^l)-{x^-j-4:X-{-4)= {x-lY-{x-^2Y = [(..•- 1)^ +(:. + 2)][(a; - 1)^ -(:. + 2)] == (x^ - x -}- 3){x^ - 3x- 1). 6. x^-{- 3x^ -\-3x~7 = {a^-{-dx^-\-3x-\-l) -8 = (a; + l)3_8 = [(a; + l)-2][(2: + l)^ + 2(:^ + l) + 4] = (x - l)(x^ -{- 4.x -\- 7). 7. a* + ^* + c* - 2a''b^ - 2a^c^ - 2W& = (ft* + ^* + c* - 2a''W - 2ftV + 2^V) - 4Jfc" = {a' -y- c'f-^Wc^^{ii^-W-&-^2hc){a^-W-&-2h^ = (a + ^ — c)(a — l-\- c){a + ^ + ^')(^ — b — c). FACTORING. 127 222. — Case XIV. A polynomial wMcJi lias a hinomial divisor is readily factored by trial. Two methods are in common use : (1) Synthetic Division (144); (2) Separation of Terms. 223. — (1) 1. To factor o? -^x'-^- 2Qx - 24 by synthetic division. The first term of the binomial divisor must be X, and the second term must be one of the divisors of — 24. It is usually best to try one of the smaller divisors, com- mencing with the one next to the square root. In this case try — 4. 1 - 9 + 2G - 24 + 4 +4-20 + 24 1 — 5 + 6 Tlie division being exact, one of the factors is x — 4, and the other is a:^ — 5a; + 6 = {x -2){x- 3). Ans. {x - 4) (a; - 2) (x - 3). 224.— (2) To factor a^ - 9.^ _^ 26a; - 24 by separation of terms, proceed as follows : (1) Arrange and grade the terms. (2) Write the ex- treme terms imchanged. (3) Write as the co-efficient of the second term a divisor of the product of the extremes. In selecting this co-efficient, try the lesser divisors, commencing with the one next to the square root. Hence in this case try — 4a;^ as the second term. (4) To find the third term: Sub- tract this second term from the second term of the proposed expression. In this example the third term is — 5a;^ (be- cause the sum of the second and third terms must be equal to — 9a;^, the second term of the proposed expression). (5) To find the fourth term : Divide the jwoduct of the second and third terms hy the first term. Hence the fourth term in this case is — 4a;^ (— 5a;^) -j- a;^ = + 20a;. (6) To find the fifth term: Subtract thisfotirth term from the third term of the proposed expression. Hence in this case the fifth term is + Qx. (7) So continue, each odd term being 128 ALGEBRA. found from subtracting the preceding even term from the corresjjonding term of the proposed expression, and each even term being found from multiplying the preceding odd term by the quotient of the second term by the first. If the last even term found as above be equal to the last term of the proposed expression, the proper second term has been selected, and the work is continued by combining the odd terms in one parenthesis and the even terms in another. If no mistake be made, the expressions in paren- theses will be alike, and the work is finished as in Case III. Thus: u?- 9a;2+ 26z-24 = r^- ^x^-bx^-^-'iOx -f 6rc-24 = x{7? -bx -\- Qi) - 4:{x' - bx -{- Q>) = (x - 4){x' - bx + G) = {x-^){x-^){x- 3). If the last even term found from the rule be not equal to the last term of the proposed expression, the wrong second term has been selected, and another trial must be made ; and so on. EXAMPLES. 2. 2x^ + 3:^2 _ 8^ _|_ 3 ^ 2t^ _ 9^2 _|_ 5,^.2 _ 5^. _ 3^ _^ 3 ^ x {^x" + 5a; - 3) -(2^2 -f bx - 3) = (a: - \){^x^J^bx-Z) = {:c-l){2x- l)(a; + 3). 3. x"" -Qx'^llx-Q = {x- 2)(x - \){x - 3). 4. x'-7x^- 6x = X {x + 2)(x + l)(x - 3). 6. x' - 21a.2 - 20 = {x^ + 4)(2;2 + l){x^ - 5). 6. 2x^ - x^ - 12 = (a; - 2)(2a;2 + 3:^4- 6). 7. x^ -3x-lS = (x- 3){x^ + 3x + 6). 8. a^-6x-{-9 = {x + 3)(x^ - 3x + 3). 9. x^-7x-irQ = {x-2){x-l){x-\-S). 10. x^-3x-4 = (x-}- l){x^ — x^ -\- X - 4:), 225. — Case XV. Certain polynomials of the general form X* + ax^ -f- bx^ + ex -f d can be decomposed into two trinomial factors. FACTORING. 129 1. To factor 2:* + ic» - 4x^ + 11a; - 3. It is evident that the first term of each trinomial is x^, that the second terms contain x with co-efficients at present unknown, and that the product of the third terms is equal to the last term of the proposed expression. Tlie required factors may, therefore, be represented by {x^ -j- ax -j- b) (a^ -{- ex -\- (I ), in which a, b, c, and d are to be found. Multiplying tliese factors together, we find that the pro- posed expression must be equal to Hence, \. a -\- c = II. b-[-ac-{-d = III. be -h ad = IV. bd = ^+ b x^-\- be ~\- ac ■i-ad 4- cl x-\-bd. 1. ^ Since bd = — 3, one 4. [of these numbers is 11. ( ± 1 and the other is 3. J =f 3. Assume^=— 1, then ^ = + 3. We now find a and c by substituting in III, and combining the resulting equation with I. Thus : 1. a-\- e = 1. ) Add. . • . 4« = 12 and a = 3, whence III.- c + 3a = 11. j e= -2. By trial, we find that the values a = 3, b= — 1, c = — 2, and 6? = 3 satisfy equation II; therefore the proper values of the letters have been found, and the required factors are (x^ -\- 3x- l)(x'' - 2x ^ 3). If the values found do not satisfy equation II, other values must be assumed for b and d, and the correspond- ing values of a and e must be found as before; and so on. 2. To factor .r* - 13.r« + 2.r^ + 34a: - 15. Herel. a-f c=:-f 2; 11. b-]-ae-\-d = -U; 111. be -\- ad = 34; IV. bd = - 15. These equations are satisfied by« = 5, b = — 3, c = — 3, d = 6; .' . the required factors are {a^-\-bx — 3)(a^—3x-{-5). 3. To factor 2a;* -6x^-{- 93^ -6x ^2 = 2 (x' - 3^;^ + ^x^ - 3x -{- 1). ]30 ALGEBRA. Here a -\- c = — 3; b -{- ac -\~ d = 1-; be -\- ad = — 3; bd=l; which are satisfied by «=— 1; b = l-; c=—2; d—2. EXAMPLES. 4. a:* + o;' - bx" - Ttx - 30 = {t? - 2a; - 5) {p? + 3:r + 6). 5. a;* + 52;^ + Ylx^Ar 272:+30= (a;^ + 2a; + 5) (a;^ + 3a; + 6). 6. a;*+ 3a;^+ 9a;2+ 10a; + 12 = (a;^ + 2a; + 4) (a;^ + a: + 3). 7. 5a;*+17a;^+ 12a;2+ 5a; + 1 = (5a;2+2a; + 1) (a.-^ + 3a; + 1). 8. ^x'' + x^ + 3a;2 - 3a; - 2 = (2a;2_|_ ^ _^ 2) (3a;2 - a; - 1). [NoTB.— Beginners may omit Arts. 226, 227.] 226.— Case XVI. A binomial of the form a" ± b^ where n is divisible by two or more odd numbers luliich are prime to each other, may be decomposed into at least four factors. Ex. a'' ± b''; a^^ ± b^^; a'' ± P; ^^ + 1; etc. The solution of one of these examples will illustrate the method for all ; hence we shall explain the method of de- composing a^^ 4- 1 into four factors. By Case XII, ^^^ + 1 = (a' + 1) {a'' - a' -\- 1) = (a + 1) (a* -a' + a^-a-\-l) {a'' - a' + 1). But a'^ + 1 = («^ + 1) (a'^ - a^ -{- a^ - a^ -}- 1) = (a + 1) («2 _ « -f 1) (ai3 _ a^^a'- a^ + 1). It is evident that a number can not be made up of two diiferent sets of prime factors; hence one factor at least in each of the above sets is composite. In the first set d^^ — a^ -\-l is the only factor which can be composite; therefore «^* — a^ -\-l is divisible by ^^ — « + 1, the prime factor of the second set not found in the first set. (^10 _ a^ _^1) -^ {a^ - a + 1) = a^^a'-a'-a'-a^ + « + 1. .-. a^° + 1 = (« + 1) («* - ft^ + «2 - « + 1) («2 - rt + 1) {a^-\-d^ — a' - a^-a^-{-a-i- 1). It is clear that the salne result would be obtained by dividing the composite factor of the second set, a^^ — a^ + FACTORING. 131 a^ — a^ -{- 1, by the prime factor of the first set not found in the second; viz., a^ — a^ -{- a^ — a -\- 1. 227. Certain other expressions are factored in like man- ner. Thus : a^^ + 64 = (a* + 4) («« - 4a* + 16). XII. But by Case X, a* + 4 = (a" -{-2a-\-2) (a^ - 2a -^ 2). . • . a^^+ 64 = {a^ + 2a + 2) {a^ - 2a + 2) (a«- 4a* + 16). But a^2 + 64 = (a« + 4a^ + 8) (a' - 4a^ + 8). X. Hence the factors of the second set must be divisible by the prime factors of the first set. By trial we find a« + 4a3 + 8 = (a^ - 2a + 2)(a* + 2a» + 2a^ + 4a + 4), and a« - 4a^ + 8 = (a^ + 2a + 2) (a* - 2a^ + 2a^ - 4a + 4). 228. Case XVII. When the paints of a polynomial are factored by previous cases, and these parts have a com- mon factor, the expression is factored by using Cases II and III in connection with the cases used in factoring the parts. 1. a''-b''-a^b={a^-y)-(a-b)^{a-b){{a^b)-\]. VL 2. a^^h'-a'^ab-})'= (a^ + b^) - (a^ - ab ^ b^) = («2 _ab^l^){a-]-b- 1). XII. 3. a« - 2ab -\- b'' - 3a -}- 3b = (a - bf - S {a - b) = (a-b)(a-b- 3). IV. 4. a^ - 6ab-\-6b^- 2a + 4i^ = (a^-5a^ + 6^^^)- 2 (a - 2^^) = (a-2^)(a-3^>)-2(a-25) = (a-2J)(a-3^>-2). VIII. 5. a'-2a^-^4:ab-4:b^-\-4:b'=(a'-]-4b')-2(a^~2ab-{-2b') = (a^ + 2a^> + 2J-) (a^ - 2ab + 2b^)-2 (a^ - 2ab + 2b'') = {a^ - 2a5 + 2b^) (a^ + 2ab + 2b^ - 2). X. 6. a^-ab-a-2b^- lb- Q = a"" - a{b-\-\)-{2b^ + 7Z> + 6). The two numbers whose sum is — (^ + 1), and whose product is - {2b^ + 7/5> + 6) are - {2b + 3) and {b + 2). (VIII and IX.) Ans. {a - 2b - Z) {a ^ b -{- 2). 7. bo" - Sab + 7a + 3J» - bb + 2 = 5a« _ a (8Z> - 7) + (3b^ - 5b -\- 2). 132 ALGEBRA. Multiply by 5, and substitute ij for 5«. (Art. 204.) . • . 1/2 - (SZ* - 7) y + 5(3^2 - 5Z> + 2) results. Now factor 5 {W -5^ + 2) (Art. 205) into 5 {U - 2) {b - 1). Since(3Z»-2) + 5(Z'-l) = (8Z>-7), .-. if-{8b-7)y-\- 5 (3^ - 2) (^ - 1) = Cv -3^ + 2) (y - 5J + 5) = (5a - 3^ + 2) (5a - 5Z> + 5). Now divide by 5. Ans. (5a - 3^ 4- 2) (a - 5 + 1). 8. a}^ + a* + 10 = {a'' + 8) + (a' + 2) = (a* + 2) [(a« - 2a* + 5)]. XII. 9. x^ + 22;'7/2 + 9?/ — 2:^2 + 4:r.?/ - Gf = (a:* + 2a;2/y2 + 9^*) - 2 (.i-^ - 2xj/ + 3i/2) = (x' ~ 2xy + 3i/2) {x" + 2i:.y + 3/) - 2 {x" - 2xy + 3"?/^) = (:?;2 - 2a:// + 3?/) {x? + 2a:?/ + 3/7^-2). X. 10. a' - 3a2^ + 3aZ>2 - Z*^ _ 3^2 _|- ^al - W- ^ {a-hf-Z{a-l)y={a-hY{a-h- 3). XIII and IV. 11. a' + a^ + 2 = (a^ + 1) + («'+!)= (a^+l)(a«-a«+ 2) = (^ + 1) («2 _ « + 1) (,«6 _ ^^3 _|. 2). XII. 12. a'+a'-10 = (a^-8) + (a^-2) = (a3-2)(a«+a3+5). XI. GENERAL REMARKS 0^ FACTORING. 229. In order to become proficient in factoring, it is necessary, first, to learn the formulas; and secondly, to be able to apply the proper formula to the proposed expression. This power can be acquired only by practice and study, but the following general rules may help the student to be- come expert in this difficult branch. Before applying any of the following rules, it is necessary to remove from the proposed expression all monomial factors common to the terms (193, 196), and to arrange the result. 230. — KuLE A. Binomials. First notice whether the proposed binomial be the differ- ence of two squares; if so, employ Case VI. If the proposed binomial be not the difference of two squares, see if its terms be both third powers, or both fifth FACTORING. 133 powers, or both seventh powers, etc.; if so, Case XI or Case XII will apply. If VI, XI, and XII fail, try Art. 210. 231.— Rule B. Trinomials. First notice if the proposed trinomial be a square (197). If it be so, factor according to Case IV. If the trinomial be not a square, then, if the first term be a square, try VIII; and if the first term be not a square, try IX. If two of the terms be squares, and twice the product of their square roots exceed the middle term by a square, then X will apply. Instead of IV, VIII, and IX, XIV may be employed, and instead of X we may use XV ; but XIV and XV may frequently be used where the others fail. Finally, in some examples. III and XVII will apply. 232. — EuLE C. Expressions having four or more terms. Try the various cases in the following order : III ; V ; VII ; XIII ; XIV ; XVII ; XV. 233. — Rule D. After resolving an expression into fac- tors, each factor must be examined for the purpose of determining if it be prime (195). 234. It is important to notice that the signs of an even number of factors may be changed (116). Thus : a^ -V' = {a - b)(a -\- b) = (- a + &)(- a - b); a?-x-(^ = (x- 3)(a; + 2) = (- a: + 3)(- x - 2); etc. It is usual to give the factors whose first terms are -j— 235. Formulas to be memorized: 1. a^-b^=(a-\-b){a — b). 2. a^-h^=(a- b)(a' + ab -\- b'). 3. ^3 ^ ^3 ^ (^ _|_ J)(^2 _ rtJ _|_ ^2)^ 4. a' + 2ab + Z>3 = (rt + b)(a + b), 6. a* - 2ab -\- b^ = {a - b)(a - b). 134 ALGEBKA. 6. 0^ -]- {a + h) X -]- ai = (x ^ a){x + b). 7. a^ — {a-\-b)x-{- ab = {x — a)(x — b). 8. x^ -{- {a — b)x — ab = (x -\- a)(x — b). 9. x^ -\- x^y^ + ^* = (^^ -\- ^y -{- y^){^^ — ^y + y'). 10. x?-\- y^-^z^—3xyz=(x-\- y-\-z){x^-\-tf-\-z^—xy—xz—yz). Exercise XIII. Resolve the following expressions into prime factors: 1. a^ - 8a* + 3^^ 6 a* - ^2 + |. 2. («^ - ^)^ - 3 (a - Z')l 7. a' - 7a' + ^a. 3. x^ — a^ -{- x^ — X. 8. a^ — 9. 4. a'-\- a^+ a^-{- a'-\- a^-{- a. 9. 1 6a* - 9. 5. (a-by-{b-aY+(a-bY. lo. (a - ^» - 6^)2 _ dK 11. a« - 2a5 -i- b^ - 6a -\- Qb -\- 9. 12. a^ - 4r/Z' + 4:b^ -\-2a-4:b -8. 13. ^2 _ 4a + 4 - ^-2 - 2^6- - cl 14. a^ + 4a - J2 _^ 6^ - 5. 15. a^ - Ua + 13. 25. 9.^*- le^;^ -f 4. 16. rc^ — 10a;?/ -f 162/1 17. a;^ — 2a^ + a;. 18. ar* - 9a;2 _|_ 8^ 28. a^* + 1. 19. 90.-^ + 3a; - 2. 29. a^^ + 27. 20. 4a.-2 - 8a; + 3. 30. a^ - 6a^ + 12a - 9. 21. 6x^ - 5a; + 1. 31. a* - ia' - da^ - 4a + L 22. Qx^ — a; — 1. 32. 2a;* — x' — 12a;. 23. ar* — 3x^-\- 1. 33. cr^ — x — 6. 24. 4a;* - 6x^-\- 1. 34. a;^ + a; + 2. CHAPTER XIV. COMMON FACTORS AND MULTIPLES. 236. A Common Factor of two or more expressions is an expression which is contained in each of them Avithout a remainder. Thus : 4a is a common factor of 12a and 20a; X -\^y is Si common factor of x^ — y^ and (x -\- yy. 237. Two expressions wliich have no common integral factor except 1 are said to be prime to each oilier . 238. The Highest Common Factor (H. C. F.) of two or more expressions consists of all the prime factors common to the expressions. Thus: ba" is the H. C. F. of \0a\ Voa\ and 25a*. The Highest Common Factor is also called the Greatest Common Divisor (G. C. D.); the Greatest Common Meas- ure (G. C. M.); the Highest Common Divisor (H. C. D.); and the Highest Common Measure (H. C. M.). 239. If two or more expressions be divided by their Highest Common Factor, the quotients will have no com- mon integral factor except 1. 240. To find the Highest Common Factor by means of factoring: (1) Resolve each expression into its prime factors, (2) Select from these the lotvest power of each common factor, and find the product of these powers, (136) 136 ALGEIJRA. 1. Find the H. C. F. of 24:a'¥c^ and 36a^(^. .-. the H. C. F. = 2^ X 31 X a' x b^ X c' = I2arbc\ 2. Find the H. C. F. of 4:0? - ^xif and 62;* + %xy\ 4:x^ - 4:xif= 2'x {x' - if) =:2\x.{x^y){x- y); ex' -i- 6xy'= 2 .^ .x.(:^-^ y^)= 2 .3.x.(x + y){x^- xy + y% .'. theH. C. F. = 2KxK(x -{- tjY = 2x{x -{- y) =2x'^2xy, 3. Find the H. C. F. of a^ — a; aJ^ J^ 2a -\- 1) and 4a2+ 4. a^ — a=:a.{(v'-l)=ia.{a-^l){a — 1). 4{a^-^l) = 2^{a-\-l){a' -a + V). a^J^2a-\-l = (a + 1)2; .-. the H. C. F. = « + 1. 4. Find the H.C.F. of ^-{a^b)x-\-ab; x'-{a-\-c)x^ac; and 01? — {a — b)x — ab. x^—{a-{-b)x-\-ab={x—a){x — h). x^~{a-\-c) x-\-ac={x — a) {x — 6"). x^—(a—b) x—ab=(x—a) (x-^b). .-.the 'EL.C.¥. = x — a, EXAMPLES. Find the H. C. F. of the following expressions : 5. Sa'bj Qa%\ A71S. 2a^l, 6. a^ - ¥; a" - dab + 2b\ Ans. a- b, 7. a^ — 8; a^ — 4a -}- 4. Ans. a — 2. 8. a' + 27; a^ -\- 2a - 3. Ans, « + 3. 9. a^ — a^; a? — 5a^ -\- 4a. Ans. c? — «. 10. iu' - 1; a;* + a:^ + 1. Ans. x?-^x-\-\, 11. ic* + 8; ic* + 4x? + 16. Ans. a^ — 2x -{- 4, 12. a' -\-l; a'- 1. Ans. 1. 241. The method of finding the H. C. F. of two or more expressions by division is similar to the corresponding case in Arithmetic, and depends upon the following principles ; 242. A factor of a7i expression is a factor of any multiple of that expression. Thus : a divides ab, and a also divides cab. COMMON FACTORS AX I) MULTIPLES. 137 243. A comm 071 factor of two expressioiis is a factor of the sum or difference of any mnltiples of the expressions. Tims : a divides ab and ac, and a also divides mab ± rac. 244. A common factor of two expressions is a factor of their remainder after division. This is merely a special case of the preceding principle, because the remainder after division is found by subtract- ing from the dividend the product of the divisor and quo- tient. Let the dividend be ab and the divisor be ac, and let the quotient be q ; then ab — acq is the remainder, which evidently contains a, the common divisor of ab and ac. 245. The H. C, F. of two or more expressions will not he altered if they be multiplied by expressions which are prime to each other. This follows from (238), since introducing factors which are prime to each other can not change the common factors. 246. Factors ichich are not part of the H. C. F. may be removed from the proposed expressions withoiU affecting the H. C. F. (238). The H. C. F. of 12, 36, and 27, is 3; if 4 be removed from 12 and 36, the resulting numbers, 3, 9, and 27, have also 3 for their H. C. F^ 247. If a common factor of two or more expressions be removed, their H. C. F. is found from multiplying the factor removed by the H. C. F. of the qnotients. To find the H. 0. F. of Sa^ («»- b^) and 3« (rt^- y"). The factor 5a^ may be removed from the first expression, and 3a from the second, and their common factor, a, is part of the H. C. F.; but the other factors may be discarded. The H. C. F. of the quotients, a^ — b^ and a"^ — b^, \^ a — b ; therefore, the H. 0. F. of the proposed expressions is a(a — b) = a^ — ab. 248. The solutions of the following examples illustrate the method of finding the H. C. F. of two expressions by division (Common and Synthetic). Before beginning the Alg.— 13. 138 ALGEBRA. operation of division, remove all monomial factors from the polynomials (246, 247). Then arrange the proposed ex- pressions according to the descending powers of some letter, and use the expression containing the highest power, as the dividend. 1. Find the H. 0. F. of Ax^ ^ 6a^ -^ x - 2x^ - 1 and ^x^ + 2x^ - 1. (1) (2) 4:c* -{-Gar" -2x'-i-x-l \ 4:X^ + 2a;^ + 4a;* + 2^;^ + —x X -}-l -^-^x" -%Q? -\-2x- 1 42;^ + 2a;2+0 -1 - Ax^ + 2a; = - 2x (2a: - 1). 4ar^ + 2x^ + - 1 | 2a; - 1 4a;» - 2a;2 2x' + 2a; + 1 + 4a;2 + + 4a;2 - 2a; + 2a;- 1 4-2a;- 1 4 -2 -0 + 1 4 + 6 -2 . 2a; -1 -2 + 1 -0 + 1 -2-0+1 2 + 1 H. 0. F. 4+2+0-1 +2+2+1 1 + 1-4 + 2 2. Find the H. C. F. of 30a;* "^ -x-\- 14. 2 + 2 + 1 2 (2 - 1) 64a;3 _ y^^ _^ 49^ _|_ ;^ (1) 30a;* - 64a;3 - 17:^^ + 49a; + 2 | lOa-^ - 23a;^ - a; + 14 30a;* 69:c3 3a.-2 42a; 3a; +i + 5a-3 - 14a;2 + 7a; + 2 bx^ - llja;2- -Ja; + 7 2^+ 7ia; - 5 = - 2 J (a;* - 3a; + 2). COMMON FACTORS A^D MULTIPLES. 139 10a:»- lOa^- - 23:r2 _ ^^ _^ 14 - 30.^2 4- 20^ 10a: + 7 7^" — 21a; + 14 7x^ - 2I2; + 14 H. C.F.=a^ Zx (2) 10 30-64 -17 +49+3 4-23 4-69 + 3 -42 + 1 +114+ 4-7 -14 3+ i - 2i+7i-5 = - 2i (1-3+2) _1 +3 -2 10-23- - 1 + 14 4-30- -20 +21- -14 10+ 3. Find the H. C. F. of 2x^ + (2a - 9) x^-(9a + Q)x + 27 and 2x^ - Ux + 18. (1) 2x^ + (2a - 9)x'' - (9a -\- C))x + 27 | 2x^ - 13.r + 18 2ar» - 13r^ + 18.c re + (a + 2) + (2a + 4)a;« - (9a + 24).t + 27 + (2a + 4^)3^ - (13rt + 26)a; + (18a + 36) + (4a + 2)a;-(18a+ 9) = 2(2a + l)x-9(2a+ 1) = (2a + 1) (2:r - 9) 2x^ - 13.r + 18 2x — 9 x-2. Aits. 2x (3) 2 + 13 -18 2 + (2a - 9) + 13 - (9a + 6) +27 - 18 + (13a + 26) - (18a + 36) 1+ (a + 2) + (4a + 2) _(l8a + 9) = 2 (2a + 1) - 9(2a + 1) = (2a + 1) (2-9) + 9 2-13 + 18 + 9-18 1- 2 140 ALGEBRA. 4. Find tlie H. C. F. of x"^ ~ ay? - aj'x' - d^x - 2a* and ?>x^ - "lax^ + Salt' - 2al Since the first term of the dividend is not divisible by the first term of the divisor, in order to avoid fractions, multiply the dividend by such a number as will render the first term divisible. The least multiplier is 3; but, since the division will be carried out to two terms in the quotient, it is preferable to multiply by 3^ = 9. If there were three terms in the quotient, multiply by 3^ = 27, etc. (1) 9:c*- 9a.^'^- 9aV- 9a^:g-18a* | 3a;^-7a2r^+3a^a:-2a^ 9:r*-21aa:^+ 9aV- Mx 3^; +4a +12ar^-28a2a;2+12A- 8a* + 10aV-15alc-10a* = ^a^{^x^-'^ax-%t^). For the same reason as before, we multiply the new divi- dend by 4 = 21 2^;^ - 3ay - 2^^^ 6a; — ha Vlx? - 28ax^ + 12a^x - Sa' 12:^;3 - 18ax^ - 12 A - lOax^ + 24.a^x - 8^^ - lOax^ + Iba^x + lOa^ + Mx -\W = 9^2 {x - 2a) {^x^ — Zax — 2a^) -h (x — 2a) =2x-\- a. Ans. x — 2a. (2) 3 H-7 -3 -f3 9- 9 +21 3+ 4 - 9- 9-18 -9+6 +28-12+ 8 Multiply previous divisor by 4 for new dividend. +10-15-10 = 5(2-3-2). 2 -1-3 +2 12-28 + 18 + 12-8 + 12 -15-10 1 + 2 9-18 = 9(1-2). 2-3-2 + 4 + 2 2 + 1 COMMON FACTORS AND MULTIPLES. 141 5. Find the H. C. F. of ^* Id? + (c — X)a^ -^ha — c and cffl^ 4- {b - ly -\-ca-b. In this case it is not necessary to go througli the form of dividing one expression by the other, because the quotient will be unity. Therefore subtract one expression from the other, use the remainder as the divisor, and one of the pro- posed expressions as the dividend. From the remainder (c — h) a? -\- {c — b) a^ — (n — b)a — (c — b), remove the factor (c — b)y which can not be part of the H. C. .F. (1) ft* - m' -\-(b-l)a^ + ca rt* + a^ — a^ — a b \a^-{-a' a -(c-\- 1) ba^-{-{c-{-l)a-b (c 4- 1)^' + (,. -^ i),,3 _i^,-\-l)a^-\-{c-\-l)a+{c + l) + (J + c + lKH-0-(^»+c + l) (6 + c-^l)(«^ + 0-l). a' + ft* - ft - 1 |a' + a« + 0— a a+l 4- n« + - 1 a'' - 1 = H. 0. F. (2) 1 l-c + (b-i) + c - 1 - 1 + 1+1 -J + 1 + {c + l)-{c + l)~(o+l) + 1 1 - (c + 1) + (b + c + l) + 0-{b + c + l) = {b + c + 1)(1 + -1 ) II - + 1 1-1-1 -0 + 1 -0+1 14-1 142 ALGEBRA. 249. Eule for finding the H. 0. F. of two polynomials. (1) Remove all monomial factors from the projjosed ex- pressioiis, and set aside the H. C. F. of these mo7iomials as a factor of the H, C. F. sought, (2) Arrange the resulting polynomials according to the descending powers of a common letter: take for tlie dividend that expression ivhicli has the highest poiver of the letter of arrangement; or, if loth have the same power, take either for the dividend. (3) Divide as in division of polynomials, until aremai^i- der is obtained lohich is of loioer degree than the divisor : if the first term of any dividend he not divisible by the first term of the divisor, multiply that dividend by such an ex- pression as ivill make it thus divisible, (4) Remove from the remainder every factor common to all the terms. Take the resulting p)oly7iomial as a divisor, and the divisor as a dividend, a7id proceed as before. (5) So contiiiue, always dividing the last divisor by the last remainder, until nothing remains. (6) The product of the final divisor, and the H. G. F. of the mono7nials set aside in the first place, is the H. C. F, sought. Verification: Divide each of the proposed expressions by the answer found. If the divisioti be exact, and the quo- tients have 710 factor in conwion, the correct answer has been found. 250. To find the H. C. F. of three expressions : Fi7'st find the H. C, F. of two of them, and then of that result a7id the third expressio7i. For the H. C. F. of two of the ex- pressions contains all the factors common to those two, and the H. 0. F. of this result and the third expression con- tains such of those factors as are found in the third. Of four expressions: First find the H. C. F. of three of them, and then of that result and the fourth. Proceed in like manner if there be more than four expressio7is. COMMOiH^ FACTOKS AND MULTIPLES. 143 EXAMPLES. Find the H. 0. F. of the following expressions : 6. Sor* — Ibx — 50; o? — Qx -\- 5. Ans, x — 5. 7. 20^ -ix- 16; 2^ - 10a; + 8. Ans, 2x - 8. 8. 2a^-{-9a^ + '7x- 3; da^ -{- bx" - Ibx + 4. Ans. x^ -{-dx — 1. 9. x^ + ^x^ - x" - 8x i- 5; a;* + 2a^ - 5a^ - 10a; + 12. Ans, a; — 1. 10. 3:?;* + 10a;' - x" -}- 4x -\- 9 ; x^ -}- dx^ - x^ + 2a; + 2. Ans, a;^ — a; -f !• 11. 4a;* - 8x^ - 20a;2 + 24a; + 20; Sa;^ + ea;^ - 24a; - 45. Ans. a^ — X — 6, 12. a^-{-23^-x'-{-x'-3;a^-aT'-{-2x'-j-x-d. A7is.x^-1, 13. a;* - 3ar» + 7a;* - 7a; - 6; a;* + a;^ - 8x^ + 7a; - 6. A71S, X — 2, 14. a;* - Sar' 4- 8a;2 - 7a; + 3; 2ar» - 9a.-* + 10.i; - 3. Ans. a;* — 4a; + 3. 15. 2a;*~ 3a;'+ 3a;2- 3a; + 1; Q>x?- 7a;*+4a;-l. Ans. 2x-\, 16. 2a;2 + 3a; - 2; 2.^' + 3a;2 + 4a; - 3; 4.^-' + 2a;2 - 1. Ans, 2x — 1. 17. x^-\- 0? - Zx- 2\x' - Zx^^x + 2; a;*- :^-hx?^^x-^\, Ans, X? — X — \. 251. A Common Multiple of two or more expressions is an expression which is exactly divisible by each of them. 252. The Lowest Common Multiple, or the Least Common Multiple (L. C. M.), of two or more expressions, is an ex- pression which is exactly divisible by each of tliem, and the quotients have no integral factor in common except unity. 253. Every common multiple of two or more expres- sions must contain all the factors of the proposed expres- sions, and may contai^i-afelie^^actors also ; but the L. C. M. 144 ALGEBRA. consists of the factors of the proposed expressions only, each factor being written with its highest exponent. Thus: 1. Find the L. C. M. of Sa^bc ; Vla^Wc ; ?>Qah\\ . •. the L. 0. M. = 2^ X 3^ a'lfd' = 72ft^Z>V. 2. Find the L. 0. M. of x"" - 2x + 1; 4^;^ - 4; a;2+22;+l. a^-2x-]-l = (x-lY; x^ ^ 2x -^ 1 = {x + 1)^; 4a;2 - 4 = 2%x - l){x + 1). .-. A71S. ^(x-^lY(x^l)\ 254. The L. C. M. of two expressions may also be found by means of their H. 0. F. For example, the L. C. M. of ab and ac is abc. This is evidently obtained from multi- plying ab by c, or ac by b. Hence: Divide one of the two pvoi^osed expressions by tlieir H. C. F., and multiply the other expression by the quotient. The product is their L. C. M, 3. Find the L. C. M. of x^ — {a — b)x — ab and x^ — {a-\- c)x-{- ac. (1) The H. 0. F. is 2: - a, (2) The quotient oi x^ — {a—b) x — abhj x — a\B> x-\-b. .• .(3) The L. C. M. = [x^ - (a -\- c) x ^ ac] [x + b\ 4. Find the L. C. M. of x^-{-x''-4:X-^ and iC«+6a;2+lla;+6. (1) The H. C. F. is x^^^x-^ 2. (2) The quotient of x^^x^-^x-4: by a;2+3:z:+2 is x-2. .-.(3) The L. C. M. = {x^ + Qt? + 11a: + 6)(cc - 2). 255. To find the L. C. M. of three expressions : First find the L. C. M. of two of the proposed expressions ; then find the L. C. M, of this result and the third ex- jjiression. 5. Find the L. C. M. of x^ -x-2', x^ -\- x - Q>', x^ - 4c. COMMON FACTORS AND 3IULTIPLES. 145 (1) The H. C. F. of ^c^ - a: - 2 and x^ -\- x - 6 is x - 2. (2) The quotient of rc^ + a; — 6 by a; - 2 is a: + 3. .'.(3) The L. C. M. of the first and second expressions is (x^-x-2){x-\-3). (4) The H. C. F. of this L. C. M. and a;^ - 4 is x - 2. (5) The quotient of 2^ — 4 by a; — 2 is ic + 2. .'.(G) The required L. 0. MAs (a^ - x — 2){x + 3)(a; + 2). 256. To find the L. C. M. of four or more expressions: Mud the L. C. M. of three of them, and then the L. C. M, of this result and the fourth. Proceed in like manner if there be more than four e.rpressions. 257. The operation of finding the L. C. M. may be ab- breviated when certain of the proposed expressions are divisors of other given expressions, by omitting such divisors. For : A miiltijjle of an expression is a multiple of every divisor of that expression. 6. Find the L. C. M. of a;* — y*; x^ -\- xy -{- y^j a^ + y\- X + y; x-yj 7? - y\ Since' 7? -\- xy -\- y^ is a divisor of or* — y^y and since a? -\- y^, X -\- y, and x — y are divisors of x^ — y^, therefore the L. C. M. of ic* — y^ and 7^ — y^ contains all the pro- posed expressions. Ans. (a;* — ?/*) (7? -\- xy -\- y"-), EXAMPLES. Find the L. C. M. of the following expressions: 7. :j^ — x; a;^ - 4a; + 3 ; a,-^ - 3a; + 2 ; x? — 5a; + 6. Ans. X*' - Qx^ + \l7? - 6a;. 8. 6a;2 + 7a; - 3 ; 3a;= + 5a; - 2 ; 2x^ -{- 7a; + 6. Ans. &7? -\- 1^3^ + 11a; - 6. 9. 6a;2 - 5a; + 1 ; 2x^ -bx-\-2', ^7? - Ix + 2. Aus. (j7^ - 17a;2+ 11a; -2. Alg.— 13. 146 ALGEBRA. 10. 1 — 0^ J 1 -\- X -{-X^J 1 — X -^ X^ — x\ Ans. 1 -\- x^ — 01^ — x^. 11. a;* - b7^ + 8:z;2 - 7ic + 3 ; 2^;^ - 9:^^ + 10:?; - 3. Ans. 2x^ - lla^ + 21a;» - 22^;^ + 13a; - 3. 12. x^-ia^ + 3x^ + 4:c - 12 ; 2a^ - x" - 18a; + 9. A71S. 2x^-da^- 17x' + 35:^ - V^x^ - 12x + 36. 13. a;2 - 4 ; a;^ - 5a; + 6 ; a.-^ - 2a; - 8 ; x^ — lx-\- 12. Ans. a;* - 7a;' + 8a;2 + 28a; - 48. Exercise XIV. Find the H. C. F. of the following expressions: 1. a;* + 2xY + if ; a;* + 2>xhf + 2yK 2. a;* + x^y^ + y^ ; a;* + a;'^ — xy^ — if. 3. X? - bx^y -\- l^xy^ - 9y^ ; a? — 2xhj + ^xy^ - Zy^. 4. a;* + 2a;y + 9?/* ; 7a;' - lla;^^ + Xbxy" + 9?/'. 5. 9a;* - 3a;?/' + 3a;2?/2 - ^x?y ; 2?/' - lOa;^^ _|_ g^^^ 6. 3a;' - 22a; - 15 ; 5a;* + a;' - 54a;2 + 18a;. 7. a;' — 8a; + 3 ; a;^ + 3a;^ + a; + 3. 8. - 5a;' - 2a;2 + 15a; + 6 ; - 7a;' + 4a;2 + 21a; - 12. 9. 20a;* ^x? -\\ 25.T* + 5a;' - a; - 1. 10. - a;* + 2a,-' - a;^ + 8a; - 8 ; 4a;' - 12a;2 _|_ 9^ _ i. Find the L. C. M. of the following expressions: 11. a;' + 1 ; a;^ - a; - 2 ; a;* + a;^ + 1. 12. ^ — X ; a;' — 1 ; a;' + 1 ; a;* + a;^ + 1. 13. a;" - 4 ; a;' + 23,-2 + 4a; + 8 ; a;' - 2a;2 + 4a; - 8. 14. a;* — 3a;' + 2a;2 + a; - 1 ; a;' — a;^ - 2a; + 2. 15. 4a;* + 9a;' + 2a;2 __ 2a; - 4 ; 3a;' + 5a;2 - a; + 2. 16. a;' - 1 ; 3a;2 - 5a; + 2 ; ^^ -\x^ -x-^\. 17. ^x^-^a? -x; 4a;' - 6a;2 - 4a; + 3. 18. 12a;2 - 15a;^ + 3/; 6a;' - ^xhj + 2xy^ - 2f. CHAPTER XT. FRACTIONS. 258. The expression 7- is employed to indicate that one unit is divided into i equal parts, and that a of these parts are taken, or that a units are divided into b equal parts, and that one of these parts is taken. 259. The expression j is called o, fraction, a is called the numerator, because it numbers the parts, b is called the denominator, because it names the parts. The numerator and denominator are called the terms of the fraction. 260. Every integer may be considered as a fraction whose denominator is unity ; thus, « = — . 261. To multiply a fraction by an integer, we may either multiply the numerator or divide the denominator by it; and, conversely, to divide a fraction by an integer, we may either divide the numerator or multiply the denominator by it. For, to multiply the numerator multiplies the number of parts taken without affecting their size; hence the fraction is multiplied. To divide the denominator increases the size of the parts, without affecting the number of parts taken; hence the frac- tion is multiplied. Again, to divide the numerator divides the number of parts taken without affecting their size; hence the fraction is divided. (147) 148 ALGEBRA. To multiply the denominator decreases the size of the parts without affecting the number taken; hence the frac- tion is divided. 262. As a result of the preceding article: If both terms of a fraction he multiplied by the same number, or if both be divided by the same number, the value of the fraction will remain unaltered. Thus: —=-- = —-, etc. DC abc 263. Since a =~ (Art. 2G0), .-. a^b = ^~-b = ~ (Art. 261). That is, a fraction represents the quotient of the numerator by the denominator (24). REDUCTION" OF FRACTIONS. 264. To reduce a fraction, multiply or divide both terms by the same number (262). 265. Case I. To reduce a fraction to lozoer terms: Divide both terms by ayiy common factor, ™ a^bc^ _ a^bc^ _ ^^^ _ ^ IWc ~ If^ ~'¥^~~^lf' To reduce a fraction to its lowest terms: Divide the numerator a7id deiiominator by tlieir H. C. F, EXAMPLES. 12a:^-3 _ 3 {^x - 1) (2a; + 1) _ 3 (2a: + 1 ) ^' 20a; - 10 ~ 10 (2a; - 1) ~ 10 ' - a;' - 8a; + 12 _ (a; - 2) (a; - 6) _ a; - 6 2- a;2 _ 4 — {x — 'l){x^'>)~ x-\- 2* a:* + ar^-a;-l _ {x + 1) (a;^ - 1) _ a; + 1 ^' x''-7?-x^\~\x-\)\x'-V)~ 'x^^ a;* + (^7? + «* a? — ax -\- (^ 4 1 1 _ ■=. , a;* + a^ — c^x — a^ x? — o? ^ x" - 5a;^ + 4 _ 3a; + 2 ^' 2x^-x^-x^2 ~ a; + 1 • 6. 7. 9. FRACTIONS. 149 Sar' + dx"" - 152; + 4 ~ 3:c - 4* iC* + 2a;2 + 9 a;2 + 2a; + 3' a:*-r»-a; + l a:2-2:z;4-l 10. 11. 12. 13. a.-* - 2a;^ - ar* - 2a: + 1 x? - '^x -{■ \ ':^ -\- (a -\- V) X -\- ab _x -\- 1) X? -\- {a -\- c) X -\- ac x -\- c' ax"" 4- hx'^^^ _ a;"""^ a^Z>a; — Z>V ~~ b(a — bx)' «' + (« + ^) f'-'^" + ^^**^^^ _ (i-\- ^ a* - ^>2a;2 ~ a^ - ^>a;* Z>a: + 2 1 2^ -I- b^x -4x- 2bx^- b — 2x (a - bf + (b- cY+(c-aY _ a? _|_ ^s ^ ^;j _ 3flj^c a-\-b-\-c 266. Case II. To reduce an integer to a fraction with a given denominator: (1) Exjjress the integer in the form of a fraction (260). (2) Multiply both terms by the given denominator, mu ^ ax a{b -\- c) Thus: a = - = — = -\—^ — -, etc. 1 x b-{- c EXAMPLES. 1. Reduce a;^ — 1 to a fraction whose denominator is a;*— 3. a:* - 4a:2 _^ 3 ^^^- c^-3 ' 2. Reduce {a— by to a fraction having a denominator {a-\-by. (a + by 3. Reduce a;* — a:^ + 1 to a fraction having a denominator a;* + a;« 4- 1. ^^15. ^^^-^44- 150 ALGEBRA. 267. Case III. To reduce a fraction to an integral or mixed expression : Arrange hoth terms according to tlie descending po2vers of a letter cominon to loth. Dimde the numerator hy the denominator. If the divis- ion le exact, the value of the fraction is integral; but if not exact, continue the division U7itil the degree of the remainder is less than that of the divisor : write the remain- der as the numerator and the divisor as the denominator of the fractional expression, reduce this fractio?i to its lowest terms and annex it to the quotient hy 7neans of the sign +. Thus: (i)^^ = a--x + l + ^. X + 1-^ x' + 2:c - 3 x—1 0? - 4a; + 3 ^ + 2 x^ — 3^; -f 2 (3) W ^.„^ 'i -1 + 15 ' x-\-% - 2x + 3 X^ — X — 1 X^ — X — 1 268. In order that a fraction may be reduced to an in- tegral or mixed expression, tlie degree of the numerator must not be less than that of the denominator; and if the degrees be equal, the co-efficient of the term containing the highest power in the numerator must not be less than that of the corresponding term in the denominator. Such fractions may be called improper fractions. 269. The signs of both terms of a fraction may be changed without altering the value of the fraction, because this is equivalent to multiplying both terms by — 1 (Art. 262). If the sign of but one term of a fraction be changed, the sign of the fraction is changed. a —a a —a FRACTIOIfS. Iso, a —a a —a iimilarly, a — h b — a a — b b — a a — c c — a c — a — c -\- a 151 270. Case IV. To reduce a mixed expression to the form of a fraction : Multiply the denominator of the fraction by the integer; to the product annex the numerator by meafis of the sign of the fraction. Under the result write the denominator, and reduce the expression to its simplest form, EXAMPLES. , b ac 4- b 1. « -f _ = '. — . c c , — b ac 4- ( — b) ac — h 2. a-\ = ^-^^ '- . c c b — ac 4- b ac — b 3. a-{ = ■ — = . — c — c c b ac — b 4. a c c — b_ac — (— b) _ac -\- b o, a — — — • c c c b — ac — ( — b) — ac 4- b ac - c — c — c c — b_— ac — { — b)_ — ac-{-b_ac — b — c~ — c ~~ — c ~~ c ' a. X -]- x-y -t- y ^, _ ^^^^ _^ y, - ^^ _ ^^i _^ ^*- *^ ^' y ~ ^ ^ — y 152 ALGEBRA. T y T ^ _ y'i x-\- y 11. x^ -}- X -]- 1 — 12. x^ — 4.^' + 1 — X X — 1 {x^-Sy _ 16x^ X"" -\- 4:X -\- 8~ X^ -{- 4:X -\- 8' 271. Case V. To reduce a complex fraction to a simple one. A complex fraction is one which has a fraction in the numerator or in the denominator, or in both. A complex fraction does not come under the definition of fimction as given in (358). It is merely the quotient of the numerator by the denominator (263). Eule: Multiply hath terms hy the L. G. M. of the de- nominators of the fractions co7itained in the numerator and denominator. Thus- m^i-iiv^--^- ^ h_ ^ h , . c _ c 1)0 _ ahc + ^ ^ ' c c be ~ abc — c^ a — J- a — -r b _1 _1^ ,„v X X xy oe^v — y (3) = X -"^ = -^. ^. ^ ^ 1 1 xy xy'' — X y — y — ^ y ^ y , b^ y a-\ a - b _ ^ a - b ^ {a - b){a -^r h) _ b- ^^^ -2 ~ ^ a" ^ (a-b)(a-^ b) a -\- b a -\- b a^-aV^b'' {a-\-b) _a^ - aV + ab"- + Z>^ _ g^ + j^^ a^l)-.l^-a\a-b) ~ a^b - Z>^ - a^ + a^b ~ - a' -{- 'Za'b-b^ FRACTIONS. 153 EXAMPLES. Eeduce the following complex fractions to simple ones: Answers. Answers, a 2ii^ a-{-b a(a — h) ^ -\- y^ 2y b b(a-]r b)' ' y x" — xy -\- y^' a — b ^-\- y a , X _ 4ab . _ X a X — a a-\-b a -{- Zb X — a ' ax ' ' Aab ^ ' b — a' a-\- b X X a -\- b " ^ ^+-U2 ^^ + ^ 10. — '■ r^ 1. Q? a — b a -\-b 272. Case VI. To reduce fractions to equivalent fractions having a common denominator: Multiply both terms of each fraction by the product of the denominators of the other fractions. Thus: 1. Reduce 7-, — , — to a common denominator. b c a a _a ac _a^c b _b fib _ ab^ 1 _ 1 be be b b ac abc' c c ab abc' a a be abc 4 3 5 2. Reduce to a common denominator X a -\- h' a — b' a' a(a — b) in^ — 4:nb a -{- b a -\- b a(a — b) a^ — ab^ ' 3 _ 3 a{a^b) _ ^a^^^ab ^ a — b~ a — b a{a-\-b) ~o* — a^ "* 5_ _ 5 {a -\-b)(a- b) _ 5a' - 6b^ a ~ a (a -\- b){a — b)~ a^ — ab^' 154 ALGEBRA. 273. Case YII. To reduce fractions to equivalent frac- tions liaving the lowest common denominator (L. C. D.): (1) Find the L. G, M. of the denominators. (2) Divide this L. C. M, hy the denomi7iator* of each fraction. (3) Multiply both terms of the first fraction hy the first quotient; both terms of the second fraction hy the second quotient; loth terms of the third fraction hy the third quotient; and so on. Note. Before reducing fractions to the L. C. D. it is usually best to reduce them to their lowest terms, as in examples 3, 4, 5, 7, helow. EXAMPLES. Eednce the following fractions to the L. C. D. : 3« n 1 a-h' a-\- h' tt^ - tf The L. C. M. of the denominators is {a^ — If), and the quotients are a -\-hy a — h, 1. 3<7. (g + Z> _ 3^^ + 3aZ> , 2h a-h lal - W X , 7 O 70 J , ^ X a-h a^h a" - b^ ' a-\-h a- b~ a" - b^ ' ^a^ -f '^a b lab - W 1 ^^^^' a^-b' ' a'- iT' a^ _ ^,2- he ac ah 2. {a - b)(a - cY {a - b)(b - c)' {a - c)(b - c)' The L. C. D. is {a — b)(a — c) {b — c), and the quotients are (b - c), (a - c), {a - h). Ans. ^—^^^—, ac {a — c) ab (a — b) (a-b)(a-c)(b-cY (a - b)(a- c)(b - c)' x^ -9 of — x-20 ^' x'-^-'x-W 0^-25 ' x^-]-S x-\- 15 x^-]-8x-\- 16 ^*^* a^ + 9rc + 20' x^ -\- 9x -\- 20' FRACTIONS. 155 x'-\-x -6' x'.-25 ' . x"^ — 4:X — 5 x^ -\- X — 6 Alls. x^-7x-\- 10' af-7x-{- 10* x-1 x-^1 11 x^-bx-{-4.' a?-^x-^' ' x-^' ic - 4* 4 1 1^_ ^* 3(a; - 2)' 2(a; - 1)' Q{x + 1)' S{x^ - 1) 3(2;^ -a; -2) a:^ -- 3a; + 2 G(a;-2)0i-2-l)' 6(a:-2)(a;2_i)^ 6(a:-2)(a,-^-l)' 1 1 a;- 1 Ans 1-x' {l-xf {l-xf J^-^ 1 -1 (1 - xf (1 - xf (1 - a:f 1 1 {a - b)(a - cY (b - a){b - c)' . b — c c — a Ans, (a -b)(a- c){b - cf (a - b)(a - c){b - c)' a -\- c b -\- c (a - b)(x - aY (a - b)(x - bY .,,, (a -{-c){x-b) (b + c){x - a) (a - b)(x - a){x - bY (a - b)(x - a){x - bY ADDITION" AND SUBTRACTION OF FRACTIONS. 274. To add or subtract fractions: (1) Reduce the fractions to equivalent fractions having the lowest common denominator (273). (2) Combine the numerators of the resulting fractions according to the rule for addition (98) or subtraction (103). (3) Write the result over the L, C. D. (4) Reduce this fraction to its simplest form. 156 ALGEBRA. EXAMPLES. 1- -^b + a-l ~ (it -^b){a-h) "^ (« + h){a -b) a" - ¥ a h _ ^ I — b _ a — b _1 ^' 2^i^b "^ W^^a ~ 2(^T) "^ 2(a - b) ~ 2(a-b) ~ 2* 2 _ 3 _ 4a; — 2 3a; _ a; — 2 ^' x" 2x~^^ ~ xi2x~^^) ~ x(2x -1)~ x(2x - 1)' x^'-i-x -6 a;2 + a;-l x^ -^ x - 5 2x^- llx + 12 20^ -^bx- 12 (2x - S){x - 4) a^-\-x-l _ g;^ + 5a:^ - rg - 20 _ {2x-^){x-fl) ~ (2x-d)(x-4)(x^) a^—3:^ — 5 x-\-4: _ 8x^ + 4a; - 24 (2a; - 3)"(¥^^4f(^T4) ~ (2a; - 3) (a; - 4) (a; + 4) ^ 4(2a;-3)(a; + 2 ) ^ 4(a; + 2 ) (2a; - 3)(a; - 4)(a; + 4) "" a;^ - 16 * b , a b , ^ 5. t:. iwT :^ + (a-b)(a-c) ' (b-a)(b-c) (a - b)(a - c) ' — a _ b{b — c) —a{a — c) _ (rt-^>)(6-6')~(^<-^)(rt-c)(^-c)"^(«-^')(a-c)(^-6')~ W—bc—a^-\-ac _ c—a—b (a-b)(a-c)(b-c)~ (a-c)(b-cy 1 ll_a;* + 4a;2 «■ ic2-4 a;3-8 ' a;^ + 8 a;« - 64 a;^ + a;- 2 a;^ - 4 _ 2a;^ + 2a; - 7 "^^ ^2 _|_ 4a; :j_ 4 + ic2 _^ 5^ 4- 6 ~ a:^ _|_ 5^ _|_ g • 8 _2__ + -_l 1_-L 4^^ - a;2 _ ^2 I (^ ^ ^)2 (^ _ yY I (3,2 _ ^2)2 ^ _ ^2' 1 1 I ^ ^ ^^^^^ a;— 3a a;-|-3aa; + a a; — a a;*— lOo^V + 9a** 48a FRACTIONS. 157 275. In many examples it is preferable not to combine all the fractions at once. 10. To simplify — - - ~-^ - -^^ + ^^^. The sum of the first and third is '— = .^•2" + a;" + 1. a;" — 1 ' The sum of the second and fourth is — z——t— =1 — x\ a;"4- 1 Ajis, a^-{-2. MULTIPLICATION OF FRACTIONS. 276. To multiply by a fraction means to take such a part of the multiplicand as the multiplier is part of unity. Thus, to multiply by J means to divide the multiplicand into four equal parts and to take three of those parts. To multiply f by f means, therefore, to divide | into five equal parts, and to take four of those parts. Now f = {^ (Art. 262). If II be divided into five equal parts, each part will be 3^. If ^5 be taken four times, the sum will be ^ + A+A+t's = W- ••■ toffor|Xi = A. ^. ., , « c no Similarly,^ X^ = ^. 277. If there be more than two fractions to be multi- plied together, the same method will apply. Thus: o A n a n Ka j tt C 6 ttCB I X t X i = t\ X i = V^, and - X ^ X ^ = ^^. Hence, to find the product of two or more fractions: 278. Multiply the numerators together for the numerator of the product, and the denominators together for the denominator of the product, EXAMPLES. {a - bf b ^ b(a - bf ^ b{a - b) a-\-b x(a — b) x{a -\- b)(a — b) ~~ x(a -\- b)' 158 ALGEBRA. x[a — x) a{a + ^) _ ax a^ + 2ax -{- x^ a^ ~ 2ax -\- x^~ {a -\- x)(a — x)' a'-b^ a-\-b f a^ - ah -\- W V _ a^ - al) -\- V ^' d'-^h^' a-h' W + a^ + b^i ~ a'^ah -{- If a^ - a-^ a^ - a -A2 a^ - 9a -{- 20 a^ + a-30 ^ a^ - 7« + 12 ^ a^ - 5a - 14 ^* — .V* ^_ 2rr?/ (--^J = -(-^ + /)- cc^ — 2xy -\- y^ \ ^ -\- y. a^-{-c'-b^-{-2ac a-\-b- c {a -\- Z>)"* (ax — bx)"" _ 1 DIVISION OF FRACTIONS. 279. Since f X f = t\, . *. y^ "^ I = f Similarly, since a c _ ac ^ ac c _a b^d~M' ''' bd'^d~b' Hence, to divide one fraction by another: Divide the nurnerator of the dividend by the numerator of the divisor for the numerator of the quotient^ and divide the denominator of the dividend by the denominator of the divisor for the denominator of the quotient, Note 1. If the terms of the dividend be not divisible by the corresponding terms of the divisor, multiply both terms of the divi- dend by an expression which will render them divisible. See Ex. 2. Note 2. If the denominators be the same, divide the numerator of the dividend by that of the divisor. See Ex. 3. FKACTIONS. 159 EXAMPLES. __?^!_ _^ y ^ "^f ■ o?-Y y^ ' x-{-y~ x^ -xy-{- y^' X ^ X _ x(a -\- x) ^ X _a-\-x ' a — X ' a -\- x {a — x){a -\- x) ' a -{- x~ a — x' f , 2x \ f 2x \ a? — X X — 1 of — bx X — 5* x^ - 3.^ y -f 3^y* - y'' _ {x + yf _ (x - y f 280. The rule in Art. 279 is the most convenient when the terms of the dividend are multiples of the correspond- ing terms of the divisor, as in Ex. 1 and 4 above. In other cases one of the following rules is preferable: 281. Second Rule. Multiply the dividend ly the re- ciprocal of the divisor. Thus: To divide y\ ^ !• Multiply y\ by | (Art. 30). Ans, |. ^. ., , «c a ac h abc c Sim.larly,^^-^=j-^X- = ^ = ^. This rule depends upon the following reasoning : Let it be required to divide j- by —. If — be divided by c, the quotient is -r- (Art. 261). But the divisor c being d times c as large as the true divisor, -^, the quotient thus obtained is d times too small; hence — must be multiplied by d for ^, . ,• , a c a d ad the true quotient. .-. — x^ = 7-X- = t~» ^ d b c be IGO ALGEBRA. 282. Third Eule. (1) Reduce the fractions to a com- mon denominalor. (2) Divide the numerator of the dividend hy the nu- merator of the divisor. Thus: . A - i = A - if = A = |. ^. ., , a c ad he ad Similarly, - + -=-^-^ = _. 283. It is evident that a complex fraction may be re- duced to a simple one by considering the numerator as the dividend and the denominator as the divisor, and finding the quotient according to one of the preceding rules. Thus: -+--2 ax x^ — 2ax + «' / . x — a 1. = ■ =- (x — a)= . X — a ax ax x^-\-{a-{'C)x-\-ac oi?-\-(b-\-c)x-[-bc _[x-[-a) {x-\-c) _^ {x—a)(x-\ a ) _ x—b ^" x'-a ^ (x-\-b)(x+cy(x-bj(x-Jrb)~x-a EXAMPLES. By each of the three methods show that : ax — x^ , a^ — 2ax + ^ _ ^^ ^' a^ -\-2ax -}-x^ ~ '^-^ ax ~ a^ - x'^ X -\- a __ X — a, ax (x a\ 2. -f- . \a x) / , 4:x \ ( 22: \ x-\-l 3. (^ + ^rir^j -^ i^ - ^373 j = ^zr5- ^6 _ 3^y ^ 3^2^4 _ y6 ^ (a; - yy _ {x + yf ^' {p^- fY '' (^ + yy ~ (^ - yf f x + 2.y 2y\ _^ f x 4- 2y _ x \ _ x -j- y ' \ y '^ xl ' \ y x-\-y FRACTIONS. 161 MISCELLANEOUS PROPOSITIONS. ft f* 284. In each of the following propositions let t- = 7 •* I. 1 -^ -^ = 1 -^ -r, Ax. V; . •. - = -. b d a c II. a b _c b . jv. .^—^ 7~ X — ~T X J ixX, i- V I . * . — -T. b c d c' ' c d III. ^+1 = ^ + 1, Ax. II;.-. X = ^ . IV. a ^ c -, A TTT a — b c — d b ^ = cl 1' Ax. III;.-. I, = a ■ V. From I - = -, whence from III - — = -^. a c ad VI. From 1 — = — , whence from IV = . a c a c JU If TTT hft (hv\(]9.c] hv TV- • . « + ^ _ c + ^ Other results may be obtained from propositions III to VII, inclusive, by using inversion (I) or alternation (II). Thus: from III, — 7 — = — j — , whence, from I, a-\-b ; also, from II, — ; — - = -; etc. Hence: c^d' ' 'c + d d' If two fractions are equal, tve may combine by addition or subtraction, in any way, the numerator and denominator of the one, provided that ive do the same ivith the other, 285. If ^= |, then ^-^ = ^ (Ax. XIV). Hence the preceding results hold with «", h^, c^, d^, instead of a, b, c, d. For example, VII would read -r-^—r^ = ' -,„ . ^ ' a" — Z>" 6" - d"" Alg.— 14. 162 ALGEBKA. 286. If a, h, and.c represent positive numbers, tlien if a>l, J—— < T-; ita = I?, y-— = -; if a -r- -\- -\- c -\-c b For, in order to compare fractions, we reduce them to a common denominator and compare tiie numerators of the resulting fractions. T.^ a -]r c ab-\-bc ^ a ab -\- ac ^-^, Now -r-. — = yTT-T— x^ ^^^ T = TTT^T—^- When a> b, -\- c b{b-\-cy b b{b-\-c) ab -{-be < ab + ac; when « = Z*, ab -\- be — ab + ac; when a < b, ab -\- be > ab -\- ac : which proves the proposition. 287. If r?, b, and c represent positive numbers, then if J a — c a .J, ^ a — c a .J, ^ a—e a b — c b b — c b b — eb ^ a — c ab — be ^ a ab — ac ^,, For , = TjT and - = -r— ^. When a> b, b — e b{b—e) b b(b — c) ab — be y ab — ae ; when a = b, ab — be = ab — ac ; when a (/^ - «)(d -c)^ c(c - a)(c - b)' a^-\-a-{-l b^-{-b + l c^ + g + 1 1^- [a - b){a ~c)'^(b- a)(b -c)'^ (c- a)(c - b)' a^-{.x' b^-^a^ c^-\-a? "• (a - b){a - c)'^ [b- a)(b -c)'^ (c- a)(c - b)' ,a (a'-hf) ( b + c) (b^-i.f){a^c) {c^ ^f ) (a^b) (a-b)(a-c) "^ (b-a)(b-c) "^ (c - a)(c - b) ' ^^' (a - b){a - c) '^ (b- a)(b - c) "^ (c - a){o - b)' 3ax a^ — a^ be -}- bx c — x 4:by 0^ — x^ a^ 4- ax a — x 21 '• (f" + « ) V J + ^) + & "■ i) [d~c)' 23. (^.±1 _ ^-y _ ^y' \ ' \x — y X -\- y x^ — y^J i(a^-ab) . 6«J ^^ b(a-\-by ' a'-b' 164 ALGEBRA. 25. 1 X 7- -^ ^^ ^ .2 26- (^ _ 1 - 2(:^ + 1) 2(2;'^ - 1)) ^^ ^)' \L ^ x^ \ - x) ' \\ - X 1 + a/ • u + ^ 1 - W "^ 1 - ^ r+^* 28 1 , 2^ 29. r— h l-|-a;l — :c*l — a; 1 + ^' la -I _ a^ - h\ ^ ( a + i , a^_-\-JA ^^' [a-i-b a' +p) ' \a-b'^ a' - bY 3j (^_±J!l _ ^ ~ y'\ ^ ( ^ + y _ ^-y \ ' W — y^ 01^ -\- yy ' \x — y x + ?// / a + ^> « -M _ /rtM:_^ _ a^ - b'\ ^^'\a-b^a-^b)' W - b' a'-V U'l' "" a'-¥ ^^' b-'-a-^ ^ a'-\- b'' X -\- a X — a X X X — a X 4- a 34. . \ • — . X — a x-\- a X -{- a X — a X — a X -\- a 1 a 35. ; . 36. CHAPTER XVI. SIMPLE EQUATIONS. CONTINUED FROM CHAPTER X bx ax — W 288. 1. Solve a a c Multiply by ac, the L. C. M. of the denominators. Then a^c — hex = a^x — aV^, Transpose. — lex — a\v = — a^c — abK Divide both sides hv—bc—a^ . • . x = \ , ^ . "^ or -{- be c. , ax -\- b a ex 4- d 2. Solve ■ T- = ■ — . e b e Multiply by be, the L. C. M. of the denominators. Then abx -{- b^ — ac = bex + bd. Transpose. bx{a — e) = ae — b"^ -\- bd, ae — b^ -{■ bd Divide both sides by b(a — e). b(a — e) 289. 3.Solve-3|l- + ^- = ^-. When an equation contains simple and compound de- nominators, it is usually best to get rid of the simple denominators first, and then of each compound denomina- tor in turn, being careful to simplify as much as possible after each step. In this example, therefore, transpose — ^- and simplify. ^ (Ix - 3) _ 8a; + 12 - 8a ; - 5 _ 1 Multiply by (6^- + 2). . • . 7a;-3 = 3a; + 1, whence x = 1. (165) 166 ALGEBRA. c, 1 6a; + 13 2x 3x -{- 4. Solve- ' ^ 15 5 5a; - 25 Multiply both sides by 15. Then 6:. + 13 -6,^ = ^^ + 25. . • . 13 = ?^±i5.. X — 6 X — b Multiply both sides by (a; — 5). Then 13a; — 65 = 9a; + 15. . • . a; = 20. ci X a -\- X 2a — b 5. Solve — ■ — = — ^ . a -f X X 2x Multiply both sides by 2a;. Then — -- = 2a -\-2x - 2a -}- b = 2x + b. a-{- X ' ' ' Now multiply both sides hj a -\- x, • . 2a;^ = 2a;^ -^ (2a -}- b) x -\- ab, whence x 2a -{-b 17 10 1 290. 6. Solve ^^_^^ 3.-10-1-2.- Multiply both sides by (6a; — 17)(3a; - 10). .■.51.-170-60. + 170 = ('^^-fHg-n 18a;2 - Ilia; + 170 .-. — 9a; = —J . 1 — 2a; . •. - 9a; + 18a;2 = ISx^ - Ilia; + 170; whence x = If. 001 o 1 ^ + '^ 2a;-16 , 2a;+5 ^, , 3a; + 7 291. 7. Solve -^t _ + __^_ 5i + -^-. Here it is convenient to clear of fractions partially, and to reduce the resulting expression. Hence, multiply by 12. Sx 4- 64 + 6a; + 15 = 64 + 3a; + 7. 12a; + 84 = 55a; — 88; whence a; = 4. ■ -(^) 292. 8. (x — a)(x — b) — {x — b)(x—c) = 2(x—a)(a—c). Simplify the first side: (x — b)[(x — a) — (x — c)] = (x - b)(- a-j-c) = {b- x)(a - c). . • . (b — x)(a — 6*) = 2(a; — a){a — c). SIMPLE EQUATION'S, CONTINUED. 167 Divide hy {a — c). ,- . b — x = 2(x — a); whence x = 2a -\-b ««« c 1 X — 1 X — 2 X — 5 293. 9. Solve x—2 x-3 x-Q x—1 Simplify each side. {7?-Ax-^^)-(x'-^x-\-^) _ {a?-12x-\-2b)-{x'-12x-\-^Q) ^ (x-2){x-^) ~ {x-Q){x-l) ' *^^* i^^ (x-2){x-^) ^ {X - ^){x - 7)- . *. - a;2 + 13a; - 42 = - x« + 5a; - 6 and a; = 4f ^ , \(2x ^ \ 7J-a; x {^ ,\ 10.Solve-(-+4)-^3- = -(--lj. Expand. \x -{- 2 — % -\- \x = ?> — \x. Transpose and simplify. Jo; = J. . • . a; = 3. on/i c 1 66a; + 1 , 4a: + 5 ^^ 294. n. Solve j^^^^^ + — --^ = 52. Multiply each term by \. 66a; + 1 , 4a; + 5 = 26. 3a; + 2 ' a; - 2 Clear of fractions. 66a;'' - 131a; - 2 + 12a;2 + 23a; + 10 = TSa^" _ ^i^^^ _ iq^^ Transpose and simplify. — 4a; = — 112. . • . a; = 28. 295. 12. Solve ^ _ -^(23; - 3) - K3a; - 1) _ 3 / a;^ - ^a; + 2 \ 2 i(^ - 1) ~ 2 \ 3a; - 2 /' X Transpose — and simplify each side. a;4-9_3ar'-a;H-6 x x -\- Q ' io{x - 1) 2(3a; - 2) 2 2(3a; - 2)* ^ _i_ 9 X I 6 Now take — — — -pr = j, oy ^^^ ^^®^^ ^^ fraction .-. 3a;2_j_25^_ 18 = 3ar' + 15a: - 18; whence a; = 0. 168 ALGEBRA. 296. 13. Solve f^- + ,f ~^, = 1 + —— L-— . 1^- 1 1(^+1) 15(1 -x-^) Multiply both terms of the first fraction by 2, of the second by 2, and of the last fraction by x^. 2x-3 2x-5 _ x^ • *• 3(a; - 1) "^ b(x-^ I) ~ + Ibi.x' - 1)' Multiply every term by the L. C. D. lb{x — l)(x -{- 1). .-. b{2x-Z){x-\-l)^'6{2x-b){x-l) = Vo(;x-l){x^l)-\-x\ Expand. lO:^^ - 5:c - 15 + Qx"- 21x-\-lb = lbx'-lb-\-xK Transpose and simplify. . • . — 26x = — 15; whence x=\^. Exercise XVI. Solve the following equations. (For the solutions of the simultaneous equations [13-24 inclusive] consult Chapter XI.) a — h a -\-'b X — c X -\-2c 3x-\- l 3x -6 _ 3x + 28 9 ^ 2a; -5 ~ 9 • 3. X a — 1 dx c = 3ah. 4. X a + I X _ — a a a-^b' 3a: - 1 6-\-x _ ^^ _3x — Q 3x -\- d ^■"^2"^ 4 "^~ 12 '^ x-j-7' 1111 6. X — 2 X — 4: X — 6 X ~ 8 2,1 6 ^ 2a; -5^0; -3 3a; - 1 a4-b a , b 8. — ' — = — . T. X — c X -\- a X — b 25 — \x 80a; + 21 _ 5a; + 28 ^* a; + l" "^ 5(3a; + 2) ~ x ■\^' 10. a; - 7 2a; - 15 1 a; 4- 7 2a; -6 2(a; + 7)* a; — 1 x-\-2 SIMPLE EQUATIONS, CONTINUED. 169 11 12. ^ o. , n. + 12x -f- 11 ' 6x- -f 5 4a; + 7 4 5 13. ^-. — = jT^-. — ; 2x-{-bi/ = 35. 54-2/ 12 -{- X ^ l^-37^ = f' 8^ = 9^,4-4. 15.^ + ^ = 5; ^ + ^ = 12. X ' y ' X y 16. {ci^ — h''){bx + Zy) = Sa^b - 2ab\' l^a-^h^ c)x + (rt^ - W)y = -^ + ab{a + 2^»). 17. 2a; — y — ^ = 12; 21. 2x + 3y — 4^ = 20; By — 2; -^ = 16; x-^2y — 3z-8; bz —x — y = 24. 3.r + y — 2z = 26. 18. a; + ?/ -[- 2 = 4; 22. a;~* + ?/~^ = ri; 3:c — ^ + 22 = 1; a;-^ +;?;-* = i; 4a; + 3^— 2; = 18. y-"^ -\- z-^ = c. 19. a;4-?/+^ = 0; 23. x-^ y-\-z-\- t-\- v = 0; x-\- z-^u= — 1; 3a;+4.y+2+5^+2^ = 0; ^-\-u-\- t = 5; 4x-2y-\-4:l+v = 0; iz + ^ •■= 7; 2a;+ y- z-2t = 23; x-\-y-z-u-\-2t =17. _6y+3z-5iJ = - 1. 20. a; 4- 2?/ + 32 + 5?t = 2; 24. a; + y + 2; — t; = «; 3a; — 4.y -f z-\- in = —9; a; + y — 2 + t; = ^>; 4a; — y — 2z-{- u = lb; x — y-\-z-{-v = c; 4:y — 2z-\- bu = —1. — x-{-y-{-z-\-v = d. Alg.-15. CHAPTER XVII. PROBLEMS INVOLVING SIMPLE EQUATIONS. 297. — 1. After losing | of my money and J of wLat was left, I gained ^ of the remainder, and then had $280. How much had I at first ? Let ^x represent the number of dollars I had at first. I lost ^ of my money, hence I had left f of 5;^; = 4a;. I lost i of what was left, so I had then left } of 4x = 3x. Finally, I gained ^ of 3a; = x, and then had 3x -{-x = 4a;. But I then had $280. .-. (Ax. L) 4a; = 280, whence a; = 70 and 6x = 350. . • . I had |350 at first. 298. — 2. A's money was -f of B's; A gained 12, and B lost $3, when A had ^ as many dollars as B. How much had each at first ? Let 7x represent the number of dollars B had at first. Since A^s money was f of B's, . * . A had f of 7a; = Qx. A gained $2; . • . A then had 6x + 2 dollars. B lost $3; . • . B then had 7a; — 3 dollars. .-. 6a; + 2 = -y>-(7x — 3), whence a; = 3. Alls, B had 21 dollars, and A had 18 dollars. 299. — 3. A lease was given for 60 years. Three fourths of the time it has already run equals -| of the time it has yet to run. How many years has it yet to run ? Let 8x represent the number of years the lease has yet to run; then 60 — 8x is the number of years it has already run. . • . 1(60 — 8a;) = 9a;, whence a; = 3, and 8a; — 24, Ans, (170) PROBLEMS IITVOLVING SIMPLE EQUATIOI^S. 171 300. — 4. E was hired for a days, at ^ cents for each day- he worked, and at a forfeit of c cents for each day he was idle. At the end of the time he received d cents. How many days did he work ? Let X represent the number of days E worked. Since he was hired for a days, and worked x days, he must have been idle a — x days. Since he worked x days at h cents a day, his total wages amounted to Ix cents. As he forfeited c cents a day for a — x days, he forfeited c(a — x) cents. The amount due him for wages, diminished by the amount forfeited, must be the amount of money actually received. . • . hx — c(a — x) = d, whence x = — r— — . 301. In solving problems involving time, rate, and dis- tance, observe that: (1) The time X the rate = the distance; Formula t X r = d. (2) — --. — -. = the rate; Formula ^ ' the time d , ,^. The distance ^j j. -r, -, d — = r. and (3) — -, = the time; Formula — = t. t ^ ' the rate r Thus, if A go 5 miles per hour for 8 liours, his distance will be 40 miles. If A go 40 miles at the rate of 5 miles per hour, his time will be 8 hours. If A go 40 miles in 8 hours, his rate will be 5 miles per hour. 5. How many miles can A walk at the rate of 3 miles an hour, so as to return to the starting-point in 11 hours, riding back at the rate of 8 miles an hour ? Let x represent the required number of miles. Since he walks X miles at the rate of 3 miles per hour, .*. his time equals \x hours. Since he rides back x miles at the rate of 8 miles per hour, . * . his time returning is \x hours. Since he takes \x hours to go, and \x hours to return, . • . his entire time = {\x -j- \x^ hours. But according to the given conditions, his entire time is 11 hours. . * . (Ax. I.) \x -\- \x =. 11, whence x = 24, Ans, 172 ALGEBRA. 302. — 6. A, who traveled at the rate of 31 J miles in 5 hours, left a certain city 8 hours before B, who traveled at the rate of 22^ miles in 3 hours. In what time did B over^ take A, and at what distance from the place of departure? Let X represent the number of hours B traveled. Then x -\-S represents the number of hours A traveled. B's rate is 22| -^ 3 = 7^ miles per hour. A's rate is 31 J -^ 5 = 6y\ miles per hour. Since d = rt, . • . B's distance is ^-^-x, and A's is \l{x -f 8). But they traveled equal distances. . • . i^-x = ^^(x -\- S), whence x = 42. A71S. B's time is 42 hours, and his distance is 315 miles. 303.— 7. A takes 5 steps while B takes 4, but 3 of B's steps are equivalent to 4 of A's. How many steps must B take to overtake A, who has a start of 80 steps ? In problems of this class it is convenient to consider a step of the pursued as the unit of length. That is, in this case, one of A's steps is the unit. Let 12a; represent the number of steps B must take in order to overtake A. Since A takes | as many steps as B, then A takes 15x steps; . * . A goes 15x units. As three of B's steps are equivalent to 4 units, then 12;:^ steps of B are equivalent to 16x units. But A's distance is 80 units less than B's distance; . • . 16x — 16x = 80, whence x = 80, and 12a; = 960, A71S. 304. — 8. Find the time between 4 and 5 o'clock when the hour and minute hands of a watch are: I. Together; II. At right angles; III. Opposite; IV. Equidistant from VIL I. Let CM and CH denote the positions of the minute and the hour hands at 4 o'clock, and let CA denote the position of both hands when together. PROBLEMS INVOLVING SIMPLE EQUATIONS. 173 Since the liour-liand goes from iv to v while the minute- hand makes a complete revolution, there- fore the hour-hand moves only ^^ as fast as the minute-hand. If, therefore, the distance the hour-hand goes in any time be X, then the distance the minute-hand ^H goes in the same time will be 12:r. . • . Let the arc HA be x minute-spaces Fig. 1. in length. Then the arc MA will be l^x minute-spaces in length. But MA = if//+ HA =20-^ X. .-. 12x = 20 + a:, whence x = l^y, and 20 + a: = 2]-^. Ans. The hands are together at 21 y\ minutes past iv. Fio. 2. II. Let CA be the position of the minute-hand, and CB be the position of the hour-hand when the hands are at right angles. Let X represent the number of minute- spaces in the arc HB, Then 12.^; represents the number of minute-spaces in the arc MA. But MA also equals b -\-x. . • . (Ax. 1.) 12x=b-\-Xy whence 3:=^^, and 12.i*=5/y. Ans. The hands form a right angle at 5^^ minutes past iv. But the hands will be again at right angles between iv and v, when the minute-hand is 15 minute-spaces ahead of the hour-hand. If, as before, the number of minute-spaces in the arc ^H HB = X, then the number of minute- spaces in the arc MHBA is 12a;=35 + a;. . • . a; = 3^ and 12a; = ZS^. A ns. The hands are at right angles a second time be- tween IV and V at 38y\ minutes past iv. 174 ALGEBRA. III. Let CB be the position of the hour-hand, and CA be the position of the minute-hand when the hands are opposite. If HB = X, then MHBA = 12^. But MHBA also equals 50 + a:. . • . \2x = bO-{-x. ,'. x = 4:^\, and 12.-?: = 54^^ Ans. The hands are opposite at 54y®Y "b minutes past iv. Fig. 4. IV. Let CB and CA be the positions of the hands when VII is midway between them; also, let HB = x. . • . DB = 1^ — x, because HD = 15, and DB = DH - BH. Since DA = DB, ,', DA = U -x, whence MDA = 35 + DA = 50 - x. But MDA = 12a:. .-.Ux^^O-x, whence x = 3|J, and 122^ = ^^^s- Fig. 5. A71S, The hands are equidistant from vii at 46^^^ min- utes past IV. PROBLEMS IJ^VOLVING TWO OR MORE UKKXOWN" QUANTITIES. 305. — 9. A and B work together for 2 hours, after which B leaves, and A completes the task in 6 hours more. Had A left, and B continued to work, he could have com- pleted the work in 3 hours more. How long would it take each alone to do the entire work ? Let X represent the number of hours A would require, and let y represent the number of hours B would require. Then A does — of the work in an hour, and B does — . X y From the first condition the work is completed when A works for 8 hours, and B for 2 hours. . * . (1) — \ — := 1. ^ ' X ' y PROBLEMS INVOLVING SIMPLE EQUATIONS. 175 From the second condition the work is completed when 2 5 A works for 2 hours and B for 5 hours. . • . (2) — I — = 1. ^ y Multiply (2) by 4, and subtract (1) from the product. 18 . • . — = 3, whence y = 6. . • . re = 12. Therefore A would take 12 hours, and B 6 hours. 306. — 10. The sum of three digits expressing a certain number, is 12. The middle digit equals the difference of the other two, and the number would be reversed by adding 396. Find the number. Let X represent the hundreds' digit, y the tens', and z the units'. From the first condition, I. x -\- y -{- z = 12, From the second condition, II. y — z — x, because we know from the third condition that z is larger than x. In order to express the third condition, we must recollect that the required number consists of x hundreds -\- y tens -\- z units; thus, 574 = 500 + 70 + 4, etc. Similarly, the reverse of the required number consists of z hundreds + y tens + x units. Hence, from the third condition, III. 100a: + 10^ + 2 + 396 = IOO2 + lOy + x. From equation III. x — z ^ — 4. From combining I and II we get 2; = 6. Therefore, from III, X = 2, and, from I, y = 4. Ans. 246. 11. What number of two digits contains the sum of its digits four times and their product twice? \0x -4- y From the first condition, I. — = 4. X + y . ' . 10;r -\- y = 4:X -\- 4y, and 2x = y. From the second condition, II. ~^^^ = 2. xy In II substitute 2x for y, whence 12a; = 4:X^, and x = 3, Ans. 36. 1 76 ALGEBRA. Exercise XVII. Solve 1 like Ex. 1, Art. 297. 1. A lost f of his money and then gained $10; he then lost $50 more than | of this amount; he next gained ^ of this remainder, but found that his total loss amounted to $233. How many dollars had A at first ? Solve 2 and 3 like Ex. 2, Art. 298. 2. A's age is | of B's; in 15 years it will be ^ of B's. How old is each ? 3. A's money equals f of B's. If A give B $12, he will have J as much as B. How much has each ? Solve 4, 5, and 6 like Ex. 3, Art. 299. 4. Divide 45 into two parts such that one part shall be I of the other. 5. A debt of $100 was paid in silver dollars and paper dollars. The number of silver dollars was -^^ of the num- ber of paper dollars. How many of each kind were used ? 6. It took A 40 hours to go from C to D and return. His rate going was f of his rate returning. How many hours did it take him to go from C to D ? Solve 7 and 8 like Ex. 16, Art. 174. 7. A's time to do a certain work is | of B's. Both to- gether can do the work in 5 hours. How long will it take each alone ? 8. A and B together do a piece of work in c days, which A alone can do in d days. How long would it take B ? Solve 9 and 10 like Ex. 4, Art. 300. 9. A was hired for 40 days, at 50 cents a day and his board. For each day he was idle he forfeited 25 cents for his board. At the end of the time he received $11. How many days was he idle ? PROBLEMS IXVOLVIXG SIMPLE EQUATIONS. 177 10. A agreed to work for a year for $200 and a suit of clothes. At the end of 5 months he left, receiving for his wages $60 and the clothes. What was the suit worth? Solve 11-1 G inclusive like Ex. 5, Art. 301, and 6, Art. 303. 11. A starts from C at the same instant that B starts from D, 60 miles from C. A's rate is 3 miles, and B's 2 miles, an hour. In how many hours will they be together: (1) If they travel toward each other? (2) If A pursue B? 12. A walks at the rate of 9 miles in 4 hours, and starts from C 2 hours before B does; but B overtakes him 45 miles from C. Find B's rate and time. 13. A and B start at the same instant from C to go to F. A, by traveling half a mile an hour faster than B, arrives at F 40 minutes before him. A's time being 16 hours, find B's rate and the distance from C to F. 14. A boatman rows with the tide 36 miles in 4 hours, and returns against a tide half as strong in 8 hours. Find the rate of the tide in each case. (See note to the 15th.) 15. A boatman rows 40 miles down stream and back again in 15 hours, and his rate down stream is twice his rate up stream. Find his time down, the rate of the cur- rent, and the rate of rowing. Note.— The rate down equals the rate of rowing plus the rate of the current; the rate up equals the rate of rowing minus the rate of the current. Formulas: (1) r-\-c — d; (2) r — cz=zu. 16. A and B set out at the same instant and tmvel to- ward each other, A at the rate of 3, and B at the rate of 4 miles, an hour. At the same time sets out with B, at the rate of 5 miles an hour, travels till he meets A, then turns about, and in 10 hours after setting out, meets B. How far apart were A and B at first ? 178 ALGEBRA. Solve 17, 18, and 19 like Ex. 7, x\rt. 303. 17. A takes 5 steps while B takes 1, but 3 of A's steps are equivalent to 5 of B's. B has a start of 80 steps: how many steps must A take to overtake him ? 18. takes 5 steps while B takes 6, but 4 of O's steps are equivalent to 5 of B's. B has a start of 30 steps: how many steps can he take before C overtakes him ? 19. A takes 2 steps while B takes 3, but A^s steps are twice as long as B's. How many steps must A take to over- take B, if A take 120 steps to reach B^s starting-point ? Solve 20, 21, 22, and 23 like Ex. 8, Art. 304. 20. At what time between 7 and 8 will the hands of a clock be: I. Together? XL At right angles? III. Opposite? IV. Equidistant from v? V. Ten minute-spaces apart? YI. Twice as far apart as they were ten minutes before ? 21. When between 3 and 4 o'clock will the minute-hand be as far past Yiii as the hour-hand is past ii ? 22. When between 5 and 6 o'clock will the hour-hand be as far past iv as the minute-hand lacks of being at xi ? 23. At what time between a and a -\-l o'clock will the hands of a clock be together ? Opposite? Solve 24, 25, and 26 like Ex. 9, Art. 305. 24. A and B work together for 3 hours, after which A alone completes the task in 5 hours more. B can do as much work in 2 hours as A can in 3. How long would it take B alone to do all the work ? 25. After A and B have worked together for 4 hours, B completes the task in 6 hours more, while A would take 8 hours to complete it. How long would it take B alone to do the entire work ? 26. An empty cistern has an inlet and an outlet pipe. Both are left open for 16 hours, when the outlet pipe is PROBLEMS INVOLVING SIMPLE EQUATIONS. 179 stopped, and the cistern is filled in 6 hours more. Had the inlet pipe been stopped instead of the outlet pipe, the cistern would have been emptied in 4 hours. How long would it take the inlet pipe to fill the cistern if the outlet pipe were closed? If both pipes be left open, how long would it take to fill the cistern ? Solve 27-31 inclusive like Ex. 10, Art. 306. 27. A number of two digits is reversed by subtracting 18, and the sum of both numbers is 110. Find the number. 28. A number is | of its reverse, and the sum of its two digits is 9. Find the number. 29. Find a number of two places which equals three times the sum of its digits, and the difference of whose digits is 5. 30. A certain number equals 12 times the difference of the two digits expressing it, and the digits will be reversed by adding 27. Find the number. 31. Find a number of three places which will be reversed by adding 297. The right-hand digit equals the sum of the middle digit and twice the left-hand one, and the sum of the three digits is 7. 32. If a rectangular field were 10 rods longer and 4 rods narrower, it would contain one acre less; if it were 20 rods wider and 36 rods shorter, it would contain one acre more. Find the length, the breadth, and the area of the field. 33. A bought a certain number of apples; had he bought 20 fewer for the same total price, each would have cost 1 cent more; had he bought 30 more for the same total price, each would have cost 1 cent less. How many apples did he buy, and at what price ? CHAPTER XVIII. INDETERMINATE SIMPLE EQUATIONS. 307. We have seen (175) that if a single equation contain two unknown quantities, the number of solutions is un- limited. Such an equation is said to be indeterminate. 308. If one solution be given of the equation ax ±I)i/ = c, all the others may be readily found. For, let X = r and y = ?i be a pair of values of the equa- tion ax -\-bif — c; then ar -j- hn = c. . • . (Ax. I. ) ax -{-by = ar -\- bn, . • . a{x — r) + b(y — n) = 0. This equation is satisfied by (x — ?•) =—bf, (y— n) = at, where t may be any quantity whatever, positive or nega- tive. Since x — r — ~ bt, .' . x = r — bt. Since y — n~ at, .' . y = n -\- at. If the equation be of the form ax — by = c, we obtain in the same way, x = r -]- bt, y = n -\- at. 309. If only integral values of x and y be required, the number of solutions will sometimes be limited. The above general values of x and y will still apply, only r, n, and t must all be integral. There can be no integral values of x and y max ± by = c, if a and b have any common factor not common also to c. For, let a = md, b = nd, while c does not contain d. Then mdx ± ndy = c. . • . mx ± ny = -^. (180) INDETERMINATE SIMPLE EQUATIONS. 181 But m and n are integers; if, therefore, x and y be also integers, iiix will be integral, so also will ny, and we shall have the sum or difference of two integers equal to a frac- tion, which is evidently impossible. Therefore no integral values of x and y can be found. 310. To solve ax ^hy = cm integers. * 1. Find the integral solutions of Zx -\- 5y = 73. Transpose the term having the larger co-efficient. . • . 3x = 73 - 57/, whence x = 24t - y -\- ^—-^. o 1 2?/ Since the value of x is integral, then — — -^, although o fractional in form, must be integral; and so also is any multiple of it. Now, multiply the numerator by such a number that the co-efficient of y shall be 1 more than a multiple of the denominator. In this case multiply by 2. 2-4?/ 2-y ... , 2-y..^ . • . — 5—- or — — ^ — «/ IS integral. . • . — — ^ is integral. 00 o 2 y Let — -^ = t, an integer. Then 2 — y = ^t, and y = o 2-^L .-. a; = 73 - 5?/ = 21 + 5L If t = Oy then X = 21, y = 2; it t = 1, then x = 2G, y = — 1; it t = — 1, then x = IQ, y = 5; etc. 311. If the solution of Zx -\- 5y = 73 be required in positive integers, there will be but five pairs of values that will satisfy the equation. For, in order that y = 2 — 3t shall be a positive integer, t can not be greater than 0. In order that a: = 21 -|- 5^ shall be a positive integer, t can not be less than — 4. .' . t may be 0, — 1, — 2, — 3, — 4, giving the following pairs of values: x = 21; 16; 11; 6; 1. y= 2; 5; 8; 11; 14. 183 ALGEBRA. 312. — 2. Solve in positive integers 39.^ — 5G?j = 11. ^ , 17?/ + 11 17v 4- 11 . . , , Here x = y -^ — . . • . —^ is mtegi-al. To find a multiple of 17 which is 1 more than a multiple of 39^ takes too much time; hence proceed as follows: 17:y 4- 11 _ 39y - 22,?/ + 11 _ _ ll(2y - 1) 39 ~ 39 ~^ '39 * Since — ^^-^r is integral, then -^-^ — must be inte*- gral, and — = 2/ + ''^m — i^^st be integral, = y-\-t. .'. y = ^dt -{-20, whence 2: = 5G/5 + 29. Here t can not be a negative integer, but it may be 0, or any positive integer. . * '. the number of values is unlimited. When / = 0, ^ =: 29, and y = 20, which are their least positive integral values. 313. From the preceding examples it appears that when only positive integral values of x and y are required, the number of them will be limited when the equation is of the form ax -\- by = c, but the number of solutions is un- limited if the equation is of the form ax — by = c. 314. The necessity for a multiplier may often be avoided by changing the form of the given equation. 3. Solve in positive integers 3x -\- 5y = 73. Add y. .'. 3x-{-6y = 73 ^y. ... ^^_2^/ = 24-f^^ 1 -\- V . • . — i—— IS an integer = t. .' . y = 3t — 1, x = 26 — 5t. o Here t may be any integer from 1 to 5, giving the same pairs of values as by the method in (310). INDETERMINATE SIMPLE EQUATIONS. 183 315. — 4. To find the least number which, wiien divided by 14 and 5, will leave remainders 1 and 3 respectively. Let X represent the quotient of the number divided by 14. . • . N=\^x^\. Let y represent the quotient of the number divided by 5. .-. N= 5^ + 3. .-. 52/ + 3 = 142; + 1; hy = Ux — 2 = 15a; — re — 2; ^/p j_ 2 X -\- 2 . • . y = 3x — ^ — - — . . • . — ;: — = an integer = t. x = 6t — 2, and y = Ut - 6. The least value of ^s 1. . • . a; = 3, and iV = 43. 316. — 5. Find the least integer which is divisible by 2, 3, 4, with remainders 1, 2, 3. Let the quotients be respectively x, y, and z. Then iV' = 2.^ + 1 = 3^ + 2 = 4;z + 3. Since2.'?;+l = 3.y+2, .*. 2.^ = 3//+1; .•,x = y-{-^^; (ft y 4-1 . • . ^^— = t; .'. y = 21 — 1; whence a; = 3/ — 1. In 2x -f 1 = 42 + 3, for x substitute its value (3/ — 1). .-. 6/f-l =42 + 3; .'. z = t + i:-l; . • . -^ = an integer = 7n, and t = 2m. Whence x = 6/y2. — 1; i/ = 4m — 1; z = 3w — 1. The least value of m is 1. Then a; = 5, and iV^= 2a; + 1 = 2 X 5 + 1 = 11. E.XERCISE XVIII. Solve the following equations in positive integers: 1. Sx + 65^ = 81. 4. bx -]-\y = 64. 2. 7. + 10, = 297. ,.- +1=-. 3. 3a; + ly = 250. 6. 3a; + 7?/ = 100. 18-A ALGEBRA. Find the least integral values of x and y wliich satisfy the following equations: 7. 7x- ^ij = 29. 9. 19:^ - ^1/ = 119. 8. 9a; - 1 l/y = 8. lo. 17a; - 49/y + 8 = 0. 11. In how many ways can 1100 be paid in dollar bills and five-dollar bills? 12. Divide 200 into two parts, such that if one of them be divided by G and the other by 11, the respective remain- ders may be 5 and 4. 13. What is the least number which divided by 28 leaves a remainder 21, and divided by 19 leaves a remainder 17? 14. What is the least number which, divided by 28, 19, and 15, leaves remainders 13, 2, and 7? * 15. Find a number of two places such that if it be divided by 19, the quotient will equal the units' digit, and the remainder will equal the tens' digit. 16. The difference between a certain multiple of ten and the sum of its digits is 99. Find it. 17. A number is expressed by three digits whose sum is 20; if 16 be taken from the required number, and the re- mainder be divided by 2, the digits will be reversed. Find the number. 18. A buys for 1100 sheep, turkeys, and chickens, 100 in all. Each sheep costs $10, each turkey $3, and each chicken 50 cents. How many of each does he buy? CHAPTER XIX. RADICAL EXPRESSIONS. 317. An indicated root that can not be exactly obtained is called a radical, surd, or irrational expression. Thus, Va^ or a^ is called a surd. An indicated root that can be exactly ascertained is said to have the form of a surd. Thus: Va^ has the form of a surd; but, since the indicated root can be exactly obtained, it is a rational expression, and not a surd. 318. It was stated in Art. 146 that there can not be an even root of a negative number. Such roots, however, may be expressed in the form of surds, and are then called im- possible or imaginary expressions. Examples : V — ^', \^ - IG; V'^^; (- 3)^; etc. 319. Surds are named quadratic, cubic, biquadratic, according as the second, third, or fourth roots are required. 320. A mixed surd is the indicated product of a rational factor and a surd factor ; as, 2 V5 ; a Vd. In this case the rational factor is called the co-efficient of the surd. 321. An entire surd is one whose co-efficient is unity. 322. A surd is in its simplest form when the surd factor is integral, and as small as possible. Alg. 16. (185) 186 ALGEBKA. 323. Sbnilar surds are those which, reduced to their simplest form, have the same surd factor; as 3 Va and — 5 Va. Otherwise the surds are dissimilar. 324. Surds are of the same order when they have a com- mon index; as, Va^ and r 5. 326. Any rational mmiher may he expressed in the form of a given surd hy raising it to a power irhose exponent is equal to the given index, and indicating the required root of the product; as, 2 = 4^ = 8^, etc. ; a = Va^ = Va^ = a^, etc. ; a + x=:(a^-\-2ax-{- a;^) ^ = (a + x)i = V(a^xy', etc. The rational numbers are here raised to certain powers, and the roots of those powers are then taken, so that the values of the expressions are unchanged. 326. The product of iivo or more surds of the same order is found iy taking the product of the iiuinhers under the radical signs, and retaining the common index. Thus: Since («nZ>«)" = ah, and ( yab^ = ah, .*. anhn=i V ah = (ah)l. Similarly: V2 . V^^ VlO; V^h. Va-b= V'd'-h^; \^a. W. VWh ^ 1^^; etc. The product of mixed surds of the same order is found hy prefixing the product of the co-efficients to the product of the surd factors. Thus: a Vb X c V d — ac V hd ; 5 1^2 X 3 4^ = 15 I^Ig = 15 X 4 = 60. 327. A mixed surd may he expressed as an entire surd hy raising the rational factor to a power whose exponent is equal to the surd-index, and placing heneath the radical sign the product of this poioer and the surd factor. Thus: 2V^=VI .V^= V\^'^ 3.3*= 1^27. 1^4= VT08; 2a VU = Vla^ . VU = Vl2^hj etc. RADICAL EXPRESSIONS. 187 328. Conversely, a factor may be removed from under the radical sign by extracting the indicated root. Tims: |/12 = V^'xd = 2 VT; \^l = i^S^<~^ = 2H; i^ = ^16X2 = 2H; Vab - U" = Vb(a - b) = l^ Va - b; etc. Hence: A71 integral surd is reduced to its simplest form by removing from under the radical sign all factors ofwliicli the indicated root can be exactly obtained. Thus: V12 = V'dQ X 2 = 6 V2 ; V76a^ — bOa^ ^ V2ba^ {;da - 27^) = 5n VSa - 2b. If the surd factor be a fraction, its numerator and denominator should both be multiplied by such a number that the indicated root of the denominator can be exactly ascertained. For example: ^b_J^bc_aV^c^ ^_//V2_'V\2^ If the index be a com^yosi/e number, and if the root in- dicated by a factor of the index can be extracted, the surd is simplified by extracting that root. Thus: VTii = 1^12 = 2 V^; y'S = V^; ^^27 = VS; etc. 329. The order of a surd may be changed by mnltiplying both exponent and index by the same number. Thus: Va= i/a^ = \/a^ = v^^^ etc.; i^Te = V7? = y"^; 8 4n 3/1 Va(a -b) = Va\a - bf = Va'{a - b)'; a' = a' = a 330. To reduce surds of different orders to the same order: (1) Find the L. C, M, of the indices. (2) Divide this L. G, M. by each index in succession. (3) Multiply both exponent and index of the first radical 188 ALGEBRA. hy the first quotient, of the second radical hy the second quotient, etc. Thus: V'Z, Vi, VS, and yo may be reduced to equiva- lent surds having an index of 30. /— 30,—- 3,— 30,—- 5,— 30,— 10,— 30,—- 331. To compare surds of the same order: Express them as entire surds, and co^npare the resulting surd-factors. Example. Compare G V^ and 4 V7. gV3 = 4/108; 4^7 = Vl02. .*. G V^ > 4 fT. To compare surds of different orders: Reduce them to entire surds of the same order, and com- pare the resulting surdf actors. Example. Compare Vb and yil. 4/5 = 1^5^ = 1^125; 1^ = t^lP = V'm. .'.Vb> 'k'n. 332. To add or subtract surds: (1) Reduce each surd to its simplest form (328). (2) If the resulti7ig surds he si7nilar, prefix the sum or difference of the co-efficients to the common surd-factor. Dissimilar surds can only he connected hy their signs. Example. 1^12 + ^^ + ^7 + ^"2 7 + 2 '^32 = 2 1^3 + 3 V^-\- 3 f3'4- 1^3 + 2 V^ = 6 V^ -\- 5 4^2. Similarly: ^^40 - J 1^320 + Vm> = 2 t's'- 1(4 V^) + 3 1^ = 2 \^- 2 V^+ 3 Vb = 3 V^. 333. To multiply surds." (1) Reduce the surds to the same order (330). (2) Prefix the product of the co-efficients to the product of the surd-factors, retaining the common sui-d-index. RADICAL EXPRESSIONS. 189 Thus: Vi X 5 V2 = 5 Vl6 = b X 4: = 20; 2i/3x3V^X4Vo'=24 V30; 2 1^ X 3 \^2 = 2 ^27 X 3 ^= 6 fl08; 3 111 1 ;2 _ /'9J\2 . 2« X 3^ = 8^ X 32 = (24y 113 2 11 1 2^ X 3' = 2« X 3« = 8« X 9« = (72)1 334. If tico or more surds have tlie same rational expres' sion under the radical sign, their product is found by making the sum of the fractional exponeiits tlie exponent of that expression. 3_ 1 1 5 Thus: Vax Va^a"" Xa"" =za^ ; 2—3-4 \ \ \ ?- + -+- H 1^2 X ^"2 X V^= 2^ X 2» X 2* = 2^ ^ * = 2^^ 335. Division of stirds is performed, when the divisor is a monomial, by a process similar to that for mullijjlication (333, 334). Thus: VS-i- V2= V4 = 2; (8i^-12i^-4i^+6)-^2 4^= 4|/f - 6 -2V2 + V^ = ^Vq - 6 -2V2-{-Vd; 3- - - i-i ^ V2 -^ V2 = 2^ -^ 2^ = 2^ ^ =z '>•) i h X3' = i ^648; 3.V6=^ = ^ = iV0.* 336. In case the dividend is not exactly divisible by the divisor, express the quotient in the form of a fraction, and * ^ V 6 is considered simpler tlian 3 -^ 1^6; because if we wish to find the approximate value, it is easier to take ^ V'C than to divide 3 by V^6, as can be ascertained by trial. 190 ALGEBRA. multiply both numerator and denominator by a factor that will render the denominator rational. The last example of the preceding article is solved in this manner. Similarly: — — = . Here both terms are multiplied by 337. If the divisor be compound, express the quotient in the form of a fraction, then multiply both numerator and denominator by a factor that will render the denominator rational. To do this, proceed as follows: 338. — First Case. If the denominator be of the form Va ± Vb, multiply both terms by {Va ^^ i^b). The de- nominator will then become a — b, a rational expression. 1 _ 1 21^+1/5" Thus: 2V2 - V5 2 V2- V^ ^ 2 V2-\- Vb 2f^-f 1 /5" _ 2^2" 4- 1^5" 8 - 5 " 3 Similarly: — —^ — = "^ — x — ~ ;=: = 5 V2-}- 2V7 5 i^2 + 2 V^7 5 ^^2 - 2 V? 20 i^4+15(2)-8(7) -eVu _ 14 VU-26 _ 7 1^14-13 25 X 2 - 4 X 7 ~ 22 ~ 11 Therefore: When the denominator is a binomial involv- ing only quadratic surds, the required multiplier ivill consist of the same terms as the given denominator, but with a diff Client sign between them. 339. — Second Case. If the denominator be of the form 2112 Va ± yb, multiply both terms by a^ =F «^ ^^ + ^''^ for the product of these factors is a ± ^, a rational expression. RADICAL EXPRESSIONS. 191 InUS. 3 3 3 3 X 3 3 3 V^3 - ^2 V3 -V2 ^^9 + ^6 + Vi o — z Therefore: Wheii the denominator is a binomial involv- ing only cubic surds, if of the form {x — y), the multiplier will be a^ -\- xy -\- y^; if of the form {x 4- y), the multiplier will be a^ — xy -{- yK As a result of this rule : If the denominator be of the form x^ -\- xy -{- y^, and involve only cubic surds, both terms should be multiplied hj x — y. If it be of the form x^ — xy -\- y^, and involve only cubic surds, the multiplier will he X -\- y. Thus: 1 i^ - 1 3-1 ~ 4 + 2 ^3-1 ^9 + ^3+1 1 2 ' t^4 + V^2 \^16 -2-{-P4: " 6 340. The following general law covers the two preceding cases, and all others involving binomial denominators: 11 1^ (1) Suppose the denominator a** + b''. Put x = a^ and y =zb^ ; let n be the least common multiple of p and q, then ic" and ?/" are both rational. If n be even, the mul- tiplier is found from dividing .t" — y^hj x -\- y. li n be odd, the required factor is found from dividing :f" + y^ by x-\-y. (Consult Articles 219 and 220.) 1 1 (2) Suppose the denominator a^ — b^. Take x, y, and n as before. The required factor is found from dividing x"" — ?/" by ^ — y- (Consult Articles 217 and 218.) 192 ALGEBRA. 341. By two operations we may rationalize the denom- inator of a fraction when that denominator consists of three quadratic surds. For, suppose the denominator to be Va -{- VJ-\- Vc : first multiply both numerator and de- nominator hy Va -{- Vb — Vc, thus the denominator be- comes a + 2 Vab -{- b — cj then multiply both numerator and denominator hy (a -\- b — c) — 2 Vab, and we obtain the rational denominator {a -\- b — cY — ^ab. Thus: VTO _ VTO ( V5 4- |/2")4- Vq _ 7^= X V6-\-V2-Vq (i/5 + V2)-l/6 (1^5+ V2)+ 4^6 5 V2-{- 2 i^+ 2 VT5 _ (5 V24- 2 ^5 + 2 V Tb)(2 VT6 -1)_ 2 fTo + 1 ~ (2 VlO + 1) (2 V16 - 1) ~ 18 V6-\- 15 1^2 -f 20 VG- 2 VfS 39 EXAMPLES. Simplify: 1. -^^63; V28; i^5; 1^; V2i6. Ans, 3 V7; 2 V'f; 5 1^1; 2 VIS; 7 i^. 2.^5^*; i^; i^^• Vm^; \^M, Ans, «2 ^5; «3 1/3- . ^2 |/f -^. 2^3 y^. 4^^4 1/3-^ 3.^; ^^; ^; ^1; ^jS- _ ^^5. lV2', }i'3; ^tTI; 1 1^3; ^^51. Ans. 2 VT; 2fl^ Ki; 3 l^^; a^ ^;^2; 2^^ l^a. 5. 2 1^80; J ^48; 2 i^; 2«2 xQi; j.« t^*. J[Wc9. 8 V^; I fe"; I i^; 2a^ i^; fa^ y^. RADICAL EXPRESSIONS. 193 e.Va^-^ d'x; Va^ - ^ct'x + ax"; V{x - y)(ij - x). Alls, a Va + x; {a — x) Vaj (x — y) V — 1. i, 6, 3a-,—- — — 6/ — 10/ Ans, \V5; iVQ; a \^dj - 2^^ y^; /H^. 8. i^ + Vf - f 1^1 4- 3 Vi. Ans. -V- ^6. 9. 3 i/- 81 - 5 i^- 24 - 3 l/- i + V - 375. J?i5. — 3 V¥. 10. |/3^K2« - ^)' - V3rt(a - 2^)'^ + i^l2^. 11. 3V^x24/fX5VT-^- f^. ^W5. 10 f'2. 12. i/63 -f- 1/14 + J^ VT8. Alls. 3 4/'2. la-Vrt - 1^^ X Va-i-Vb. Ans. Va!" - b. ^^*3V^' i^s' '^^' t^' 1^-4' Ans. |t^3; i^; ^3; f 1^; - V^. 1 15. 16. Vb - V2' V? + 1' i^i - 4/7 ' ^?^5. i( V5 + 4^); - i( 1^7 - 1); i( fH + i^). ry^ — — — . Ans(i(V6-2-{-V2).) 17 -. — --Y- . Ans. i( 1^ + t^lO + 1^). ^^'^-^^2 ^v -r -r ; 18. 1 . Ans. i(\^ - f^+ 1). 1/54-1 19. -3— 3^= sF- ^^^^- i( ^+ ^^)- 1/25 - 1/20 + 2 1/2 Alg.-17. 194 ALGEBRA. IMAGIKARY EXPRESSIONS. 342. All imaginary square roots may be reduced to one of the two forms, ^ i^- 1 or h^ V- 1. Thus : V-^- 9 = V9(- 1) = 3 |/^=n[; V^^ = V6(-~ 1) = Vq V^^. It is frequently convenient to use aV — 1 instead of V — a^, also h^ V — 1 instead of V— b. 343. Since the square of the square root of any number will produce that number, .' . {V — ly = — 1. Since the third power of any number is equal to the product of the square and the first power of that number, . • . ( V— Vf = ( V^^Y V^^ = { - 1) ^^^^ = - ^^'^^- Since the fourth power is equal to the square of the square, .*. ( |/zn:)* = ( v^iriY(inriY = (_ i) (_ i) 3= + 1. in like manner, we find the fifth power of V— 1 = V— 1; the sixth power of V— 1 = — 1; the seventh power of V^^ = - \/^^\\ etc. Thus all the powers of V— 1 may be expressed by four formulas: 1. ( V^^)*™ + i = i/~i; II. ( i/Zri)4n + 2 ^_ i. III. ( /:r-i)*«+3 = - |/:ri; iv. ( ^^=T)*" = + 1. For example : ( ^^1)*» = ^^^H. ; ( V^^)^^ = - 1 ; ( 1/371)51 = _ |/— 1- ( i/Zri)52 = -I- 1. 344. The rules for addition and subtraction of imaginary expressions are the same as those for ordinary surds; but before applying the rules for multiplication and division (333, 335), the factor V— 1 must be removed as explained in (343). RADICAL EXPIIESSIONS. 195 EXAMPLES. 1. 2 V^=^ + 3 /^=^ + 5 V^^S - 8 V-2b = 4 V^^ + 9 V^l + 20 V^l. - 40 /^i =- 7 2. 2 V- 12 + -3 l/- 18 = 4 /^=^ - i |/- 3 - 3 /^^ = JgJL V^^ - 3 V^-2, 3. -/^^^ X i^^^ X i^^^ = 8 ( vq v^)( Vs v-i)( V2 i^-i)=6(-iy=-6 v-1. 4. V-ab -h t^^^ = V^, 5. I^«^ -^ V— a = 4/:iT. 3 - 3 V-^ _ 3(1 - i^- 1) 2 - 2i/" !(1 - V- 1) 7. 4^^^ _ (3 - /^^)(2 - /^^) 2 + |/- 1 (2 + V- 1)(2 - V- 1) = l->/^:^. 8. (I - ^- 1)*^ = [(1 - /^=^fP(l - '*^^^) = (- 2 V^T)2(1 - V^^) = _ 4(1 - V^^), 4^-f-34/-iU|i/^=^.) 9. 2 V- 24 10. 3 /^ X 4 V^ X 5 V^r -=- 2 V^t - 10 V2A SQUARE ROOT OF A BINOMIAL SURD. 345. The product or quotient of two dissimilar quadratic surds will be a quadratic surd. Thus: (1) Vab X Vabc = ab Vc; (2) Vabc -^ Vab :^ Vc; (3) Vab X V^ = a Vbc. For, since the surds are dissimilar, one or more of the factors under the two radical signs must be unlike (323); 196 ALGEBRA. hence in the product or quotient these unlike factors re- main under the radical sign {326, 333, 335). 346. The sum or difference of tivo dissimilar quadratic surds can he neither a rational number nor a single surd. For, if possible, let Va ± Vb = Vc, in which Va and Vb are dissimilar surds, and Vc either rational or a single surd. Then, by squaring, a ± 2 Vab -\- h = c. That is, ±2V^ = c-a-b. But Vab is not rational (345) ; . • . ±2 Vab can not be equal to c — a — b, b. rational expression. Hence the sup- position that Va ± Vb = V^ is false. Therefore Va ± Vb can not be equal to a rational expression nor to a single surd. 347. A quadratic surd can not equal the sum of a ra- tional expression and a surd. If possible, let Va = b± Vc; then a = b"" ± 2b Vc -\- c. . *. a — b^ — c = ± 2b Vc; that is, a rational number is equal to a surd, which is impossible. Therefore Va can not equal b ± Vc. 348. It a -]- Vb = X -\- Vy, in which a and x are rational and Vb and Vy are quadratic surds, then will a = x and Vb=Vy. For, by transposing, Vb — Vy z= x — a. Now if Z' and y were unequal, we would have the difference of two unequal surds equal to a rational expression, which is impossible (346). . • . b = y, and therefore a = x. 349. IfVrt-+ Vb =x-{- Vy, then\/« — Vi = x - Vy, For, by squaring, a -\- Vb = x^ -^ 2x Vy -{- y. .' , a = x" -\- y Sind Vb = 2x Vy. (Art. 348.) RADICAL EXPRESSIONS. 197 By subtracting, a — Vb = of — 2x \^ + V- Extract square root, . * . y a — Vb =^ x — \fy. Similarly it may be shown that \iy a-\- ^Tb — Vx -^ Vy, then Va — Vb = Vx — Vy, 350. To extract the square root of a binomial surd, a -\- Vb, Let I.Va-\-Vb = i^-\-Vy; tlien (349) II. Va~ Vb = Vx- Vy, By multiplication, III. Va^ — b — x — y. By squaring I., IV. a -{- Vb = x -\-2 Vxy -\- y, . • . V. a = x -\- y. VI = V + III. 2x = a-\- V¥^. .',x= l{a^ V~^^^), Nil = Y - III. 2y = a- V^^^. .'.y = ^{a- V^^), Therefore x and y are known; .' , Vx -\- Vy \q known, and Vx — Vy is known. . * . \ a -\- Vb and Va — Vb are known. For example, extract the square root of 11 + 6 V2. Here x -^ y = 11 and x - y = Vl21 - 72 =Vl9 = 7. ,'. x = 9 and Vx = d; also, y = 2 and Vy = V2. . • . V^ll + 6 ^2 = 3 + 1^. Or thus: 11 --[- 6 t^2 = 9 + 6 1^2 -f 2 = (3 + V2Y. . • . Vll -\-6V2 = 3-\-V2. 351. From (350) it appears that Vx=Vi{a-\- VcF^) and Vy = y \{ci — Vc? — b); hence unless c^ —b be a per- fect square, the values of V^and l^^^will be complex surds, and the expression Vx ■\- Vy will not be as simple as V a-\-Vb itself. , 198 ALGEBRA. 352. A binomial surd of the form V^ -j- Vdc may be written thus : Vc{a -{- t^). If then a^ — b he a perfect square, the square root ot a -\- Vb may be expressed in the fx)rm Vx -\- Vy. . • . the square root of Va^c -\- Vbc is of the foj-m {Vx -\- Vy) Vc, For example, find the square root of 4 ^2 -|- 2 Vq. 4 1^ + 2 1^ = 1/2(4 + 2 1^) = V2( V3 + ly. r,V4:V2 -^2 V6= t^(4 -\-2V^y= ^^2 ( t^ + 1). Also, V" 11 V2 + 12 = V 1^2(11 + 6 1/2) = ^"2 X Vll + 6 i^2 = t^(3 + V2). 353. The square root of a binomial surd may frequently be found thus : Fiiid ttvo numbers whose sum is the ra- tional term^ and whose 2^Toduct is the square of half the radical term. The square roots of these numbers, con- nected by the sign of the radical term, is the required root. For example, to determine by inspection the square root of 4 — 2 1/3. The two numbers whose sum is 4 and whose product is 3, are 3 and 1. . • . v 4 — 2 4/3 = t^ — 1. Also, a/i8 + 8|/5 = i/l0 + |/8 = |/l0 + 2l/2, because 10 and 8 are the numbers whose sum is 18, and whose product is (4 Vlf = 80. 354. The square root of a binomial surd may also be found by solving the equation x^ — (rational term) X ^ = — ^ (square of radical term), and connecting the square roots of the values of x by the sign of the radical term. Thus: Extract the square root of 7 + 2 VTo. Solve the equation x^ — 7x = — 10; .-. a:2-7.T + -V- = l; .'. x-i= ±1; x = 6oy2. .'.V7-\-2Vi^ = Vb-^ V2, RADICAL EXPRESSION'S. 199. In like manner the square root of 16 — 5 4^7 is found ])y solving the equation a^ — 16a: = — ij^. Here x = ^^- or |. .-.Vie - bVl = V^-- Vl- = :^{bV^- Vli). EXAMPLES. Extract the square root of each of the following expres- sions : 1. 13 - 4 VTa Ans. 2 ^2 - V5. 2. 10-2 V^, A71S. Vl - 4/3. 3. 9 + 4V'5^ Ans, 1^+2. 4. 21 - 4 Vb, Ans, 2 Vb - 1. 5. 11-4V7. Ans. ^7-2. 6. 2i- - f V^, Ans. i{VQ- f3). 7. 8 VTS - 30. ^7/5. \^b{ Vb - V3). 8. 16 V7- 42. ^7i5. >k^(3 - V7). 9. 11 V^ - 60. Ans. V'30( V^- Vb). 10. 8 1^+2i^. /^^7^.9. H{Vb-{-V^). 11. 9f^+20. I^w5. V^(V^+2). 12. 15 i^+ 12 V3". y W5. V^(3 + Vg). EQUATIONS INVOLVING RADICALS. 355. To solve an equation containing a single radical: Arrange the terms so as to have the radical alone on one side, and then raise both sides to a power corresponding to the order of the radical. Thus : I. Solve I. Vx" - Qx 4- 24 = a; + 2. II. Square I. o;^ — 6a; + 24 = a.-^ + 4a; + 4. . • . a; = 2. 200 ALGEBRA. Solve 2x— Vx^ — 1 = 3. 11. Transpose I. 2x - 3 = VaF-1. III. Square II. Ax^ - I2x -\- 9 = x" - 1. IV. Transpose III. Sa:^ - 12x = - 10. V. Multiply IV by 12. 3Qx^ - lUx = - 120. VI. Complete the square. SGa;^ - 144.r + 144 = 24. VII. Extract square root. Qx-12=±2 Vg. .'.x=2±i V6. 3. Solve 2x - V8x^ + 26 + 2 = 0. II. Transpose I. 2x + 2 = ^Sx^ + 20. III. Cube II. 8x^ + 242;^ -f 24a; + 8 = 8x^ + 26. IV. Transpose III. 24a;« + 24a; = 18. V. Divide IV by 0. 4:X^ + 4^; = 3. VI. Add 1. 4:x^ -^ 4a; + 1 = 4. VII. Extract root. 2a; -j- 1 = ± 2. . • . a; = J or — |. 356. If an equation contain two radicals, two steps may be necessary in order to clear the equation of radicals. It is usually best to have the larger radical term alone on one side; but if the radicals are reciprocals, both are placed on one side, and the remaining terms on the other. As before, the equation is raised to a power corresponding to the order of the radicals. 4. Solve Va; + 4 + V2x - 1 = 6. Transpose I. Square II. Transpose III. Square IV. Transpose V. II. y2x — 1 =. 6 - y^ + 4. III. 2a; - 1 = 36-12 ^/a; + 4 + a; -f 4. IV. a; - 41 = - 12 Vx^^. V. x^ - 82a; + 1681 = 144a; + 576. VI. a;2-226a;=-1105. . •.a;=221or5. 367. If we attempt to verify the above values, we find that 5 satisfies the given equation, but that 221 does not. We find further that the four equations, I. Vx-i-4:-{- V'2a;- 1 = 6; IL- Va; + 4 + V2x^ = 6; III. Vx^- V2^^1 = 6; IV.- t^^+T- 1/2^^ = 6, RADICAL EXPRESSIOIS^S. 201 all reduce to the quadratic a^ — 226x = — 1105, whose roots are 221 and 5. Now 5 satisfies the first of these forms and 221 the second, whilst neither 5 nor 21 satisfies the third or fourth forms. These examples show that when an equation has been reduced to a rational form by squaring, the roots found may or may not satisfy the equation in the form originally given. Both 221 and 5 would satisfy all the forms, how- ever, if we agi-ee that every indicated square root may be either + or — , and that we take whichever sign produces the proper result. oeo o 1 ax — P , Vax — b 358. — 5. Solve -— -: = c -\ . Vax -\-b ^ XL Simplify each side of I. Vox — b = c -\ ; . c c III. Transpose II and clear. i^(c — 1) = c^ -{-he — b. IV. Square III. ax(c - If = {c' -{-be- bf. V. Divide IV by a{c - If. x = (i-t-^il'. •^ ^ ' «(c — 1)^ 359.-6. Solve V2x - 1 + Vdx^2 = V^x^ + V5x-4. II. Transpose I. V2x-\- Vbx-l = V4x-3 - V3x-2, III. Square II. 2a; - 1 - 2 V(2x - l)(bx — 4) -\-6x-4 = 4:X-3-2 V(4:X - 3) (3x - 2) + 3a; - 2. . • . IV. - 2 V{2x-l)(5x-l) = - 2 |/(4i - d)(dx - 2). V. Divide IV by — 2 and square the result. 10a;2 _ 13^ _^ 4 ::::: 12^^^ _ 17^; + 6. .'. x = l. EXAMPLES. Find the value of x in the following: 7. Vx-\-2 - Vx — 3 = 1. Ans. x = l, 8. V^x-{-l — V2x — 1 = 1. Ans. a; = 5 or 1. 9. Vbx — 3 4-2 = 5. Ans, x = Q, 202 ALGEBRA. 5x- 1 ^ V6x + 1 . . 10. -t=z = 4 -] ^ — . Ans. X = D. Vbx - 1 3 11. \^8a^ + 61 - 1 = 2a;. ^?i5. cc = 2 or - 2J. 12. H , = o • ^^^^- i?^ = 1 ± '^^^. a - Va^ - a^ ^ 2rtc 13. - = (?. Ans. x= i-^-—-. a -f |/^2-Zlc2 6-2 + 1 -j o 14. a: + 1^9 4- 2;^ = -^==. ^?^s. cc = ± V^. Exercise XIX. Simplify (328): 1. V8; V48; 1^5; fTs"^^- ^72^?/; 4^45 (a -^'f; ^96^^^ 2. \^bi^^; V'iOtt*; 1^686; 1^16(0; - ?/)^- l^8r(^^^"pl 3. 4 VTS; 3 VT2; f 1^; J V^; - V^^; (a - h) V(a + bf, 4. 4 1^27^*; (« - ^^) -' P{a^^; \^(d' - a')(a - h)\ 5. 4^; V^; 4 VY-; (« - ^)/^^- (^ + ^)(^ " ^)~^"- 6. 3t^; hi', {a-^l)V-^^- 'v{a-^h){a-h)-\ 7. t^; Hi) a H2a^j Pa' - a%' Vl^¥^; VcF¥+^, 8. t^; ^8; ^6; V'«2-2rt^>+Z»^- t'(a - bf; \^-S{a^bY. 9. Pa^ - 3a^ + Ub^ - ^V t^81^*Z^=^V t^^* 3 1^?^. Keduce to the same order (330): 10. I^andl^; 1^ and V2; l^and f^; t^andl^. RADICAL EXPKESSIONS. 203 Compare the following surds (331): 11. 3 /3 and 2 V7; 3 H and 2 1^; 4 ^2 and 3 H; 12. i V2 and J Hi; Vb and 2 l^f ; 2 1^ and t^SS". Simplify (332): 13. 4 f 24 + Vbl - |/96 + 3 ^50 - 2 l/l)8 - 5 V72. 14. V^+ i^-V- - i^ - i i^+ i vTs + V^. 15. 5 4/a^ - a^b + ^'^cT^^Wh - 8 ValF^~h\ /^ 16. Va«(« - 3<{>) + ^'-^(Sft - ^) - i/(a + ^)(««- J«) 17. H^ - \y/\\ - ^+ ^"40 - V"- 625. Simplify (333): 18. 5 VT2 X 2 1^; 3 V'^'x i Vl6; 2 fT2 X i V20. '/ 19. J ^2 X 6 Vj; ( l^+ i^) X 1^6 ; ( /3 - -^^2)1 '/ 20. i^ X V'S ; ^"4 X VS X t^; 3 t^ X 2 f^ X i f^. 21. l/(a - h) X -/(^"^y^; (a - V) '^{cT^h) X V'(7r=T)"* 22. ( V^ - V^) ( V3 + 4/2); t^i~+y X V'a^ + y. Simplify (335): 23. (2 VZ- 3 V%- V^O) ^ l^e"; {\ l^f ^ iy ^) X 2 V^. 24. ( '^M + 6«¥'+ Vu^~^fUa^') -^ 6 l^r+T^^ 25. ( y'si -^ V'S) X t'G + t^ X V^ + i V^2 ^ i ^. Rationalize the denominators (336-341, inclusive): 3 2 4 3 V^-}-Vc 2 , , _. . 4/2 2 1^2 1^ 4^8 V9 |/, rtc 27. 4^6 4^5 + /3 2 + 1/3 1/3 + 1/2' 1/5-1/3' 2-1/3' V3 + 2 1^ 204 ALGEBKA. 28. 29. 30, 1 1^ 1 y4 — 1 Vi—V2 vTe + 2 + f4 3 1^ VT04-2 Simplify (342, 343, 344): 31. V^^s - V^^-^- i V^ - i V^^- 2 i^'^. 32. 2 V- 3 X 3 |/- 2 X |/- 8 X 1^- j. 33. (3 V^Xi ^^2)'; (-1- V^^'; ( 34. V- 4 ^ Vi; J -7)1 ^; 6 1/- 10 -^ i i/- 5. 35. yC:i6+ ^-81+ 3 V2 V^3 + 2 X V2 V^l - 2. 36. Vg + V^^3 xV6- /^n:3 - 35 V^. Simplify (350, 352, 353, 354): 37. ^5 - 2 1^ + Vu -4 1/6- V^O - 12 V6. 38. -\/l7 - 12 V2 + -v/49 + 20 V 6 - V28 - 16 VS. 39 ;-i-f (4 - 2 i/3)i VlO + 4 4^ V7 - 4 1/3 * 40. V4 + 3 V2 4-^/2 l^+i^-V^4 4^-2 1^6'. 41. a/3 - 2 V^; \/6 - 4 4^; V"? - 2 VlO; V 9 - 4 V2. 42. (8+2 ^15)^; a/7 - 2 1/6; Vll - 4 l/?; (2 - V^)\ Solve (355-359, inclusive): 43. Vx^^ = - 3. 45. V2x' - a; + 1 = a; + 1. 44. Vx^-6x-f7 = x + 2. 46. |/^12 - Vx'-U = 2. RADICAL EXPRESSIONS. ' 205 47. Vx-\-5 + Vx — 3 = 4. 48. Vx-\-2 ^l-{-x = Vl- Vl i-x, 49. Vbx - la + Vbx = 7«i 2 50. Vx^Vl-\-x= , 51. a Vx — l ^ = (a -{- 'b){a — h), 52. i i/2 + a; + i 1/2 + a; = ^ V2x, 53. iz=z + 1==-- = 6. / x-{- Va^ -1 X- Vx" -1 a; + 2 ic — 2 55. 4/« + a; = |/^- Va-o;. 57. 3a; - 2 VSa: + 7 = 8. 56. a; + i^a; + 5 = 7. 58. 2a; + 3 Vbx = 25. 59. - VSx + 7rt + 7 i^a = l^3i. >. 60. V2a; + 7 + V3a; - 18 = Vlx + 1. 62. X V6x-'-\-x = ^^^^. j^ xi ^ 63. Vx-\- \/{x - |/[1 - a:]( = 1, a;^ + l 64. ar^ - - a; + V'6a;-\ ^ ^'^'y CHAPTER XX. QUADRATIC EQUATIONS. CONTINUED FROM CHAPTER XII. 3G0. The General Method of solving affected quadratics (187, 190) is as follows: (1) Reduce the equation {hy clearing of fractions^ trans- position, and combining like te7'ms) to the form ax^-\- l)x=^c. (2) Multiply or divide both sides by such a number that ihe co-efficient of :^ shall be a square. Thus: a^x^-\-abx=ac, (3) Complete the square of the first side by adding to each member the square of the quotient obtained from dividing the second term, by twice the square root of the (bV P first term. Thus: aV -|- abx + («"] = ^^ + t- (4) Fxti^act the square root of each side. Thus: ax + - = ±y ac-i--. (Place the double sign before one side, as explained in Art. 189.) (5) Transpose the known term to the second side, and divide both sides by the co-efficient of x. Thus: 361. Two modifications of the general method are often employed, — the Coinmon Method and the Hmdoo Method. (206) QUADRATIC EQUATIONS. 207 COMMON^ METHOD OF SOLVING QUADRATICS. 362. Rule. (1) Reduce the equation to the form ax^-\-bx=c. (2) Divide both sides by the co-efficient of xK Thus: a a (3) Add to each side the square of half the co-efficient ofx, Thus"^ + ^^4-fAV-'- + ^=^^^^±^' ilius. .'^ + ^ ^ -i- (^2^J - a^ 4.a' 4a^ (4) Extract the square root of each side. Thus: (5) Transpose the hnown term to the second side* Thus: -b ± V^ac + b^ (3) 2a 363.— 1. Solve 2x' - Ix + 3=0. Transpose + 3. . • . 2a:* - 7:r = - 3. Divide by 2. . • . a;* - |a: = - |. Add (J)Ho each side. .-. ^ - J:?: + (J)' = H - f = !i- Extract square root. . * . a; — J= ± J. . * . a; = 3 or J. Verification: (1) 2(3)* - 7(3) + 3 = 18 - 21 + 3 = 0. (2) 2(jr-7a) + 3 = i-f + 3=0. 364.-2. Solve —5 = --. a; — 1 X -\- 6 35 Clear of fractions. . • . 35a;+105-35a;4-35=a:«+2a;-3. Transpose. . • . — ar^ — 2a; = — 143. Divide by - 1. . • . ic^ + 2a: = 143. Add 12=1 to each side. . • . a.-^ + 2a: + 1 = 144. Extract square root. . * . a:+l= ± 12. . • . a:=ll or —13. Verification: (1) ^- - ^-^ = ^- - ^^ = ^. 1 1 11111 _13_1 -13 + 3 -14 -16 10 14 35' 308 ALGEBRA. 365.-3. Solve —^ + --- = 3 + — --. 2x + 9 Transpose -^—- — and multiply each term by 18. 79^ 54 . • . -^ — — ^- = 54:-^^x-16 -4:X -lS = 20-x, 4:X -\- d Clear of fractions. . • . 72x - 54= —4:X^-\- 77a; + 60. Transpose and simplify. . • . 4a;^ — 5a;r=114. Divide by 4. ,' . x^ — ^x^^l^. Add (|)< . • . a,-^ - f a: + (1)^ ^ 11 + li* ,= 18 j«. Extract root. r , x — ^ = ± ^^K . • . a; = 6 or — y-. HINDOO METHOD OF SOLVING QUADRATICS. 366. EuLE. (1) Reduce the equatio7i to tlie form ax^ -{- bx = c. (2) Multiply both sides by four times the co-efficient of x^. Thus: 46iV + 4(2^2; = 4ac. (3) Add to each side the square of the co-efficient of x in the equation ax^ -j- bx = c; that is, add V\ Thus: 4ft2a;2 + ^abx ^^^ ^ac + W, (4)- Extract the square root of each side. Thus : 2ax-\-l= ± V4:ac + b\ (5) Transpose the known term to the second side, and divide both sides by the co-efficient of x. Thus: 2ax= —b ± V4:ac + bK — b± V4:ac + b^ 2a 367.-4. Solve (2x - 3)^ = Sx, Expand. Transpose. Multiply by 16. Add (20)1 Extract root. . Ax^ - 12a; + 9 = 8a;. . 4a;2 _ 20a; = - 9. . {8xy - 320a; = - 144. . (Sxf - 320a; + 20^= 400 - 144 = 256. . 8a; - 20 = ± 16. . • . a; = 4J or J. QUADRATIC EQUATIONS. 209 368.-5. Solve - + | = 5-(a; - 1). Multiply by 12a;. . • . 72 + 22;^ = IS.^^ - 15x. Transpose and simplify. . • . — 13.r^ + 15a; = — 72. Multiply by 4(- 13). . •. (2Gxy-( ) = 3744. Add (15)'^ to each side. . • . (26xY-( )+15*=3969. Extract the square root. . • . 26a; — 15 = ± 63. . • . a; = 3 or - f |. 369. It will be observed that, after multiplying by four times the co-efficient of x^, (1) The co-eflBcient of the first term is the square of twice the co-efficient of x^ in the original equation, since 4rf^ is (2^)^; (2) The second term of the product is not used in the work. Advantage is taken of these truths to save multiplication, as in Ex. 5 above. SPECIAL METHOD OF SOLVING QUADRATICS. 370.— 6. I. Solve ar^ + 4 = 5a;. II. Divide by x. III. Square each side. IV. Subtract 2x8 from each side. V. Extract the square root. VI. Add V and II. 371.-7. I. Solve 0^ + 24: = 10a;. II. Divide by a;. a; -f 24a; "^ = 10. III. Square II. a;^ -|_ 48 + ( ) = 100. IV. Subtract 2 X48. a;^ - 48 + ( ) = 100 -96 = 4. V. Extract square root, x — 24a; ~^ = ±2. VI. Add V and II. 2a; = 10 ± 2 = 12 or 8. . • . a; = 6 or 4. Alg.— 18. X + x = 5. x" + 8 + 16 25. x" -8 + 16 9. X _4 _ X ■■ ±3. 2x: = 5±2 1 = 8 or 2, . X = 4 or 1. 210 ALGEBRA. X — Q a; - 12 5 372.-8. I. Solve 12 X - Q 6 11. Square. (^-^^ j - 2 + (^—J = -. III. Subtract - 2X2. (^S) +^+ (S') =^M- TTT -1-111 1 "^ — t) ■ , X — L/i . Lo IV. Extract square root. -A = ±—-. ^ a; — 12x — 6 6 V. Add IV and I. ^i^^^] = | ± t? = 3 or - |. \x — 12/ GO 3 . •. 2a: - 12 = 3x - 3G, Avhence x = 24; or 6:r — 36 = — 4^: -f 48, whence a; = 8.4. 373.-9. I. Solve x" ^ (a^ - h^) = 2ax. II. Divide by ic. x -f ( j = 2a. III. Square. a^ -\- 2(a^ - b'') + { )= ^a\ IV. Subtract 2 X 2(^^2_ ^2)^ ^2 _ 2(^2 _ ^2>) _^ ( ^ ^ 4^2^ V. Extract square root, x — { ) = ± 2Z'. VI. Add V and II. 2:?: = 2a ± 2^'. .',x = a±'b, 374. Kule for solving quadratics by the special metliod. (1) Reduce the equation to the form ax^ — c= — Ix. (2) Divide loth sides ly x. Thus: ax = — b. X (3) Square the resulting equation. Thus: {axY-2ac-^[^'=h\ (4) Subtract tivice the second term from each side; that is, add Aac, Thus: {axy + ^^^ + (— ) = b^ -\- Aac. (5) Extract the square root. ax-\ — — ± Vb^ -\- 4ac. (6) Add the equations fourid by usi?ig the second and fifth steps. Thus: 2ax = - b ± Vb^ + 4:ac. QUADRATIC EQUATIONS. 211 (7) Solve the resnUing simple equaiion. Thus: -1)± Vb^ + 4ac "^ 2l. • The last term on the first side being the same in each step, need not be expressed, as in Examples 7 and 9 above. In Example 8, above, an expression takes the place of x. 375. The general equation ao(^ -\-dx = c has been solved by each of the preceding methods, and it is found that X = . We may make use of this formula, therefore, instead of going through-the wocess of finding x, 10. Find X in the equation, —x^-{-lix = 33. Herea=-1, ^'=+14, 6--h33. JH4ac=190-132=64. -U±Vm -14 ±8 , _ r,x = — ^ = _2 - = + 3 or + 11. In Example 1, Art. 3G3 (solve 2x^ — 7x -{- ^ = 0), a = 2, b= -7, c= -d. 4ac + ^ = - 24 4- 49 = 25. -b± V4ac + b^ + 7 ± 5 2a = 3 or J. 376. In the examples heretofore considered we have found two different roots of a quadratic equation; in some cases, however, we shall find really only one root. For example: 11. If a^ — 6x -]- 9 = 0, by extracting the square root X — 3 = 0, and therefore x = 3. It is, however, convenient in such cases to say that the equation has two equal roots. 377. If the quadratic equation be represented by aa^4-bx=c, we have found that the roots are ^-^r Za -b - V4ac + b^ and 2a 212 ALGEBRA. (1) These roots are different, unless iac -\- b^ = 0, and then each of them is a (2) Since the two roots have the same expression. Viae -f b'\ both will be real or both will be imaginary: real when 4^6' + b^ is positive; imaginary when 4«c -|- b^ is negative. (3) Both roots will be rational if Aac + Z*^ be a perfect square. Both roots will be surds if Aac + b^ be not a per- fect square. Exercise XX. Solve by the General Method (360): 1. 4^2 _ 3(^ - 1) = 10.r. 3. 15x^ - Ux = - 3. 2. 2x^ + ^ = 3. 4. {3x - 2)(x -l) = 14. 5. (x - '7){x - 4) + (2x - ^){x - 5) = 103. 6. 8^ + 11 + - ^ . x-\-2 _ A-x _ 7 x-\-4: 2x - 12 _ 10 ''' x-i 2^ ~ 3 • ^' X - 4 "^ re + 4 ~ T* Solve by the Common Method (362): 9. rr" + 7rc = 8. 11. 2a;2 - 3:?: = 54. 10. ^ -\x-= 34. 12. 22^ + 1 = !!(:?: + 2). 1 1 1_ _ 13- 3 + 3-q:^ + 3 + 2^ ~ 3.1 8 14 ^ — X A — X X -\-2' 0? — ^ ^ X 'i . X — \ QUADRATIC EQUATIONS. 213 Solve by the Hindoo Method (366): X -\-22 4 Ore - 6 17. (2x -\-3y=- 8x. 21. S X 2 5,3 14 a:4-4 7-x 4:r+7 , 1 o ^^^ oo ' ' 1 x-i-2^x x-\-^' 3 x-3~ 9 19. (5a:- 3)2-7 = 44:^ + 5.23. g^H^ . ^'^ ~ Y^ = ]-. X ^ x -\- b 2 ^20.|(^-3)=|(.-3). 24.i^+-^ + |^^ = f. . Solve by the Special Method (374): 25. 2x^-^~ = 7. 29. x^ -'^-^= -Z Vb, x^ x^ X 56 26. a:T = _ 1. 30. X -\- c(a—h)x-^ — a—b-\-c, 2 7,2 J 27. Vx-3x-^= -2, 31. 2a; H ^ = 3^-5. ^ a: O 2 28. 62:-^ — a;i = — 1. 32. Sx(b -{- c) jy--^ — r = a. ^ ' x(b -\- c) Solve by Formula (375): 33. ^3^-\-^xz=- ^^. 37. 23^2 = ^4"^ + 14. 34. 780^^ _ 73x = - 1. 38. — ^ + 2 72 35. a^x" + ^) = 2a3a: + 1. 39. 4.T + - - - = 4r?. a; + 2^a; + 4 a;-|-6* = 4 /2<72 -f 2h\ 36. cc^ — a;(2a — h) — 2ab. 40. x^ — rc P^g ]^ ) ~ ~ "'• CHAPTER XXL PROBLEMS INVOLVING QUADRATICS. 378. — 1. A SELLS a box for |24, and by so doing he gains as much per cent as the box cost. AVhat was the cost of the box ? Let X denote the number of dollars the box cost. Then — — - denotes one per cent of the price, and x times •—— = -^ denotes x per cent of the price. Q^ X SincQ the cost is x, and the gain is —r--, therefore x + --— r is the selling price. But 24 equals the selling price. . • . (Ax. 1) re + — - = 24, whence a; = 20 or — 120. Only the positive value of x is admissible; therefore the box cost $20. 379.-2. A rows 7 miles down stream and back again in 3 hours 20 minutes. Supposing the river to have a current of 2 miles per hour, find the rate of rowing. Let X denote the rate of rowing; that is, the number of miles A could row per hour in still water. Then a; + 2 is the rate down stream, and .t — 2 is the rate up stream. Since the distance down is 7, and the rate 7 down is a; + 2, . • . — — - is the number of hours required a; -f 2 (214) PROBLEMS INVOLVIIfG QUADRATICS. 215 to go 7 miles down stream. Similarly, ^ is the time re- quired to go 7 miles up stream. 7 . 7 The time down being — — -^, and the time up being , 7 7 therefore the entire time is — -— - A -. But the entire a; -f- 2 X — 2 7 7 10 time is 3^. . • . (Ax. 1) — -^ + ^— -^ = — , whence x = b or - f The negative answer to the quadratic, — ^, does not apply to the question proposed, hence it is rejected. Ans. The rate of rowing is 5 miles an hour. ^^^380^3. A left C for D at the same instant that B left ^D for U." A reached D 9 hours, and B reached C 1 6 hours, after they met on the road. Find the time A required for the journey. Let X denote the number of hours A required to perform the journey. Since B evidently required 7 hours longer, . • . B's time was x -^1 hours. Let d denote the distance from C to D; then A's rate was — , and B's rate was . From the time they started until X X —J— / they met was a; — 9 hours, hence A traveled in that time f^(x — 9) -^^ '- miles. But B performed this distance in 16 X hours; . • . -^^ was B's rate. We now have two expres- 16a; (j(x — 9) d sions for B's rate: . • . -^- — - = — — ^, whence a^ — 2x — ' 16a; a; + 7 63 = 16a: and x = 2$ or — 3. As before, reject the nega- tive root; therefore A's time was 25 hours. 381. — 4. A worked x days, and received as wages $96; B worked X — 6 days for $54. Had A worked a; — 6 days 216 ALGEBRA. and B x days, they would have received equal sums. Find X ; also find the daily wages of each. 96 Since A received $96 for x days' work, he received — X dollars per day, and for :?; — 6 days he would have received 95(2: — 6) — ^^^ dollars. Since B received $54 for x — ^ davs' X 54 work, he received dollars per day, and for x days he X o , , , . , 54:r 3 „ 96(.T - 6) 54x would nave received dollars. .*. — ^ — -, X — Ki X X—Ki whence \^{x — 6)^ = ^x^ and 4(2; — 6) = ± 3a*, from which a; = 24 or ^-. But ^'- is not admissible, because B's time would then be (2^- — 6) days = — ^- days. Ans. ^ = 24; A received $4 a day and B $3 a day. 382. In solving problems, as in the preceding examples of this chapter, results will sometimes be obtained which do not apply to the question actually proposed. This is owing to the fact that the algebraic language is more gen- eral than ordinary language; hence the equation not only expresses the conditions of the given problem, but it will sometimes apply to other conditions. When this is the case, the conditions of the second problem are usually the contrary of those given in the proposed problem. Fre- quently, however, it will be found that only one root applies to the problem proposed, and that no obvious interpi-eta- tion occurs for the other. To illustrate the method of in- terpreting negative answers, take the following problem. -^ 383. — 5. A bought a certain number of apples for 80 cents; had he bought 4 more for the same sum, each apple would have cost 1 cent less. How many apples did he buy? 80 Let X denote the number he bought, then — is the price X of each; if he had bought x -J- 4, the price of each would PROBLEMS II^^VOLVIJn'G QUADRATICS. 217 have been — -— . . * . — —7 = 1, whence x = IQ or ic + 4 X -\- 4: X -20. If A bought — 20 apples, then he must have sold 20 apples, hence the answer, 20, applies to the following problem : 6. A sold a certain number of apples for 80 cents; had he sold 4 fewer for the same sum, each apple would have sold for 1 cent more. How many apples did he sell? Let X denote the number he sold: then, reasoning as in 80 80 Problem 5, we have — -— = 1- 1, whence a; = 20 and X -\- 4 X — 16; thus the number 20, which appeared with a negative sign as a result in Example 5 and was in that case rejected, is here the admissible result. Exercise XXI. 1. Find two numbers whose sum is 10 and product 21. 2. What number added to its square gives 132 ? 3. Find two numbers whose sum is 16, and the sum of whose squares is 146. 4. A sold a horse for $24, and by so doing he lost as much per cent as the horse cost. Required the price of the horse. 5. A bought a certain number of oxen for $240, and after losing 3 sold the remainder for $8 apiece more than they cost him, thus gaining $59. How many did he buy ? 6. Divide a into two parts such that their product shall be equal to the difference of their squares. 7. A field which is 6 rods longer and 2 rods narrower than a certain square field, contains 128 square rods. How long is each field ? 8. A banker had two kinds of money with which he paid a bill of $40, giving 60 of the more valuable coins and 100 Alg.-19. 218 ALGEBRA. of the less valuable. It required 8 more of the latter than of the former to make one dollar. What was the value of each coin? 9. A receives from a banker 60 francs and 140 marks for $47, and it requires one more of the former than of the latter to make one dollar. What is the value of each coin? 10. A sold 16 pounds of mace and 20 pounds of cloves for $65. He sold 12 pounds more of cloves for $20 than he did of mace for $10. What was the price of a pound of cloves? 11. A and B had jointly $1000 in business. A's money was in trade 9 months, and B's 6 months. When they shared stock and gain, A received $1140 and B $640. What was each man's investment ? --. A's gain for n months _ A's capital B's gain for n months "" B's capital" 12. A cistern has two pipes, one of which will fill it in 4 hours less than the other, and both together can fill it in 6f hours. How long will it take each separately to fill it? 13. A pasture was rented at a fixed weekly price by A and B, who agreed to share the rent in proportion to the number of animals put to pasture. The first week A put in 4 animals and B paid $20. The next week B put in 2 additional animals and paid $4 more. At what price was the pasture hired? 14. Find a number of two places, such that the units' digit, increased by 2, shall equal three times the tens' digit, and the square of the units' digit is 12 more than the re- quired number. 15. Find the number of two places whose units' digit is 3 more than the tens', and 125 times the units' digit equals the square of the required number. 16. The square of the time past noon, increased by the PROBLEMS INVOLVING QUADRATICS. 219 time to midnight, equals the time past midnight. What o'clock is it ? 17. The hands of a clock are together between two con- secutive numbers, the difference of whose squares is 1 more than the square of the less. What o'clock is it? 18. A traveled 48 miles in a certain time. Had he traveled 2 miles an hour faster, his time would have been 4 hours less. Find his rate. 19. A set out from C toward D at the same instant that \ B left D for C. They met 18 miles nearer to D than to C. • A arrived in D in 8 hours, and B in D in 18 hours, after they met. Find the distance from to D. 20. A set out from C, and B from D, distant 60 miles, at the same instant. A arrived in D 2 hours, and B in C 8 hours, after they met. How far from D did they meet? 21. A rowed 20 miles and back in 14 hours, going out with a tide of 2 miles an hour, and back against a tide of 1 mile per hour. How long would it have taken him to row the stime distance in still water? 22. A steamer went 40 miles down a river and back in 16 hours, the current being 3 miles per hour. Owing to an accident to her machinery, she could steam back with only half the power she had when going down. How long did she take to return? 23. The height of a room is two thirds of the breadth, and the breadth is two yards less than the length. The plastering at 50 cents per square yard cost $80. Find the dimensions of the room. 24. A loaned two sums of money, which differed by |200. His rate on each was one per cent of its principal, and the interest on both was $52. Find each principal. CHAPTER XXII. PROPERTIES OF QUADRATICS. MAXIMA AND MINIMA. 384. Feoh (362) it follows that every quadratic equation i h^ c h^ can be reduced to the form ic^ H — x-\- -—7, = — 1 -„. a ^a^ a ^a^ 1) c c For convenience, let — = 2», and - = q • also, let V- a ^ a ^ a —2 = m^. Then the general quadratic reduces to the form a? -f ^p^ ~\- P^ ^^ ^ -\- P^' .'. (x-\- pY = q -\-p^ = m^ ; . • . {x -{- pY — m^ = 0, and, from (199) and (200), {x -\- p -\- i})){x -{-p -m) = 0. 385. Every quadratic equation has two roots, and 07ily two. For, {x -\- p -\- m){x -\- p — m) = can be satisfied by- putting each factor equal to zero, and it can be satisfied in no other way. From x -\- p -[- 7)1 — ^, x^ — p — m ; from X -\-p — ill =2 0, x = — 2) -{- ni. Hence every quadratic has two roots, and only two. 386. In every quadratic equation, reduced to the form o? + ^P^ = g, the sum of the roots is equal to the co-efficient of the second term with its sign changed, and the product of the roots is equal to the knoiun term loith its sign changed. For the roots are — p — 7n and — p -\- 7n, whose sum is — 2p, and product p^ — m^ = — q. 387. If the two roots be represented by r and s, then it follows from (386) that a;^ — (r -j- .v) x = — rs, (220) PROPERTIES OF QUADRATICS. 221 . . • . ar' — (/• + ^^)-^ -^ rs = 0; .' . {x — r)(x — 5) = 0, a result which agrees with that obtained in (384). Whence: b G Every quadratic reduced to the form x^ -\ — x = can he decomposed into two linomial factors, the first term of each ieing x, and the second terms the tivo roots with their signs changed. 388. An expression of the form ax^ -\- bx -\- c is called a quadratic expression. A method of factoring such expres- sions is given in (205), but it is evident from (387) that the factors are readily found from placing ^a^ -\-bx -\- c=^0, and finding tlie roots of the resulting quadratic. If these roots be represented by r and s, the required factors will be a(x — r)(x — s). If r and s be not rational, the proposed expression is prime. 389. If both sides of an equation be divisible by a commo7i factor involving x, the equation is satisfied by putting that factor equal to zero. For, it {x—a)k = {x—a)mf then {x—a)k—(x~a)ni= 0. . • . (x — a)(k — m) = 0. Whence x — a = 0, and x =a. In like manner, if k — m involve x, then h — m = 0. 390. If r, 5, and t represent three values of Xy then: I. x-r = 0-, II. X — s = ()\ III. x-t = 0. .'. {x - r){x - s){x - t) =0. This is a cubic equation, as may be seen by performing the indicated multiplication. Hence we may infer that a cubic equation has three roots, and only three. In like manner, the number of roots of any equation is equal to the degree of the equation. Similarly, it may be inferred that if r is a root of an equation whose second side is zero, then a: — r is a factor of the first side. 391. From (389) and (390) it follows that quadratics and many equations of higher degree may be solved by factoring. 222 ALGEBRA. EXAMPLES. 1. Solve a;* + 2a;« - ic = 2. By transposing, x^ + 2:c^ — a; — 2 = 0. .-. 2:3(:c+2)-(a:+2) = 0, and (2:+2)(;r-l)(^+a;+l)=0. By putting each factor in succession equal to 0, we find :c=: -2, 1, -J(-l ± |/^=^). The equation is of the fourth degree; it has four roots. 2. Solve 3a;3 + 5a;2 - a; + 2 = 0. 32;^ + 6a;2 - a;2 - 2rc + ic + 2 = 0; .-. (:?; + 2)(32;2-.'r + l) = 0. Since a; + 2 = 0, .-.0;= — 2. Since Sx^ — cc + 1 = 0, .-. ii^ = i(l ± V- 11). 3. If a:^ - 8^2 + 192; = 12, then {x - 3)(:c-l)(a:-4)=0. . • . a; = 3, 1, or 4. 4. li^x" — :i^ — x — ^, then :r(22: — l){3a; + 1) = 0. .*. ^=0,i, -1 392. From the preced ng articles it is clear that any equation may be formed if its roots be known. Thus: If the roots be a' and r, the equation is (x — 8){x — ?•) = 0; that is, x? — {s -\- r)x = — sr. If the roots be s, r, and t, the equation is {x — s)(x — r){x — t) =0', and so on. EXAMPLES. 1. The equation whose roots are 5 and — 2 is a;2 _ (5 _ 2).r = - (5) (—2); that is, x^ - ?,x = 10. 2. If the roots are — 1 and |, the equation is ^' - (- 1 + f)^ = - (- 1)1, ovx^-{-ix = ^. 3. If the roots are 3 ± Vb, the equation is a^— 6x = —4, because the sum of 3 + '/S and 3 — 1^5 is 6, and their product is (9 — 5) = 4. 4. Form the equation having twojwts, whose sum is s, and their product jt?. i jt7is, '^ — sx = --^^ PROPERTIES OF QUADRATICS. 223 5. If the roots are 2, 1, and — 3, the equation is (x - 2)(x - l){x + 3) = 0; that \s, a^ - 7x -^ Q =0. 6. The equation whose roots are 0, 2,-3, — 4, is (x-0)(x-2)(x-{-^){x+^)=0; thatis, x'-\-b3^-2x^-24:X=0. 393. From (384) it follows that every quadratic equation may be written in the form x^ -j- 2px = q, in which p and q may represent any numbers, positive or negative, integral or fractional. Let r represent the first value of x, and s the second; then we have : (1) a? + 2px = + Q^ whence \ s = -p-\-Vp'-\-q, =.-p-Vp'^q, (2) x^-2px = + Q^ whence \ s = -\-p-Vp^ + q. (3) x> + 2px = -Q> whence \ = — p — Vp"^ — q. (t) x^ — 2px = -Q> whence •! = j^p^Vp^-q, = -^p-Vp'-q, Since p"^ -\- q > } i\ then i^/ 4- g > P- Since ;>^ — j/, tlieu in the third and fourth forms r and s both become impossible quantities. MAXIMA AN"D MIN^IMA. 394. It is often useful to determine the greatest or least values which a quadratic expr(?ssion can have. A quadratic expressio?i is of the general form ax^-^dx-\-c, in which a, h, and c are fixed or constant quantities, and x may have any value we please to assign to it (388). 395. To determine the maximum or minimum value of a quadratic expression, put it equal to ni, and solve this equa- tion ; by this means x will be expressed in terms of 7n, and it will be easy to see what will be the greatest or least values allowable for m, so that x shall be a possible quantity. EXAMPLES. Find the maximum or minimum value of the following: 1. Of 3a;2 - ?,x + 2. Zx^-^x^2 = m; .'.x = \{^± Vl2m - 15). In order for x to represent a possible number, the ex- pression 12in — 15 can not be negative ; hence the least value of 12;;? — 15 is zero, in which case m — f. Hence the minimum value is |. 2. Of x^-^x^ 16. x^-^x^\a;" -^ b^ = b^ -\- 4ac. IV. Extract square root, divide Dt. -4. x'- x" - X = 2ax'' + b = x^ = - -5^^+(fr - f = ± f . = ± 2 or ± ± Vb^-\-^ac. V. Transpose and -b±Vb^^ 4:ac 2a VI. Extract nth. ro( 2. Solve x^ — 5:^2 = Add ay. Extract square root. whence (236) - b ± Vb"" + Uc 2a = ^/ - 4 = f . .-. a;2 = 4 or 1; 1. EQUATIONS WHICH MAY BE SOLVED LIKE QUADKATICS. 227 398.-3. Solve x -}- 6 Vx = 16, II. Add 9. X -\-6 Vx-i- 9 = 25. III. Extract square root, x^ -\- 3 = ± 5. . • . rr^ = 2 or — 8, and a; = 4 or 64. 399.-4. Solve x-^ - 4:X-^ = 21. II. Add 4. x-^ — 4:x-^ -f 4 = 25. III. Extract square root. x~^ — 2 = ± 5. . *. a:"^ = 7or 3; whence a;~* = 49 or 9, and x = ^ ot ^. 400.— 5. Solve x^ -dx^ = —8. II. Add (1)1 a^-9a^^ (|)2 = »^t. -8 = -^-. III. Extract square root, ic* — f = ± }• IV. Transpose. ar»= f±J = 8orl. Since a:^ = 8, . • . .t' - 8 = 0. .'.{x- 2)(x'-{-2x + 4)= 0. Therefore (Art. 389) x —2 = 0, whence a: = 2; and x^-\-2x -f- 4 = 0, whence x = —1 ± V— 3. In like manner, since a;^— 1=0, . *. (.t— l)(a;*+ic+l)=0. If 2;-l = 0, a: = l;if a;2 + a; + l = 0, x = i(-l± V"^). 401. — 6. Find six sixth roots of 64. Let X represent a sixth root of 64. . • . a:' = 64, whence a;' — 64 = 0. .-. {x - 2){x^ + 2x-i- 4)(a: + 2)(x'' - 2x + 4) = 0. Ans, x = 2, - 1 ± t^^, - 2, 1 ± V^'Z, 402.— 7. Solve X + Vx-^2 = 10. II. Add 2. (a; + 2) + V^^ = 12. This is plainly in the form of a quadratic, because if Vx -\-2 — y, then ic + 2 = y\ III. Add J. (a; + 2) + l^a; + 2 + J = 12J. IV. Extract square root. Vx-\-2 -f J = ± 3J. V. Transpose. I^a; + 2 = 3 or — 4. VI. Square. a; + 2 = 9 or 16. . • . a; = 7 or 14. 228 ALGEBRA. 8. Solve 2x + i^lx + 8 = 1 II. Multiply by 2. 4^; + 2 VIx~^8 = 7. III. Add 8. (4:x + 8) + 2 ^4a; + 8 = 15. Add 1 and continue as in Ex. 7. Ans. a; = :J or 4 J. 9. Solve ^x^ + 15r?; - 2 Vx^ + 5.^ + 1 = 2. II. Divide by 3. a;^ + 5:r - | VaJ^- 5a; + 1 = |. III. Add 1. {x^ + 5a; + 1) - I i/^^qrs^TjTl = 5, IV. Add (4)1 (^2 4- 5a; + 1) - f 1^^ + 5a; + 1 + i = -V- V. Extract square root. Va^ + 5a; + l — ^ = ±1^. A71S, x = ^, — J/, 0, — 5. 403. An equation which will remain unaltered when — X is substituted for x, is called a reciprocal equation. Ex- ample : x^ — x^ — o(^ — X -{-1=0. Every reciprocal equation of odd degree is divisible by a; — 1 or cc + 1, according as the last term is negative or positive. Every reciprocal equation of even degree with its last term negative, is divisible by a? — 1. In every case the reduced equation after the division will be reciprocal, of an even degree, and with its last term positive. By this means a reciprocal cubic may be reduced to a quadratic, and one of the fifth or sixth degree to a biquad- ratic, which latter may be solved as follows : 404.— 10. Solve x^-\-^ -4:X^-\-x-{-l = 0. II. Divide by a;l x" j^ x - 4: ^- ^\ = 0. X X III. Add 6. a;2 -I- 3 + 1 _|_ ^ 4_ 1 = 6. XJ^ X IV. Combine terms. [x -\ — ) + (^ H — 1=6. EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 229 V. Complete the square. Ix -| — ) + (a; H — J -f ;^ = 6^. VI. Extract square root, x -\ \- i = ± 2|^. X ,' . X -\ — = 2 or — 3, wliich is best solved by the Special X ' Method (374). Ans. a; = 1, 1, or i(- 3 ± Vb). 11. Solve x^-{-a^ -2x^-x-]-l = 0. Ans. x = l, - 1, K- 1 ± Vb). 405. Cubic equations with the second power of x missing may sometimes be solved as follow : 12. Solve 8a:» + I62; = 9. II. Multiply by 2x. 16a;* + 32x^ = 18x. III. Add (2xY. 16a;* + 36a:* = ^a^ + 18a;. IV. Add (if. 16a;* -{- 36a;* + (|)2zzz4a;2+18a;+(|)l V. Extract root. 4a;2 + f = ± (2.^ + |). From 4ar^ + I = + (2a; + f ), a; = 0, or f From 4a;2 + f = - (2a; + f ), a; = i(- 1 ± 4^^^^^). The value of a; = satisfies the equation after being mul- tiplied by 2a;, but is not a root of the given equation. 406. Rule for solving cubics of the form ax^ -\- bx = c. (1) Multiply each side hy sUch a mimber that tlie first term shall be a perfect square. Thus: a^x^ -j- bx^ = ex, (2) Add to each side x^ with a co-efficient winch is the square of a divisor of ac, a7id such that the number lohich completes the square on one side also completes the square on the other side. (3) Extract the square root of each side {prpfix ± to the square root of the second side), and solve the resulting quadratics. 13. Solve a;^ — 3a; = 2. II. Multiply by a;. a;* - Zx" = 2x, III. Add (Ixf. x*^-27? = x?-\- 2x. 230 ALGEBRA. IV. Complete the square, x^ — 2x^ -\- 1 = x^ -|- 2:r -f- 1. V. Extract square root. a;^ — 1 = ± (:c -j- !)• Ans. X = 2, — 1, — 1. No. Ill may also be formed by adding {^xf = 4a;^ 14. Solve .T^ — 6a; = 9. II. Multiply by x. x'' - Qx? = ^x. in. Add (Zxy. x"^ + Zx^ = 9a:2 _|_ 9^^ IV. Add (1)1 2;* + 3a;2 + {^f = ^x'J^^x-{- {ff. Ans. :c = 3, i(- 3 ± V^^). 15. Solve a? - nx = 12. II. Multiply by x. x^ — Ux^ — 12x. III. Add {Ixf. a;* - 12^;^ == x^ + 12a:. IV. Add 36. a;* - 12a;2 + 36 = :t? + 12a; + 36. Whence a; = 4, — 3, — 1. Ill may be formed by adding either (3a;)^ or (4a;)^ 407. Certain biquadratics, with a? missing, may be solved in a similar manner. Thus: 16. Solve a;* - 6:^^ - 8a; - 3 = 0. 11. Transpose, x^ — Qx^ = 8a; -|- 3. III. Add (2xy. x^-2x^ = 4a;^ + 8a; + 3. IV. Add 1. a;* - 2a;2 + 1 = Ax^ + 8a; + 4. Whence a; = 3, — 1, — 1, — 1. 408.— 17. Solve a;* - lOa;^ + 35a;^ - 50a; = - 24. II. Complete the square of the first two terms by writing 25a;^ as the third term, and the equation becomes a;* - 10a;3 + 25a;2 + lOa;^ - 50x = - 24. III. . • . (a;* - lOa;^ + 25a:2) + 10(a;2 - 5x) = - 24. IV. Add 25. {a^ - bxf + 10(a;2 - 5a;) + 25 = 1. V. Extract square root. (x^ — 5a;) -|- 5 = ± 1. Solve a;^ — 5a; 4- 5 = + 1, whence a; = 4 or 1. Solve a;^ — 5a; + 5 = — 1, whence a; = 3 or 2. 18. Solve a;* - 2a;^ + a; = 132. II. a;* - 2a;» + ar^ - a;2 + a; = 132. III. Combine, {a^ - xf - {x" - x) = 132. EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 231 IV: Add i. 0^2 - xY -{x^-x)-\-\ = 1321 V. Extract root. {t? — x) - ^ = ± 11 1. Ans. a; = 4 or - 3, or ^(1 ± V- 43). 19. Solve 4a;* — 4:X^ + bx^ — 2:^; = 8 in a similar manner. Ans. x = l{l± Vll) or i(l ± V- 31). The solutions of Examples 17, 18, 19, illustrate the method of solving biquadratics, such that if the square of the first two terms be completed, the fourth and fifth terms will contain the square root of the first three terms. 409.— 20. Solve x' - lOo;^ + 120:?; = 144. Transpose 120^ and complete the square of the first two terms. .-. x^- lOa:^ + 2bx' = 26x^ - 120:c + 144. Both sides being perfect squares, extract the square roots, x^ — 5x = ± {5x — 12). It x'-6x= -{-(5x- 12), x = 5 ± VU. It x'-bx= - (5rc - 12), x= ±2 VZ, 21. Solve x*" — 10a:'-}- 9^^+ 40a; = 25 in a similar manner. A71S, a; = i(9 ± Voi) or |(1 ± ^21). Equations (20) and (21) are examples of biquadratics such that if the first and second terms be on one side and the remaining terms on the other, the number which com- pletes the square on one side will complete the square on the other side also. 410.— 22. Solve V¥T~8 + V^^~^^S = Vvf -\- 1. II. Take tlie identical equation (a^ -|- 8) — (x^ — 8) = 16 = 17 — 1, and divide this equation by the given equation. The given equation is of the form a -\- b = m -]- n, and the assumed identity is of the form a^ — b^ = m^ — n\ . • . III. V^~^S - V^^^^S = Vl7-l. IV. Subtract III from I. 2 Vx^-8 = 2. . • . a; = ± 3. This equation is introduced for the sake of illustrating the artifice employed in the solution. Similar methods may often be employed with advantage. 232 ALGEBRA. 411. The principles explained in Art. 284 may frequently be used to abbreviate the solutions of equations. Thus: x^ 4- Qx" + 1 _ 81.g* + 16 23. bolve ^^^^^ + 1) - 8Lt* ^"16' 11. From Art. 284, Case VII, III. a;* - 4.^3 + Qx"" _ 4a; + 1 32 (x + 1)* _ Slx"^ \x-lf~ 16 • IV. Extract the fourth root. ^ = ± -^-. a; — 1 2 When £±| = + |:, ^ = 3or-i. When |±1 = _ 1*^, ^ = j(l ± V323). 412. When none of the methods heretofore explained will apply, if the equation have rational roots they may be found by factoring, as follows: Transpose all the term.s to one side; resolve this expression into its prime factors, and place each factor equal to zero. The values found from solving these equations are the required roots (391). 413. Conversely: In order to find the factors of an ex- pression, place it equal to zero, and solve the resulting equation. Thus: The factors of a;* — 2x^ + re — 132 may be found as in Ex. 18. One root being 4, . • . a; — 4 is one factor; another root being — 3, .*. a; + 3 is a factor; the other two roots being irrational, the quadratic, x^— x -f- 11, of which they are the roots, is prime. Hence the required factors are {x — 4)(a; + 3)(a;^ — ^ + H). In like manner, 4a;* — 4x^ -\- ^x^ — 2x — S may be fac- tored as in Ex. 19. The four roots being irrational, the two quadratics, 2z^ — a; — 2 and 2x^ — x -\- 4, are the re- quired prime factors. EQUATIONS WHICH MAY BE SOLVED LIKE QUADRATICS. 233 Exercise XXIII. Solve like Example 2, Art. 397 : 1. X* - 6a^ = - 8. 3. x-^ - 5x^ = - 4. 2. a;* - (rt + Ij)x^ = - ab. 4. x^ - 33a;t = - 32. Solve like Example 3, Art. 398 : 5. X — 10^^ = — 16. 7. x^ — 2x^ = 3. 6. x^ — 3x^ = —2. 8. 32- + 2 1^ = 1. Solve like Example 4, Art. 399 : 9. x-i + 5x-^ = 22. 11. a;-2 - 35x-^ = - 216. 8 3 -1 _^ 10. 3X~T^ — iX 4 = 7. 12. X n — X n =z — 2. Solve like Example 5, Art. 400 : 13. xf' -7x^ = 8, 15. x^ - 7x^ = - 10. 14. ar' + Ux^ = 1107. 16. a^^x' = 2. Solve like Examples 7, 8, and 9, Art. 402 : 17. X - Vx-{-5 = 1. 19. x-\-3 Vbx = 20. 18. X + V5x 4- 10 = 8. 20. 2x + V2a;+ 3 = 9. 21. a;2 + 3 = 2 ^^-2x-\-2 -f 2a:. 22. a.-2 + 5a; + 4 = 5 Va^ -\- bx + 28. 23. Vr^ — 2x -f 9 - - = 3 - a:. 24. a;^ - a; + 3 V2x' - 3a: + 2 = - + 7^ Solve like Example 10, Art. 404 : 25. ar + a: + ^"^ + ^~^ = 10. 26. a;2 + 3a- - 3x-^ + a.--^ = 6. 28. x' - x^ - x-'' + a:-* = 18. Alg.— 20. 234 , ALGEBRA. Solve like Examples 13, 14, and 15, Art. 406: 29. 4^^ - 39^; = - ^5. 31. 9a^-4x-^ = U, 30. x« — 32^ = 18. 32. X — x-i = 2. Solve like Example 16, Art. 407 : 33. x'-{-^x'-2x + 8 = 0. 35. 4^;* - 5^x^ -{- 8x = 15. 34. x!^-^3a^-^6x = 5. 36. x^ — 5x + 6^:^ + 3 = 0. Solve like Examples 17 and 18, Art. 408 : 37. x' - 4x^ + 8x^ — Sx = 5. 38. a;* + 8.c3 + 130;^ - 12^: = 10. 39. a^ - lO.'c^ + 35./;* - SOa;^ ^^ 24. 40. 4.y* - 4^;^ - 52;^ + 3:^ = 4. Solve like Example 20, Art. 409 : 41. 2;* - 6x^ 4- Sx^ -8x = 16. 42. 3^ - 4:^3 _ 5^2 _|_ 6^^ 3:, 1^ 43. 4:c* - 82;^ + 3:^2 _ g^, ^^ iq^ 4.4.. ^x' - U:^ - 117cc = 243. Solve like Example 22, Art. 410 : 45. Vx^' + 21 - ^x^ + 5" = 2. 4^ + 2 V3rta; — X = 3a Vl — Ax. Solve like Example 23, Art. 411 : 49. — ^ , = -. 51. ■ -zz^=:^ = 2. 50. p^ ; = ,7. 52. - -^ 7-=^— = ^• «a; + 1 - |/^2^2 _ 1 2 |/7^2_|_4_2 4/3^._1 CHAPTEE XXIV. SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 414. Quadratic equations involving two unknown quantities require different methods for their solutions, according to the form of the proposed equations; hence no general rule can be given. There are, however, three cases of frequent occurrence, for which the following observa- tions will be useful: 415. — Case I. When one of the unknown quantities can be eliminated, and the resulting equation solved according to preceding methods. This can usually be done when one of the equations is of the first degree, or is capable of being changed into an equation of the first degree, and it can frequently be done in other classes of problems. Elimination by substitution is the method commonly employed (180). 416.— 1. Solve: I. x^ ^ y^ ^^ 25. II. 3a; + 4?/ = 24. III. From II, y = i(24 - 3x). IV. .'. I becomes a^-\- ^(24 — 32;)^=: 25, whence a; = 4 or ||. From III, when a; = 4, y = 3, and when a: = ||, ?/ = 4||. 417.— 2. Solve: I. x^ + xy -^ y^ = 7. IL 2x + 3?y = 8. IIL From 11, x = ^{8 - 3y), IV. ... I become. (^^J+ {^l) y + f = ,. V. Expand IV, and simplify, iy^— 8y = — 9. Whence ^ = 24^ or 2 ; and from III, x = \ or 1. (235) 236 ALGEBRA. 418.— 3. Solve: I. x -^ y z=. x\ II. 3?/ - a: = if, III. From \, y = :f? — X. IV. By substituting in II, 3(.t2 - x) -x=^{a? — xf. V. Transpose and simplify IV. x^ — 2x^ — 2x^ -\~ 4:X = 0. VI. By factoring V, x{x - 2)(x'' - 2) = 0. • • • cc = 0, 2, ± V2, From III, when x = 0, y = 0; when x = 2, y = 2; when xz^ ± V2, y = 2 ^ V2. 419.— 4. Solve: I. x^-\-y''-x-y = G — 2xy. II. x^ + 3/ = 19. The first equation is of the form (x + yy — (x + y) = G. .-. (•c + yY-(x^y)-\-^ = C)i, whencex-{-y = 3or -2, and x=^ ~ y or ~2 — y. In II for a: put 3 - ?/. . • . HI. (3 - yf + 3?/^ = 19. . • . 4?/2 — 6?/ = 10, whence y = 2^ or — 1. Shice .-?; = 3 - //, . • . a^ = 3 - 2^ = i or 3 - (- 1) = 4. Similarly, in II substitute the value ~2—y for x, and solve. Ans. x = i, 4, -31 'i; y = 21, - 1, 11, _ 2i. 420,-5. Given I. a^y^ j^ xy = 10. To find x and z/. II. 2x + 3?/ = 7. By solving I according to Art. 406, III. xy z=2 or — I ± V— 4. From II, X = 4(7 - 3?/). Substitute in III. yl ~ ^] = 2 or — 1 ± V^^^, .'.3y'-7y= -4. or 2T-4 V^^, whence ?/ = f , 1, or -^(7 ± VrS^lsV^), and a: = f, 2, or = J(7 T \^73 q= 48 4/^1). The impossible values of x and y (318) are usually lejected. SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 237 421. — Case II. When the two equations are symmetrical with respect to x and y, put « + 5 for x, and a — h for y. An expression is said to be symmetrical with respect to a: and y when these quantities are similarly involved in it. ll\m^'.'x^-x^y-xy^^y^; 2a;^+3a; + 3!/- 5; x^^y^-x-y. In like manner, a:'+ x^y— xy^— y^, — 2xy-^ 3x — 3y — l, 22_[_^2_ ^ _j_ y^ ^YQ symmetrical with respect to x and — y. Also, 4:X^ + 9y^ — 2x — 3y is symmetrical with respect to 2x and 3y. 422.-6. Solve: I. r' + y-« = 27^. Let a; = o^ + ^. II. x + y~^ = 3J. Let y-^ = a — i. III. Substitute in I. {a + J)» + (a - hf = 27^. .-. 2a^ -{- 6ab^ = 27^, IV. Substitute in II. (a + b) -j- {a - b) = Si. .-.« = }. V. Substitute ^ for a in III. . • . 2^)' + HiW = ^"i- VI. Simplify V. . ' . W + ¥^ = 27^, whence ^^ = ± f . a: =a-\-b = i±i = 3ori. y~^ = a — b = -} ^ i = i or 3. .*. y = 2 or |. 423.-7. Solve: 1. x^ -\- y^ = 10. Substitute for x, a-\-b, II. ar* -f- y'' = 28, and for y, a — b. III. (a + by + (« - ^')^ = 10, . • . 2«« + 2^*^ = lb. IV. (a + Z')^ -f (-^ - by = 28, . • . 2a^ + 6rt^>2 = 28. V. Multiply III by 3a. . • . 6(i^ -\- 6ab^ = 30rt. VI. From V take IV. . • . 4«« = dOa - 28. VII. Multiply VI by a. 4«* - 30^?^ = - 28a. VIII. Add (4:ay. 4:a^ - 14.0^ = 16rt« - 28«. IX. Add (J)^ 4.a' - 14^^ + (|)'^ = 16«^- 28rt + (J)l X. Extract square root. 2a^ — ^— ± (4a — |), whence fl^ = 2 or — 1 ± f i^. From III, when a = 2, ^ = ± 1. . • . a: = 3 or 1. When az=z -l±^V2, b= ± V"- i ± 3 V^. ^W5. a: = - 1 ± I 1^2 ± V-i ± 3 f2, 3, 1; y = - 1 ± I 4^ If /-I- ±3 V^, 1, 3. 238 ALGEBRA. 424.-8. Solve: I. {x + y)(x - yf = 160. 11. {x-^ry){x' + f)=^SO, Let X — a -{-!), and y ■= a — I. Then x -{- y = '^a, x - y = 2b, x'^y^= W + W. in. Substitute in I. 2a{%hY = 160. . • . «Z»« = 20. IV. Substitute in II. 2a{2a''+ 2¥)= 580. r . a^-{- aW=l^b. Subtract III from IV. . • . a^ = 125, and « = 5. From III, h = ± 2; whence x = 1 or ?,, y = ?> or 7. Having a^ = 125, three values of a may be found (401). 425. Many equations which are not symmetrical with respect to x and y, may be solved in like manner. 9. Solve: I. a; — ^ = 1. Substitute a -]- b for x, II. a^-\-y^= 35. and a — b for y. III. I becomes 2^* = 1, whence b = ^, IV. II becomes 2a^ + 6ab^ = 35. V. In IV put iforb. . • . 2a^ + |« = 35. Solve V as explained in Art. 406. r. a = 2i or i(- 5 ± V^^7), .', x = 3 or J(- 3 ± V^^7). y = 2 or i(- 7 ± V^^^87). 426. — Case III. When the equations are homogeneous with respect to x and ?/, and of the second degree, substitute tx for y, and divide one equation by the other. ^ 10. Solve : 1. 0^ + xy = 15. . • . III. x^ + tx^ = 15. / II. xy -y^ = 2. . • . IV. tx" - fx'^ 2. V. Divide III by IV. ^-J = ~, ^ t — f 2 From Chapter XX. t = % or \. In III put I for L a;^ + frc^ = 15. .-. x= ±d. y=lx. .'. y= ±2. In III put i for if. . • . 1:^2 = 15. r.x= ±^V2, y = lx=±^V2, SIMULTANEOUS EQUATION'S INVOLVING QUADRATICS. 239 427. Equations of this class may also be solved by elim- inating the known quantity, then finding the value of one of the unknown quantities in terms of the other, and sub- stituting this value in one of the given equations. Thus, to solve Ex. 10, multiply I by 2, and II by 15. III = I X 2. 2a^ + 2x1/ = 30. IV = II X 15. 15x1/ - 15/= 30. Subtract IV from III. 2x^ - Uxy + 15/ = 0. 16.^2 _ ( ) _^ 169^2 ^ _ 120/ -}- 169/ = 49/. 4iX — 13// = ±ly. .' . X = by ov fy. Substitute in II, and solve the resulting quadratics. 428. It frequently happens that a new equation, simpler in form than the two given equations, can be formed by combining them. Especially is this the case when one equation is divisible by the other. Three examples are ap- pended illustrating this artifice, but skill in the use of such methods must be acquired by experience. 429.— 11. Solve: I. a^ - rry + / = 7. II. x" -^ x'y^ -\- y' = 133. / III = II -^ I. x^-{- xy + / = 19. IV = III - I. 2xy = 12. . • . xy = 6. V = III + (-^7/ = G). {x + yY = 25. .'.x-^y=±b. VI = l-(xy = Q). (x-yY = l. ,',x-y=±l. .-. a;= ± 3, ± 2. y = ± 2, ± 3, 430.— 12. Solve: I. x^ -{- y^ -\- x -}- y = IS. II. xy = 6. / III. Add twice the second equation to the first. . • . (x -\- y)'^ -{- {x -\- y) = 30, whence a; -f- y = 5 or — 6. Now take x -}- y = 5 or — 6, and xy = 6. Solving this pair of equations (421), we find a; = 3, 2, - 3 ± VS; ?/ = 2, 3, - 3 :f V^. 240 ALGEBRA. 431.— 13. Solve: I. x" + xij - (jif = 21. II. x^— 2y^ = 4. 111 = 1^11. Mi^2/^21 y 4 IV. From III, x = ^y. In II substitute for x its value, f ?/. . • . \y^ = 4, whence y =z ± 4:y and, from IV, x — ±^. 432. Simultaneous equations involving three or more un- known quantities usually require special methods for their solutions. In addition to the artifices heretofore explained, two others are submitted for the consideration of advanced students. 433.— 14. Solve: I, x -{- y -\- z = 2. II. xy -\- xz -\- yz = — 5. III. xyz = - 6. Assume a new unknown quantity, v, whose values shall be X, y, and z. .'.TV. V — x = ^. Y. V — y — 0. Yl. v — z = 0. VII. Multiplying IV, V, and' VI together, v^ — (x -\- y -\- z)v^ + (xy -\- xz -\- yz)v — xyz = 0. But X -]- y -\- z = 2; xy -{- xz -\- yz = — 5; xyz = — 6. Substituting in VII, . • . VIII. v^ - 2v^ - bv + 6 = 0. .-. IX. (v-l){v-^)(v-\-2)=0. .'. v = l, 3, - 2. The given equations being symmetrical with respect to ic, y, and z, it is evident that each unknown quantity may have any of the values of v. When :?; = 1, ?/ z= 3 or — 2, and ^ = — 2 or 3, etc. 434.-15. Solve: I. -^ = 2. .-. IV. ^±i^ = i X -j- y xyz 2 ^ xyz _18 ^ ^J ^ -\- ^ _ '^ ' X -\- z~ 1' ' ' ' xyz ~ 18' III. ^-^^ =.- • VI '^'^^ = - ' y -i- z 2' ' ' ' xyz 9 * SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 241 x-\-y^z _ 5 VII = (IV+V+VI)-f-2. xyz From VII take IV, V, and VI, respectively. .-. VIII. -=^.. IX. - = \. X. 1 = 1 xy 18 xz Q yz ^ Divide each of these equations by the product of the other two. ±6, 2/ = ± 3, ±1. Solve: Exercise XXIV 1. 3^-2 + 5/ = 47, 2a;2 - Zy^ = 6. 2. x-y = 3, ^x" - bif = 27. 3. x^ -{- xy = 15, 3xy + 4y^ = 34. 4. 2x 4- 3.y = 14, ic"^ + ?/-i = f. 6. x^ - 4!/^ = 9, xy-{- 2;/ = 18. X — y x -\- y 3 -^ 7. a: + .y = 9, a:' + ?/' = 189. 8. .'c — 2/ = 4, a:* — ?/* = 124. 9. X y 19, rc^- ?/i= 1. 10. a; + ?/ = 126, a;^4- y^ = 6. 11. a:4-,iy= 10, Vxy-''-{- Vx^/ = f. 12. 3:' + ?/- a; + ?/ = 2a:^ + 2, af*+ y^= 26. 13. rr + Vxy-\- y = 14, a:^ -f-^2-f- a;?/ = 84. 14. (a: + ?/)3 - a: — ?/ = 60, xy = 8, 15. a; + ?/ = 1, a^-tf = Q, 16. x-y = d, x'-{-y^ = 27. 17. a:" - «/' = 3, a:^ _^ ^^ ^ g. ^^^^ ^^^ ^^.^ ^^ values.) 18. a:^ + ?/2 = 17, a.-^ + y^ = 63; (find two pairs of values.) 19. x" + y' = 20, ic»;^,,2^1^=4g: (find two pairs of values.) Aig-21. /^v^^^^r^F^ 242 ' ALGEBRA. 20. X — y -\- xy{x — y) -{- x^y^ = 17, ^y + (-^ - yf + ^y{^ -y) = i3. 21. x^-i-y''^ 33, 5x-i-if=: 35. 22. x{x-\-y-}-z) = bO, y(x-\-y-{-z) = 30, z{x-^ y-\-z) = 20. 23. ic^ + ^^ + ^^ z= x^ -{- y^ -{- z^ = X -\- y -]- z = 1, 24. a: +2/ + ^ =-+-+- =^-, .T^. = l. 25. X -{-y -^ z =7, rr^ + / + ;22 = 21, a:?/^; = 8. 26. Find two numbers whose difference is 4 and whose product is 45. 27. The sum of two numbers multiplied by their product is 70, and their difference multiplied by their product is 30. Find the numbers. 28. The product of two required numbers is 12 times their difference, and the sum of their squares is 52. 29. If A had worked 5 days less and had received $4 less per day, he would have earned $9; if he had worked 4 days less and had daily earned $5 less, he would have earned |8. How much did he earn? 30. A number is equal to 3 times the product of its two digits, and is 18 less than its reverse. Find the number. 31. A went from to H, a distance of 24 miles, in 5 hours, riding half the way and walking the other half. Ee turning, he rode half the way at a rate 2 miles faster than when he went out, and walked the other half at a rate 2 miles slower than before. He reached in 7^ hours. Find his rates of riding and walkiug. 32. A and B gained by trading, |100. Half of A's in- vestment was less than B^s by |100, and A's gain was ^g- of B's investment. Find the investment and gain of each. 33. Find three numbers whose sum is 10, whose product is 18, and the sum of whose squares is 46. CHAPTER XXV. INEQUALITIES. 435. Two expressions containing the same letter will have their values changed when different values are assigned to that letter, and the expressions may be so related that one will be larger than, equal to, or less than the other, according to the values of the given letter. For example, when x = ± 1, t^ -{- 1 = a^ -{- x ; when X < — 1, a^ -\- 1 < of -\- X. For all other values of x, x'+l> a^-{-x. In other cases, however, the relations may be such that one of the two can not be greater than the other. Thus: 2x can not be greater than of -^ 1, whatever value be given to x. 436. The axioms employed in the consideration of in- equalities have been already given in Art. 62 (numbered from 8 to 13 inclusive). In using the axioms for multiplication and division (Axioms X and XI), it must be observed that if the multi- plier or divisor be negative, the inequality will be reversed. Thus: 6 beiug greater than 2, if both be multiplied by — 3, we shall have — 18 less than — 6. Similarly, if both be divided by — 2, we shall have — 3 less than — 1. 437. From Axioms VIII and IX it is evident that a term may he transposed from one side of an inequality to the other, provided its sign he changed. Thus: 16 -4 > 11. .-. 16 > 11-f 4. Also 16 + 4 > 11. .•.lG>ll-4. (243) 244 ALGEBRA. 438. From (436) it follows that the signs of all the terins of an inequalUy may he changed, provided the sign of in- equality bo reversed; that is, > changed to <, or < changed to >. Thus: 6 > 4, but — 6 < — 4. 439. Two inequalities are said to subsist in the same sense when both are read is greater than, or both ^9 less than. Thus : 6 > 4 and 5 > 2 subsist in the same sense. 440. Two inequalities are said to subsist in a contrary sense when one is read is greater than, and the other is less than. Thus: 6 > 4 and 2 < 5 subsist in a contrary sense. 441. Both members of an inequality may be raised to the same odd power without changing the sense. Thus: 4 > 2. . • . 64 > 8. Also, - 2 > - 3. . • . - 8 > - 27. 442. If both members of an inequality be raised to the same even poiver, the sense will be unchanged if both be positive, but the sense luill be changed if both be negative. Thus: 5 > 3. .-. 25 > 9. Also, - 5 < - 3, but (- 5)2 > (- 3)1 If one member be positive and the other negative, both members may be raised to the same even power without changing the sense, if the positive member be numerically the greater. If the negative member be numerically the greater, the sense will be changed. If the two members be numerically equal, an equation will result. Thus: 5 > - 3, then 25 > 9; 3 > - 5, but 9 < (- 5)^; 3 > - 3, but 9 = (- 3)1 443. For finding which of two given expressions is the greater, the following is a fundamental theorem : I. The sum of the squares of tivo unequal ^lumbers is greater than twice their product. Formula: If a — b is not zero, a^ -\- b^ > 2ab. For, {a — by must be positive, whatever the values of a and b (114). INEQUALITIES. 245 Since every positive number is greater than zero, . • . (a — by > 0. Expanding, a^ — 2ab -\- b^ > 0, Add 2ab to each member (Ax. VIII). . • . a"" -i- b"^ > 2ab. Ua = b, then a^ -\- b^ = 2ab. 444. The following are evidently special forms of the above general case, and result from substituting particular values for a and b in the formula a^ -\- P > 'Zab: II. fl"H-^»" > 2rt"^l 11 a A- b 11 III. a -^b > 2«J bi. .', ^-^ > a2b2. 1111 IV. fl52 + ^2 > 2ai bi. 445. If ((b be positive, both members of I, a^ -\-b^ > 2ab, may be divided by ab without changing the sense. .-.v. ^ + *>2. b a Whence: Tbe sum of any fraction and its reciprocal is grvafer than two, unless the value of the fraction is unity. 446. By adding 2ab to both sides of the inequality a^ -{-W > 2ab, we have : VI. {a + by > 4.ab. Whence: The square of the sitm of tioo tmeqnal numbers is greater than four times their 2)roduct. 447. In order to prove any proposed inequality, the fol- lowing method is recommended: In the first place, analyze the proposed statement as follows: (!) Assume the statement to be true. (2) Simplify both members as far as possible by adding equals to both, subtracting equals from both, etc.; the object being to reduce the proposed inequality to one whose truth is evident. To prove the truth of the proposed inequality, begin 246 ALGEBRA. the analysis ended, and proceed ly steps the reverse of those taken in the analysis. For example: 448.— VII. To show that a^-\-W> a% + ah\ Analysis : If a^ -{-¥ > a^h + alf, then by factoring each member we find {a -\- b)(a^ — ah -\- b^) > al?(a -{- b). Divide both members by the common factor a-\-b. .'. a^-ab-i-b^> ab. Subtract ab from each member. . • . a''-2ab-\-b^> 0. That is, {a — bf > 0, a known truth (443). By reversing this analysis, we show that the proposed inequality is true, as follows: Since {a - bf > 0, r . a" - "lab + Z'' > 0. Add ab to each member. . * . (Ax. VIII) a^—ab~\-b^>ab. Multiply each member by {a -\- b). . ' . (Ax. X) a^ -\- b^ > a^b + ab^, which proves the prop- osition. 449. JVo inequality involving letters is true under all circumstances. Thus: 6« > \a if a be any positive num- ber, but ^a < 4« if « be any negative number. In order to determine in what cases a proposed inequality is untrue, it is necessary to examine carefully every step of the proof. Take, for instance, the proof of formula VII (448). (1) {a — by > for all unequal values of a and b, (2) a^ — ab -\- P > ab for all unequal values of a and b, because adding equals to unequals never changes the sense. (3) To derive a^ + W from a^ — ab -{- b^, it is necessary to multiply by « + ^. If, therefore, a -\-b\iQ any positive number, the sense is unchanged (436); if a-\-b be equal to 0, an equation is formed; if « + ^ be negative, the sense is changed (436). Whence: I. d'^b^ = a% + aW, \ia=±b. II. i^ + ^^ < c^b + aV^y a a ■\-b be a negative number. In all other cases ci" -\-b^> a^b + ab^. INEQUALITIES. 247 450. If three or more letters occur symmetrically (421), the proof of the proposed inequality usually depends upon a corresponding inequality with two letters. Thus: VIII. To show that ci" + Z*' + ^'^ > al) + rtc + Ic. From I (443), I. (v" -\- ^ > 2ah. Similarly, II. a^ + '^^ > ^'''<"j and " III. ^'^ + 6-2 > 2bc. Adding I, II, and III (Ax. XII), IV. 2a^ + 2b^ + 26-2 > 2ab + 2ac + 2hc, Divide IV by 2 (Ax. XI). . • . a^' + b^-i- a" > (lb + ac -\- be, 461.— IX. To show that ^3 4- Z/^ -f (^ > l(a'b + ab^ + rt^c + ac^ -\- Pc + br). From VII (448), I. a^ + Z*' > ^^^ + «^'. Similarly, 11. a^ -{- c^ > a^c + ac^, and III. Z/^ + ^' > b^o + Z»6'l Adding I, II, and III (Ax. XII), IV. 2a^ + 2^' + 2c* > (i^b + aZ'^ + a=^c + ac^ + Z/^c + bo\ Divide IV by 2 (Ax. XI). ... a^j^V^c^> ^{a^b + ab^ + ^'^ + «^' + ^'^ + ^^'). Formulas VIII and IX are also true when one of the three fundamental inequalities becomes an equation, and the same remark may be made in regard to all formulas similarly derived. 452. If both members of two or more inequalities be positive, and if the inequalities subsist in the same sense, the product of the corresponding members will form an inequality in the same sense. If 6 > 5, and 9 > 8, then 6 X 9 > 5 X 8. 453.— X. To show that {a + b){a + c)(J + c) > Sabc. From formula III (444), I. a -\- b > 2aH\ Similarly, II. a -\- c> 2a^c^, and III. b -{-€> 2b^c\ 348 ALGE13RA. Multiply together I, II, and III. r. (a-}- h){a + c){h -\- c) > Saic, Exercise XX Y. Show that each of the following propositions is true for all unequal positive values of the various letters employed. Advanced students should also state the conditions under which the propositions are untrue. 1. a'c + hh > ?.ahc; u^h + c^h > ?aiIc; IM + chi > ^ahr, 2. a^' > 3(2rt — 3), unless rt = 3. a-\-h 2al) 4. a^h + al'' > 2(ru\ ^' 2 ^ a + 0' 5. a* + ^^ > 2a^\ 6. a^c + m + arl) + c^b + l^a + 6'^^ > Q>al)G. 7. r/.3 + Z/^ + t^ > B^rZ/c. 0. a^ y a'^ -\- a — ^, unless a — \. 9. a^ > 3 (a'- + 3rt — 9), unless a = 3. 10. a + b-{-c> 3aU^ci 11. {a + b-^ cf > 27rt5c. 12. r^* + Z/^ + a' + ^Z* > 2««Z'^ + 2chl\ 13. rt -f Z> + c + ^Z > 4:aH^cidk 14. (ri -r Z> + c + tZ)* > 4*aZ'CfZ. 15. 2(r^Z>+rt6'+Z>6-)>rt^+Z^2+6-^ if a+L>c, a+c>b, b^Oa, 16. a' -f Z/* 4- c* > rtZ'6-(« + Z* + c). . Z>-|-c^ + / a Z> / a a -\- c -\- e a ^ a e c 18. , ; , 7 , . < y , if y > - > -T. b -\-d -\- f b b t d 19. (« + Z* + •6')(a2 + Z'^ + c^) > 9«Z>c. CHAPTER XXVI. LOGARITHMS. 454. The Logarithm of a number is the exponent of the power to which ii fixed number, called the basey must be raised to be equal to the number : that is, if U = n, then I is called ilie logarillim of n to the base b. 455. The logarithm of n to the base b is written log^, n; thus, logft 11 — 1 expresses the same relation as b^ = n. For example, 4^ = 64 and log^ 64 = 3 may both be translated, 3 is the logarithm of 64 to the base 4. COMMON LOGARITHMS. 45,6. In practical calculations the only base that is used is 10; logarithms to the base 10 are called common loga- rithms. Thus, since 10- = 100, therefore the common log- arithm of 100 is 2, written thus, logig 100 = 2, or more briefly, log 100 = 2. 457. Since 10^ = 1, . • . the log of 1 is 0. Since 10^ = 10, . • . the log of 10 is 1. Since 10=^ = 1000, . • . the log of 1000 is 3. Since 10~^= yV = -1, .*. the log of .1 is — 1. Since 10-2= _^y =: .01, .-. the log of .01 is - 2. Since 10"=^ = y^W = -001, . *. the log of .001 is - 3, and so on. It is evident that the logarithm of any numbers between 1 and 10 is -|- a fraction ; 10 and 100 is 1 -f-; a fraction ; 100 and 1000 is 2 + a fraction; (249) 250 ALGEBRA. 1 and .1 is — 1 -j- a fraction; .1 and .01 is — 2 -|- a fraction; .01 and .001 is — 3 -f a fraction; etc. 458. The fractional part of a logarithm can not be ex- pressed exactly, either by common or decimal fractions, but the approxnnate value may be found, true to as many decimal places as may be desired. For instance: Let it be required to find the logarithm of 2; that is, to find X in the equation 10^ = 2. Since 10' = 1 and 10^ = 1 10, . •. a; > and < 1. Let a; = + y-^; then lO^^ =: 2, whence, by raising each side to the y power, 10 = 2^. Since 2^= 8 and 2*= 16, .-.?/> 3 and < 4. Let ?/ = 3 + 2;-'; then 2' + ^' =: 10, or 8X2^ = 10, or 2^ == |. . *. 2 = (|)^ Since (i)=^ = -Vr < 2, and since (f)* = ||f > 2, therefore ;2>3and<4. Let^ = 3+?;-^ .•.2 = (f)'+i .-.2 = By trial v is found to be greater than 9 and less than 10. If we take v = 9, then ;z == 3 + v"^ = 3 + 1 = 2/-; y =^ 3 + 2-1 = 3 -f /^ = If ; a; = ?/-! = || = .30107 + . If we take v = 10, then z = 3-\- v'-' = 3 + ^V = li; y = 3 + ^-1 = 3 + M = W; ^ = y-' = i%h = -30097 + . Now the true value of x must be between .30107 and .30097. By sliorter methods of Higher Mathematics, the logarithm of 2 is found to be .3010300, true to the seventh place. 459. The logarithm of a number consists of an integral part, called the characteristic, and a fractional part, called the mantissa. Thus, log 2 = 0.30103; here the character- istic is 0, and the mantissa is .30103: log 200 = 2.30103; here the characteristic is 2, and the mantissa is »30103. 460. TJie mantissa is always positive; hence a negative logarithm consists of a negative characteristic and a posi- LOGARITHMS. 251 five mantissa. Thus, the log ot -^% = — 1 -{- .3010300; the log of .03 = - 2 + .477121 (Art. 457). 461. When the characteristic is negative, it is usual to write the minus sign over the characteristic, and to omit the plus sign before the mantissa. Thus: log .2 = T. 301030 ; log. .03 = 2.477121. A negative logarithm may also be expressed by adding 10 to the charactei-istic, and indicating the subtraction of 10 from the resulting logarithm. Thus : log .2 = 0.301030 - 10; log .02 = 8.301030 - 10. 462. T/ie characteristic of a logarithm of an inte(/er, or of that of a mixed number, is one less than the number of integral digits. Thus, the chaiacteristic of the logarithm of 50 is 1, because 5() > 10 and < 100; hence the logarithm of 50 is between 1 and 2 (Art. 457). Similarly, the char- acteristic of the logarithm of 320.G24 is 2, and of 5476 J is 3. 463. The characteristic of a logarithm of a decimal frac- tion is negative, and is numerically one more than the num- ber of cipJiers between the decimal point and the left-hand significant figure of the decimal. Thus (457): The characteristic of the logarithm of any number between .1 and 1 is — 1; of any number between .01 and .1 is — 2; of any number between .001 and .01 is — 3; etc. 464. Rule for determining the characteristic of a loga- rithm of an integer, of v. mixed number, or of a decimal. If the left-hand significant figure be n places to tlie left of units* place, the characteristic of the logarithm will be + ^J if the left-hand significant figure be n places to the right of units' place, the characteristic of the loga7'ithm will be — n. The characteristic of the logarithm of 6435 is -\- 3, because the left-hand significant figure, 6, is three places to the left of units' place. The characteristic of the logarithm of 252 ALGEBRA. .00G5 is — 3, because the left-hand significant figure, 6, is the third place to the right of units. If the left-hand figure be in units' place, the characteristic will be 0. 465. Let n represent any numler, and let zv represent any whole mnnber, eitlter positive or negative; then will the logarithms of n aiid 10"' x n have equal mantissas. For, let c be the cliaracteristic and m the mantissa of the logarithm of a; then, from the definition of a common logarithm (45G), 10'' + '" = ??. Multiply both sides of this equation by 10"'; . •. (Art. 117) io^ + «' + »« =^ lO'^xn. Whence c -\-w -\- m is the logarithm of lO"' X n, in which (c -f w) is the characteristic and m is the mantissa. 466. From (465) it follows that changing the position of the decimal point in any number does not affect the mantissa of the logarithm. For, moving the decimal point to the right is equivalent to multiplying by 10.\ 10^, 10^, etc., according as the point is moved one place, two places, three places, etc. In like manner, moving the decimal point to the left is equivalent to multiplying by 10 ~^ = ^^, 10 ~^ ~ f Jo, 10 ~^ = YirVo' ^'^c., according as the point is moved one place, two places, three places, etc. Thus, the logarithm of 3570 being 3.552G68, then log 35.7 = 1.552G68, log .357 = 1.552668, log .'0357 = 3.552668, etc. 467. Log {ab) = log a -{- log h; i.e. : The logarithm of any product equals the s^imoftlie logarithms of tlie factors. For, if 10^ = a and 10^ = b, then 10^ + " = ab (Art. 117). But X = log a, y = log b, and x -\- y = \og (ab) (Art. 456). . • . log (ab) = log « + log /;. Similarly, log (abc) = \og a-\- log b + log cj etc. EXAMPLES. Given log 2 = 0.301, log 3 = 0.477, log 7 = 0.845, and log 10* = X (Art. 457); find the logarithms of the following numbers by adding the logarithms of the factors: LOGARITHMS. 253 1. 6. A?is. .778. 8. 24. Ans. 1.380. 2. 8. Alls. .903. 9. 27. Af/s.lAU. 3. 9. ^;^5. .954. 10. 3.2. A}is. 0.505. 4. 12. ^y^s. 1.079. 11. .30. Ans, 1.556. 5. 14. Ans. 1.14G. 12. .042. ^?^5. 2.623. 6. 16. ^^^5. 1.204. 13. 490. Ans. 2.690. 7.18. Jy^s. 1.255. 14. .0056. Ans. 3.74:8. 468. Log a^ =p log «; i.e.: The logarithm of any power of a number is obtained from 7nuUiplyi?ig the logarithm of the number by the exponent of the power. For, if 10^ = «, tlien (10^)^ = a". . ' . 10^ = a^ (Art. 127); . • . log a^ =px =p log a. EXAMPLES. Given log 2 = 0.301, log 3 = 0.477, log 7 = 0.845, and log 10^ = x; find the logarithms of the following numbers: 1. 64. Ans. 1.806. 5. 196. Ans. 2.29->. 2. 81. Ans. 1.908. 6. .036. Ans. 2.556. 3. 343. Ans. 2.535. 7. .000128. Ans. 4.107. 4. 72. A71S. 1.857. 8. 25600. Ans. 4.408. 469. Log a-7-b = log a — log b ; i.e. : 77ie logarithm of a qnotient is obtained from subtracting the logarithm of the divisor from the logarithm of the dividend. For, if 10^ = a and 10'' = b, then 10="-^ =za^b (Art. 139). . • . (x — y), which is the logarithm of a -r- b, equals log a — log b. Similarly, log Z>"~^ = log 1 — log J = — log ^ = -^log b. Log 4 =^ log 3 - log 7 = (- 1 + 1.477) -.845 = 1.632. 254 ALGEBRA. EXAMPLES. Given logarithms of 2, 3, and 7 as before, also log 10 = 1^ find the logarithms of the following numbers : 1. 5. Ans. .699. 5. f. A)is. IAd6. 2. J. Ans. 1.699. 6. f. Ans. 1.854. 3. |. Alls. 1.824. 7. 1.25. yl?;s. 0.097. 4. 3|. ^W6\ .544. 8. jV v4?i6'. 2.912. 470. Log V(( = -^; i.e.: 77ie logarithm of any root of a numher is equal to the logarithm of the nnmher divided hy the index of the root. For, if 10^ = a, then 10^ = Va. . • . log V?^ = - = ^^. r r Thus, if log 5^ = 4.893, then log W = |(4.893) = .699. Similarly, if log 2 = .301, then log V^ = \ (.301) = .060. Also, log V."0u3 = |(- 5 + 2.477) = -^ (3.477) = 1.495. 471. If a negative characteristic be not exactly divisible by the index of the root, add to the characteristic the least negative number which will render it so divisible, and prefix an equal positive number to the mantissa. Thus: i(3.624) =r i(- 4 + 1.624) = 1.406; ■i(- 7 + .835) = \{- 10 + 3.835) = 2.767. EXAMPLES. Given log 2 ■= 0.3010, log 3*= 0.4771, log 7 = 0.8451, log 10 = 1 ; find the logarithms of the following • 1. V§. Ans. 0.1505. 3. 3i Ant^. 0.1193. . 2. Vt". J ?zs. 0.2817. 4. 3-^ ^?i5. 1.5229. LOGARITHMS. 255 5. ei Ans. 0.5187. 9. VAm. ^^is. 2.9226. 6. 5^. Ans. 0.1398. lo. 'Vjm. Ans, 1.7603. 7. 15 -i ^;/«. L6080. n. (.0084)1 Ans. 2.3394. ' 8. .7i ^yi.«. 0.8838. 12. 7^ x 3i ^W5. 0.4010. 472. The remainder obtained by subtracting the logarithm of a number from 10 is called the cologaritliin of the num- ber, or tlie arithmetical complement of the logarithm of the number (abbreviated into colog or a.c. log). Thus: log 3 — 0.4771 and colog 3 = 9.5229 ; log 7 = 0.8451 and a.c. log 7 = 9.1549; log 200 = 2.3010 and colog 200 = 7.6990; log .005 = 3.6990 and colog .005 = 12.3010 ; log .0009 = 4^9542 and a. c. log .0009 = 13.0458. To find the cologaritbm: Begin at the left, siihtrad each figure from 9 down to the last significant figure, and sub- tract that figure from 10. 473. Let a and h represent any two logarithms, then it is evident that a — h =^ a-\-{V1, and I will be negative when n < 1. If ^ be less than unity, I LOGARITHMS. 2G1 will be negative when ii > 1, and I will be positive when n <\. For every base the logarithm of 1 is 0. 479. In every system of logarithms the rules for I, mul- tiplication (4G7); II, powers (468); III, division (469); and IV, roots (470), are the same as those given for common logarithms. Thus: I. If h"" = vi and h^ =in, r . h^^^^ ~ mn, II. If h'' = n, then ^'^^ = n^, 7)1 III. If ^^ = m audby = n, .'.b''-'' = -. IV. If ^^ = n, then bl = Yii, 480. TV/e difference between the logarithms oftioo consecu- tive numbers becomes less as the numbers themselves become greater. For, log {x -f- 1) — log X = log ' . As x increases, • *^ X -\- 1 the value of the fraction decreases, hence the loga- X X 4- 1 rithm of decreases; therefore its equal, log {x -\-l) — log X, decreases. EXPONENTIAL EQUATIONS. 481. An Exponential Equation is one in Avhich the un- known quantity is an exponent or index. Thus, a'^ = b; Va = b. Such equations usually require logarithms for their solution. EXAMPLES. 1. If 2^ = 32, find X. Since 2^ = 32 = 2^ .'.x = 5, 2. If 2^ = 12, find X. Since 2^ = 12, then log 2^ = log 12; that is, ^ log 2 = log 12; .'.x= M ^ ^^_ = 3.58538 + . 262 ALGEBRA. 3. Given 20* = 100, find x. log 20^ = log 100; . • . :r log 20 = log 100; _ log 100 _ 2 _ • ' • ^ - 1^2"0" " LSOl - ^'^^^ + • 4. Find log 144 in the system whose base is 2 Vs. Let X represent the required logarithm. Then, from Art. 454, (2 I^S)* = 144 = 12^; hence by squaring each side 12^ = 144^ = 12*; .' . x=z 4,, 5. Find logs 2. Given common logarithm of 2 = .3010, Let X represent the required logarithm. Then 5* = 2 ; . • . rclog 5 = log 2 ; . • . a; = |-^. logs Since log 10 = 1 and log 2 = .3010, . • . log (5 = J/) = 1-.3010 = .6990. .•..= §!« = |J-=. 4306 + . 6. Find logs J ; that is, find x in the equation (f )* = J. ^ ^logl^ 18751 ^-^1249^ 1249 _.^g log I 1.8239 ' -.1761 l^^l 7. Given 4^ + 2* = 72, find x. If 2* = y, then 4* = y\ .•.y'-^'y = 72,.'.tj = 8ov-9. Discarding the negative value oty, we have 2'"=8, . • . x=3, 8. Given 4* + 2* = 12, find x. As before, ^^ -|- ?/ = 12 and ;/ = 3 or — 4. . • . 2"" = 3, whence x = ,— ^ = '-w7:7-r. = 1.585, nearly, log 2 .3010 -^ 9. What is the logarithm of 3125 to the base 5 ? Ans. 5. 10. What is the logarithm of 8 to the base ^ ? Ans. — 3. 11. Given log 2 = .301, find the number of digits in the sixty-fourth power of 2. A7is. 20. 12. Find loga 243. Ans. 5. 13. Given (|)^ = .5, find x. Ans. 3.103. 14. 25^" + S'^ = 650, find x. Ans. 2. 15. 3^.+ 3* = 42, find X. Atis. 1.631. LOGARITHMS. 263 Exercise XXVI. 1. 3.1416 X 672 = ? 6. (If^ = ? 2. 3.1416 X (2.7)* = ? 7. (2#=? 3. 70642 ^ 82.3 = ? 8. (6.7)^* = ? 4. 21.06 ^ 6.9 X .541 = ? 9. Find log, 125, 5. If 10-^^=2, find log .0625. lo. 2lt=? Find the value of x in each of the following: 11. 6^ + 6l = 30. 12. a^*^*' = d. 13. a^^'c^ = n, 14. I. 2^.3^^ = 2000; II. hx = 3z. 15. a^ - a'' = b^ - i. 16. {a' - ^>2)*(x-i) =(a- h)"^ -^ {« + hy. 17. 3^'-*^ + * = 1200. 19. (V-)'' - (tr = 2. 18= (^Y ~ (I)* = 6. 20. 8^ - 3(2^) = 2, CHAPTER XXVII. RATIO, PROPORTION, AND VARIATION. 482. The Ratio of two numbers is their relative magni- tude expressed by the fraction of which the first is the numerator and the second is the denominator. Thus the ratio of 6 to 2 is indicated by the fraction f ; the ratio of 2 to 6 is indicated by the fraction f ; the ratio of f to f is indicated by the fraction f = f -^ }. 483. The ratio ot a to b is usually written a : h (read a is to h). The first term of the ratio is called the antecedent , and the second term is called the consequent. When the antecedent is greater than the consequent, the ratio is called a ratio of greater inequality; when the ante- cedent is equal to the consequent, the ratio is called a ratio of equality ; when the antecedent is less than the consequent, the ratio is called a ratio of less inequality, 484. When two quantities are commensuralle (that is, can be expressed in integers in terms of a common unit), their ratio is equal to the ratio of the two numbers by which they are expressed. Thus, the ratio of $9 to $12 is y%; of 2^ ft. to 31- ft. is 2J -f- 31 = J-o. 485. Tivo quantities of different hinds can have no ratio. Thus, 2 hours and 3 inches have no relative magnitude. (364) KATIO. 265 486. When two quantities of the same kind are iyicom- mensurable, it is not possible to express their ratio by a fraction both of whose terms are integers; yet a fraction of this kind may be found which shall express the ratio of the two quantities to any required degree of accuracy. For, let a and h represent two incommensurable numbers; also, let b = 7ix, where n is an integer, and let a be greater ft 171 X than mx and less than (rn + 1)^« Then — > — ^; that is, — > — ; also, T < — — — . Thus the difference between -z- b n b n b 1)1 1 and - is less than — = ii-^. Since 7ix = b, when x is n n diminished n is increased and ?i~^ is diminished. Hence, by taking x small enough, n~'^ can be made less than any assigned magnitude, and therefore the difference between - and - can be made less than any assigned magnitude. 487. A ratio will not be altered if both its terms be mul- tiplied or divided by the same number. For, a : b = rrt and ma : 7nb = — r. Since - = — r, b mb b mb .', a : b = ma : mb. 488. A ratio of greater ineqicality tvill be diminished, and a ratio of less inequality tvill be increased, by adding the same number to both terms of the ratio. For, let -r represent any ratio, and let a new ratio be formed by adding any positive number, x, to both terms of the original ratio ; then , , > or < r> according as -\- X b ab -\-bx ^ ^ (tb -{■ ax ,, . . , . , , , iTT— i — \ > or < Y77— ; — ^ ; that IS, according as ab + bx b(b -i-x) b{b -\-x) ' * ' > or < ab -{- ax; that is, according as bx > or < ax; that is, according as 6 > or < a. Alg.-23. 266 ALGEBKA. 489. In like manner it may be shown that a ratio of greater inequality will M increased, and a ratio of less in- equality will he diminished, hy taking from both terms of the ratio any number which is less than each of these terms. 490. Ratios are compounded by multiplying together the fractions that represent them. Thus, ac : hd is said to be compounded of the two ratios a : h and c : d. 491. When the ratio « : ^ is compounded with itself, the resulting ratio, a^ : h^, is called the duplicate ratio of a : h. Similarly, a^ : h^, the ratio compounded oi a : h, a '. b, a \ b, is called the triplicate ratio of a -.b. The ratio Va : Vb i^ called the subduplicate ratio of a : b; and Va : Vb is called the subtriplicate ratio ot a : b. 492. A ratio of greater inequality compounded with an- other increases it, and a ratio of less inequality compounded with another diminishes it. Let the ratio x : y \>q compounded with the ratio a : bj the compound ratio ax : by is greater or less than a : b, ax a according as j— is greater or less than -j-; that is, according as X is greater or less than y. 493. Eatios are compared by reducing the fractions which represent them to a common denominator, and comparing the numerators of the resulting fractions. Thus, 4 : 5 is compared with 7 : 9 by comparing f with ^. Now f = |^, and J = Jl ; . • . 4 : 5 is greater than 7 : 9. EXAMPLES, 1. Compare the ratios 5 : 7 and 2 : 3. Ans, 5 : 7 > 2 : 3. 2. Find the value of f : |. Ans. 15 : 14 = |f. 3. Compound 2 : 3, 3 : 4, 6 : 7. Ans, 4 = 3:7. PROPORTION". 267 4. Compound the ratios a -]- b : a — b, a^ -\- b^ : {a -\- iy, (^2 _ h^Y : «* - bK Ans. 1. 5. Write the duplicate ratio of 3 : 2, and the subdupli- cate ratio of 100 : 169. Ans. 9 : 4, 10 : 13. 6. Two numbers are in the ratio of 6 : 7, and if 4 be sub- tracted from each, the remainders will be in the ratio of 4 : 5. Find the numbers. Ans. 12 and 14. 7. Compound the duplicate ratio of 3:4, the triplicate ratio of 2 : 3, the subduplicate ratio of 16 : 9, and the sub- triplicate ratio of 27 : 64. Ans, 1 : 6. 8. Show that 1^2 : VT> /3 : fo". PROPORTION. 494. Proportion is an equality of ratios. Thus, if CL C a, b, c, d are four numbers such that t- = ;t, these four numbers are called proportionals, or are said to be in pro- portion. This proportion may be written a : b = c \ d (read, the ratio of a to b is equal to the ratio of c to d), or a '. b :: c : d (read, a is to b as c is to d). 495. When a : b : : c : d, a and d are called the extremes; b and c are called the means ; a and c are called the ante- cedents ; b and d are called the consequents; a and b are to- geth^ called the first couplet; c and d are together called the second couplet, 496. When four numbers are in proportion, the product of the extremes is equal to the product of the means. a c For, \i a '. b '.'. c \ d, then - = -. . • . (Ax. IV) ad = be, 497. Hence, ii a : b :: b : d, then ad = b\ In this case b is said to be a 7nean proportional between a and d, and d is called a third proportional to a and b. 268 ALGEBRA. 498. If any three terms in a proportion are given, the fourth may be determined from the equation ad == be. 499. If the product of tivo nnmlers he equal to the product of two others, the four are in proportion; the two factors of either product heiiig taken for the means, and the two factors of the other product for the extremes. For, if ad = be, by dividing by bd, j-j = -- ; a at IS, T c ■=d- •-■" :b'.:c:d. 500. If ad = be ', then it follows from (499) that I. a \b \: c '. d, ad be II. a '. c :: b : d. ad be since -7- = --. dc dc III. d\b \\ c\ a, ad be smce Y- = T-» ba ba IV. d '. c '.'. b : a, ad be since — = — . ac ao V. b : aw d : c. be ad since— = — . ae ac VI. b '. d '.'. a : c, be ad since — , = — ,. ca eel VII. c : a '.: d : b. be ad since —^ = —- . ab ab VIII. e : d :: a : b, since -— = =-:. bd bd It is evident, therefore, that if four numbers be propor- tionals, the relations between them may be expressed in eight ways. To change one of the above forms into another: Take the product of the extremes equal to the product of the 7neans, and divide both sides of the restdting equation by the product of the consequents of the reqnired proportion. PROPORTION. . 269 601. If four numhcrs are propo7iionaIs, they are in pro- portion toll en taken by ALTERNATiOiq^; that is, the first is to the third as the second is to the fourth, li a '.h :: c \ d, then a : c w b : d. ^ a c li- 1 1 ^ a b - , For, - = -; multiply by - ; . • . - = -; or, a : c :: b : d. a c U/ 502. If four numbers are j^roportionals, they are in pro- portion wlien taken by inversiojst; that is, tlie second is to the first as the fourth is to the third. It a : b :: c : d, then b : a :: d : c. ^ a c ^ a ^ c ,, . . b d For, -=-; .'. 1 -^ - = I -^ - ; that is, - = -, b d b d a c or b : a :: d : c. 603. If four numbers are in proportion, they are in pro- portion by COMPOSITION"; that is, the stun of the first and second is to the second as the sum of the third and fourth is to the fourth. It a : b :: c : d, then a -\- b : b :: c -\- d : d. or a -}- b : b :: c -\- d : d. If the given proportion be taken by inversion, it can be shown similarly that a -{- b : a :: c -\- d : c. In like manner, by alternation and composition, a -\- c : c :: b -\- d : d. 604. If four numbers are in proportion, they are in pro- portion by DIVISION; that is, the difference betiveen the first and second is to the second as the difference between the third and fourth is to the fourth. It a : b :: c : d, then a — b : b :: c — d : d. ^ a c «-,C-,i a — b c — d ^"^'1 = 1' ■'■b-^ = d~^' ^^"""^^ ir^-u-' or a — b : b :: c — d : d, Similarly, hy inversion and division, b — a : a :: d—c : cj also, by alternation and division, a — c : c :: b — d : d. 270 ALGEBRA. 505. If four numhers are in proportion, they are in pro- portion by coinposition a?id division. If a : b :: c : d, then a -\- b : a — b :\ c -\- d : c — d. From (503), —^ = —j-J and from (504), —^- = ——. . • . (Ax. V) — —T = — ■ — jj or, a-\-b : a—b : : c-\-d : c—d. Or — c — a 606. In a series of equal ratios, the sum of the antecedents is to the sum of the conseqiients as any antecedent is to its consequent. a-\-c^e-^(j'.b^d-^f-{-h'.', : c : d. Since -=-;,.'. ad = be. d p p Since -^ = ~, .' . cd = dc. a a p p Since y- = ^^ .'. ed= fc. Since j- — 1' .' . gd = he. It a Whence, by addition, d{a -\- c -\- e -\- g) =^ c(b-\-d +/+/0' .-. (499), « + c + e + ^: b -{- d -\- f -]- h :: c : d. 507. If a, b, c, d be in co7itinued proportion, that is, ita : b :: b : c :: c : d, then a : c :: a^ : b^, and a : d :: a^ : b\ a _b J a _a fl^. .^_«^. ' b c c b b b c b^ that is, a\ c \\ a^ \ V. ^. .^^abcaaa a a^ Similarly, S >< ^^^ d = b ^ b ^ P '' ' 1 = V' that is, a: cl :: a? : b\ PROPORTION. 271 508. If four numbers are proportionals, the first couplet may he multiplied or divided by any number, as also the seco7id co}(plet, and the resulting numbers will be propor- tionals. It a : b :: c : d, then ma : 7nb : : nc : nd. ^ a c ma nc r or, — = —; . • . — - = — -, or ma : mb '.: nc \ nd, b d mb nd 609. If fo\ir numbers are proportionals, the antecedents may be multiplied or divided by any number, as also the consequents, and the resulting members will be proportionals. It a : b w c\ d, then ma \ nb : : mc : nd, I-, « c ^, . ma mc ma mc For, T- = 7; therefore -r- = —j-; .•.—- = -—; b d b d nb nd or, ma : nb :: mc : nd, 610. If tivo proportions have a common couplet, the other couplets form, a proportion. If a : b :: c : d, and it a : b :: e : f, then c : d :: e : f, ^ a c ^ a e c e For,^ =^, and^ = -. .■.^=j.; or, c : d :: e : f. 611. If the antecedents in tivo proportions be the same, the consequents will be in proportion. It a : b :: c : d, and it a : e :: c : f, then b : d :: e : f. It a : b :: c :d, then (Art. 501) a : c :: b : d. It a : e :: c : f, then (Art. 501) a : c :: e : f. Whence (Art. 510) b : d :: e :f Similarly: If the consequents in tivo proportions be the same, the antecedents will be in proportion, 612. The products of the corresponding terms of two or more proportions are in pro^jorfion. It a . b :: c : d, e : f :: g : h, h : I :: m : n, ,, a c e a k m then T=i^ -7= Ti T=-' b d f h I n The product of the first sides is equal to the product of the second sides; .*. -j-^j =: -j—-; .•. aek : bfl :: cgm : dhn. 272 ALGEBRA. 613. If four numbers he i7i proportion, their liTce potvers, or like roots, will be in proportion. Itaib ::c: d, then a"" : b"" :: 6" : d\ (I c oP' (j^ For, J = —; .•.—- = — -, where n maybe integral or frac- tional; .-. a" : Z^" :: c" : ^". 614. If two quantities be increased or diminished by like parts of each, the results luill be in the same ratio as the quantities themselves. For, a ma r . a :b :: a ±-Ta : b ± -,b. d d 616. If a and b are incommensurable, and if c and d are also incommensurable; also, if when — lies between — and b n ^-, then - also lies between — and , however the 71 d n n numbers m and n are increased, then ^ = -r. b d ft Q For, if J- and -^ are not equal, they must have some assignable difference, and since each of them lies between — and , this difference must be less than — = « - 1. n n n But since n may, by supposition, be increased without limit, n~^ may be diminished without limit; that is, n~'^ may be made less than any assignable magnitude. Therefore, j- and ft (1, c -7 have no assignable difference, so that we may say — = -, Hence all the propositions respecting proportionals are true of the four numbers a, b, c, d. VARIATION". 273 EXAMPLES. 1. If 4 : 6 : : a; : 36, find x. Ans. x = 24. 2. If 3, X, aud 867 are in continued proportion, find x. Ans. X = bl. 3. Find a third proportional to 25 and 40. Ans. 64. 4. Find a mean proportional to 16 and 9. Ans. 12. 5. If 5^ — 8a; : 7a; — 5^ :: 6 : 1, find the ratio of x to y. Ans. 7 : 10. 6. Ila^ -b^:a-\-I} ::{a- hf : x, find x. A ns. x^= a — l. 7. What number added to each of the numbers 4, 8, 10, 16 will make the results proportional? Ans. 8. 8. Vx -\- 5 : Vx - 5 : : 5 : 3. Find x. A ns. x = WK 9. Given x -\- y : x :: 5 : 3, and x : 2 :: S : y, to find x and y. Ans. x = ± 3, y = ± 2. 10. Given x -\- y : x — y :: 3 : 1, and r* — «/^ = 56, to find X and y. Ans. x = 4:, y ^ 2. VARIATION. 516. Variation is an abridged form of proportion. It is employed when two quantities are so related that if the value of one be changed, the value of the other is necessarily changed. Thus, if A work for $4 a day, his total wages will de- pend upon the number of days he works. If he work twenty days, he will receive $80; if he work twice as many days, he will receive twice as many dollars; etc. That is, as one day is to 7i days, so is A's daily salary to A's total salary for n days. Hence we say that if A^s salary for one day be fixed, his total salary varies as the number of days he works. 274 ALGEBRA. 617. Variable duantities are such as admit of various values in the same computation. In this chapter variables are represented by capital letters. 518. Constant or Invariable Quantities are such as have only one fixed value. In this chapter constants are repre- sented by small letters. 619. Two variable quantities are said to vary directly when their ratio is constant. Thus, the length of a shadow varies directly as the height of the object which casts it. If ^- = c, a constant quantity, we say that A cc B, the symbol a being used instead of varies or varies as. Since ~ = c, .' . A = cB, 520. Two variable quantities are said to vary inversely when their product is constant. Thus, the distance being fixed, the rate and time vary inversely; for, as the one is increased, the other is diminished in the same ratio. c 1 1 If AB = c, then A = -^ = c X -j^, and A cc-jr. B B B 521. If J = cBD, where c is constant, then A is said to vary as B and D jointly, written thus: A a BD. For example, A^s total wages will vary jointly as the number of days he works, and his wages per day. cB 522. If ^ = -T,-, where c is constant, then A is said to vary directly as B, and inversely as D, written thus: A a yr. For example, the time occupied in a journey varies directly as the distance, and inversely as the rate of travel. 523. Besides the four kinds of variation heretofore men- tioned, many other forms are to be noted; thus, A may VARIATION. 275 vary as the square or the cube of B; inversely as the square or the cube; directly as the square and inversely as the cube; etc. For instance, the intensity of the light shed by any luminous body upon an object varies directly as the mass of the luminous body, and inversely as the square of M its distance from the object. Thus, A = ^ + ^, in which A cc C, and B oc—, and if when Z) = 4, (7=1; when D = 6, C =2; find the equa- tion between C and i). 30. If iS^ a ^^ when / is constant, and S a /when t is con- stant, and when t = ^ff= 2S, find the equation between /S^,/, and t. CHAPTER XXVIIL SERIES. 531. A Series is a succession of numbers, called the terms, each of which is derived from one or more of the preceding ones by a fixed law, called the htw of the series. Thus, 1, 3, 5, 7, 9, etc., is a series, and each term is found by adding 2 to the term immediately before it. 532. An Infinite Series is one that consists of an iin- limited number of terms. A Finite Series contains a limited number of terms. 633. A Converging Series is an infinite series in which the sum of the first n terms can not numerically exceed some finite number, however great n may be. Thus, 1, I, ^y ^, yig-, ^^5, etc., is a converging series, be- cause the sum of any number of its terms is less than 2. 534. The Sum of a converging series is the limit to which we approach by adding together more terms, but which can not be exceeded by adding together any number of terms whatever. By taking n large enough, the sum of the first 71 terms of the series can be made to differ from the limit by less than any assigned magnitude. 635. A Diverging Series is an infinite series in which, by taking n large enough, the sum of the first n terms can be made larger than any finite number. Thus, 1, 2, 3, 4, 5, 6, etc., is a diverging series. (279) 280 ALGEBRA. 536. If we divide I hj 1 — x, the quotient is the infinite series i -\- x -{- x^ -\- cc^ -^ x*' -^ . . . . If 2; be any positive number less than unity, this series is convergent. Thus, if x = ^, then 1 -^ (1 — :c) = 1 -f- ^ = 2, and the series is 1 + i +, 4 + i + iV + • • • } whose sum can be made to diifer from 2 by as little as we please, by increasing in- definitely the number of terms taken. By taking x small enough, we can make 1 -i- (1 — re) ap- proach as near unity as we please. Thus, it x = .01, 1 — (1-^) = W5 if* = -001,l^(l-^) = ^^; etc. By taking x nearly equal to 1, the sum of the series can be made very great. Thus, if x = .99, 1 -^ (1 — a:) = 100; ilx = .999, 1-ir (1-x) = 1000; etc. If x = \, = l-{-x-^x^-\-x^-\- . . . becomes 1 — X ' ' J-— ^ = 1 + 1 + 1 + 1+ . . . That is, ^ = an unlimited number, expressed thus: i = ex. The only meaning to be attached to this expression is, that , by taking a denominator S7)iall enough, we can make the value of a fraction as large as we please, without limit. If a; > 1, the series is divergent. Thus, if a: = 4, j-^ = -i-; butl+2: + a;« + a;« + a;* + .. , =1 + 4 + 16 + 64 + 256 + . . . , and the more terms of the series we take, the more does their sum diverge from — ^. 637. There are certain fractions which, on a certain sup- position, assume the form |. Such fractions are called vanisliing fractions. Thus, if x = 1, ~ ^, = -. 1 — Or The symbol % denotes indetermination, and may be in- terpreted in one of two ways: (1) The expression which equals f is really indeterminate, and may be satisfied by any value. SERIES. 281 (2) The fraction may have a factor common to both terms, and, by the supposition made, this factor becomes 0, 1 — x^ whence each term becomes 0. Thus, 7, becomes - 1 — x^ when X = 1. It x = 1, then .t — 1 = 0. Now see if .t — 1 or 1 — a; is a factor of each term. If so, divide each term by the common factor, and then the real vahie of the frac- tion is readily found. Thus: 638. The number of different series is unlimited, but three kinds are of primary importance, called Arithmetical, Geometrical, and Harmonical Progressions. ARITHMETICAL PROGRESSION. 639. An Arithmetical Progression is a series each term of which is derived from the preceding, by adding a con- stant number. Thus, 1, 3, 5, 7, 9, . . . is an arithmetical progression. 540. Let a represent the first term, and d the common difference; then the terms of the arithmetical progression are a, a -\- d, a -\- 2d, a + 3d, a + 4fZ, . . , , in which the series will be increasing or decreasing according as d is positive or negative. Since the co-efficient of d in each term is less by unity than the number of the term, . • . the nth term is « + (n — l)d. If I represent the ?^th term, then I = a-\-{n — l)d. (I) 641. Let s represent the sum of the series. Then s = a -{- (a -{- d) -}- (a -{- 2d) -{- , . . -\- I. By writing the terms in reverse order, we have also s = l-\.{l-d) + (l-2d) + . . . + «. Therefore, by addition, 2s=i(a-{-l) -{-(a-\-l)-{-{a-{-l) -\- . . . to 71 terms. . • . 2s = n{a -\- 1), whence *• = ^ [a -\- l). (II) Alg.-24. ^ 282 ALGEBRA. 642. In an arithmetical progressio7i, the sum of any two terms equidistant from the beginning and the end, is equal to the sum of the first and last terms. For, the rth term from the beginning is « -j- (?• — 1)^7, and the rth term from the end is I — {r — \)d, and the sum of these two terms is {a -f- I). 543. If the numler of terms in an arithmetical progres- sion be odd, the middle term is equal to half the sum of the extreme terms. Every odd number is of the form 2r + 1 ; therefore, let n = 2r + 1. Thus the middle term, m, has r terms before it and r terms after it; hence m is the (r + 1) term, count- ing from each end. Therefore, m = a -\- rd, and m — l — rd. (Art. 540,1); whence, by addition, 2m = a -{- 1; . • . m = :^(a -}- 1). 544. The Arithmetical Mean between two numbers is the number which, when placed between them, will form with the given numbers an Arithmetical Progression. Thus, 7 is the arithmetical mean between 5 and 9, and a -\- d is the arithmetical mean between a and a -\- 2d. From Art. 543 it follows that the arithmetical mean between two numbers is half their sum. 545. To insert a given number of arithmetical means between two numbers. Let a and I be the two given numbers, and let m. repre- sent the number of terms to be inserted. Then the mean- ing of the problem is, that we are to find m + 2 terms in arithmetical progression, a being the first term and Z the last. Since Z = « + (m + \)d, (540, I,) .-. tZ = -^-— ^-. This finds d, and the m required means are a-\- d, a-\-2d, a-\-Zd, , , . , a-\- md. SERIES. 283 546. From the two equations, (I)l = a-{- {?i - l)cl; (II) 5 = I (a + 0; any two of the quantities a, d, I, n, s may be found when the other three are given. For example, — To find I when a, d, s, are given. Since n is neither given nor required, eliminate n. From I, n = — \- 1. From II, n = — — ^ . d a-\-l . • . — J h 1 = — 7—1 i ill which I is required. d a-\-l ^ Clear of fractions. P — a^ -\- ad -\- dl = 2ds, Transpose and add --. 1!^ -\- dl -{- — = 2ds -\- a^ — ad -\- -j-. Extract square root. I -\- ^d = ± V2ds -\- (a — ^dy\ ,'.l= -id± V2ds + (a - id)\ EXAMPLES. 1. Given a = 8, d = 2, s = 44, find n, Ans, n = 4. 2. Sum 2, 5, 8, 11, . . . to 20 terms. Ans. 610. 3. Find d when a = 2, Z = 50, 5 = 520. A7is, d = 2{^. 4. Find the sum of the first u odd numbers. Ans. n^. 5. The first term of an arithmetical progression being 2, and the fifth term being 10, how many terms must be taken that the sum may be 90? A7is. 9. 6. If s = dQ, a = 12, d = - 2, find u. Ans. n = 9 or 4. 7. Insert six arithmetical means between — 1 and 20. Ans. 2, 5, 8, 11, 14, 17. 8. The sum of the first two terms of an A. P. is 4, and the fifth term is 9; find the series. A7is. 1, 3, 5, 7, 9, etc. 9. In the series 1, 3, 5, 7, . . . the sum of r terms : the sum of 2r terms ::1 : x; determine the value of x. Ans. 4. 10. Divide 3 into 5 parts whose common difference is ^. Ans. i, I, I, I, 1. 284 ALGEBKA. GEOMETRICAL PROGRESSION". 547. A Geometrical Progression is a series of terms each of which is derived from the preceding, by multiplying it by a constant factor, called the ratio. Thus, 1, 2, 4, 8, 16, . . . is an mcreasing geometrical progression whose ratio is 2; also, 18, 6, 2, f, ... is a decreasing geometrical progression whose ratio is J. If a be the first term and r the ratio, then a, ar, ar"^, ar^, ... is a geometrical progression which is an increasing series when r > 1, but a decreasing series when r < 1. 648. Since the exponent of r is less by unity than the number of the term, therefore the nth term = ar^~^. If / denote the wth term, then I = a7'^~^. (I) If s denote the sum of the series, . • . s = a -\- ar -{- ar^ -\- ar^ + . . . + «?'"" V . • . sr = ar -{- ar"^ -j- ar^ + • • • + dv"^'^ -\- «?'". Hence, by subtraction, sr — s = ar'^ — a. Whence .9 = a{r--l )^ ^^^. r — 1 ^ ^ Since ar^ z=z rl, ,' . s = --. (HI) When r < 1, formulas II and III will be more conven- ient if written : II. 5 = —; -. 1 — r 549. The Geometrical Mean between two numbers is that number which when placed between them will form witli the given numbers a geometrical progression. Let the two numbers be a and b, and let G represent their geometrical mean; then, a, G, and b being in geomet- rical progression, ~ = — . ,' . G^ = ab, whence G = Vab. That is : The geometrical mean between tivo numbers is equal to the square root of their product. SERIES. 285 650. To insert a given numher of geometrical means he- tween tioo given terms. Let a and I be the two given terms, r the ratio, and rn the number of terms to be inserted. Then the meaning of the problem is, to find m-\-2 terms in geometrical progression, a being the first term and I the last. From (548, I), I = ar^^'j .'. r = (J^^^^, This finds r, and the required terms are ar, ar^, ar^, . . . , «r*". 551. If r < 1, then the larger n is the smaller will r" be, and by taking n large enough ?" can be made as small as we please. If 7i be taken so large that r" may be neg- a(l — r") lected in comparison with unity, the value ot s = — ff(l - 0) a ^ reduces to s = — ^^ = r . Hence: 1 — r 1 — r By taking n large enough, the sum of n terms of a decreas- ing geometrical series can he made to differ as little as we please from a -^ {I — r). This statement is sometimes abbreviated into the follow- ing : The sum of an i7ifnite number of terms of a decreas- ing geometrical ^progression is « -i- (1 — r). 552. In (548) we have five quantities occurring, namely, a, r, /, n, s; and these are connected by the equations I and II, or II and III, there given. We might, there- fore, propose to find any two of these quantities when the other three are given; it will, however, be found that some of the cases of this problem are too difficult to be solved. The following four cases present no difficulty: (1) Given a, r, n, (3) Given r, n, I. (2) Given a, n, I. (4) Given r, n, s. In the four cases in which 7i is required, the unknown quantity is an exponent, and may be found by the use of logarithms. ^S6 ALGEBRA. EXAMPLES. 1. Sum to n terms 3 + 2 + f + • • • 2. Sum to infinity 3+2-^1+. . . 3. Sum to infinity | — f + ^t ~ • • • 4. Given a, r, s, to find L Ans. I ns. 9 j 1 — (1)"}. A7IS. 9. Alls. If a-h{r- 1). 7' 5. Given a, r, I, to find n. Ans. n = — ^= [- 1. log r 6. Eind two numbers whose sum is 13, and geometrical mean 6. Ans. 9 and 4. 7. Find three numbers in geometrical progression whose sum is 26, and the sum of their squares 364. Ans. 2, 6, 18. 8. The sum of two numbers is 10, and their geometrical mean is | of their difference ; find the numbers. Ans. 9 and 1. 9. Find three numbers in geometrical progression whose sum is 21, and the sum of the first and second : the sum of the second and third :: 1 : 4. Ans. 1, 4, 16. 10. The sum of three numbers in G. P. is 7, and the sum of their reciprocals is J ; find the num.bers. Ans. 1, 2, 4. 11. There are four numbers in G. P., the sum of the ex- tremes being 18 and of the means 12 ; find them. Ans, 2, 4, 8, 16. 12. There are four numbers in A. P., and if 2 be added to the third and 8 to the fourth, the first, second, and the two sums will be in geometrical progression; find the num- bers. A71S, 2, 4, 6, 8. HARMON^ICAL PROGRESSION^. 653. A Harmonical Progression is a series of numbers whose reciprocals are in arithmetical progression. Hence the general representative of such a series will be 1 1 1 1 1_ a' a + cV a + 2d' a-^dd' ' ' ' ' a-{-(n-l)d' SERIES. 287 554. Questions relating to harmonical progression are readily solved by writing the reciprocals of the terms so as to form an arithmetical progression. monic mean, then 555. If a and I denote two numbers, and H their har- H a~ b H' 2 _1 1_ 0', .-. d'-\-2ab-^¥>4.ab; .'. ^> -M . .-. A> H. 559. The three numbers, a, b, and c are in arithmetic- al, geometrical, or harmonical progression according as a — b a a a ^. , ^ = — or =- or = -, respectively. b — c a b c ^ ^ b^ = ac. For, (1) If a — b- c a , .*. « l-b. = b (2) If a — b- b c a then ab — b^ = ab- ac, (3) If a — b- b c a 1 1 •** b a _ 1 ~ C 1 EXAMPLES. 1. Insert five harmonical means between ^ and ^. ^ns. i, \, \, ^, i^. 2. Find two numbers whose arithmetical mean is 3, and whose harmonical mean is f. Ans. 2 and 4. 3. Given the first two terms, a and b, of a harmonical progression, to find the nth term. A 7 "^ Ans. I [n - l)a - (n - 2)b. 4. The first term of a harmonical progression is ^, and the sixth term is ^; find the intermediate terms. ^^^s. J, i, i, ^^. 5. What number added to each of the numbers a, b, c, will give results in harmonical progression ? ab — 2ac -\- be Ans. — . a — 2b -{- c SEKIES. 289 Exercise XXVIII. AVheii X = a, find the value of : x*^ — a*^ x^ — rt* x'^ — «" 1. . 2. -3 3. 3. . X — a of — or X — a 4. Find the sum of n terms of the series 1, 2, 3, 4, 5, . . . 5. The sum of five numbers in A. P. is 35, and the sum of their squares is 335; find the numbers. Note. Let the numbers be represented hy x — 2y,x — y, x, x-\-y, x-\-2y. In general, if the number of terms in an A. P. be odd, it is convenient to represent the middle term by x. If the number of terms be even, it is convenient to represent the two middle terms by X — y and x-{-y. 6. Find four numbers in A. P. such that the sum of the squares of the extremes is 50, and of the means is 34. 7. Find four numbers in A. P. such that the product of the extremes is 22, and of the means 40. 8. The sum of the fourth powers of three consecutive natural numbers is 962 ; find them. 9. The sum of a certain number of terms of the series 21 -f 19 + 17 + . . . is 120 ; find the number of terms. 10. Sum to n terms the series 5, 9, 13, . . . 11. Sum to 71 terms the series whose 7-th term is 2r — 1. 12. Sum to 71 terms the A. P. whose first term is 71^ — ?i + 1, and the common difference 2. 13. Find the sum of five terms of a geometrical progres- sion whose second term is 3 and fourth term 27. 14. The sum of four terms in geometrical progression is 700, and the difference of the extremes is f J of the differ- ence of the means ; find the numbers. 15. In a G. P. show that the product of any two terms equidistant from a given term is constant. Alg -25. 290 ALGEBRA. 16. In a G. P. show that if each term be subtracted from the succeeding term, the successive differences are in Gr. P. 17. The sum of the first three terms of a G. P. is 28, and of the first four terms is 60; find the seventh term. 18. Three numbers whose sum is 21 are in A. P. ; if 2, 3, and 9 be added to them respectively, the sums are in G. P. ; find the numbers. 19. The fifth term of a G. P. is 8 times the second, and the third term is 12 ; find the series. 20. The sum of an infinite G. P. is 3, and the sum of the first two terms is 2|; find the series. 21. Show that b^ is greater than, equal to, or less than ac, according as a, h, c are in A. P. , G. P., or H. P. 22. If X is the harmonic mean between a and h, show that 7 = - + r- X — a X — a 23. If a, h, c are in H. P., show that ^ — ; — , — ; — , — —7, -\- c a -{- c a-\-h are in H. P. 24. Find two numbers whose difference is 8, and whose harmonic mean is If. 25. If tty h, c are in arithmetical progression, and h, c, d in harmonical progression, prove that a :h :: c : d. 26. A sets out from a certain place and travels one mile the first day, two miles the second day, three miles the third day, and so on. B sets out five days after A, and travels the same road at the rate of 12 miles a day. How far will A travel before he is overtaken by B ? 27. Prom 256 gallons of wine a certain number are drawn and replaced with water ; this is done a second, a third, and a fourth time, and 81 gallons of wine are then left. How many gallons are drawn out each time ? CHAPTER XXIX. PERMUTATIONS, COMBINATIONS, AND CHANCE. 560. The Permutations of quantities are the different orders in which they can be arranged. Quantities may be arranged singly, in pairs, in groups of three, and so on. Thus the permutations of the letters a, b, c, taken two at a time, are ab, ba, ac, ca, be, cb. Permutations are also called variaiions, or arrangements. 561. The number of permutntiojis of n letters taken r at a time is n (u — l){n — 2){}i — 3) . . . (n — {r — 1]), Let a, b, c, d, . . . , be the n letters; and let P^, F^, Ps, Pij . . ' f Pry respectively denote the whole number of permutations where the letters are taken singly, in pairSf three together, four together, . . . , r together. The number of permutations of n things taken singly is evidently equal to the number of things ; . * . Pi = n. The number of permutations of n letters, taken in pairs, is n{?i — 1). For, from the n letters, a, b, c, . . , , if we remove a, there will remain (^i — 1) letters. Writing a before each of these we shall have ab, ac, ad, . . . ; that is, there will be {71 — 1) permutations in which a stands first. In like manner, if a be replaced and b be removed, it is readily shown that there are (n — 1) permutations in which b stands first; similarly, {n — 1) permutations in which c stands first; and so on for the n letters. That is: Pg = ()i _ 1) -|_ (;i _ 1) -j_ (/t _ 1) -|_ . , . to n terms; r, Pfi^nin — 1). (291) 292 ALGEBRA. The number of permutations of n letters, taken three at a time, is n{ii — l)(w — 2). For, if we remove «, there will remain {ii — 1) letters, Ij, c, d, . . . If these {n — 1) letters be taken two at a time, according to the preceding para- graph {)i — l){n — 2) permutations can be formed. Put a before each of these, and we have (71 — l){n — 2) per- mutations, each of three letters, in which a stands first. Similarly, there are {n — 1){h — 2) permutations, each of three letters, in which i stands first; similarly, there are (n — l){n — 2) permutations, in which c stands first; and so on. .' . P,= (n-l)(7i-2)-{-{ji-l)(7i-2) + {n-l){7i-2) + ... to ^ terms. Whence, P3 = n(n — l)(n — 2). In like manner, P^ = ?i(?i — l)(n — 2)(n — 3), P^ = n{a - l)(7i - 2){n - 3)(n - 4), and so on. Hence we may conclude that Pr = n(n - l){u - 2)(?i - 3) . . . {n - r + 1). 562. The general proof of the formula in (561) depends upon mathematical induction (216). We have shown in (561) that this formula holds if r = 1, 2, or 3, and similarly, by taking successive values for r, we can show that it is true in any selected case. We shall now show that if this formula be true for any selected value of r, it will be ti'ue for the next higher value; that is, if Ps = n(n - l)(n - 2) . . . (n - [s - 1]), then P, + i = n(7i - l){u - 2) . . . (n - [s - l])(n - s). For, in the value of P^, substitute u — 1 for 71, then out of the (n — 1) letters, b, c, d, . . . , we can form {n — l)(n — 2)(?i — 3) . . . {n — 1 — [s — 1]) permuta- tions, each of s letters. Put a before each of these, and we obtain as many permutations, each of {s + 1) letters, in which a stands first. Similarly, we have as many in which b stands first, as many in which c stands first, and so on for the n letters. P^ f 1 is therefore equal to the sum PERMUTATIONS. 293 of n terms, each equal to {n — l)(^i — 2) . . . (t^ — s), .'.P,^, = n{n-l){n-2). . . {n - s). Since this formula holds when the letters are taken three at a time, therefore it holds when they are taken four at a time, therefore it holds when they are taken five at a time, and so on; thus it holds for every value of r. 563. If r = n, then P„ = n{n-l){n-2) . . . {n-7i^\)', that is, Pn = n{)i - l){}i - 2) ... 1. Whence: The 7iumher of permutations of n things, taken all together, is equal to the 2^1'oduct of the natural numbers from 1 up to n. For the sake of brevity, 1 . 2 . 3 . . . (?z — l)y^ is denoted by hi, which is read, factorial n. Thus, 1.2.3.4. = [4. 564. It is evident that w(?i-l) (/^- 2) . . . {n-r^l) = [n{n-l) . . . (;^-7'+l)] [{n-r){n-r-\) . . . 1] _ \± ^ (u — r){n — ?• — 1) ... 1 ~" \n—r ^ Pr \n — r 565. The number of permutations of n letters, taken all together, of which p are a's, q are b's, r are c's, . . . , is t^ilEklL- • • ) For, suppose iV^ to represent the entire number of per- mutations. If in any one of these permutations the p a's were changed into p new letters different from any of the rest, then these p letters could be arranged in \p different ways; hence, without altering the situation of any of the remaining letters, we could from the single permutation produce l;j different permutations. The same would be true for each of the N permutations; hence, if the p a's were different letters, the whole number of permutations would be iV X 1^, of which still q are b's, r are r's, etc. Similarly, if the q b'B were also changed into different letters, the whole number of permutations would he N X \p X '\q; and 294 ALGEBRA. if the»r c's were also changed, the whole number would be N X [pX \qX\r; and so on until all the n letters are dif- ferent. But when this is the case we know that their whole number of permutations equals |^. (563.) .'. JVx[pX\qX\r_.. . . y = \ii; ,'.]V=\n-^(\p\q^. . . ) 566. The nu7nber of loeriinitationn of n different letters, when each may occur once, twice, thrice, . . . , r times, is li^. Let the n letters be a, h, c, . . . First take them singly: this gives n permutations. Next take them two at a time: here a may stand before itself, or before any one of the remaining letters, as aa, ah, ac, . . . ; that is, there are n permutations of this kind in which a stands first. Shnilarly, there are n permutations of the form hh, ha, he, hd, . . . , and so on for the oi letters: thus there are nX n ^ r? permutations of the letters taken two together. Similarly, by putting successively a, h, c, . . . before each of the permutations of the letters taken two at a time, we obtain n X n^ = n^ permutations of the letters taken three at a time, and so on for all values of r. 667. The sum of such permutations of n letters, taken 1, 2, 3, . . . , r together, equals n -{- n^ -\- 7i^ -\- , . , rf = n{l-\-ii-\-it^-\- , . . 7^^-l) = n ('— \ IJ EXAMPLES. 1. How many permutations can be formed with the letters a, h, c, d, e? F, = 5, P, = 20, Ps = 60, P, = 120, P, = 120. Ans. 325. 2. How many permutations can be formed out of h o rn e 9 Alls. 4 4- 12 + 24 + 24 ^ 64. PERMUTATIONS. 295 3. In how many ways, taken all together, can the seven prismatic colors be arranged ? Ans. [^=5040. 4. How many numbers of two figures each can be ex- pressed by the ten digits? A)is. 90. 5. How many words, each containing two consonants and one vowel, can be formed from the alphabet ? The 20 consonants can be arranged in 20^ = 400 ways. Since a may be placed before, between, or after the con- sonants, therefore 3 X 400 words can be formed contain- ing a. Similarly, 1200 words can be formed containing each of the other vowels. Therefore the required number of words is 7200. If repetitions be not allowed, the number of words will be 7200 - 3G0 = 6840. 6. Out of 10 consonants and 4 vowels, how many words can be formed, each containing 3 different consonants and 2 different vowels? The 10 consonants can be arranged in 10 X 9 X 8 ways. The 4 vowels can be arranged in 4 X 3 ways. Each pair of vowels can form 10 different words with each group of consonants. Ans. (10 X 9 X 8) X (10) x (4 X 3) = 86,400. 7. How many different words, each containing eight letters, can be formed from the letters of mainmalia ? Ans. [8-^- ([3[3) =1120. 8. The number of permutations of n things taken 3 to- gether : the number of permutations of {n + 2) things taken 3 together :: 5 : 12; find n. Ans. n = 7. 9. The number of permutations of 7i letters taken 3 to- gether is 20n ; find n. Ans. n — 6. 10. How many different words can be formed out of the letters of the word Mississijjpi, taken all together? Ans. [U -^ ( (£14(2) = 34,650. 296 ALGEBRA. 11. How mauy words of six letters might be made out of the first ten letters of the alphabet, no letter being used more than once in any word, and each word containing two vowels ? Ans. (7 . 6 . 5 . 4) X (15) x (3 . 2) = 75,600. COMBI]^ATIONS. 568. The Combinations of things are the different collec- tions that can be formed out of them, without regarding the order in which the things are placed. Thus, the combinations of the three letters a, b, c, taken two at a time, are ah, ac, he ; ah and ha, though different permutations, forming the same combination. 669. The numher of comhinations of n Ihings, taken r at a time, is 1, each product will be larger than the multiplicand, and Cr^x> Or- For some value of r, how- ever, the multiplier changes from being greater than 1, to being either equal to 1, in which case 6V + 1 = CV, or less than 1, when C^^i < C^. (1) When ^^-^=1, then r=i(/i-l) and r^l = \{n^l). Since r is integral, its value, \{u — 1), must be integral; .' . n must\)e an odd number. Hence: When n is an odd integer, the greatest number of combinations is obtained when the n things are taken \(n — \) at a time or \(n -\- 1) at a time, the results being the same in these ttuo cases. Thus, if n=5, then (7^=5; C^=10; C's^lO; C,=5; C,=l. If n = 7, then Ci=7; 6^2=21; 6'3=35; C,=35; C,=21; etc. (2) If ^J-^ < 1, then r > U^i - 1). (7, will be the greatest when r is the first integer greater than ^{n — 1). When n is even, this integer is -. Hence : Whe7i n is even, the greatest number of combinations is obtained tvhen the n n things are taken - at a time. Thus: If ?i = G, then 6^=6; ^2=15; r3=20; C^=U; C,=6; C\=l. If w = 8, then 0^ = 8; C^ = 2S; C^^bQ', (\=-.'70; C\=56; etc. 298 ALGEBRA. EXAMPLES. 1. How many combinations can be formed of nine things taken five at a time? three at a time ? Ans. 126; 84. 2. Eour persons are chosen by lot out of ten; in how many ways can this be done ? A7is. 210. 3. Find the whole number of combinations of six things. Ans. 6 + 15 -h 20 + 15 + 6 + 1 == 63. 4. The number of combinations of ^n things taken two at a time is 28; find n. Ans. 7i = 24. 5. The number of combinations of n things taken three at a time is y^^ of the number taken five at a time. Find n. Ans. n = 12. 6. In how many ways can 12 men be divided into two classes of 8 and 4 ? A us. [1^ -^ ([8 [4 ) = 495. 7. How many words of six letters might be formed out of the first ten letters of the alphabet, if each .word differ from every other in at least one letter ? Ans. 210. 8. How many parties of 8 men each can be formed from a company of 40 men ? Ans. [4 -^ ( [8 |32). 9. The number of combinations of {n -)- 1) things, 4 to- gether, is 9 times the number of combinations of n things, 2 together; find n. Ans. n = 11. 10. A person wishes to make up as many different dinner- parties as he can out of 20 friends. How many should he invite at a time? A7is. 10. 11. How many combinations can be formed from the letters in the word notatifxn, taken three together ? Ten combinations can be (formed, in each of which the three letters are different. Four combinations contain two 7i's, four contain two o's, and four contain two /'s. Ans. 22. CHANCE. 299 CHANCE. 573. If an event may happen in a wa3'S, and fail in h ways, and all these ways are equally likely to occur, then the probability or chance of the event's happening is — — r, (I — j— and the probability or chance of its failing is . This may be regarded as the definition of the word chance in mathematical works. This definition may be further illustrated as follows : If an event may happen in a ways and fail in h ways, it is evident that the chance of its happening is to the chance of its failing as a is to b ; therefore, the chance of its happening is to the sum of the chances of its happening and failing as a is to {a + b). But the event must either liappen or fail; lience the sum of the chances of its happening and failing is certainty. There- fore, the chance of its happening is to certainty as a is to {) is the chance of its failing. 575. When the probability of the happening of an event is to the probability of its failing as a is to b, the fact is expressed in ordinary language thus: the odds are aio b for the event, or ^ to « against the event. When a = b, then the odds are evc7ij that is, the event is as likely to happen as not. 576. If there be any number of events A, B, C, etc., such that 07ie 7nust happen and only one can happen; and if a, b, Cf etc., be the number of ways in which these events 300 ALGEBRA. can respectively happen, and each way be equally likely to occur, then the chances of the events are proportional to a, by c, etc., respectively. Consider, for example, three events: then A can happen in a ways out of {a -^-b -^ c) ways, and fail in {b + c) ways; therefore, by Art. 573, the chance of A's happening is ^ . Similarly, the chance of B's happening is — ■ — ; — ; — , and the chance of ^^ ^ a^h -\- c C^s happening is a -\-b -\- c 577. If four white balls, five black balls, and six red balls are thrown promiscuously into a bag, and a person draws out one of them, the chance that this will be a white ball is /^, that it will be a black ball is -f-^ = ^-, and that it will be a red ball is j\ = f . The chance that the ball drawn will be either white or black is y\ = f , that it will be either white or red is |, and that it will be either black or red is |^. Now let us consider the chances of the different cases if two balls be drawn. The number of pairs that can be . 15 X 14 formed out of 15 things is — ~ = 105. The number of pairs that can be formed out of the four white balls 4X3 5x4 is —^ — = 6; out of the five black balls is — - — = 10; and 6x5 out of the six red balls is — - — = 15. Therefore, the chance that the two balls drawn will be both white is yf^ = -^j; that they will be both black is -f^^ = ^\, and that they will be both red is ^^ = ^. Also, since each white ball might be associated with each black ball, the number of pairs consisting of one white ball and one black ball is 4 X 5 = 20; hence the chance of drawing such a pair is y^y^ = j^. Similarly, the chance of drawing a white ball and a red one CHANCE. 301 is ^§, and the chance of drawing a black ball and a red one is f. The sum of these six chances is, of course, equal to unity. The chance that at least one of the two balls drawn will be white is /„\ = |^. For this may occur in any one of three ways: (1) Both may be white, the chance being yf^; (2) One may be white and one red, the chance being -^q\; (3) One may be white and one black, the chance being ^^^. Therefore, out of the 105 ways, there are 6 + 24 -f 20 = 50 ways in which a white ball may be drawn. The chance that neither ball is white is (1 — ^^) = J|. In like manner, the chance that at least one of the two balls dmwn is black is y% = 4? ^^^ *^^^^ ^* ^^^st one is red is -jW = |f . If three balls be drawn, the chances of the various cases may be calculated similarly. The chance that the three balls will all be white is jf^; that all will be black is :^^j; that all will be red is -^j; and so on for the other cases. 578. In a bag there are four white and six black balls. Find the chance that, out of five drawn, two only shall be white. The number of combinations of ten things taken five at a time is 252. The four white balls may be taken two at a time in 6 ways, and the six black balls may be taken three at a time in 20 ways. Since each pair of white balls may be drawn with any three of the black ball^, therefore, out of the 252 possible ways in which five balls may be drawn, there are 6 X 20 = 120 ways in which two are white and three are black. Hence, the required chance is |^| = ^^. Suppose that we are required to find the chance that at least two shall be white. The chance that two will be white and three black is ||^; that three will be white and two black is ^f^; and tliat four will be white and one black is ^f^. The sum of these chances is the required chance, i.e., ill — ii' ^^^^^ chance of not drawing as many as two white balls is 1 — J^ = J-|. 302 ALGEBEA. 679. A's cliance of winning a prize is J, and B's |: what is the chance tliat neither will obtain a prize? The chance that one will obtain a prize is (i -f -i) = f^-^ hence the chance that neither will win is 1 — -f^ — j\. 580. A series of events such that only one of them can happen, may be called a series of exclusive or dependent events. Two or more events such that both or all may happen, are called non-exclusive or independent events. For example, if a copper be tossed twice, it may fall head up both times; if it be tossed six times, it is possible for it to fall head up six times. 581. If there are tw^o or more independent events, the occurrence of all of them simultaneously or in succession may be regarded as a single compound event. Thus, in tossing a copper twice, the event of its falling head up at both trials may be regarded as an event compounded of iivo simpile events; namely, with head up at the first trial, and with head up at the second trial. The chance that the coin will fall head up at each trial is \. If these separate chances were added the sum w^ould be 1, that is, certainty; a result obviously false. A little consideration will show that the chances should be multi- plied instead of added. For, in the double fall there are four possibilities equally likely to occur : 1. Both times head up. 2. First time head up, second time head down. 3. First time head down, second time head up. 4. Both times head down. Only one of these four ways gives '' heads" both times; hence the chance of heads both times is J = ^ X J; that is. The chance that two independent events loth happen is the product of their separate chances of happening. CHANCE. 303 In general, suppose that there are two independent events, C and Dj and let C happen in a ways and fail in b ways; also, let D happen in a' ways and fail in b' ways, all of these ways being equally likely to occur. Each of the {a -f b) ways may be associated with each of the {a' -f ^') ways; thus there are {a -f- b)((i' + b') compound cases which are equally likely to occur. In aa' of these compound cases both events happen; in ab' of them the first liappens and the second fails; in a'b of them the first fails and the second happens; and in bb' of them both fail. Thus: a' X -> T} is the chance that both events happen. a -\- b a' -Y b X —, 71 is the chance that C happens and D fails. a ^ b a' ^ b, X , Ti is the chance that C fails and D happens. a^b a' -\-b' - X —, 77 is the chance that both fail. a-^b a' -{- b' 582. In like manner, if there be any number of inde- pendent events, tlie chance that they will all happen is the product of their respective chances of hapj)ening, and the chance that all fail is the product of their respect ire chances of failing. For example : The chance that A can solve a certain problem is ^; the chance that B can solve it is^; and the chance that C can solve it is f. What is the chance that the problem will be solved if they all try? The problem will be solved unless they all fail. A's chance of failing is f ; B's is J; and C's is |. The chance of all failing is } X |^ X ^ = J. Therefore, the chance that the problem will be solved is 1 — ^ = -J. The chance that all will succeed is J X |^ X | = yV* 304 ALGEBRA. 583. Suppose that four white balls, five red balls, aud six black balls be thrown promiscuously into a bag; required the chance that in two successive trials two red balls will be drawn, the ball first drawn being replaced before the second trial. Here the chance of drawing a red ball at the first trial is ^y and the chance at the second trial is again J; hence the chance of drawing two red balls is {\Y = |. Let us now consider the case when the ball first drawn is not replaced before the second trial. Here the chance of drawing a red ball at the first trial is J; if a red ball be drawn at first, out of the 14 balls remaining 4 are red; hence the chance that a red ball will be drawn at the second trial is f ; therefore, the chance of drawing two red balls is 1 X f = ^\. This result corresponds with the one obtained in (577), when two balls are drawn simultaneously, and a little consideration will show that the two processes are really identical. 684. A bag contains five white and six black balls. If five balls be drawn in succession, and no one of them be replaced, what is the chance that the first three will be white and the last two black ? The chances for the successive trials are -f^, ^^o, f, |, f . Hence the chance for the compound event is y\ XiV X f X « X, f = tI T- If the three white and two black balls were drawn in any order, instead of being drawn in an assigned order, the chance would be 10 X yfr == -ff- ^^^'^ ^^® fi^^ things of which three are alike and the other two are alike, may be |5 arranged in f^==-- = 10 ways (Art. 565). As in (583), this [3|^ result corresponds with the one obtained when the five balls are drawn simultaneously. 585. One bag, A, contains three white balls and four black balls; another bag, B, contains two white balls and three CHANCE. 305 black balls; required the probability of obtaining a white ball by a single drawing from one of the bags taken at random. Since each bag is equally likely to be taken, the chance of taking A is J, and the chance then of drawing a white ball from it is f ; hence the chance of obtaining a white ball so far as it depends upon A is ^ X ^ = -^^. Similarly, the chance of obtaining a white ball so far as it depends on B is |. Therefore, the chance of obtaining a white ball is (ft + i) = «• 686. If a person is to receive a sum of money m. in case a particular event happen, and if p represent the chance that the event will happen, then jom is called his expectation. Thus, suppose that there is a lottery with 20 tickets, and one prize worth $100. If a person own three tickets, his chance of drawing the prize is -^^^ and his expectation is worth /^ of 1100 = $15. 587. A and B draw from a bag containing three white balls and three black balls. A is to draw a ball, then B, and so on alternately; and whichever draws a white ball first is to receive $60. Find A's expectation : (I) if the ball drawn be replaced; (II) if it be not replaced. I. A's chance of drawing a white ball on the first trial is h B's chance of having a trial is the same as A's chance of drawing a black ball, that is, J; and since the ball drawn by A is replaced, B's chance of drawing a white ball is ^ X J = }. If B draw a black ball, of w^hich the chance is J, then A will have a second trial, and his chance of draw- ing a white ball on this trial is :^ X i = ^. B's chance of having a second trial is J, and his chance of drawing a white ball on the second trial is y^, etc. Evidently A's chance of drawing a white ball is the sum of the infinite series i + i + yV + - • •=! (Art. 551). Hence A's ex- pectation is f of $G0 = $40. Alg.-2G. 306 ALGEBRA. II. A's chance on tlie first trial is J. B's cliauce of having a trial is -|-; and if A drew a black ball, B's chance of draw- ing a white ball is |; therefore his chance of having a trial and drawing a white ball is J x | = yV? ^^^^t his chance of drawing a black ball is -fjj. A^s chance of having a second trial is the same as B's chance of drawing a black ball, i.e., ■f-^] and if two black balls have been drawn, A's chance of drawing a white ball is f ; hence A's chance of having a second trial and drawing a wliite ball in that trial is -f^ X f = ^%; also, A^s chance of drawing a black ball is -f^ X J ^ -^^^ B^s chance of having a second trial is equal to A's chance of drawing a black ball, i.e., ^^, and then there are only white balls left; hence B^s chance of drawing a white ball, if he have a second trial, is certainty = 1, and his chance of having a second trial and drawing a white ball is 2VX 1= iV- A's chance, therefore, is the sum of ^ and -^^ = JJ. and his expectation is Jf of |G0 = $39. EXAMPLES. 1. In a bag are three white and five red balls: find the chance that, o/ie being drawn, it shall be (I) white, (II) red; hco being drawn, they shall be (III) both white, (IV) both red. Ans. I. |; II. |; III. A; IV. 3V 2. In a bag are three white, four red, and five black balls : if one be drawn, find the chance of its being (I) white, (II) red, (III) black; if two be drawn, of their being (IV) white and red, (V) both black, (VI) one at least red; if three be drawn, of their being (VII) one of each color, (VIII) two of them black, (IX) one of them white. Ans. Li; 11.^; 111.^^; IV. ft; V. ^V; VI. H; VII. ft; viii. A; IX. If 3. The odds against a certain event are 3 to 2, and the odds in favor of another event independent of the former are 4 to 3. What are the chances (I) that both happen ; (II) that the first happens and the second fails; (III) that CHANCE. 307 the first fails and the second happens; (IV) that neither happens ? I. -i-, ; II. ^; III. ^; IV. A- 4. If from a lottery of 30 tickets marked 1, 2, 3, 4, ... , four tickets be drawn, what is the chance that 3 and 4 will be among them? A7is. yfj. 5. A has three shares in a lottery where there are three prizes and six blanks; B has one share in another lottery where there is but one prize and two blanks. What are their comparative chances of drawing a prize ? A^s chance of drawing three blanks is -^^ ; . • . A's chance of drawing at least one prize is ^. B's chance of drawing one prize is ^ = ^\. . * . A's chance : B's : : 16 : 7. 6. When n coins are tossed up, what is the chance that one, and only one, will turn up head? • A7is, n -j- 2". 7. The skill of A is double that of B; what is the chance that A will win four games before B wins two? A's chance of winning any one game is f . His chance of winning four games in succession is (f )* = Jf . His chance of winning four out of the first five games is ^^. In order to win four before B wins two, it is evident that A must either win four straight or four out of the first five. His chance of doing one or the other is ^ -f if = JJf. Ans. |^. 8. A draws five times from a bag containing three white and seven black balls, replacing tlie ball di-awn after each trial. Every time he draws a white ball he is to receive $2, and every time he draws a black ball he is to pay $1. What is his expectation ? A's chance of drawing a white ball the first trial is ^; hence his chance of receiving $2 is -^, and his expectation is ^ of $2 = .60. A's chance of drawing a black ball on the first trial is y\; hence his chance oi paying $1 is -^, and liis expectation is — ^^ of $1 = — .70. Hence his expectation is to lose 10 cents each trial. Ans. — 50 cents. 9. It is 3 to 1 that A speaks the truth, 4 to 1 that B does, and 6 to 1 that C does; what is the probability of the 308 ALGEBRA. happening of an event, if both A and B assert that it hap- pened, and C denies it ? If the event happened, A and B tell the truth and C is mistaken, the separate chances being respectively f, f, and \; hence the compound chance is f X | X \ = ^o- If the event did noi happen, both A and B are mistaken and C tells the truth, of which the separate chances are J, ^, f; hence the compound chance is J X^^ X f = ^fV Now, the event either did or did not happen, and there are six chances in favor of its happening to three chances against it; hence the chance of the happening of the event is f = f . Exercise XXIX. 1. How many permutations can be formed out of the letters in the word MiUo?i 9 2. How many different words can be formed out of the letters in the word CincinnaHf taken all together ? 3. The number of permutations of 2n things taken three at a time is twenty times as great as the number of permu- tations of n things taken two at a time; find n, 4. How many different words can be formed from the letters in the expression ab^(?d^ 9 5. The number of permutations of n things taken three together is J the number taken four together; find n. 6. If the number of permutations of 7i things taken three together be 12 times the number of permutations of |?i things taken three together, what is the number of per- mutations of n things taken all together? 7. The number of permutations of n things taken all together is 720; find n. 8. If 210 different words can be formed from seven letters, of which a certain number are a'& and the others are all different, how many r^'s are there? CHANCE. 309 9. How many combinations can be made of 11 things taken four at a time ? five at a time ? ten at a time ? 10. How many combinations can be made out of the letters in the word Homer ? 11. The number of combinations of n things taken two at a time is 15; find ??. 12. The number of combinations of 3?i things taken four at a time is -^- of the number of combinations of 2ii things taken three at a time; find n. 13. How many different sums of money can be formed from a three-cent piece, a half -dime, a dime, and a dollar 9 14. From a company of soldiers numbering 96, a picket of 10 is to be selected; determine in how many ways it can be done, (I) so as always to i7iclude a particular man; (II) so as to exclude the same man. 15. If the number of combinations of r things taken {a — b) together be equal to the number of combinations of r things taken (a + b) together, find r, 16. A boat's crew consists of eight men, three of whom can row only on one side and two only on the other; find the number of ways in which the crew can be arranged. 17. There are ten tickets, five of which are blanks, and the others are marked 1, 2, 3, 4, 5. What is the probability of drawing 10 in three trials, the tickets being replaced? 18. If the tickets be not replaced? 19. Two players of equal skill, A and B, are playing a set of games for a prize of ^80. A needs two games to win the set, and B needs three games. Find the expectation of each. 20. A bag contains eight tickets numbered 1, 2, 3, . . . , 8. A draws two tickets: what is his expectation, if each odd number be worth $1, and each even number be worth 5 cents? CHAPTER XXX. CONTINUED FRACTIONS. INDETERMINATE CO-EFFICIENTS. BINOMIAL THEOREM. 588. A Continued Fraction is one of the general form 1 c + etc. , in which a,h,c,o, . represent positive integers AVhen the number of temis a, h, c, . , . is finite, the fraction is said to be terminating. For the sake of abbreviation the continued fraction is sometimes written, — -, -7 — -, — —, etc. In this chapter we shall consider m3im\y proper continued fractions, as indicated above; but an integer plus a con- tinued fraction may also be regarded as a continued fraction and treated accordingly. 589. To convert any proper fraction into a continued fraction : Proceed as in finding the H. C. F. of the terms of the fraction. Thus: to convert ^\ into a continued frac- 9 1 tion, divide both terms by 9 ; whence — = 7 7 1 Similarly, divide both terms of - by 7: .*. - — ^ . 2 1 Now divide both terms of — by 2; . * . 7 -^ ^ 7 3 + f _9 _ J^ 1__ _ L_ 1 ^^^^' 25 ~ 2+^1 +^3 4-^2 (810) CONTINUED FRACTIONS. 311 It is evident that every proper fraction may be converted jjito a terminating continued fraction; for, by continuing the process indicated above, we must finally arrive at a point where the remainder is zero, and the operation termi- nates. In like manner, an improper fraction may be con- verted into an integer plus a terminating continued fraction. 590. To find the value of a terminating continued fraction: Perform the operations itidicated, beginning at the right, ■^"•o.l "3 + 1 -3 + i+-163- ■ ^ + —1 2-+A 591. To find the approximate value of an infinite con- tinued fraction: Omit all its terms heyond any assumed term, and oUain the value of the resulting fraction as in the preceding article. Let the fraction be — -, -^ — -, — - , , , a -{- b -{- c -\- If we omit all the terms after the first, the approximate result thus obtained, — , is greater than the continued frac- a tion, because the denominator a is less than the true denominator a -\- -: — r — —- b -\- etc. If we omit all the terms after the second, the approxi- mate result thus obtained, = — -, is too small; for b m; also that Vb < ||f|. 6. If 8' = 32, find X, Ans. |. 7. If 3* = 15, find X. Ans. 2.465. 8. Show that V¥+l = . -I- ^, ^, ^, etc. 9. Show that i/l = l + A^,J-, A-^ J_, etc. 10. I md the value of — , ^^, — , -. Ans. — ^-. Alg.-27. 314 ALGEBRA. INDETERMINATE CO-EFFICIEN^TS. S94.* The method of developing algebraic expressions, by assuming a series with unknown co-efficients, and finding the values of the assumed co-efficients, is termed the method of Indeterminate or Undetermined Co-efficients. 595. li A-^ Bx-^ Cx'^D:^ ■\- , , , = A' -\- B'x + C'x^ -\- D'a? + . . .for all possible values of x (the co-effi- cients of the variable, x, being finite quantities and inde- pendent of x), then A=A',B = B\C=G',D = D', etc. Since the equation is true for all possible values of x, it must be true when x = 0', then A =:^ A'. If ^ = ^' be subtracted from both sides, then Bx -\- W -]- Da? -^ . . . = B'x-{- C V + D':x? -f . . . J)WidiQhj x; .' , B-{-Cx-\-Da? -\- , . . =5'+(7'a:+i>V+ . . . This equation is true for all values of x, because the various co-efficients are independent of x; whence, as be- fore, put 2: = 0, and B = B', In a similar manner, C = C, D •= D' , etc. 596. \iA-\-Bx-^Cx^^Do?-^. . . = 0, for all possi- ble values of x (the co-efficients A, B, C, D, etc., being finite quantities and independent of ic), then A =0, B =0, C=0, I) = 0, etc. Since the equation is true for all values of x, it must be true when x = 0; ,' . A = 0. As before, divide by x, and then put x = in the re- sulting equation; .' . B = 0. In like manner, C = 0, j9 = 0, etc. 597. To develop into a series by means of inde- ct ~j~ ox terminate co-efficients. The series will consist of the powers of x, with certain co-efficients depending upon a or h, and it is evident that I2^DETERMIl!fATE CO-EFFICIEin:S. 315 X will not occur in the first term. Therefore, assume a-\-bx Clear of fractions by multiplying both sides by a + hx. ,', a = aA -\- aB x -\- aC rc^ + aD I ar* + . . . -i-Ab -\-hB -^hcl 4- . . . I. ,' , aA = a, from which ^ = 1. 11. aB-{- Ah = 0; that is, a^ + ^> = 0; . *. ^ = . a 2 "2* III. aC + ^'^ = 0; that is, a(7 + Z'f- ^Wo;.-. (7=+^ IV. aD-\-})C = 0; that is, aD + h{^^ =0;.'.D=- Whence — — r- = 1 x-^ -^a^ ■,^+' • • a-{-dx a a^ a* ' If additional terms be required, they are readily supplied by observing the law of the series; viz., Each successive term is found from multiplying the next preceding term by Xy the general term being ( xj 598. To develop ^ into a series. oX — X The first term will evidently contain x~^. Hence put ^^^ = Ax-^ -\- Bx"" -{- Cx -}- Dx"" -\- , . . Clearing of fractions, 1 = 3.1+3^ x+3C x^-{-dD \x^-{- ,,. B -C -A I. 1 = 3J; .-. A =i. 11. = 3^- A; that is, = 35 - ^-; . • . ^ = i. III. = 3C - 5; that is, = 3C - i; . • . C = ^, etc. general term being (iYx"-\ 316 ALGEBRA- PARTIAL FRACTIONS. 599. An algebraic fraction may be sometimes decomposed into two or more simpler fractions. 2x 3 To decompose -^ — — — -—- into two fractions whose de- X — oX ~\~ /i nominators shall be the factors of o:^ — 3x -j- 2. The factors of ic^ — 3:?; -f ^ ^I'e x — 1 and x — 2. 2a; -3 A , B Assume -^ — -— - = 4 x^ — 3x-\-2 X — 1 ' x — 'Z Clear of fractions. . • . 2^: — 3 = ^(^ — 2) + B(x — 1). Transpose. (2^ + ^ - 3) -j- a;(2 - ^ - 5) = 0. From (596), I. 2^ + ^ - 3 == 0. II. 2 - .4 - ^ = 0. 111 = I + n. ^ -1 =:0; .-. J = 1. .-.^ = 1. 22^-3 1.1 ' ' Q?-^x-]-'Z x-\ ' x-2' EXAMPLKR. Prove the following identities: 1. i^| = l + 5^ + 15a^ + 45a?+. . . + {^x){ZxY-\ 1 1,2 ,42,8,, 2- 3-2-^=3+9^ + 37 •^+8i^+- ■ - + 1 (H"-'- = 1 + 2ic + 3a;2 + 42:^ + . . . + 7ix' 1 - 2:c + ^ x"" 4.1 1 * {x + l)(r^ - 3:?: + 2) 3(:r - 2) ' Q{x + 1) 2(2;-l)* ,/-^rn — 2 I ^ a;* . 0^ 2a Sa^ ' 16«^ 6.vr+7+^ = i + | + ^^-g + ^ 52; - 14 . 3,3 7. Decompose ^-^gi+S- ^'''- J^ + ^ BINOMIAL THEOREM. 317 THE BINOMIAL THEOREM. Peck's Proof. 600. Lemma. —] = na;""^, for all values of 7i. x — yjy^^ In other words, if y =^ x, then, for all values of n, ic" — ?/" _ * nx" \ x-y I. When n is a positive integer. rr^ .»/'* By division, ——x'^~'^-\-x'^~*y-\-x^~'^y'^-\- . . . +«/"~^ X — y We have shown that the division is exact (Art. 217). The number of terms in the quotient is n, because there are ?i — 1 terms containing y, and one term containing x only. li x = y, each of these terms becomes x'^~'^, and their sum \x — y J^-' II. When 11 = T-, a positive fraction. a a a — b Let x= r^; .'. x^ = r", and x^~ = x~^~ = r"~^ a Let y = s^; . *. «/& = 5", and it y = x, s = r. xb — y^ _ ^''* — s** _ J r — fi x-y ~ r" - 6-^ "" 1 r'^ - s" r — s ar"- III. When n is a negative integer = — m. x-y ^ \ x-y ly^^ 318 ALGEBRA. IV. When 71= — —, a negative fraction. X ^ ^x ^\—x^ — T-X = nx""-^. 601. To expand {ct + xy. I. Assume (« + ^)" = ^ + 5a; + Cx^ ^ Dx" -\- , , Making 2: = 0, we find a" = A; hence II. (a + x)"" = a''-i-Bx-i- Cx^-\-Da^-i- , . . Substitute y for x, then III. (a + yr = a^ + By-i-Cy'-i-I)f+.., IV = II - III. {a + xY - (a + yY = B{x - ?/) + C(x' - y^) + i){x^ -y') + . Divide both members by the identity {a-\-x) - {a-\-y) = x- (a-{-x) - {a-\ry) B -{- C{x -{- y) + D(x' ^ xy + f) -^ , . But it X = y, then, for all values of n, V becomes VL 7i(a JrxY-'' = B-i- 2Cx + Wx^ + 4^;?:^+ . . . Multiply both members of VI by (a + x)j hence VII. 7i{a-\-xY = aB-\-2aC\x-\- 3aD I x"" + 4«^ a;^ + . + -5 I +3C I -f3Z> Multiply both members of II by n; hence VIII. 7i(a-{-xY = na"" + ?i5:r -^-nOx" -\-nDx? + ^^^2;* + . Equating the second members of VII and VIII, IX. aB-\-2aC X + '6aD 4-2(7 :^-{- x'^^UE + 3Z) = na"" + ^i^rc + ^^(7:^2 _|_ ^^^^ + . . . Hence, by the principle of indeterminate co-efficients, X. aB^.iia''; .-. B = na^-\ BINOMIAL THEOREM. 319 Xl.UO+B = nB; .-.0 = 3 ^^^ = '^^a'-\ XII. 3«i) + 2(7=^i6'; . n - pin - 2) _ n{n - 1) {n - 2) and so on. Substituting these values in II, we have, for any exponent, {a + xY = a« + 7icC'-''x + ^^v^^~" •'"V -^a;^ + 2 3 "^ .-^ + . . . . . . + iC". 602. If a; be negative, then the terms containing odd powers of x are negative; .'. (a - xy = «"- 7i««-^r + !^MLlI_ii^n-2^_ ^ ^ ^ 603. 7)) Jind the numerically greatest term in the ex- pansion of (a + ^)". I. If u be a positive integer. An examination of the series will show that the (r + l)th term is derived from the ?'th term by multiplying the latter by '. — . _ = f — L 1 j_. Evidently this multi- plier diminishes as r increases, but as long as it is greater than unity, the (r + l)tli term is greater than the'rth term. The rth term will be the greatest when the multiplier first becomes less than unity. When f — — Ij- < 1, then r > ^^ — TTT"-'' ^i^^<^^ tl^6 ^'tli t^i'^ is the greatest when r is. a ~f~ the first integer greater than i~\ -. 320 ALGEBRA. If ~^ j- be an integer, then the (r4-l)th term is equal to the rth, and each of these is greater than any other term. II. If u be a positive fraction. If — > 1, there is no greatest term, for /the series will a jaiiest tei evidently diverge. If — < 1, the greaiiest term or terms may be ascertained as in I. ^0r III. If n be negative, either integral or a fraction greater than 1. The multiplier that changes the rth term into the (r+l)th , — n — r-\-l b 71 -\-r — 1 b ^,, IS . — = . -. As the numeric- r a r a ally greatest term is sought, reject the negative sign be- fore the multiplier; then, as in I, the rth term will be the greatest when . - is first less than 1 ; that isX^ ^ when r is first greater than -^= --. oJr / a — b ^ As in I, if _j^ be an integer, there are two equal terms each greater than any other; and if, as in II, - > 1, there is no greatest term. IV. If n be negative and < 1, and - < 1, the first term lY) I n* 1 7i is the greatest ; for in this case -— ^- . - < 1 for all r a values of r; that is, each term is less than the preceding. The same rules apply to the expansion of (« — bY, be- cause the terms differ from those of {a -\- b)^ only in the signs, and the sign does not affect the arithmetical value. ^W^^^ BINOMIAL THEOREM. 321 EXAMPLES. Expand to four terms each of the following expressions: 1. (1 _|_ x)^. Ans, l-^ix-ix^-i- -^a^ - . .' . 2. (1 + ^)"^. ■ ^ns. l-ix + ^x^- tV^ + . . . . „ .1 . X x^ 6o^ 4. Va^ — X?. Ans, a — -^ 5-^ — tf~6 — • • • Find the grea^st term in each of the following : 6- (3 + 1)^. ^ns. Second term = 810. ^ 7. (2 - i)-*A Ans. Y^[ 8. (I+I)". Ans, Fifth and sixth, 9. (1 + i)-^^ Ans. Thij:d. 10. (l-7\)-t. ^W5. Thipd. Exercise XXX. Reduce each of the following to a continued fraction, , and find the successive convergents: r-^ 1. ,WA. 2. iV^-A- 3. if J. 4. iff. « Find the first four convergents in the following: V 5. Vs. 6. vTr. 7. V53. 8. i^i. 9. 1^. Y if 10. Simplify ^, 3I-, 5^:, 6^, ^ . 2 11. Expand (l + a:)~^ 13. Expand 3-1- 4a; 1 4- a: 12. Expand (1 - xY\ 14. Expand ^_^g^,_j_3^ « 823 ALGEBRA. 3.r 15. Decompose x" - ^x + 15' 16. Decompose -^--^-^. 17. Decompose ^^—^^-. 18. Decompose -j. 20. Expand (1 — a:*)^. 2 19. Expand (.T + y)-\ 21. Expand ^-^— -^^jl 22. Write the 49tli term of {a — x)^. 23. Find the greatest term of (1 + xy, when x = \ and w — 4. 24. Extract the square root of 24 approximately. 25. Eind the number of terms in the expansion of (« + ^ + c + dY\ 26. What is the sum of the co-efficients in the expansion of (2«-5 + 36f«? 27. In the expansion of (^a — J — 2c)^^ what is the sum of the co-efficients of the terms containing odd powers of a^ 28. Simplify -, 7^—7 + 7 — r-JY^^ ^ -" ((t — b) ^ (a -f b) ^ CHAPTER XXXI. BUSINESS FORMULAS. 604. Interest is money paid for the use of money. The principal is the sum lent. The aviount is the sum of the principal and interest at the end of any time. The rate is the interest of one dollar for one year. 605. Interest is of two kinds, simple and compound. Simple Interest is interest charged upon the principal only. Compound Interest is interest charged upon the sum of the principal and the interest due. SIMPLE INTEREST. 606. Let the principal be represented by P; the interest on $1 for one year by r; the amount of $1 for one year by R; the number of years by n; the amount of P dollars for n years by A. Then: I. R = l-\-r. II. Interest on P for one year = Pr. III. Amount of P for one year = P(l + r) = PP. IV. Interest on P for n years = Pur. *V. Amount of P for 7i years = P(l -}- 7ir), From (V), A = P{1 -{- nr). 607. If any three of the quantities A, P, n, r, be given, the fourth may be found by substituting the given quanti- ties in the equation (V), A = P(l -f nr). Thus: 324 ALGEBRA. \ Ex. 1. At what rate will $500 amount to $650 in 3 years 4 months ? Substituting in Formula V, 650 = 500(1 + ^r). A71S. The rate is 9^. Ex. 2. At 6^ per annum, what principal will amount to $650 in 5 years ? As before, 650 = pfl + 5 X ^). r.P = ^- = 500. Ans, $500, 608. Since P will, in n years, amount to ^, it is evident that F at the present time may be considered equivalent in value to A due in n years; thus F may be regarded as the present worth of a sum A due in n years. Thus, from Ex. 2 in the preceding article, $500 is the present worth of $650 due in 5 years, if money be worth 6^ per annum. 609. Discount is an allowance made for the payment of a sum of money before it is due. From the definition of present worth, it follows thai a debt due at some future time is equitably discharged by pay- ing the present value at once; hence: The teue discount is equal to the amount clue diminished hy its present loorth. Thus: N\.D = A-F^A- --^ = ^i!i!L. 1 -|- nr 1 -\- nr For examples in simple interest consult any arithmetic. COMPOUI^D INTEREST. 610. When the interest is compounded annually^ The amount of F in one year is FR; in two years is FR{\. -j- r) = FR^; in three years Is F R^{\.-{-r) — F R^ ; etc. • *. the amount of F in n years is FR^, That is: VII. A = FR^ = F(l + r)». BUSINESS FORMULAS. 325 611. Having given any three of the quantities A, P, r, n, the fourth may be found by substituting the given quanti- ties in VII. A = P(l^ r)\ Thus: VIII. F = jY^n^ IX. r = V~ - 1. Since R- = i-..', X. n = ^^^-M^. F log li 612. The interest is equal to^-P=Pi^-P=P(i2»-l). .'. D = A — F = FiR"" — 1); or, since F = -^-, therefore, 613. When the interest is due more frequently than once a year. (1) When due semi-annually. The amount of P in ^ year is P f 1 -f 9-) 5 in lyear isP^l + ^j; in n years is P (1 -f ^) • .•.{XII)^=p(l + 0". (2) AVhen due quarterly (XIII), ^ =P ^1 + ^^ . / ^. \ 12n (3) When due monthly (XIV), ^ = P ( 1 + -- 1 . (4) When due q times a year (XV), ^ = P f 1 + - j . XVI. As before, D = A- F = F XVII. The present worth, P [('+?)■■-'] (-?)■ 326 ALGEBRA. AN^NUITIBS. 614. An Annuity is a sum of money payable annually, or at fixed periods in»the year. 615. To find the amount of an annuity left unpaid for any mimher of years, alloiving compound interest. The sum due at the end of first year = 8; of second year = 8 -[- 8R; of third year =:8^8R-{-8I^; and so on for n years. .-. A = 8 + 8R -^ 8R' -\- SB' + , . . +8R''-K r. A = 8{l-{-R + R'-{-R'-j- . . . + 7?"-i). ...XYIII. ^=^--_l) = ^(^-I). R — 1 r 616. By means of Formula XVIII, joroblems relating to sinking funds may be solved. Thus: A city owes $300,000, payable at the end of 10 years, without interest. What sum must be set apart annually, as a sinking fund, to pay this debt, money being worth 6^, compounded annually ? . - v^. V } ^ ^TTTTT c ^r 1300,000 X .06 , ,^ ^^ /< From XVIII, ^S' = -^^p;^ = __A___^__ =, |22,670. 617. To find the present ivorth of an annuity to continue a certain number of years, allowing compound interest. Let P denote the present worth ; then the amount of P in n years is equal to the amount of the annuity in the same time; . • . Pk^ — — ^^ — ^—. h — 1 . • . XIX. P = '^(^ - ^'"^ = S\l-{l + r)-' }_ R — 1 r 618. If the annuity is perpetual, then, in Formula XIX, (l + ^")~" differs from zero by less than any assignable quantity. . • . XX. P = — = -. BUSINESS FORMULAS. 337 619. To find the present worth of an annuity to com- mence at the end of jy years, and then to continue q years. Subtract the present worth of the annuity for p years from the present worth of the annuity for p -\- q years. Thus: ^ " R-\ R-1 - R-l^^' ^ f' or, XXI,P = '-^R-^-R-^-^) = -j X-^\^ , 620. If the annuity is to commence at the end of p years and then to continue forever, we must suppose q to be infinite; hence, in XXI, R~p~^ differs from zero by less than any assignable limits; hence, XXII. F = ^{R-^ -0) = '^"^ ^ R^{R - 1)* 621. To find the annuity when the present worth, the time, and the rate per cent are given, S(l- R-"") From XIX, P = R _ P(R-l) ^ PR^{R-l) _ PrR- 1 - R-^ R^ -1 ^-1* EXAMPLES. In all the following examples days of grace are not con- sidered, and one mouth is counted equivalent to ^^j of a year: 1. The sum of $300 amounts in 25 years to $900. What is the rate: (1) at simple interest? (2) If the interest be compounded annually ? Ans, (1) 8^; (2) 4^^. 2. The interest on a certain sum is $18, and the discount on the same sum for the same time and at the same rate is 115; find the sum. Ans, $90. 3. In how many years will $10 amount to $105 at 5^ compounded annually? Ans. 48.17. 328 ALGEBRA. 4. In how many years will any sum double itself at com- pound interest, at 5^ ? Aus. 14.2066. 5. In how many years will a sum of money treble itself, if compounded annually at 3|^? Ans. Nearly 32. 6. What is the amount of an annuity of $120 for 20 years at 6^ per annum ? Ans. $4412.80. 7. At 5^ per annum, what is the present worth of an annuity of 1500 for 30 years ? A7is, 17687.86. 8. What is the present worth of a perpetual annuity of $600 at 6^ per annum? Ajis, $10,000. 9. At 4^, what is the present worth of an annuity of $450, to commence at the end of 10 years, and to continue 20 years? Ans. $4130.10. 10. At 6^ per annum, what is the present worth of a perpetual annuity of $5000, commencing at the end of 15 years? Ans. $34,780. 11. At 8^ per annum, compounded quarterly, what will be the amount of $1000 in 2 years 7 months ? Ans. $1225.11. Exercise XXXI. 1. What is the amount of $1 for 200 years at 5^ per annum, compound interest ? 2. Find the logarithm corresponding to the amount of $1 for 1000 years, compounded at 5^ per annum. 3. In what time will any sum double itself at 8^ per annum, if the interest be compounded quarterly ? 4. In how many years, at 5^ compounded annually, will $1 amount to $10 ? 5. What is the compound discount on $1000, due in 20 years, at 5fo ? BUSINESS FORMULAS. 329 6. What is the present worth of an annuity of $1000, at 5^, for 30 years? 7. What is the present worth of a perpetual annuity of $1000, at 5^ per annum ? 8. A debt of $40,000, at 6^ compound interest, is dis- charged by eight equal annual payments; required the annual payment. 9. Find the present worth of an annual pension of $90, at 5^ compound interest, to continue forever. 10. What is the present worth of a perpetual annuity of $1000, at 4^, payments to begin at the end of 10 years? 11. A debt of $3700 is discliarged by two payments of 12000 each, at the end of one and two years; find the rate of interest paid. 12. At 4,^, what annual sum should be paid for 30 years, at the beginning of each year, to secure the payment of $10,000 at the end of that time? 13. An annual premium of $300 is paid to a life insur- ance company for insuring $10,000. For how many years must the premium be paid in order that the company shall sustain no loss, money being worth 4 per cent compounded annually ? 14. A debt of %cl, compounded at rji^ per annum, is discharged in n years by equal annual payments of $Z>; find n. Alg.-28. CHAPTER XXXII. TEST EXAMPLES. Time, 2 hours for each set. I. 1. When « = 5, 5 = 4, c = 2, find the value of 3« -2[b-2c-3(a-b + 2c)-\-(- a-2b-{- c)] - — . c 2. Simplify 4.a-3{a-b-[2{2a - db)-{-a-2b)]-3c\, and verify the result when a = 4, b = ~ 2, c = d. 3. Simplify {a - 2b) {2a - 3b) - (2b - a)(b - 4«). 4. Expand (x + l)(2a; - l){3x - 2). 5. Expand {— c — 2d)K 6. Divide {- a - b)' hy - (a -{- bf. 7. Divide — a^ — ¥ — c^ -\- 3abc by a + 5 + )" by (a + b)\ 7. Simplify (a - by - {b - af + (J - af. 8. Simplify Vd' -{- b^ -{- c" - 2ab - 2ac + 2bc. 9. Divide the cube of (a^ — b^) by the cube of {a + b). 10. A can do in 7 days what would take B 11 days; how long would it take both, working together, to do a piece of work which A alone can do in 21 days ? III. 1. Interpret 6* = 36, and find the value of x, 2. Simplify 3(« - b)a'' - a^ -\- a^b. 3. Expand {a — b)(a—c) — (n — b){a -\- c) — (a—c)(a + c), 4. Expand (a^ - ab + by + (a^ + ab -i- J^f, 5. Divide {a — b- cf^-ia -b- c)'^-(a — b-cY by (a-J-c)"^-". 6. Divide {—a— b —c){ab -{- ac-\-bc) -\- abc by {—a—b). 7. Given 4«a; — 2b = 3ab — 6«*a; ; find x, 8. The sum of a certain number of quarter-dollars and an equal number of dimes is $21. How many dimes are there ? 9. A is three times as old as B; 10 years ago he was four times as old. How old is A ? 10. A bought a certain number of oranges at two for five cents, and an equal number at five cents each: by selling the lot at three for a dime he lost 15 cents. How many did he buy ? IV. 1. Interpret x^ = -^, and find the value of x, 2. Simplify 2a - 2{a - 2 - 2[a - 2 - 2{a - 2)]}. 3. Expand {x — 2)(x — b){x + c). 332 ALGEBRA. 4. Show that (a' - ^^ + l)(a* - 1)+ a\a'-l)(a' + 1) = 5. Divide — a;' — ?/' + 3a;i/ + 2a; + 2^ + 1 by —1—x—y. 6. Divide (:^+?/^)(a;^ —2/^) hy (2;^+ ic^y + 2/^)(a;^— ^y+ «/^). 7. Simplify (« -f- Z* - c)^ + (- a - ^ + c)l 8. Given i{7x + 9) - 2(a; - i[2a; - 1]) = 14; find x. 9. A has as much money as B and together; B has twice as much as 0; and A has $40 more than 0. How much has A? 10. A can walk from to D in 20 hours, and he can ride the same distance in 6 hours. He goes from to D, walking half the time and riding half the time. How long does it take him to go from to D ? V. 1. Expand (1 + 3x -\- ^x" -\- x^){l + 2^ + a^). 2. Expand and simplify {a^ -\- x -\- ly — {a^ — x -{- 1)^ 3. Divide - a^ - b^ - 1 ^ 3abhj - a - b -1, 4. Show that {a^ + b^)((y' + d'') = {ac ± bdf+(ad =F bcf, 5. If — ^ ~— = dx — 7)1, what must be the value of m in order that x = 7? 6. A can do twice as much as B, and B twice as much as C ; working together they complete a piece of work in 7 days; how long would it take A alone to do the work ? 7. The time past noon equals one third the time past midnight. What o'clock is it? 8. If each side of a square field were four rods longer, the area would be one acre more; find the area of the field. 9. Given I +1 = 7, 4"~9-^' find x and y. TEST EXAMPLES. 333 10. Given x-^2y-\-dz= 8, "i 2x— y — z = 15, Vfind x, y, and z, 3x-4:y-2z = 16; ) VI. 1. Reduce [(ar* — 2abf + a^b^] -i- {3^ — ah) to its simplest form. 2. From the cube of (2a — b) take (a — 2by, 3. V^ - Gx^y + 15a:y - 203^y^ + lbxY-6xy^ +«/*' = ? 4. («i - 1)5 z::: ? 5. A's age is three times the united ages of his three sons, whose ages differ by two years respectively. In four years A's age will be | their united ages. How old is A? 6. Given - - -^ = 1, find X and y. 7. If x^ — 3x = 0, find X, 8. Ux^ — Sx= — 15, find x, 9. The square of A's age is 75 more than 10 times his age. How old is A ? 10. A's age is now equal to the square of B's, and in four years it will be four times B's. How old is A ? VII. 1. Expand (-2a- dby. 2. Extract the fourth root of 5 to 2 dccjmal places. 3. A travels three times as fast as B, and requires 5 hours longer to go 60 miles than B does to go 10 miles; find A's rate of travel. 4. A number expressed by two digits equals 7 times the sum of its digits, and if 18 be subtracted from it, the digits will change places; find the number. 334 ALGEBRA. 5. If a rectangular field were 2 rods longer and 4 rods wider, it would be a square containing 64 sq. rods more; find the area. 6. Solve 2x^ — 3x = 20. 8. Factor x^ — xK 7. 2:^-\-x = 21; find x. 9. Factor — a^ — 3x -}- c(^ -\- d. 10. Factor x^ — lice + 18. VIII. 1. Find the G. C. M. ot x^ - da^ — 5c(f -\- 3x -\- A and 2x* _ 6^:3 - 8x^ -\-6x-{-6. 2. Find the L. 0. M. of {a - by, a^ - h\ a^ _ ^ab + l\ 3. Solve 25^2 _x^x= -2, . 4. Form the quadratic whose roots are 5 ± V^. 5. Form the cubic whose roots are 1, — 2, 2. 6. Find three numbers whose sum is 11 : the sum of the first and third is one less than twice the second, and the smallest number is the positive root of the equation 2x^ 4- ic = 3. 7. Expand (1 + fT) (2 - '/S) ( Vs - 1). 8. Solve x" ■\-2x? -x = 2, 9. Simplify —— -—. 10. Find four fourth roots of 4. IX. 1. Factor a^ + 3a^ -4a- 12. 2. Find the H. 0. F. of a'' + 1 and a^ - a' - a^ - l, 3. Find two numbers whose sum is 4, and the sum of whose squares is 10. 4. The square of f of a certain number is one more than the number itself; find the number, 5. A's rate in going 80 miles exceeds his time by 2; find his rate. TEST EXAMPLES. 335 6. Solve a^ -^ 5x = 12. 7. Form the equation whose roots are the factors of a^-x-2, 8. The time past noon is the square root of the time past midnight; what time is it ? 9. 2VE-Vs6-{-Vi5- '/125 = ? 10. If x^ -{-y^ = 10 and xy = 3, find x and y, X. 1. The log of 5 to the base a; is J; find x. 2. The log of 2 is .3010; what is the log of 2J? 3. If 2^ - 4 : a; - 3 :: 2a; + 1 : 2, find a;. 4. V^~^fU - Vx^-9 = 2; find x, 5. Show that a^ + 3^^ ^ 2b{a + b), unless a = 5. 6. Simplify 5 h X 2 h'^ -r- f^l08. 7. Simplify (5 + V^^)(5 - V^^)( Vd- 2 V^^K V3-^ 2 i^^). 8. If « : ^ : : c : <7, show that a^-\- ab : ab — i^ :: c^ -{- cd : cd — d\ 9. Given a; + a;-^ = 2.9, find the value of x, 10. A sold a horse for $144, and gained as much per cent as the horse cost him; find the cost. XL 1. Show that a^ -}- b"^ -]- 1 > ab -{- a -}- b unless a = b=l. 2. Extract the square root of 20 VG — 48. 3^ Given log 2 = .3010; how many figures express 2^^ ? 4. Given 25^ + 5^ = 650; find x. 5. If a; + 5 : 2a; — 3 :: 5a; + 1 : 3(a; - 1), find a;. 6. Ua:b::b:c, prove a : c :: {a -^ bf : (b -\- cf. 7. The eighth term of an A. P. whose common difference is 3, is 14; find the third term. 336 ALGEBRA. 8. Find two numbers whose sum is 10 and whose geomet- ric mean is 3. 9. Sum the infinite series i — i + f — •/t + ' • • 10. Insert three harmonical means between 2 and 10. XII. 3/— 1. Simplify V'lO - \^'Z 2. Find X when x^ - 4:X -}- 2 : x^ - 4:X :: x^ - 2x - 1 : x^ ~ 2x - 2. 3. In an A. P., given d, I, s; find a. 4. Find three numbers in G. P. whose sum is 13, and the sum of whose squares is 91. 5. If 3^3^ = 20^ and 5a; = ?>y, find x and y; having given log 2 = .3010 and log 3 = .4771. 6. How many different sums of money can be formed with 2 cents, a three-cent piece, a half-dime, a dime, and a quarter-dollar ? 7. If log xy = 1.4982 and log (x 4- y) = 1.6543, find log x and log y. 8. How many figures will express the amount of $1 for 1000 years at 6^ per annum, compound interest? (Use table of logarithms.) 9. What is the present worth of a perpetual annuity of $1200 at 4^ per annum ? 10. Of the same annuity, beginning after 10 years ? XIII. 1. Divide a -{-b -\- c — 3 Vahc by ^^ -|- ^^ + c^* 2. Multiply .-ija + 1 + ^~^ hy x^ ^ 1 -\- x~K 3. li a = — S, b = — 5, c = ^, find the value of (- af - 2{b - Ac - [a ^ b -\- 12c - (a - b -}- Sc^)]]. 4. Resolve into prime factors x^ — 7x^ — 8. TEST EXAMPLES. 337 5. Vl6x-^ - 322^-3 + 24a;-2 - 8x-^ + 1 = ? 6. Find two consecutive numbers, the difference of whose cubes is 61. 7. Expand (a^ — 3^>)*. 8. Expand (2a-^ - 3b-fy. 9. Factor rt» — 256 "^ 10. Simplify VbXi^Xf^. XIV. 1. Divide ic* — ?y* by a:~^ + ?/~^ ^. , .^ a; + V2 - x^ 4 2. Find a:, if ^ = -. ^^_- ^2 -2^2 3 3. ^28 - 5 4/12 = ? 4. Expand to four terms Vl ± dx, 6. Divide 21 - 5 4^ by ^3 + 1. 6. Sum 3, 3^, 3f , . . . , to fifteen terms. 7. Form the equation whose roots are double those of 3^-2x = 3. 8. Insert six arithmetical means between 2 and 23. 9. Itx-\-i/ = a and x^ -\- xy ~{- y^ = h, find x and y. 10. Find the fifth term of {x + «)"". XV. 1. Solve Vrc + 22 — Va; + 3 = 1. 2. Find three cube roots of 27. 3. If 2* = H^, find X. 4. If the base is 25, of what number is —^ the logarithm 5. In the common system, what is the log of ? 6. If (|)« = I, find X, 7. Simplify— , —,—,-. Alg.-29. 338 ALGEBEA. 8. By indeterminate co-efficients, find four terms of the quotient of----, 9. Expand (a — bfK 10. If . ] , = a-' 4- ^"' + ^"S find X. a -^ -\- X XVI. 1. Given log 2 = .3010; find log 2"^ 2. Expand to four terms (a"" — b'')~^, 3. If 6^ = 36, find x, 4. Form the equation whose roots are the reciprocals of the roots of a;^ — 7:?; = — 10. 5. Divide a^ + 3a'b + Sab^ -{- b^ - c^ hj a -}- b - c, 6. If 7(x — y) = S(x -\- y), vrhat is the ratio ot xto y 9 ^. ,.„ 1 2,1 7. Simplify h X — 1 X X -\- 1 8. If a-\-b : a—b w c -\- d '. c — d, show that a \b \\c \ d. 9. Solve a^ + («-^ - ^-> = {ab)-\ 10. If c? is the least term in the proportion a : b :: c : d, prove that a^ -{- d^ > b^ -\- c^. XVII. 1. Extract the fourth root of 7 — 4 1^. 2. Solve Vl4 - X + fl4 + a; = 4. 3. Simplify {a^ - b') X («' - b') -^ {a - bf. 4. Simplify (a;^-3a:2-15a:+25)-i-(a;3+7a;2+5a:-25)-^ 5. What must be added to 9:r* + 12^:^ + 20:?; + 25 to make it a perfect square ? 6. Expand (2a; — «)^ TEST EXAMPLES. 339 7. Given 6 (x^ + ^i) + ^ [^ + ^) ^ ^^' ^^^ ^' 8. The sum of five numbers in A. P. is 15, and the sum of their squares is 65; find them. 9. If a^(c(^-y^) = 25 and y^x" + «/')= 19^, find x and y, 10. Divide a + aH^ + b by a^ + t^^^ + ^>i XVIII .t of 16 2. Find the maximum or minimum value of 1. Extract the cube root of 16.3131 ... to two places. 2x2-2a;+5 2;2 2x -4- 3 3. Solve: I. {x ^ y){x' ^ y^) = 203; II. {x'-y^){x-y) = 63. 4. Solve: I. x -\- y -\- z = \0\ II. 7?-^y^-\-z^=z 38; III. xy — xz — yz— — 1. 5. Simplify (1 - of X (« - l)"'. 6. Divide x-^-\-x~^y~^ -^y-'^h^x'^ + ^~i?/~^ +y"i 7. Find a fourth proportional to a~S ^"^ c~^ 8. Sum to infinity 9, — 3, + 1, — J, -|- . . . 9. In a bag 9 white and 3 red balls are placed ; if 5 balls be drawn, find the chance that all the red balls w^ll be among those removed. 10. Solve 15^:^ —9x-4. = 2x Vdx + 4. XIX. 1. Solve x' + lOa; = 57. 2. Given (2 5)^ = ^, find x. 3. Simplify (4a;-t)-i 4. (x--^ -y-') -^ {x^ ~y^) = ? 1 3 5- ^ ^^; find X. X — 14- X 340 ALGEBRA. 6. A's rate exceeds B's hj^ of a mile per hour, and he goes 39 miles in 45 minutes less than B does; find A^s rate. 7. Find the fourteenth term in the expansion of (2~^—by^. 8. Solve IV+T -V-x'-{-i=Vl- xK 1 _ '%x 9. Resolve into partial fractions - - a; - 20 10. Find an A. P. such that the sum of n terms shall be equal to n^, XX. 1. Sum the A. P. whose first and twelfth terms are G and 61, and the entire number of terms is sixteen. 2. Find the A. P. such that the first, third, and thir- teenth terms are in G. P. 3. Insert five harmonic means between — J and — f^. X^ 11^ 4. Solve 1- -^- = 9, and a; + ?/ = 6. y ' X 5. Expand (1 + r - 2)^ 6. Solve 20^ = 100. 7. If — I — = 5 and —„ + — „- ==13, find x and ?/. X y X? ' if •' 8. How many words, each containing one vowel and three consonants, can be formed out of the first thn-teen letters of the alphabet? 9. If ax^ -\- bx -{- c becomes 24, 37, 60, respectively, when X becomes 2, 3, 4, what will it become when x = — 3? 10. Prove that the difference of the squares of any two odd numbers is exactly divisible by 8. ANSWERS. Exercise I. 1. A, $175; B, $225; C, $360. 2. 16; 4. 3. 252; 336. 4. 38. 5. 14, 15, 16. 6. 100. 7. 5. 8. 28. 9. 18 years. 10. 5. 11. 60. 12. 20. Exercise II. 1. 1; - 11; 11; 2. 2. - 5; 4; - 8. 3. 2; 4; 4. 4. 1; - 1; 4; 1; - 2. 5. -J; -^; |; j\; f 6. 57 a.d. 7. 33 B.C. 8. 125 B.C. 9. 55. 10. + 1266; - 322; - 623; - 1375; - 840. ii. $3823. 12. - $1200. 13. 268; 22 steps. 14. 75 steps; 675 steps. Exercise III. 1. 22. 2. 17. 3. 42. 4. 256. 5. 3. 6. 28. 7. 0. 8. 16. 9. 64. 10. 0. 11. 3. 12. 2. > Exercise IV. 1. 6a. 2. a(db~c). 3. «(«— 3). 4. 3(b--c). 5. 6{a—'t). 6. b{a + 3). 7. a(a^ - 1). 8. a(b + 1) + c(b + 1) = {a + c){b + 1). 9. a\a + 1) + (a + 1) = {a^ + !)(« + 1). 10. (x'-{-l)(x-l). 11. {a'-{.l)(a^-a+l), 12. x[(a+b) + {a—b)]=2ax. 13. {« — ^)(«/ + 2;). 14. b{a—c){d—a). 15. [(a + ^) + (r? - b)] ^x- 2a V^, 16. {a^ + Z'')(a - ^'). 17. (3a + 2^)a;i 18. Sx{a - b), 19. (4 + c)(a - b). 20. 3(2a - b){c + (?). (841) 342 ALGEBRA. Exercise V. 1. (2b-^c)x. 2. i2b-{-c-\-l)x. 3. 10(a-b). 4. (2x-y) {a + 3b). 5. 3a -7b- 2c, 6. a - a^ = a(l-a^). 7. 2a. 8. 4^> - 8 = 4(6-2). 9. ^+ (a-b)a^ + (a-b)x-i-(a-l). 10. x^-^ x{y + 2;)+(^2+?/2;+22). ii. a?J^x\y-z^l)-\- xy. 12. (« - b)x' -(a- b)^? -(a-\- b)x + (a - b). Exercise VI. 1. 6a;* — 96. 2. x^ - 6:^3 - 24a;2 + 121a; - 120. Z. x? — 4:ix — 120. 4. 4a;« — 6a^ -\- Sx'' — lOa;-^ — 8x^ - 5x - 4. b. ^ + 3xy + ^» - 1. 6. a:* - 6a;» + 13^:^ - 12a; + 4. 7. a;^ + (6 — a + c)a-^ -j- (— ^^ ~ ^'* + ^^)^ ~ ^^^' 8. a;^ + 6a:^ + 15a;* + 20a;3 + ISa;^ _|_ g^. _|_ j^ 9. ^ _ 6a;^ _|_ is^^i _ 20cc' + 15a;2 - 6a; + 1. 10. 2x' - 8a;« - Zaf' + 12a;* - la? + 28a;^ + 3a; - 12. 11. a^-2^a-\ 12. a'-« + 2 — «-i. 13. fl^^^"^ + a^b^ + fl^~iZ'i 14. a^ + <7Z>^ — a^Z' — b^. 15. 6a;2y+2i/'=:2i/(3a;2+?/^). 16. 8a-(a-2+l). 17. 16a;(a;2+4). 18. W - d\ 19. ^» + Z*^ + cl 20. 2^2 + 2^,2 + 2c^ = 2(a^ + 6' + c'). 21. 4a;^(a;2 + y^). 22. 4:xy(:^ - ?/^). 23. a;* + y. 24. [2a;(a;2 _|_ 3^2) -|2 ^ 4^6 _j_ 24:^*^!* _|_ 36a:y. 25. y* + 7y^ + 16^^ + 12.y + 4. Exercise VII. 1. (^ - 9a^y + 36a;y - 84aY + 126a^y' - 126a;*^'^ -f 84ar^?/«-36a;2^^ + 9xy^-y\ 2. {a^-6x'y+10xY- 10a;y+ bxt/ - y')~\ 3. a* - Sft^Z^ + 2ia^^ - 32aP + 16Z^*. 4. (81a;* — 216a;»y + 216a;y - 9Qxy^ + I61/*) -\ 5. 8a^?/+ 8a;y^ = 8a;t/(a;2 _|_ ^2)^ g_ ^2_2xJ^^-2x-^-i-x-\ 7. a;- 42'^+ 6 - 4a;-^ + x-\ s. a^-6a^ + 15a* - 20ft^+15rt2_ 6a + 1. 9. a^-\-y^-l + 3x^y -~ 3x^ + 3a;.?/« - Si/^ + 3a; + 3y - Qxy. 10. ^2 + 62 + c^ -{- d^ - 2ab '- 2ac'+ 2ad + 2i^6; - 26^ - 2cd. 11. [2(a2 -j- b')f = 4a* + Sa^Z^^ + 4:bK 12. (4a'^+24a*+36a2)-^ 13. a«-12at+54a-108an81. ANSWERS. 343 14. 4«*-4«^+a^H-12a-6+9«-^ 15. a:«+8a;^«/+282:y+ 66a^y^ -\- lOx^y*^ + SGa:'/ + 2Sxhf + 8a;/ + y\ i6. :c^ - 6a;t + 15a;t- 202:^+15 - Gaj-t+a^-i 17. i^a^ ± ^Wx'ay-{- U^x'a^y^ ± 4:bxaY + ay. 18. 1 ± 10a; + 402;^ ± SOa;^ + 80a;* ± 32a;5. 19. a^ + b^x^ + c-^a;* + 2rt^>a; + 2acx' -h 2Z>ca;». 20. «' - 8^>^ + c^ - 6r«2^> + Sa^c + UaU" + 12^>2c + dac" - 6/^c2 — 12a5c. 21. (a*a;* + 4«V + 6aV + 4aa;7 + a;«)-^ 22. a;(2 - a;)(2 + dx") or 4a; - 2ar5 + 6ar» - 3a;*. 23. - 7 + 20x-Ux' + 4:x^. 24. 1 + 4a; - 2a;2-4a;3 + 25a;* - 24a;^ + 16x\ Exercise VIII. 1. -2a(^, 2. -2a -]- ib^ - Qa^. 3. -{7n-nY{7n-\-n). 4. a' + 2a^ + 3^2 +4^ + 5. 5. a;^ _ 2a; + 3. 6. a^ + 2fl2J + ^ab^ + 4^^ 7. a;" - Ja; + c. 8. a;* - (« - l)a;' - {a- b- l)x^ -{a- l)x + 1. 9. a"" -\- b^ -}- c^ -^ ab -\- ac - be. 10. x" + x{2y - 1) + (4?/ + 2?/ -f- 1). 11. a* + a^J^ + b\ 12. {« + by - 2c(a + J) + 4^^. 13. - 2 + 3«j - tt-i+rt-l 14. (a + lf -2(« + l) + 1 =a^ 15. 1 - cd{a-b)^ + (a- J)i 16. d\a - bf-^\a-bY - ^\a-b)\ 17. {a -\-b - c)-^ - {a^b - c)-^ - {a -\- b - c) -^. 18. x^ + a;t-a;^ + a:^ +1. 19. a^-2a'bc + 4Z*V. 20. «' + 2^2^, -f 2aW + Z>1 21. « - a^ - a-^ + «-^ 22, a ^b-\-c. 23. a' - 2aJ + b\ 24. a"' + a~\b - c)~^ ^ {b - c)-\ Exercise IX. 1. 3a;y; 5a'Z»«; M{b - c) ; <6a\x - y)\ 2. {a - bf; {2x - If ; {a - 3)^. 3. {a - b -{- cY ; a'^-\b - c)^""*; x^'{a - 2^)*». 4. (x - yY; (3a - 1)^"; {a - b - V)\ 5. 3:^2 _ ^ _ 1, 6. 4a;3 __ 5^2 _ 3^'^ 2. 7. 421; 737; 3789; 4.164. 8. 1.5811 +; 3.7947+; .4110+ ; .5477+; .1732+. 9. 2a;2 + 4a; — 3. 10. a;^ — 3;=^ + a; — 1. 11. a + b -^ c. 12. 42; 48; 4.7; 9.6. 13.2.34; 32.1; 4.68. 344 ALGEBRA. 14. 1.07 +; .43 +; 2.51 +; 2.33 +. 15. «"' - 1. 16. 3; .4; 6a^-\ Exercise X. 1.7.' 2.f. 3.-3. 4. ^-±-^. B.-h 6.cl-^(b-{-c-a), c 7. T^TTx- Q. M^iad-\-'bc). 9.-77. 10. i a— h)-^i 4,— c), c{a-\-o) ^ 2 \ / V / 11. - |. 12. b{a + c) -^ {2b -i-c- a). 13. ff 14. 30. 15. — 2J. 16. « — 4. 17. 6, 4. 18. 36. 19. 12 minutes. 20. 12 hours. 21. 1|^ days; 2f days; 2f days; 3f days. 22. 400. 23. A, $90; B, $150; 0, $200. 24, $2400. 25. 8. 26. 8, 4. 27. 6. 28. A, 18 years; B, 6 years. 29. A, 40 years; B, 25 years. 30. A, 20 years; B, 5 years. 31. $30. 32. 39 rods; 41 rods. 33. $200. 34. $400. 35. Wine, 85; water, 35. 36. 60,84. 37. 3 a.m. 38. 3 p.m. 39. $5. 40. 300. Exercise XI. 1. X = Si, y ^ 4. 2. X — 56, y = 35. 3. x = 2, y = i, 0=5. 4. X = 2, y ~ 1, z = 3. 5. X = 4:, y = 5, z = 2, V = 3. 6. A in 20 days, B in 80 days, in 240 days. 7. A has $232, B has $332, 8. A, 49 years ; B, 21 years. 9. ^. 10. 72 apples and 60 pears, ii. A has $180; B, $480. 12. A, $360; B, $280. 13.30,10. 14.2,4,7,9. 15. ^±^, a — h -\- c — a-\-b-\- c 2 ' 2 ' Exercise XII. 1. 4, 1. 2. 3, 1. 3. 3 f 4. - 1, - 9. 5. 9, 5. 6. 4, - 8. 7. 4, 5. 8. 14, - 3. 9. 6, - 1. 10. 5, - J. 11. 4, J. 12. 3, - f. 13. 3, - ^. 14. I, - |. 15. a ± b, 16. —a ±b. ANSWERS. 345 Exercise XIII. 1. a.a.a.{a^-8a-\-3). 2. (a-b){a - b)[{a-I)Y-3']. 3. x(x^ + 1)(^ - 1). 4. ala" + a + l)(a + l){a^ - a -\- 1). 5. (a - bYlia - by + (a-b)-^ 1]. 6. (a' - i){a' - \). 7. a{a^-\){ci^-\). 8. (a*-3)(a*+3). 9. (4«^2-3)(4aH3). 10. {a—b—c—d)(a—b—c-\-d). ii. (« — ^— 3)(« — 5 — 3). 12. (a-2Z'+4)(a-2^>-2). 13. (rt-2-J-c)(a-2 +5+4. 14. (a -{-2Y - {b - dy = (a + 2 - b -^3)(a-\-2 -\- b - 3) = (a-b-^6)(a-^b-l). 15. («-13)(«-l). 16. (^-8?/)(ic-27/). 17. x(x-l){x-Jfl)(x-i)(x-\-l). 18. (:c^-8)(3;-l)(a: + l). 19. (3a; + 2)(3a; - 1). 20. (2a; - 3)(2a; - 1). 21. (dx — 1) (22:-l) 22. (3a;+l)(2a;-l). 23. (x'-x-^ix' -^-x-l). 24. (2x'-x-l){'Z3^-{-x-l) = (2x-Jrl)(x-l){2x-l){x-i-l). 25. (3a;«-2a;-2)(3r' + 2a;-2). 26. (a*-2)(a» + 2a*+4). 27. («-l)(a2+a + l)(rt + l)(«2-rt+l)(«2 4. 1)(«*_a2 4_l). 28. («' + l){a'' - a« + 1). 29. (a* + 3)(rt« - 3a* + 9) = (a' + 3) (a* - 3«2 + 3)(«* + 3rf2 + 3). 30. (a - 2)^ - 1 = {n - 3)(«2 _ 3^ ^ 3). 31. (a - 1)* - 9a^ = (a* -5^5 + 1) (ri^ + « + 1). 32. x{x — 2)(2.c2 + 3a; + 6). 33. (x - 2) (x^ + 2a; + 3). 34. (x + l)(a;* - a;^ + a;'' - a; + 2). Exercise XIV. 1. x^ + y\ 2. x^ -\- xy + y^. 3. x — y. 4. x^ — 2xy + 3/. b. x — y, 6. a; — 3. 7. a; + 3. 8. a;^ — 3. 9. Sa;^— 1. 10. a; — 1. 11. a;* — a;® — a;* — a;^ — a;^ — a; — 2. 12. a;^ — a;. 13. a;* - 16. 14. a;^-3a;5 + 7a;»-5a;«-2a; + 2. 15. 12a;^ + 23a;« + a;* + a;^ - 8ar^ + 2a; - 4. 16. 12a;« - Sa;^ - 3a;* - 10ar» + Sa;^ + 3a; - 2. 17. 12a;« - lOa;^ - 20a;* - 6x^ + 2x^ + 3a;. 18. 72a;* - 90:i^y + 42a;y - 30a;2^ + 6f. Exercise XV. 1 1 3a;-l 1 X 1. — ^. 2. — ; — . 3. ^ T- 4. a;-3* x-\-y' 2a;-l' a:+l* a;2-2a; + 2 346 ALGEBRA. 1 1,1 1 ^' 6.x . n.x — l-{-x~\ 8. X ' x^-l-^x-"" x^-a^-^l' 2a; - 3 9 1 10. 7T— TT ^^^o. 11. (x^ - l)(2:r + 3)' (1 - a;)(;^ + 2)^* a: + 2' 2:r?/ y^ — ax 1 12. 0. 13. , •' . . 14. T^ ^. 15. —r- 16. 1. 17. 1. XT — y 0^ — x^ aoc 18. 0. 19. 7n^. 20. -r- . 21. \ j^ \—. 4y c^ W a^ lac , M\ ^ 2(fl5 - Vf a\a + l) 22. 2U-;.-f— . 23. 2. 24. ijT^T f-^-f- 25. - ^ ^ ^ l-a;+a;2 i_|_^ ^aUa-W 26. 0. 27. 1. 28. ., , ' . 29. .-^. 30. , , ^.^ , ^, . 31. 2 ■ 2 - 32. - ^ ' g ' -. 33. «. 34. ^^ -.. 4 a{df^e) 35. 777 T-TT. 36. ^ -' ^ ' Z{x + 1) ^'^Z + ^e + c/ Exercise XVI. c(3^ - h ) ac(3ab + 1) <^^(^> - a ) ^' 2b • ^' ^' c-ad ' ^- b(b-\-a)' 5. 5. 6. 5. 7. 2|i. 8. «^ -f- (a + Z* — c). 9. 3|. 10. 8. 11. — 107. 12. — |. 13. a: = 2, ?/ = -V- 14. a; = 5, 1/ = 4. a^ — Z>^ «^ — Z>^ «5 a^ 15. X — -TTT. y — T^^ ^;. 16. X = — — r, y = -. 17. :c = 20, y = 16, 2; = 12. iQ. x := 2, y = ?,, z = - 1. 19. a; = - 15, ?/ = — 16, 2 = — 8, ?^ = 31, ?f = 22. 20. a: = 6, ?y = 4, 2; = 1, ^^ = — 3. 21. a; = 10, y = S, 2 2 2 = 6. 22. a: = — —7 , ?/ = 7— — , z = -^-- — . 23. rr = 5, y — 2, z = — d, t = — 4:, v = 0, 24. x = i(a + b + c-d), 2/ = J(« + ^-c -\-d),z = ^(a-b^c^d), V = J(— a -\- b -\- c-{- d). It ^ (a -\- b -\- c -\- d) he repre- sented by A", then x =^ j^ (s — d) ; y = ^ {s — c)j z = ^ (s—b)S V = I {s — a), ' ANSWERS. 347 Exercise XVIT. 1. 287. 2. A, 15 years; B, 20 years. 3. A, $48; B, $60. 4. 9, 36. 5. 36, 64. 6. 24. 7. A, 8J hours; B, 12J hours. cd 8. days. 9. 12. lo. $40. ii. 12, 60. 12. rate, ^ d '~~~ c miles an hour; time, 18 hours. 13. B's rate is 12 miles an hour, and OF is 200 miles. 14. 3 miles an hour and IJ miles an hour. 15. Time doivn, 5 hours; rate of current, 2 miles an hour; rate of rotving, 6 miles an hour. 16. 72 miles. 17. 300. 18. 720. 19. 480. 20. I. 38y\ minutes past 7. II. 21^j minutes past 7 and at 54^^j- minutes past 7. III. 5/y minutes past 7. IV. 13^^ minutes past 7. V. 27^1 minutes past 7 and 49yV minutes past 7. VI. 44|f minutes past 7 and 58^ minutes past 7. 21. 49^^ minutes past 3 o'clock. 22. 46^ minutes past 5. 23. Together at ff a minutes past a o'clock; opposite at ff {a ± 6) minutes past a o'clock. 24. 8J hours. 25. 13 hours. 26. 9-J hours; 46 hours. 27. 64. 28. 45. 29. 27. 30. 36. 31. 124. 32. Length, 80 rods; width, 20 rods; area, 10 acres. 33. 120 @ 5 cents each. Exercise XVIII. 1, X = 2, y = 1. 2. Formula: a: = 41 + 10^, y — 1 — 7t, in which jf=b, -1, -2, -3, -4. r,x=n; 31; 21; 11; 1. g = l', 8; 15; 22; 29. 3. Formula: x = 81 -\-l!t, g = 1 ~ 3t, in which t may be 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11. .-. a; = 81; 74; 67; 60; 53; 46; 39; 32; 25; 18; 11; 4. y= I; 4; 7; 10; 13; 16; 19; 22; 25; 28; 31; 34. 4. Formula: x = 12 -}- 1, y = 8 — 10/, in which t may be 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11. .•.a;=: 12; 11; 10; 9; 8; 7; 6; 5; 4; 3; 2; 1. y= 8; 18; 28; 38; 48; 58; 68; 78; 88; 98; 108; 118. 5. a; = 12 - 7/ = 5. 6. x = n-^lt = 31; 24; 17; 10; 3. y=5t-l =i. y= l-3t= 1; 4; 7; 10; 13. 348 ALGEBRA. 7. 2: = 8, «/ = 3. 8. X- 7, y = 6. 9. x = 11, y = 18. 10. x = ^7,y= 13. n. 19. 12. 185 and 15; 119 and 81; 53 and 147. 13. 245. 14. 97. 15. 21; 42; 63; 84. 16. 100. 17. 974. 18. 5 sheep, 1 turkey, 94 chickens. Exercise XIX. 1. 2 i^; 4 i^; 5 Vb; 3a V'Z; 6x V^ij; 3(a-b) Vo(a-b) 2. Sxif h; 2a H^i; 7 H; 2(x ~ y) H; 3{x-y)-' i^. 3. 12 y^; 6 ^; 2 V3 ; 3 V2; V^; (a' - W) l/JTf^. 4. a-^ K^'"- = a-^ Pa"; V^~^j {b - a) PT+a. 5. f t^; f /6; f V36; Vc b'j a + b a — b Va- b. 6. 4^3; i V30; Va + b; ^-^7^ Va^ - b\ 7. 2 t^; 2 ^2"; 2a« \/2a; a f^ - ab; aV Vl; a^b Vb. 8. V^', V2; V2; V^T^; VT^H); V- 2{a -fjj, 9. i^a - b; b V3a{a - b); ^Ho; | V2. 10. vTg and ^27; \^6 and ^"8; t^4and \^^; t^and t^9. 11. 2 V7> 3 V3; 3 H> 2 1^6; 3 H> 4 1^2. 12. 4 ^"27 > i 1^; 2 t^ > 1^; 2 i/5 > 1^88. 13. 7 1^6 - 29 V2. 14. 4/3 + I V2, 15. 8(« - J) V^^TT. 16. - 2& |/« - b. 17. J^a 1^. 18. 20 V6; ^2; 4 Vl5. 19. V^; 10 V^; 5 - 2 1^. 20. 3t; 4 1^54; ^24. 21. (a — b) Va — b —{a — b)^~', {a — b^Va — b. a + & 22. 1; {x-^y) ^ ' 23. V2- V'6~ 1^; 3 V2. Ct ./TT 24. I VG. 25. 4 1^. 26. V%; 1 1^2; v'^"; 2 t'2; 'k'27; " ^^ + " ^. ANSWERS. 349 27. 3 V2 - 2 1^; 4 + VT5; 7 + 4 1^; i(2 V^ - i^). 28. 1^4- 1; ^ (29 + 4 fT); 10 + 4 Vg"; V2 ^ 1. 29. K2 V'^' + V^4 + 1)^ t/4 +J^+ 1; iCi'I - V^). 30. J(- 3 1^2 + 2 V'S + |/30) ; ^( |/Io + 2 - t^li"); |( V5 + 1/3 + 1^2). 31. - J V~^. 32. 18. 33. - 24 V3; 2 /^=^- 2; -7. 34. 2 4^=^; J V6; 12 V2. 35. 11 t^^n. 36. - 4 V^ 37. 5( i^S"- V^). 38. 2 l/2. 39. i( Ve"- V^- 5). 40. i [3 i^+ 4 + f^-2 ^J t^- 41. i^2 - 1;_2 - V^={ V^-1) V2; VE- V2; 2 f^-1. 42. V^ 4- i^S"; 4^-1; Vl - 2; i (V6 - V2). 43. 16. 44. y^o- 45. 3 or 0. 46. 37. 47. 4. 48. — |. 49. -^. 25 ^ 50. i- 51. (a + Z*)'. 52. g-^IT^- 53. 3. 54. ± | r5. 55. ± i ^4:ab — b\ 56. 4. 57. 6. 58. 5. 59. 3a. 60. 9. 61. 5. 62. ± f 63. i|. 64. |. Exercise XX. 1. 3, i. 2. 1, - |. 3. I, i. 4. 3, - f . 5. 10, - 2. 6. 7, - f 7. 3, - f 8. 8, - 44. 9. 1, - 8. 10. 6, - ^^. 11. 6, - f. 12. 7, - li. 13. |(- 3 ± V3). 14. i(27 ± V67). 15. 2, i. 16. 2, i. 17. - i or - 4^. 18. 3, - 4. 19. 3, - ^. 20. i - f. 21. 2, if. 22. 21, 5. 23. 7, - -V^. 24. 3, - ^. 25. 9, J. 26. 64^ 49^ 27. 1, 9. 28. 9, 4. 29. 5, 80. 30. a — b, c, 31. a — b, a 4- b a 2a „ 00 -^- ^^-bT-c'-W+TY 33. -|,-3|. 34. A, ^g. 35. a±-. 36. 2«, — b. 37. 3, — Jji^. 38. 1, — ^. a ± b a -\- b a — b 39. —7^—. 40. — -^, — -^. 2 a — b a -\- b 350 ALGEBRA. EXEKCISE XXI. 1. 7, 3. 2. 11, - 12. 3. 11, 5. 4. $60 or HO. 5. 16. 6. 1(3 ± V^) and - ^(1 ± 1^5). 7. The square, 10 rods; the rectangle, 16 rods. 8. 50 cents, 10 cents. 9. A franc, ;^0 cents; a mark, 25 cents, lo. Hi- li. A's, $600; B's, $400. 12. 12 hours, 16 hours. 13. $40 per week. 14. 37. 15. 25. 16. 2 P.M., or 12 M. 17. IOJ4 minutes past 2. 18. 4 miles an hour. 19. 90 miles. 20. 20 miles. 21. 13^ hours. 22. 13 J hours. 23. Height 4 yards, breadth 6 yards, length 8 yards. 24. $600, $400. Exercise XXII. 1. (^ — '7x = - 12. 2. x^ -x = 6. 3. x'^ — p= — |. 4. x^ ^ ^x = ^. 5. x^ = 25. 6. x^ — 4tX = — 1. 7. x^ — 32; = - }. 8. 2^ - 2x = - 4. 9. x^— 2ax = i^ — a\ 10. a^—2x Va — h—a. 11. a? + ^xJ^— lOx = 24. 12. a^— 5a^ -i- 2x = - 12. 13. ^* - 2a;» - 6^ + 8a; = - 8. 14. x^ - (2r^ + l)x = (I - a)(a + 2). 15. a^ - 3^:^^ _^ i?;(3«2 - 1) = a^ - «. 16. a;* - 4«a;2 + x%6a^ — 5) — 2«2;(2«^ - 5) = - «* 4- 5^2 _ 4. 17. (:?; - 4)(:i; - 2) = 0, .*. ic = 4 or 2. 18. 4, — 2. 19. 4, — 3. 20. — 1, i(l ± |/^~3). 21. 1, |. 22. -i, ± i 1^3. 23. a, i(3-5«). 24. 1 ± V— 2, — 1 ± |/— 2". 25. Minimum J, when x=^. 26. Minimum — J, when a^ = — |. 27. Maximum j\j when ic— 7. 28. Minimum positive value 1, when x = 0. EXEllCISE XXIII. 1. ± 2, ± V2. 2. ± V7h ± Vb. 3. ± 1, ± 8. 4. 4, 1. 5. 64, 4. 6. 16, 1. 7. 729, 1. 8. h (- 1)' = L 9- tV^ (- ^y. 10. 1, (f)i 11. i ^. 12. i (- 1)^ 13. 2, ANSWERS. 351 - 1. 14. 9, (- 41)1 15. 2 h, 5i 16. ±P-2, ± 1. 17. 4, - 1. 18. 18, 3. 19. 5, 80. 20. 3, V-. 21. 1. 22. 4, - 9, i(- 5 ± V^^. 23. 2, 0, 1 ± |/3y. 24. 2, - 4, i(3± t^505). 25. i(3± 4/5), -2± V3. 26. -2± ^,i(l ± V5), 27. i(8 ± 3 4/7"), i(15 T 4 Vu)- 28. i a/10 ± 2 |/21 r:r ± -K 1/7 ± t'f), (- 2 ±. 1^3)^ = 30. ± Vd, ± \/i(- ^±V'- i (1 T ^3) i/- 2. 29. 1, 2J, - 3f h -1. 15). 31. 32. 4(3 ± V E),1. 33. i(l ± V^^), l(- 1 ± V^n^), 34. i(l ± 4/^^19), 4(- 1^V5). 35. 1(1 ± V^^^2d), j(_l±|/4i). 36. 1(1 ± 4/- 3), i(7 q= 3 4^). 37. 1± V^, 1 ± 2 V~l. 38. 3, -5, -2± V2. 39. ±4 V'lO±2 1^33, ± i \/lO ± 2 1/^=^. 40. i(l ± ^^). i(l ± i^^^). 41. 2(1 ± i^), 1 ± 4/^=T. 42. 4(- 1 ± V5)^(5 ± 1^21). 43. i(3± 4/41), i(l± 4/=^). 44. i(13± |/493),±3 t^^. 45. ± 2. 46. Kl ± ^^)- 47. 3, |(- 1 ± V^^), 3a -1 . .^ ^ . 2 48. 12(^+1)" 52. M, |. 49. ± r2rtC— ?, 0. 50. ± Exercise XXIV. V4a-l 51. 2. 1. a;= ± 3; y = ±2. 2. rr = 6, 4; f/ = 3, 1. 3. a: = ±10, ±3; y=T^h ±2. 4. a: = 4, J; 1/ = 2, 34. 5. a; = ± 5; y = ±2. 6. x= ±4; y = ± 2. i. x = 5, i; y = 4, 5. 8. a; = 5, — 1; ?/= 1, - 5. 9. a: = 27, — 8; y = 8, — 27. 10. X = 125, 1; y = 1, 125. 11. a; = 8, 2 ; «/ = 2, 8. 12. a: = 1 ± 2 1/3, i( - 1 ± Vs'i) ; y = - I ± 2 V3, 4(1 ± ^51). 13. X = 8,2; y = 2,8. 14. a: = 3, 1; ?/ = 1, 3. 15. 2; = 2, 4(- 1 ±V^j9y, y= -1, i(6T V^W). 16. a; = 3, i(l ± 1/- 15); 2/ = 0, i(- 3 ± V^H^). 353 ALGEBKA. 17. x = 2; y = 1. 18. x = 4, — 1; y= —1, 4. 19. x = i, -2; y = 2, -4. 20. a^ = 3, - 1, j\(- 19 ± V-283); 2/ = 1, - 3, tV(19 ± i/-^283). 21. X = 2, - 1 ± 1^; 12/ = ± 5, (40 =F 5 V2)k 22. X = ± 6; y = ± 3; z := ± 2. 23. tc = 1, y = 0, z = 0, and these values may be inter- changed. 24. oj = 3, «/ = 1, 2 = ^; and these values may be interchanged. 25. x = 4, y :=z2, z = Ij and inter- changed. 26. ± 9, ± 5. 27. 5, 2. 28. 6, 4. 29. |56. 30. 24. 31. Eates of riding, 6, 8; of walking, 4, 2. 32. A's investment, 1600; B's, $400; A'sgain, $60; B's, $40. 33. 1, 3, 6. Exercise XXVI. 1. 2111. 2. 167. 3. 858.4. 4. 5.6417. 5. 2.796. 6. 1169.72. 7. 1.387. 8. 96.08. 9. 3.483. lo. 7.6117. 11. 1.7963. 12. l^---li^. 13. ,, ''^ '^^ . biog a b\og a -\- d log c 14. .- = .8483. 15.^-^^^. 16.|5^fe|. 17.4.54 log a log (a -\- b) or - .54. 18. - 1.7092. 19. 2.408. 20. ' i/' - 3?/ = 2; .-. 2/ = 2 or - 1; .-. 2^ = 2 or - 1, x = l. Exercise XXVII. I. H>i,ii>M. 2. If«>*,^, ^ > . Z. x" - 11a; + 28 : o?. 4. 32 : 27. 5. 4. 6. 27, 18. 7. 20, 4i. 8. H. 9. 18, 13i. 14. ^^-^d a — — c -\~ d 23.B = 7A. 24.. 5AB:^2(A^-{-B'). 25. C=- VB^^\ 26. 2. 27. ^^---g^-- 29. Z>=2(7+|. 30. /S^Jft^ ATS^SWERS. 353 Exercise XXVIII. 1. 4a\ 2. |r?. 3. na^'-K 4. !^i!^J_). 5. 1^ 4^ 7^ 10, 13. 6. ± 1, ± 3, ± 5, ± 7. 7. ± 2, ± 5, ± 8, ± 11. 8. 3, 4, 5. 9. 10 or 12. 10. n(2?i + 3). 11. n'. 12. ^{n^ -j- n- 1). 13. 121. 14. 108, 144, 192, 256. 17. 256. 18. 3, 7, 11, or 18, 7, - 4. 19. 3, 6, 12, 24, 48, . . . 20. 2, f , f , . . . , or 4, - f, + f , - /^, . . . 24. 9, 1 or ^, — 7y2^. 26. 36 miles. (B will be over- taken afterwards by A at a distance of 120 miles from the starting-point.) 27. 64, 48, 36, 27, wine; 0, 16, 28, 37, water. Exercise XXIX. 1. 1956. 2. 50400. 3. 3. 4. 12600. 5. 9. 6. 11?- 7. 6. 8. 4. 9. 330, 462, 11. 10. 31. 11. 6. 12. 2. 195 195 "•^^- ^^- ^6' [ikjry ''•''- ''•''''- ''--'''- 18. Vo- 19. A's expectation, S55; B's, $25. 20. $1.05. Exercise XXX. 1. j^^^^s^'"^ ^'^'^'''''^ 2T' IT' 4T' W ^' 2T' 11^' ^• [Convergents, 4, i, t-«_, j|, ^^t^, ^(^, m. Integral fractions, ^, 3I-, 3^, 3I-, ^-, ^-, ^-, 1 + ' l-f'2' Convergents, i, f, JJ, f|, -j'V'V, iSVt. t%V. Hfi HH- Alg.— 30. 354 ALGEBRA. Integral fractions, — -, — -, — -, — -, -. Convergents, i ^\, j%\, ^%\, Integral fractions, ^, A_, _L, _L, _L, ^_. ^ Convergents, 1, f , }, -f , fj. 5. Convergents, 2, 3, J^^, -y., |2^ etc. 6. Convergents, 4, ¥. -¥/. -VAS etc. 7. Convergents, 7, -^^^.^ -2^, -V-, etc. 8. Convergents, 10, -%%^, VA-"^ VirVV-^ ^^c. 9. Convergents, 6, 7, \% -V^,etc. 10. tWo- 11. 1 - a; + ^- ^'+ ^'- . . . 12. 1+ a; + a;2+ x^-\- . . . 13. -|-fa; + ff^^- Vr:z:'+ • • • 15 9 14. 1 — X — X^ ~\- ^:i^ — 7x^ — . . . 15, ^77 fT — .u o ~x' 2(:2: — 5) 2(2;— 3) 2 1_ _7 5_ 1 ^^" a; + 1 "^ a;^-a;-r ^"^^ a; - 7 ic - 5* ^^' 2(1 -x)'^ 19. o;"^ — 22;"^?/ -f 3a; ~y — 2(1 + :?:) ' 1 + a;2 , ^ , X x^ bx^ ^ _2 4a;-y + . . . 20. 1 - - - g - g^ - . . . 21. a 3 - 1^-1^2 _^ 5^-V^4_4^a—'^°Z*«+ . . . 22. + 1225a V«. 23. 2d and 3d terms, each f. 24. 4.89898 -. 25. 286. 26. 4^^ 27. 3(6)^ 28. 2a^ + ^Wb^ + 70fl^Z>* + 14.ab^ = 2a{a' + 21fl*Z>^ + 35^^^^* + '7b'). Exercise XXXI. 1. $17380. 2. 21.2000. 3. 8 J years. 4. 47 years. 5. $623.35. 6. $15375.73. 7. $20000. 8. $6440.40. 9. $1800. 10. $16903.31. 11. 5.451. 12. $172. 13. 22. log b — log {b — rd) ^^' log (1 4- r) • ANSWEKS TO TEST EXAMPLES. I. 1. 57. 2:16a - 9{b-c); 109. 3. - 2{a -\- b)(a - 2b), 4. 6ar^ - 2^2 - 5:c + 2. 5. c* + 8o^d + 2^chP + d2c(P + 16^*. 6. {a + bf, 7. - a^ - ^>2 _ ^^ _|_ ^^ _|_ ac + ^c. 8. 2a^o^. 9. Four cents. 10. llj. II. 1. 13. 2. {a - b)(4:a - b). 3. G3. 4. «* - (i - 6')V - 2abc{b - c) - bh\ 5. - 8a^ - 27^^ - 0^ - ^6a^b - Ua'c - 54^*2^ - 27b^c - 6ac^ - Qbc" - 36abc. 6. (- 1)". 7. {a - bY[{a - by + {a - b)-\- 1]. 8. ± (a - b - c). 9. (a — by. 10. 12|- days. III. 1. 2. 2. 2a\a - b). 3. 2bc - a" - 2ac + c". 4. 2^* + 6a2^2_|_2j4, 6^ la-b-c)'^ + ''-{a-b-cY-{a-b-c)^''-'^. 6. rt J + ac + ^c + c^. 7. ^-. 8. 60. 9. 90 years. 10. 36. IV. 1. A. 2. - 4(a - 3). 3. a^ -^ (c - b - 2)x^ + (2b — 2c — bc)x + 2bc. 4. Each equals a^ — 1. b. x^ — x{y + !)-(- y' + «/+!). 6. a;^ - ?/^ 7. 0. 8. 5. 9. 160. 10. 9^5 hours. V. 1. 1 + 53: 4- \W + lO.r' + 5.^* + x\ 2. 2^(3.r* + "Ix" -f 3). 3. «^ + J^ + 1 — «^ — « - ^. 5. 14. 6. 12J days. 7. 6 p.m. 8. 324 square rods. 9. a; = 8, «/ = 9. 10. .t = 8, y = 3, z^-2, (355) 356 ALGEBRA. VL 1. a;* - babx" + "laW. 2. 7a^ - Mb - QaW + W. 3. x" - 2xy + y\ 4. a^ - 5a^ + 10^ - lOa^ + 5a* - 1. 5. 36 years. 6. x = 16, y = 8. 7. 3, 0. 8. 5, 3. 9. 15 years, lo. 36 years. VII. 1. 16r^* + 96a^ -f 21GaW + 21QaP + 81^*. 2. 1.49 +. 3. 6 miles an hour. 4. 42. 5. 80 square rods. 6. 4, - 24. 7. 3, - 3^. 8. x,x.(x - l)(x + l)(a;2 _|_ ;l)^ 9. (^:' + 3)(1 - x). 10. (a; -2)(x- 9). VIII. 1. :r2 _ 1. 2. (a^ - ¥){a - hf. 3. I or \. 4. x" -- 10.2; =-22. 5. a^-x'-4x = -4:. 6. 1, 4, 6. 7. 4-2 1^3. 8. - 2, ± 1. 9. 2 + i^. 10. ± 4/2, ± 4/^=^. IX. 1. (a4-3)(r? + 2)(rt -2). 2. «* + l. 3.3,1. 4.3, -f. 5. 10 miles. 6. 3, i(- 3 ± -Z^^). 7. ?/' ^F (2a; — 1)?/ = — :^ + rr + 2. 8. 4 p.m. or 9 a.m. 9.-44/5. 10. a; = ± 3 or ± 1, ?/ = ± 1 or ± 3. 1. 25. 2. .3980. 3. 1. 4. ± 5. 6. ^ t^. 7. 308. X ;980. 3. 1. 9. 24, |. 10. $80. XI. 2. 2( 1^ - V2) t^. 3. 31. 4. 2. 5. 3 or f 7. - 1. 8. 9, 1. 9. tV 10. 3i, 3i, 5. XII. 1. i( V^5 + 4^+ !)• 2. ±2. 3. id ± V{l-\-idy-2ds. 4. 1, 3, 9. 5. x = 3.5606, y = 5.9343. 6. 4h 7. log a; = .57625, log 2/ = .92195. 8.26. 9.130,000. 10. $20,284 -. ANSWERS. 357 XIII. 1, ^f _^ jl _j- ct — Pab — Vac — \^c. 2. x^ -\-l-{- x'^- 3. 11. 4. {x - ^){x' + 2a; + 4)(:z: + l){x^ " ^ + ^)- 5. 4.x-'' - 4:X-' + 1. 6. ± 5, ± 4. 7. «' - 12at^> + 54:ab^ - 108aH^ + 81Z/*. 8. IQa-^ - ma-H'^ + 2l6a-'b-^ - 216a-^b-^ + 81^"! 9. («* + A)(«' + i)(« + i)(« - i). 10. ^^1^3500. XIV. 1. x'y - xY + '-^Y - ^y'- 2. ± f 3. ± (5 - i^). 4. 1 ± - - ^a;« ± ^^f - . . . 5. 13 i/3"- 18. 6. 80.- 7. x" -4.x = 12. 8. 5, 8, 11, 14, 17, 20. 9.x= \[a ± VU - ?>a% y = i[« =F i^T^="3^]. ,,(;z+l )(n + 2)(7^ + 3) 10. ^ 3" i"""^ "^ • XV. 1. 5, - 30. 2. 3, i(- 1 ± V^^). 3. f 4. f 5. - CO. 6. .449 +. 7. ff. 8. I - «•>; + tW^ - ff|2^ 9. «" - 11^'"^ + 55rt»^2 - 165rt«^^ + 330r?^^»* - 4G2««^^ + 4G2a«d« - 330a*^' + 165a^Z»« - SSa'-^^^ + lla¥'' - ¥K 10. — a, — b, XVI. 1. 1.8997. 2. r^-^ + ^rt-^J-.f |rt-i"^»^+3^V«-^"^^+. 3. 1. 4. 10.c2 - 7a; = - 1. 5. {a + 5)^ + (« + ^)c + ^''. 2 6. .T : ?/ : : 5 : 2. 7. -3 ^.. 9. ^ ~ S — a ~ ^ XVII. 1. ± i( ^6 - V2). 2. ± 13. 3. («^ + ^^ + ^')(« + ^)- (a;^ — 25)(2:'' -)- 2^ — 5) 358 ALGEBRA. SOa^a^ - 40a;2«3 4. lOxa"^ - a\ 7. x = 2, J, - \, - 3. 8. 3 - 2 i^, 3 - 1^2; 3, 3 + i^, 3 + 2 V2. 9. x=±i, y = ±i. 10. a^ — Vab + bk XVIII. i. 2.53. 2. Minimum |. 3. ri; = 5 or 2, ?/ ~ 2 or 5. 4. a; = 5 or 3 ; ?/ = 3 or 5 ; ;2; = 2. 5. — (1 — af. 6. x~^ —x~^y~^ -\- y~i. 7. -t--. 8. 6}. 9. ^. XIX. 1. 3, i(-3 ± V-67). 2. -2. 3. Jo:. 4. -a;-i^-i- x~^y~'^ — x~^y~\ 5. 4, f. 6. 4J miles an hour. 1 1 5 — a; 4 -f a; 10. 1, 3, 5, 7, . . . XX. 1. 696. 2. OJ, 3«^, 5a, la, . , . 3. — |, — |, - f, — -f^, — yV- 4. a; = 4 or 2, ?/ = 2 or 4. 5. 23 - 10 ^^^, 6. 1.5372 +. i.x = \QY\,y = \Qx\. ( 8640, if no repetitions of consonants be allowed; I 12000, if repetitions of consonants be allowed. 9. 109. OF THE ^NfVfiRsiTY OF ECLECTIC EDUCATIONAL SERIES. Published by VAN ANTWERP, BRAGG & CO., Cincinnati and New York. By mail, post-paid, on receipt of the price. RAY^S NE^A/' ARITHMETICS, SPECIAL TERMS will be made by the publishers on supplies for first introduction into schools. Ray's Arithmetics have a more extended use and a more wide- spread popularity than any other series. Teachers, everywhere, throughout the length and breadth of the land, are familiar with their pages, and millions of pupils have gained their arithmetical knowledge from the study of their principles. More than ten thousand editions of these books have gone forth from the press. RAY'S NEW PRIMARY ARITHMETIC, l6mo, 94 pp., price, .15 RAY'S NEW INTELLECTUAL ARITHMETIC, 104 pp., price, .25 RAY'S NEW PRACTICAL ARITHMETIC, l6mo, 336 pp., price, .50 Two Book Series : RAY'S NEW ELEMENTARY ARITHMETIC, i6mo, 192 pp., price, .35 RAY'S NEW PRACTICAL ARITHMETIC, i6mo, 336 pp., price, .50 RAY'S NEW HIGHER ARITHMETIC, l2mo, cloth, 212 pp., price, .85 RAY'S TEST EXAMPLES IN ARITHMETIC, Published with and without answers. Five thousand Test Examples, carefully graded. Can be used with any arithmetic. RAY'S ARITHMETICAL TABLETS, Test Examples in Arithmetic in Tablet form. From 32 to 48 pp. each, each leaflet having 5 to 10 problems at the top, and sufficient paper to solve all. Carefully graded. Nos. i, 2, 3, 4, 5, 6, 7, 8. By mail, single tablet, 10 cents. Per dozen, i.oo RAY'S NEW ALGEBRAS. RAY'S NEW ELEMENTARY ALGEBRA, Primary Elements of Algebra. i2mo, 240 pp., price, .80 RAY'S NEW HIGHER ALGEBRA, Elements of Algebra for High Schools, College, and private students. By Dr. Joseph Ray, Revised by Del. Kemper, A. M., Hampden Sidney College. i2mo, 406 pp., price, I.oo RAY'S TEST PROBLEMS IN ALGEBRA, Twenty-five hundred Test Problems in Algebra. Prepared for Ray's series by Teachers in the Cincinnati High Schools. l2mo, cloth, 153 pp., price, .50 ECLECTIC EDUCATIONAL SERIES. VAN ANTWERP, BRAGG & CO., Publishers, Cincinnati and New York. RAY'S HIGHER MATHEMATICS. RAY'S PLANE AND SOLID GEOMETRY. Identical with the first part of Geometry and Trigonometry following. •i2mo, 276 pp. Price 70 cents; RAY'S GEOMETRY AND TRIGONOMETRY. Written for the Mathematical Course of Jos. Ray. M. D.j by E. T. Tappan, A. M. 8vo, 420 pp. Price $1.20; RAY'S NEW ASTRONOMY. Revised Edition. The Elements of Astronomy, with numerous Engravings and Star Maps. By S. H. Peabody, A. M., Regent of the University qf III, i2mo, half roan, cloth sides, 350 pp. Price $1.20; RAY'S ANALYTIC GEOMETRY. A Treatise on Analytic Geometry, especially as applied to the properties of Conies : including the Modern Methods of Abridged Notation. Written for Ray's Mathematical Course, by Geo. H. Howison, M. A. A treatise thoroughly serving the purposes of liberal education as a college text-book, and furnishing the in- creasing body of special mathematical students in schools of technology with an introduction to the works of the great masters, and a complete Manual of Conies and Quadrics. 8vo, sheep, 574 pp. Price, $1.75; RAY'S SURVEYING AND NAVIGATION. Surveying and Navigation, with a preliminary treatise on Trigonometry and Mensuration. By A. Schuyler, LL.D. 8vo, cloth, 404 pp., and Tables of Loga- rithms, Natural Sines and Co-sines, Natural Tangents and Co-tangents, Logarith- mic Sines and Tangents, Traverse Table, Miscellaneous Tables, Meridional Parts, Corrections to Middle Latitudes. Price, $1.20; RAY'S CALCULUS. Elenients of the Infinitesimal Calculus, with numerous examples and applica- tions to Analysis and Geometry. By James G. Clark, A. M., Willia^n Jewell College. 8vo, sheep, 441 pp. Price, $1.50; ECLECTIC SCHOOL GEOMETRY. Primary Elements of Plane and Solid Geometry. By E. W. Evans, M. A., late Professor of Mathematics in Cornell University; Revised by J. J. Burns, A. M., Supt, of Schools, Dayton, O. i2mo, half roan, 155 pp. Price, 60 cents; The Eclectic School Geometry will be found the best and most economical brief course on this subject yet offered. It is a concise treatise adapted to the use of students who can not spare the requisite time for mastering the larger works. It will be found useful, also, as a course of first lessons for those who intend to pursue the study more at length. Contains practical illustrations, and exercises and examples in the application of Algebra to Geometry. In the revision, the work is made to conform to the " New Geometry," and is especially adapted to High Schools by the addition of numerous exercises and original demonstrations. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW j AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. HOV 8 ^y''^* ftpb 5 IbJb nro 1 Pi fry-^':; «^t:o lo iw^iii i)EC 2t ^U< LD21-100w-7,'33 1'S'?989