;,'7/. Names of the Series 33 7 TABLE OF CONTEXTS. 24. Two Views of the Signs + and — 38 25. Positive and Negative Numbers. Plus and Minus .... 34 26. Like and Unlike Signs 34 Section II. — The Fundamental Rules Extended to apply to Opposite Numbers. 27. Rules for the Fundamental Operations applicable to Opposite Numbers 34 . . I. — ADDITION. 28-30. Addition of Positive and Negative Numbers. Examples, Rule, Exercise . . . . • 34 II. — SUBTRACTION. 31-33. Subtraction of Positive and Negative Numbers. Exam- ples, Rule, Exercise • 38, 39 III. — MULTIPLICATION. 34. Def. of Multiplication. Illustration, a. Explanation ... 40 35. Investigation of the Reasons for the Rule of Signs in Multi- plication. Two Cases 41 36-37. Rule and Exercises in the Multiplication of two Factors, 42 38. The Commutative (two cases) and Associative Laws in Mul- tiplication 42 39. The Sign of the Product when three or more Factors are multiplied together 43 , .1. The Product of any Number of Positive Factors .... 43 2. The Product of an Even Number of Negative Factors . 43 3. The Product of an Odd Number of Negative Factors . . 44 40-41. Rule and Exercises in the Multiplication of more than two Factors 44 42. Powers of Algebraic Numbers. Definition 46 a. Index of Power 46 1. Rule for Signs. 2. Exercise 46,47 IV. — DIVISION. 43-44. Definition, Rule of Signs, Exercise 47, 48 45. Roots of Opposite Numbers 48 a. Indices and Radical Signs . . » ; 48 b. The Double Sign ± 49 1. Square Root (1) of + Number \ .^ (2) of - Number (.:::: ^* Cube Root • ... (1) of + Number \. . . .^ (2) of - Number ( *^ 3. Other Roots j (1) Index a Prime Number } .q I (2) Index a Composite Number j 4. Rules for Signs of Roots, (1. (2. (3 49 46. Exercise in Extracting Roots 50 TAI5M-: OF CONTKNTS. (HAI'lKi: III. — LkIIKUS used to SniNlFY NlMBEUS.— Definitions and Explanations. AKTUI.i;. I'AiiK. 47. Use of Letters the Second Distinguishing Characteristic of Algehra. Problems to illustrate this 51-53 Section I. — Letters loith the Signs. 48. Definition of Algebraic Expression. Illustrations .... .");} 49. List of Signs used in Arithmetic and Algebra .');{ 50. The Sign +. Illustration :>4 51. The Sign — .'>4 52. The Sign X 54 a. The Period as a Sign 54 b. No Mark. Explanation 54 bH. The Sign -f. a. F'ractional Fonn 55 54. The Parentheses 55 55. The Equality Sign 55 a. Explanation. Example 55 50. An Exponent .56 a. No Exponent written, 1 understood 56 b. Meaning and Treatment of Fractional Exponents . . 56 57. Power of a Number. Definition . 5(J a. An Exponent is called the Index of a Power .... 56 b. Second Power is called the Square ; third Power is called the Cube 5() 58. The Radical Sign, y/ 5<} 59. Index of Radical 56 a. No Index written, 2 understood 56 60. Definition of lioot 57 a. Square Root and Cube Root 57 Section II. — Classification of Symbols. 61. Definition of -Si/m6o/.s. a. Kin 70-72. Binomial, Trinomial, Polynomial 59 ! 10 TABLE OF CONTENTS. ARTICLE. Section IV. — On the Term. page. 73. Coefficient and Literal Parts of a Term 60 74. The Coefficient 60 a. Coefficient and Exponent contrasted 60 b. Tlie Coefficient 1 not written 60 75. The Literal Part 60 a. Dimensions of a Term 60 76. Similar Terms 61 a. Addition and Subtraction of Similar Terms .... 61 b. Like and Unlike Signs of Terms 61 77. The Degree of a Term 61 a. The Exponent 1 understood must be counted .... 61 78. Homogeneous Terms 61 a. A Polynomial Homogeneous 61 CHAPTER lY. — Exercises in Notation. Section L — Exercise in Beading Algebraical Expressions. 70. Exercise 62 Skction it. — Exercise in Writing Algebraical Expressions. 80. Exercise . 63 Section III. — Numerical Values. 81-82. Definition of Numerical Values. Exercise 64 83. Numerical Values of Similar Terms. Exercise 64 84. Numerical Values used to verify Equations 6.5 8."). Vo find the Numerical Value of a Letter in an Equation . . 65 CHAPTER v. — Addition. 8(;. Definition of Addition 69 a. Meaning (59 87. Examples of Addition 61) 1. Terms Similar and with Like Signs (l.-(2 69 2. Terms Similar with Unlike Signs (1. -(2 70 3. Not Similar, or Mixed 71 4. Wlieii Terms to be added constitute a Single J^olynomial . 71 88-89. Uule (1-5). Exercise "..■.... 72 CHAPTER VI. — Sir.THACTiON. J)0-93. Definition, Examples (1, 2), Rule (1, 2), Exercise . . 74-7() CHAPTER VII. — Symuoi.s of Acjoheoation. Ar.so Exercises in Addition and Si utijaction. 114. Symbols of Aggregation. Definition 77 95. (^lantity in Algebra. Definition 77 a. Quantity, Algebraic Expression, and Number .... 77 Section I. — "On Compound Terms. 96. Addition and Subtraction of Similar Com])ound Terms with Numerical Coefficients 78 TAHLK <»K «<>N PKNTS. 1 1 ARTICLK. I'AtiK. J)7. Formation of rompound rooflieients for Common Literal Part 7S OS. Ff)nnati(m of Conipoiiiiil ('ootti<-i«Mits for Similar ('onj])oiiiiil IVruis 7'' Skc'TIon II. — Principles and liulis votinectetl wif/i the I'sc of the St/tnhol.s of Ax IV. — r^es of the S//nihnlfi. 10:5. General Exercise S:J C'HAPTElt Vlli. Ml I.TIIM.K ATIMN. 104. Definition a. Literal Proof of Kiilc for Si«:ns S.") 105. Rule for Signs SO KKi. Multii)lication of Monomials S(» 1. The Signs (1. -(4 S<» 2. Numeric'al Cot'tticient SO :^,. The Literal Part, (1.(2. (', so. ST 4. Examples ST 107-108. Rule, 1-:*.. Exercise SS 101>. Multiplication of Polynomials SO 1. Multiplication of Sum by a Single Ttiiii SO 2. Multiplication of Tenu by a Sum SO :). Multiplication of two Sums SO 4. Mtdtiplication of m — »i by / — )/ 00 .'>. Distributive Law 00 110-112. Examples, ifule. (Monomials, Polynomial ;in I .Mo- nomial, Polynomials (1 and (2. Exen-ise IM) 0:i 11:1. Tse of one L«'tter to stand for a (Quantity 04 rt. A Fonnula 0.') 114. Three Theorems in Multiplication. Proofs and Examples. 0."» 07 11'). Exerci.se in the I7se of the Theorems 07 lie*. Other Noteworthy Examples in .Midlipli'-.itio '.»S 1. S(juare of a Trinomial Sum OS 2. Sijuare of a Polyiunuiai OS '•). Tlieon'm W OS 4. Cube of a Hinondal OS r>. Theorem V OS (J. Fourtli Power of a Pinomial OS 7. Simple Multiplications 00 8. Tln'onnn VI 00 0. Exercise 00 12 TABLE OF CONTENTS. ARTICLE. PAGE. 117. Simple Powers. Definition 09 1. Monomials, Examples, Rule, Exercises 09 2. Polynomials 100 (1. Square of Binomials, Trinomials, Polynomials . . . 100 (2. Cubes, Fourth Powers, etc. References 101 CHAPTER IX. — Divisiox. 118. Definition 102 119. Monomials, a. Two Ways of Writing 102 1. The Signs, Reason, Rule 102 2. The Coefficient 102 3. The Literal Part. Particular Cases 102 (1. Letter in Dividend and not in Divisor 102 (2. Letter having same Exponent in both Terms .... 102 (3. Letter with Greater Exponent in Divisor 103 4. Examples 103 120-121. Rule for Monomials. Exercise 103-104 122. Leading Letter in Polynomial ...... ."^ ... . 104 123. A Polynomial arranged 10.5 124. Division of Polynomials 10.5 1. Polynomial by Monomial. Example and Explanation . 105 2. Polynomial by Polynomial. Example and Explanation, 105 a. Nature of Long Division in Algebra 106 b. Necessity of Dividing First Term of Dividend by First Term of Divisor, etc. 107 125-127. Examples, a, b, c, d. Rule, 1, 2, 3. Exercise (1, (2, (3, (4, 107-109 128. Zero and Negative Exponents. Theorems of the Notation, 110 1. Zero Exponents. 2. Negative Exponents. 3. Exercises, 111-113 129. Simple Roots. Definition (45) 113 1. Monomials, (l-(3. Example, Rule, Exercises. (4. Dif- ferent Ways of Regarding the Same Power 114 2. Polynomials. (See Chap, xviii.) 114 130. Four Theorems in Division. Divisibility of Binomials, 1-4 115-116 5. Examples in the Use of the Theorems of Division, with Particular Reference to Powers of Quantities which are themselves Powers 117 131. Exercise on the Theorems of Divisibility 118 CHAPTER X.— Factoring. 132. Definitions of Terms used 120 1. Divisor. 2. Multiple. 3. Prime Quantity. 4. Com- posite Quantity 120 1.33. Monomials, (1. Standing alone, (2 120 134. Binomials 121 1. Difference of two Squares 121 2. Difference of the same Powers 122 3. Difference of the same Even Powers 122 4. Sum of the same Odd Powers 123 5. Binomials separable into two Trinomial Factors . . . 123 TABLK OF C(>M1:NT.S. ^ 1 H ARTK I.K. 1A<.I;. 135. Trinomials 124 1. TriiiouiiHl Squares 124 2. Siiiiph' Trimmiials not Sciuart's 125 ;5. Trinomials th«» Product of any two Binomials. 1, 2, :J, Examples, Kule, Exercise 127-128 4. Trinomials of the Form A^+aA^B^+B^ 12« 5. Trinomials the Proiluct of a Binomial and Trinomial . 128 136. Quadrinomials 121) 1. The Cube of a Binomial. Explanation, Rule, Exerciser, 121)-130 2. Quadrinonnals the Difference of two tSiiuares .... 130 3. Quadrinomials the l*roduct of two Binomials .... 131 4. Quadrinomials the Product of BiiHjmials and Trinoimals, llHi 137. Polynomials of more than Four Terms . » 134 1. Expressions which are Perfect Powers. Note . . . ,1:54 2. Polynomials the Difference of two Squares 1:54 3. Polynomials the Square of a Trinomial I:i5 4. Other Polynomials IHT) General Remark. — This Classification requires that all expressions must be considered from the standpoint of their Developed Forms. 138. Promiscuous Exercise in Factoring 130 CHAPTER XI. — Common Factoks. 139. Common Factors, II. C. F 138 1. Common Factors, Definition 138 2. Highest C'ommon Factor 138 a. g. c. d. and h. c. f 138 Skction i. — First Method. — The h. c. f. by factoring. 140-142. Principles involved in finding the h. c. f. by factoring. Rule (1, 2, 3), Exercise, Note, Miien adapted . . . 138-140 Section II. — Second Method. — By Continued Division. 143. Principles involved in finding the h. c. f. by Continued Division 141 1. Principle of First Metluxl 141 2. A Divisor of a Quantity is a Divisor of any Multiple of it, 141 3. A Common Divisor of two Numbers is a Divisor of their Sum or Difference 141 144. Application of Principles to justify Metho.s. 154. Definition, a. Illustration, h. An Indicated Division . . 156 SECTION I. — Classification and Principles. 155. Classification of Fractions 157 1. Witli respect to their Origin. (l.-(o. Simple, Complex, Mixed Quantity 157 2. With respect to their Capability of Reduction to a Mixed Quantity. (1. Proper. (2. Improper 157 156. Fundamental Principle in Fractions 158 157. The Three Signs 158 1. Effect upon the Fraction of changing these Signs. (l.-(5. Cases 159 2. Rules. (1. (2 159 Sectiox II. — deduction. 158. Definition. Kinds 100 I. — TO LOWKST TKHMS. 159-160. Reduction to Lowest Terms, Explanation and Exercise 160 II. — TO MIXED OK EXTIKK QlAXTrrV. 1()1-162. Reduction to Mixed or Entire Quantity .... 161, 162 16:^-164. Dissection of Fractions 162 165-166. Fractions written as Entire Quantities 163 III. — MIXED Ql'AXTITIES TO nil'HorEIi FHA( I'lONS. 167-168. Reduction of Mixed Quantities to Iiiipiojx'r Frac- tions 168, 164 IV. — i;Ki)r< Tiox or fi!A(ti()Xs ojt ixteoeks to kqi'iva- I.KXT 1 JLVCTIOXS IlAVIXCx (HVEX DKXOMIXATOIiS. 169-170. To reduce a given Fraction or Integer to an Equiva- lent Fraction having a given Denominator 165 171-172. To reduce two or more Fractions to Equivalent Frac- tions having a Common Denominator. L'^sually the 1. c. d 166, 167 Sectiox hi. — The Fundaxicntal lialcs in Fractions. Ho. Definition. Explanation 168 I. ADDITIOX. 174-175. Addition, Rule, Exercise Ji'.s, KiO ir. SIIJ'il{A< TIOX. 17<5-177. Subtraction, Rule, Exercise 170, 171 111. — Ml I/!1IM.[( A riox. 178-179. Multiplication, Rule. ^^ />. Ivxercise 173 TABLE OF CONTENTS. ARTin.K. IV. — i)lVISIi>N. l'A«iK. 180-181. Division, Kule, Exercise 17'. 182-18:3. Complex Fractions, Definition, ExjUanation, Exercise . 17<> 184. Promiscuous Exercise ITS SECOND GENERAL SUBJECT. — SIMPLE EQUATIONS. CHAPTER XIV. — SiMPLK Equations avitii Onk ^^K^<>^v^. Skctiox I. — General Definitions. lHr>. Definition of an Equation IM* 18(J. Identical and Conditional Equations 1S2 187. Identical Equations 1S2 a. Character, h. Identity Sign, rrr 18:J 188. Conditional Equations 188 180. Explanation of Conditional Equations 183 a. Verifying, h. Simplest Case. c. Importance of the Equation in Algebra 188, 184 190. Members of an Equation 1S4 191-15)2. Numerical and Literal Equations 184 198. To Solve an Equation 184 194. Known and I^nknown Quantities in Equations 185 195. The Degree of an Equation 185 190. Simple Equations, a. Remark 185 Section II. — Algebraical Method of Treatment. 197. Nature of the Treatment 185 I. — LOGICAL TKKM.S. 198-200. Proposition, Problem, Theorem, I)«M)i«>n^f?;tti<)ii. ( or<>l- lary, Scholium, Hyjwthesis, Axiom Ix". II. — AXIOMS. 207. Axioms 18() a. Addition and Subtraction 180 1. Equals added to, etc 180 2. Subtraction 18<> 6. Multiplication and Division 180 8. Multiplication. 4. Division 1S(> c. Powers and Koots 187 6. Powers. 0. Kwts 187 7. Quantities equal to the same Quantity, etc . . .187 8. General Axiom 1ST III. — SOLITION OF EQUATIONS. 20^209. Solution of the Ecjuation, ax = b. Exercise . . . 187, 188 210-211. Solution of Integral Equations. Theorem of Transr position. Exercise 188-190 212-218. Solution of Fractional Equations. Theorem. Clearing of Fractions 191, 192 ( 'on. I. Signs of all Terms may be changed 191 SciioL. I. Any Common Multiple may be used . . .101 ScHOL. II. General Kemark. Examples 191 16 TABLE OF CONTENTS. AKTICLP:. I'AGK. 214. Examples of Complete Solution UK] 21.5. General Rules for Solving Simple Equations 104 1. Normal Process, (l-(4 11)4 2. General Process, (1 (2. llemark 11)4 210. Exercise 104 217. General Kemarks on Solution of Equations ... . . . 100 Section 111. — Froblenis. — One Unknown. 218. Explanation of Problems of Algebra. Analytic JVletliod . 200 219. Solution of Problems. Two Operations 200 1. Statement. 2. Solution 200, 201 220. In the Enunciation we may see 201 1. Description of an Unknown 201 2. Assertion of Equality 201 221. Function of a Quantity 201 222. General Rule for solving Problems 201 1-5. a, 6 201, 202 223. Proportions turned into Equations 202 224. Exercise 202 CHAPTER XV. — Simple Equations Containin(; Two oil Moke Unknowns. 225. Simultaneous Equations 213 a. Conditional Problems solved with one or with more Unknowns 213 b. Indeterminate Equations . 214 c. Any Proper Axiomatic Operation does not change the Identity of an Equation 214 226. Elimination, a. Kinds 215 Section I. — Simple Equations with Two Unknowns. 227. Method of Solution 215 I. — SUBSTITUTION. 228. Elimination by Substitution 215 229. Substitution to find Value of other u^nknown 215 230. Exercise 216 II. — COMPAKISGN. 231-232. Elimination by Comparison. Exercise .... 218, 219 III. — ADDITION AND SUIiTH ACTION. 233. Elimination by Addition and Subtraction. Definition. Ex- planation 220 1. To make Coefficient the same (l-(3 220 2. Adding or Subtracting 221 234. • Exercise 221 IV. — SPECIAL METHODS OF ELIMINATION. 235. Special Methods 224 A. Elimination by Undetermined Multipliers 224 TAliLK OF CoNTKNTS. 17 AKTK I.K. I'A«ii;. •_';'.»;. < it'iieral Process by Uiuletenuiued Multipliers. Bezout's M«*thc)(l. Exercise 224 IJ. Eliinliiation by means of one or more Derived E 238. Methoil and Principle by finding tlie li. c. f 22(5 231). (ieneral Exercise 227 240. Problems 22i) Section II. — Shnultaneous Equations contniniiKj Three or more Unknowns. 241. Metliod of Solution. Choice of Plan 2;W 242-245. Examples, Kule 1, 2, 3, a, 6, Exercise, Problems . 233, 238 CHAPTER XVI. — Indeterminate and Redundant E. Kule 1-4. Exercise 242, 243 250. Problems. Diophantine Analysis 244 Skctiox II. — Redundant Equaiions. 251. Definition. Two Kinds 245 1. Compatible. 2. Incompatible, a. Reference .... 245 Section HI. — Negative and Inconsistent Solutiojis in Simple Equations. 252. Problems involving; Arithmetical Inconsistencies .... 245 2.53. Lessons from the Examples of the Preceding Article . . . 240 1. Significance of Negative Results 249 2. Positive Results are Impossible in some Problems, due to the Presence of Negatives in some Stage of the Solution . 24t) Section IV. — Literal Problems. — Generalization. 254. Definition, a. Explanation 24l> 255. Exercise, a, b. Explanations 241> THIRD GENERAL SUBJECT. — NOTATION CON- CLUDED.— POWERS, ROOTS, AND RADICALS. CIIAPTEK XVII.— Of Powkrs. 25(5. Involution 25(> Section I. — Monomial Poivcrs. 257. Exercise, Reference, a. Fractions 250 Section II. — Binomial Powers. 258. Investigation of Binomial Powers. Newton's Theorem, 1-5 257, 2(50 a. Meinoriter Rule. b. Remark 2(J0 259-200. Examples of use of Binomial Theorem. Exercise . . 200 18 TABLE OF CONTENTS. ARTICLE. Section III. — Polynomial Powers. t'aok. 261. Development of Polynomial Powers . . r 262 a. Polynomial Squares 262 CHAPTER XVIII. — Of Exact Roots. 262. Evolution, Definition, a. Division of Chapter 265 S PACTION I. — Monomial Boots. 263. Roots of Monomials 26.5 264. Signs of the Roots. Two Square Roots 265 265. Imaginary Quantities 265 266. Exercise in Extracting Real Roots of Monomials .... 266 Section II. — Square Root of Polynomials. 267. Square Root of Polynomials. How effected, a. Explana- tory 266 268-269. Rule 1-3. Exercise 268 270. Extraction of the Square Root of Arithmetical Numbers . 269 271-272. Rule 1, 2, 3. a, 6, c. Exercise 271, 272 Section III. — Cube Hoot of Polynomials. 273. Extraction of Cube Root of Polynomials, 1,2 272 274-275. Rule 1, 2, 3. Exercise 273, 274 276-278. Extraction of the Cube Root of Arithmetical Numbers, 1, 2. Rule 1-5. Exercise 274, 276 Section TV. — Extraction of other Boots of Polynomials. 279. Other Roots by Square and Cube Root Processes .... 276 1. Fourth Root. 2. Sixth. 3. Eighth. 4. Ninth, etc. . . 276 280. Prime Roots other than the Square and C'ube Roots. The Fifth Root as an Example . . . . 277 281. Exercise 277 CHAPTER XIX. ^ — Of FifACTioNAi. Exponents. 282. Origin, a. Illustration 278 283. Meaning of Terms 278 284. Fundamental Principle Governing use. — Root of Product = Product of Roots 278 a. Order Indifferent 279 b. Other Terms of Exponent 279 285. Fractional Exponents and Radical Sign 279 286. Object of Separate Treatment 280 1. To show same Rules apply 280 2. As an Exercise in their use 280' 287. Fractional Exponents and the Fundamental Rules . . . 280 1. Addition and Subtraction 280 2. Multiplication and Division 280 3. Fractional Powers and Roots of Quantities with Frac- tional Exponents, a. Remark 281 288. Exercise 281 TAHLK OK CONTENTS. 10 AKTULK. CIIAPTEK XX. — Of Kaou ALS. ia<;k. 281). Definition. n = Irrationals (►r Sunis 2S4 ((. Distinction between Onlinary Decimal Fra<'tions and Irrational Decimal Fractions l'S4 2i)l. Treatment of Radicals, 1. Reductions. 2. Fundamental Rules. :]. E Skction I. — Heduction of liadinils. 2'.)_*. Kinds of Keduction . 285 I. — SIMI'LIFICA rioNS. 293. Reduction of Radicals to Efiuivalcnt ones liavin*,' a Lower Index 28."> 2{)4-21)o. Rule. Exercise 285 29t>-298. Simi)lification by Removing: a Factor from tbe Radical. Rule 1-2. Exercise 280, 287 299-301. Simplification of Fractions under tlic Si<;n. a. Reason. Rule l-:i. Exercise 287, 288 II. — KKDl'CTION TO STHD F()«M. CONVKIJSK OI'KIJATIOX. 302-304. Reduction of EIntire Quantities to the Form of Surds. Rule. Exercise 289 305-307. Reduction of Ra0. Pro1. Solving? a (Quadratic Equation by Factorin}; 347 352. Solvini^E(piationsof any Degree by Factoring ...... 348 Ii53-;io4. Rule for Solving Equations of any Degree by Factor- ing. 1, 2, 3. a, b, c. Exercise 349, 350 355. Conversely, to construct an Equation, having given its l{(K)ts. a. Iniaginaries 350, 351 Section II. — Discussion of Problems. — Failure of Processes of Solution. 'M>(). Discussion of the Equation of the Second Degree. Great- est and Least Values of the Coefficients. Two Classes. 1. Real Roots, 2. Imaginary Roots :^'>2, 353 357. Maxima and Minima Values :i53-357 a.. P^xplicit and Implicit Functions 353 358. Discussion of the Equation x=^ for limitiii<; Values of b a and 6. Infinites and Infinitesimals. 1-5 . . . 357-359 .359. Problem of the Lights .359-362 3»K). Validity of Processes of Solution 362-1^66 1. In the Extraction of Roots (1. (2. (3 362 2. An Equation obtained by Squaring or Cubing, etc. (1. (2 3(52-.3(J4 3. An Equation may not be multij)lied or divided through by a Function of the I'^nknown which becomes either Zero or Infinity. (l.-(5 364-366 Section III. — Equations of Higher Degrees solved like Quadratics. 361. Equations containing but one Unknown solved like Incom- plete Quadratics 3(MJ-368 362. Equations containing but one Unknown solved like Com- l)lete Quadratics. Two TyjHi Forms, (ieneral Remark, 3(W-373 3(^3. Simidtaneous E 2a/>. 2 — 110. Exer- cises and Problems 381 , 382 CHAPTER XXV. —Ratio and Pkoportiox. 369. Definition of Ratio, a. Denoted by Colon 383 h. Other Way of Writing 383 370. Reduction of Ratios . 383 371. Comparison of Ratios 384 372. Compounding Ratios 384 a. Duplicate, Triplicate, etc 384 h. May be seen in Arithmetic 384 373. Definition of Proportion, a. The Colon Form read, etc. . 384 374. Method of Treatment 384 375. Transformations of a Proportion 385 1. Theorem. Test. 2. Converse. Different Forms . . . 385 3. General Theorem 385-386 a. Inversion 386 h. Alternation 386 Queries (l.)-(5.) 386 376. Xew Proportions from a Given one 386 1. The Composition Theorem, a. Direct Derivation . 38(), 387 2. The Division Theorem 387 3. The Composition and Division Theorem 388 4. The Like Powers Theorem 388 5. The Like Roots Theorem 388 377. Combinations of Two or more Proportions 388 1. In a. Continued Proportion, etc ,388 a. Shorter Notation 389 h. Application in Geometry 389 2. Multiplication and Division of the Corresponding Terms of Two or more Proportions ,.,,,. 389 TAIiLK <>l CONTENTS. -j;} :i7S. Si»ecial F'onns :;;>0 1. A Moan l*roiK)rtioual ;J5)0 !<• in ;>;>0 2. Two Mean Proportionals :^90 ;170. Exercise in Hatio and Proportion ....... :V.)0, 301 ■.\<(y Variation. 1. Direct. 2. Inverse. ;;. otiirr. . . :',1)2. :}i>:5 (lIAPTKn XXVI. — Kxi'ONKXTiAi. Im/i ath>\s a\i> LiXiAltlTIIMS, ".-^i. Detinition of Exponential E<|iiations 394 a. Failnre except in Ca.se of Exact Powers 3JM :)S2. Lojiaritluns. a. Three Nnnibers considi'n-d HU4 h. Logarithms approxituate :5!)4 '.\K\. Systems of Logarithms. Tables :Ji)5 a. Logarithm of 1 = ;Jl)5 \. 2 as the Base of a System :}S)5 b. Logaritlims are Fractional Ex])onents :>{).■> 2. Unity cannot he the Base of a System. Other iJascs . :Ji)r> c. Base taken Positive ami Greater than 1 :W($ 384. Tenns and Notation 35)6 :i85. The Briggsian or ( 'ommon System 3!M$ \. Logarithms of Fractions. 2. Of Numbers between 1 and 10. 3. Of Numbers Greater than 10 :5!><'), 3!>7 3M»5. Briggsian Mantissas 31>7 (I. Mantissas always taken Positive 400 Tables of Logarithms :50{), 399 387. Explanation of accompanying 4-Place Table, 400 ;i88. Rules to tind the Logarithm of any Number 4(X) 1. Characteristics 400 2. Manti.s.sas 402 (1. Mantissas of Numbers of Three Figures 402 (2. Mantissas of Numbers of more than'I'hree Figures . 402 (3. Exercise 402 389. Conversely. To tind a Number from its Logaritlnns. To find the Significant Figures from the Mantissa .... 404 1 . When Mantissa is same as in the Table 404 2. When Mantissa is not the same as one in Table . . . 404 3iK). Exercise 40."> 391. Uses of Systems of Logarithms 405 392. To Multiply by Logarithms. Examj)!*'. Kule, Exercise . . 40<5 393. To Divide by Logarithms. 1. Example. 2. Rule. 3. Exercise. 4. Evaluation of ( 'omi)ound Exi)ressions. 5. ?^\ercise. 407, 408 394. To Hai.se to I'owers. 1, 2, 3 408, 401> 39.5. To Extract Hoots. 1, 2, 3 40i», 410 3(m. Accuracy of Results 410,411 397. Logarithms to Other Bases can be derived from the Briggs- ian Logarithms 411, 412 398. Solution of Ex]K)nential Equations by Logarithms. Exam- ples. Exercise 412 24 TABLE OF CONTENTS. ARTICLE. CHAPTER XXVII. —The Progresstoxs. page. :]99. Arithmetical Progression 414 400. Fonimla for finding the ?i"' Term in a. j). 414 401. Formula for finding the Sum of n Terms in a. p. ... 415 402. Other Problems in a. p 416 408. Exercise 417-419 404. Definition of Geometrical Progression 420 405. Formula for finding the n'" Term in *y. j> 420 406. Formula for finding the Sum of n Terms in g. p 420 407. Other Problems in g. p. 1. Solved without Logarithms 421 2. Solved with Logarithms 422 3. Problems which lead to the Solution of Equations of Higher Degrees than Simple or Quadratic 423 408. Summation of a Decreasing g. p. which extends to Infinity, 423 409. Repeating Decimals as Examples of f/. p.'s 423 410. Exercise 424, 425 CHAPTER XXVIIL — Interest, Annuities, and Bonds. 411. Interest and Annuities as Furnishing Exercise in Loga- rithms and the Progressions 426 Section I. — Interest. 412. Definition of Interest. Kinds 426 413. Simple Interest. Problems 426 414. Annual Interest 427 415. Compound Interest, Definition and Problems 427 1. To find the Amount when the Principal, Rate, and Time are given 428 2. To find the Present Worth of a Sum Payable at a Criven Time at a Given Rate of Discount 428 3. To find the Time when the Amount, Principal, and Rate are given 429 4. To find the Rate when the Amount, Principal, and Time are given 429 Section II. — Annuities. 416. Annuities Certain, Definitions, and Problems 429 1 . To find the Amoimt of an Unpaid Annuity 430 (1. Overdue 430 (2. Sinking Fund 431 2. To find the Present Cost of an Annuity 431 (1. Annuity Perpetual 432 (2. Deferred Annuity 432 (3. Repaying a Loan in Annual Instalments .... 432 Section III. — Bonds. 417. Definition and Problems in Bonds 433 1. To calculate Price at Time of Issue to realize given % . 433 2. When Interest Payments are made q Times a Year . . 434 4. Value when Payments have been made. Also at Intervals. 436 5. To calculate Rate of Interest for Given Cost .... 436 418. Exercise in Interest, Annuities, and Bonds 437 TEXT-BOOK OF ALGEBRA, INTRODUCTION. CHAPTER I. AI/JKBRA AS A i:i:AN("H (»F M ATHKM ATICS. J>Y way ot preparation lor the study of alj^ehra, it will be helpful to show its nature as one of the mathematical sciences, and to point out its relation to two other hraiuhes of mathematies, arithmetic and geometry. 1. Mathematics is the science of the ex;i«f relations of quantity as to magnitude and form. a. The word " quantity " comes from a Latin adjective which means "how much," or "how many." Anything that lias size or can he measured is a quantity. Any area, as 100 acres, any content, as 25 bushels, any length of time, as 10 liours, any number, as 1.'), is a quantity. Quantity appears under one or other of two fonns, luunber or extent. Thus we may say that a l)asket containing; apples has iV2 in it; or we may si^ak of the number of inhabitants in a t6wn as, e.g., 2000. Or, on the other hand, we may s]M>ak of a square mile, or a cord of wood, and in this way denote the size of the obj<'ct named. Now, many quantities have not only size but also shajKN e.g., a house or a field; and so mathematics treats both of tlie size of objects and of their 8hai>es. * This chapter may be entirely omitted at tlie discretion of tlie teacher. It is too difficult for younp pupils. 25 •20 TEXT-BOOK OF ALGEBRA. The following are intended as suggestive rather than exhaustive definitions of arithmetic and geometry. 2. Geometry treats of quantities in respect to their posi- tion, size, and shape. 3. Arithmetic treats of numbers with reference to the art of computation. a. Arithmetic shows how to write numbers in the shortest and most convenient way; how to multiply and divide them; how to extract roots, and the like. Algebra, on the other hand, treats of numbers in the way of finding general truths in regard to them. 4. Algebra is that branch of mathematics which employs general characters as well as figures in the study of numbers, marking its numbers off into two opposite kinds. 5. The General Characters used. — In arithmetic we study numbers by using the arable system of writing them. In algebra not only figures stand for numbers, but letters regarded as general characters are used for the same pur- pose, each letter standing for some number. However, when a letter is used to stand for a number, it is not like a figure, as 5 (which stands for 5 only, and can not mean any- thing else), but may stand for any number. This peculiarity the student will find one of the principal advantages algebra has over arithmetic. Arithmetic teaches how to add and subtract, multiply and divide, extract roots, and the like, when figures stand for the numbers; algebra teaches how to perform the same operations when letters stand for the numbers. 6. The Two Kinds of Numbers. — The definition gives an- other difference between arithmetic and algebra. Arith- metic employs only one kind of numbers, using that kind to stand in one place for a debt, in another for a credit ; some- times for a gain, sometimes for a loss, and so on. Now, when both gain and loss, for example, appear in the same INTKODlCTloN. 27 problem, contradictions may arise in the arithmetical lan- guage. Thus, if a man in Imsiness gain $i)()0 in the first part of a year, and lose ^2(K) in the latter part, the amount of his year's gain is found by subtracthiff $200 from $500. Couti-ariwise, if two men start in business with the same sum, and one loses $20(X), while the other gains $3000, the difference in their fortunes will be obtained by adiluKj $2000 to $,'5000. The same peculiarity presents itself in examples in longitude and time. We ask what is the difference in longitude between Xew York and Berlin, and expect the student to add to get the result. Algebra makes such (pies- tions clear by pointing out the opposite nature of the num- bers and marking them in such a way as to show tliis. It also enables one to solve more difficult ])roblenis containing such numbers. 7. Opposite numbers will be investigated in the next chapter, before taking up the literal notation. All the laws governing the simultaneous use of such lunubers will be develoi)ed while still using the familiar arabic system. 28 TEXT-BOOK OF ALGEBRA. FIRST GENERAL SUBJECT.— THE ALGE- BRAIC NOTATION. CHAPTER II. OPPOSITE NUMBERS. 8. Ill Algebra numbers are separated into two opposite kinds. We will study the nature of opposite numbers first, and afterwards consider the changes necessary in the funda- mental operations of addition, subtraction, multiplication, and division. SECTION I. Nature of Opposite Numbers. 9. Arithmetical and Opposite Numbers. — The science of arithmetic recognizes only real objects and will admit no element of unreality. It would say, for example, that a man can not lose more than he already has ; that there is no such thing as a number less than zero. This may be described as a failure in arithmetical analysis. For it fre- quently happens that a man's debts exceed his credits, and he is worse off than if he had nothing at all. So stocks vary from premiums to discounts, and latitudes from north of the equator to south of tlie equator, and so on. As was pointed out and illustrated in Art. 6, the operations of addition and subtraction often become confused in the solution of ])roblems, because no distinctions were made at the outset between the two kinds of numbers. AL(J K131 : AlC N OTATIUN . 29 10. The Arithmetical Series. — Arithmetic uses only one set of numbers, commencing with 1 and increasing without limit. We may write down an arithmetical series as follows: 1 2 S 4 o 6 7 8 9 10 11 12 IS 14 . , . ^ . (See Art. 66 for dots of continuation, and 62, b for the meaning of oo , which is used to denote an indefinitely large number.) Fractional and irrational numbers are included between the whole numbers. 11. An Opposite Series. — Let us now commence at zero and write a similar series extending in the opposite direction, understanding that every num- ber on the left has a signification opposite to that of the same number on the right. If numbers to the right hand mean credits, then numbers to the left hand mean debts; if numbers to the right mean north, then numbers to the left mean south ; and so on. Combining the two, we have oD .... 7, 6, 5, 4, 3, , 0, 1, 2, 3, 4, 5, 6, 7 .... 00 12. The Series to be used. — For convenience merely we will imagine our double series of num- bers to extend vertically. In order to distinguish between "above" and "below" numbers when re- moved from their places, some marks will be necessary. 13. Marking the Two Kinds of Numbers.' — The letter a (for above) might be written over all the upper numbers, and tlie letter b (for below) over all the lower ones. However, the notation adopted I The student i.<; asked to prepare such a scale as is found in the mar/?ln on stiff pasteboard, and to make const.int reference to it, v«rifyin>f hv the scale all the addMiOD? ftnd subtraction given, until quite Iftuiiliar w)th it, -. + 00 13 12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 ^0 TEXT-BOOK OF ALGEBKA. is immaterial so long as two convenient marks are chosen and used consistently. We shall emjjloy small j^Zv^*- and minus signs, not to denote addition and subtraction, but simply as marks for distinguishing the two kinds of num- bers. A dagger and a dash might just as well be chosen. 14. Instances of Opposite Numbers. — This double series of numbers finds an application to problems involving (1) A merchant's gains or losses. (2) Income and outlay. (3) Latitude and longitude. (4) Scales, as the thermometer, etc. (5) Time, a.d. and n.c. (6) Attractive and repulsive forces, as positive and negative electricity, etc. 15. Definitions of Addition and Subtraction in Arithmetic. — Let us now examine addition and subtraction for a single series of numbers. Addition means literally ^i^^^m^ together, and is performed by counting the total number of units in the different numbers added ; and subtraction means finding a third number which added to the smaller of two numbers, will give the larger. These definitions are sub- stantially the same as those given in the arithmetic. Now, how shall they be changed when both series are in use ? 16. Algebraic Addition. — Kemembering the opposite na- ture of the two kinds of numbers (11), let us understand that adding an " above " number to another means counting upward that many units, while adding a " below " number means the opposite, counting doivmvard that many units. To illustrate, let us take four simple examples : +Q -{.-^^ = +11 Explanation. —When +5 is added to +6, we count 5 units up from +6 and have +11. -7 -f -9 = -16 Explanation. — When "9 is added to "7, we count 9 units down from "7 and have "16. Both these cases are just as in arithmetic. But, now, if adding +5 carries us 5 units upward in the series, there is al(;i:bkaic notation. -M no ivasoii why the starting-}K)int (i.e., the minil>er to wliich the other is julded) shoiihl be one i)oiiit in the double series rather than another. Let us suppose the st:iitini,'-p()int is "3, to which we are to add "'"5. -3 -f- +5 = -. Exi'LAXATioN. — Starting at ~3 and count- ing 5 units ujjwardy we have "'^2. Tiikewise, adding a '' below " number to an •• above," we write, +6 4- ~9 = ~3. Explanation. — Startingat "♦'6 and count- ing 9 units downward, we come to ~3. 17. Algebraic Subtraction — Hy the definition, Art. 15, the subtrahend and remainder added must etpial the minuend. We shall extend the application of this principle to algebraic numbei-s. +7 -^ +8 = ^IT). \\\ the rule for addition. ^IT) — +8 ="^7. C'onse(iuently, when '•"8 is subtracted from +15, in order to get "•'7, one must count downward 8 units. +13 -f -11 = +2. Hy the rule for atldition. '*J — "11 = +13. Hence to .subtract "11 from +2 we must count 11 units upward to get +13. Therefore to subtract an ^^aboce^^ number count from the minuend downward as mani/ units as there are in the sub- trahend, white to subtract a " belour^ number count from the )uinuend upward as many uuits as there are in the subt ra- it end. Evidently this is Just the opposite of the rule for addition. a. To subtract one number from another, or to find their difference, is the same as to find the didance between them on the scale. If the minuend is above the subtraliend, tlie difference is marked "above," -f ; if it is below, the difference is marked " below," — . 32 TEXT-BOOK OF ALGEBRA. 18. Exercise in Addition and Subtraction. — The student is expected to write down the answers and give the reasons as in example 1. 1. +s -\-+S = +16. To add +8 to +8 count 8 units up from +8. 2. -11 -\r +16 = ? 7. +8 - +4 = ? 3. -9 + +2 = ? 8. +6 - +9 = ? 4. -19 + -1 = ? 9. -3 - -5 = ? 5. +32 + -17 = ? 10. -17 - -12 = ? 6. +42 -{--55 = ? 11. +26 - "15 = ? 12. -17 - +29 = ? 19. Meaning attached to Zero (0) in Algebra. — Zero has a real place in the double series, (12) — as real, for ex- ample, as the zero point in the thermometric or any other similar scale. It has a real place in the latitude scale; viz., the equator : and so in other instances. Consequently it can appear just like other numbers in algebra. Thus adding +5 to means going up from to +5. Adding "13 to means passing down from to ~ 13. We have 1. -}- +6 = +6. Starting at and moving up 6 units gives +6. 2. — +o = "5. Starting at and moving down 5 units gives "5. 3. -|- "3 = "3. Starting at and moving down 3 units gives "3. 4. — "4 = +4. Starting at and moving up 4 units gives +4. 20. The Use of Other Signs to Mark the Series. — From the first and second equations of the last article we see that instead of the sign " + ", " ^ + "^ " ii^^y be used ; and instead of the sign '' - ", " — + " may be used. ALGEBRAIC NOTATION. o3 IJut the latter notation lias "^ before both kinds oi num- bers. Consequently this mark need not be retained. Hence instead of the sign " "•" ", " + " may be used, and instead of the sign " ~ '', " — " may be used. Note. — llatl the thirtl and fourth equations Ixmmi selected, the results would simply duplicate those found, the + and - merely being interchanged. 21. The in the Series Sign just obtained is dropped. — Since adds no value and can be readily supplied, it is plain that it need not be written, and that its absence will make no difference in either our operations or results. As in the first e(puition of Art. 19, no sign, before a number and a -|- sign signify tlie same thing. To illustrate, -h 21 = + lil = 21 ; - 21 = _ 21. 22. Conclusion. — 15y the last two articles it appears that we nuiy distinguish between the two series by + and — signs. A -|- sign prefixed to a number shows that it belongs to one, and a — sign that it belongs to the other series. 23. From what we know of the nature of addition and subtniction, the sign of addition (-[-) refers to what may be called the direct series, and the sign of subtraction ( — ), to the other, which may be called the reverse series. Naturally no sign at all is direct, so that both + and no sign refer to this series. 24. Two Views of the Signs. — The two signs -{- and — seem to have a double meaning : + to denote the operation of addition, and to mark direcrt nund)ers ; and — to denote subtraction, and to mark reverse numbers. In the thermometer scale -|- means above zero. In ac- counts -f would ap[)ropriately refer to income, — to outlay. But it woidd be better to regard them as always indi- cating the o})erations of addition and subtraction, being understood when no other quantity precedes either of them. 34 TEXT-BOOK OF ALGEBRA. 25. Positive and Negative Numbers. — lumbers from tne direct series are called positive or plus iiuinbers, while those from the reverse are called negative or minus numbers. 26. Like and Unlike Signs. — When numbers or terms have the same sign preceding them they are said to have like signs ; otherwise they are said to have unlike signs. Thus 5, +7, +13 have like signs. Also —4, —3, —41, —50 have like signs. Again — 28 and + 4 have unlike signs; and 3, —6, +11, —23, —24 have unlike signs. SECTION 11. The FuNnAMEXTAL Rules fou Opposite Nimhehs. 27. Rules for all of the fundamental operations will now have to be investigated. I. -ADDITION. 28. Addition of Positive and Negative Numbers. — There are two cases. 1. When the numbers added have like signs. The oper- ations now to be considered are the same as those in 18, but large + and — signs take the place of those there used to mark the series. Explanation of (2. — By 16 when — 7 is added to — 3, we count 7 units down to — 10 ; (-'■• ~\' ^ (2- — 2 — 5 added to this result gives — 15; and — 12 added to — 15 gives the answer, — 27. Or, more simply, the numbers are added as in Example (1, and the — sign prefixed to the result. + 4 (2. - 3 + 6 — 7 + 7 — 5 + ^) -12 + 26 - 27 al<.i:i;i;aic NoiwrioN. 35 (.3. Set down as tlie above iind add -|-".), -j-7, -}-<), -\-\i>, and +81. (4. A«ld -(•)!, -L':;. -1>7. -."). -1. and -IT)!. (.\ Add -s;;, -8.S, -s;;, -s:{, -8:i, -8.s, -8;5, -S'A. 2. W'lien the numbers added have unlike signs. (1. Add 5, - 4, + IT). - I), and + 1. ;-, KxpLAXATiox. — Following the method given — 4 in 1(), we have, -|- 15 — 4 added to + 5 gives 4- 1? — 9 +15 atlded to + 1 gives + 16, 1 — added to + 1(5 gives + 7. 8 + 1 added to + 7 gives + 8. Ans. (2. Add - 11. - :i. 4- ir>. - (;. and - 14. — 11 Exri-AN ATIO.N. — o — .S added to — 11 gives — 14. + IT) + 15 added to — 14 gives + 1. — 6 — added to + 1 gives — 5. — 14 — 14 added to — 5 gives — 19. Ans. -^19 (;i Add in, - 0, ;^, - 49, - LV), 14, and 2. a Add ()02. 27. - 1903. 1292, - 3r)9, - 1. ll i> a well-known })rinei|)le in aiitlmietie that the sum is the same in whatever order the numbers are added. If this be true in algebra there will be an obvious advantage in changing the order of the numbers added, so that all the l)ositive numl^ers come first, followed by all the negative numbers. In the examjdes just given it is j)lain that start- ing with zero we count in the i)ositive direction in all a certain distance, denoted by the sum of the positive numbei-s, and in the negative direction in all a certain distance denoted S6 TEXT-BOOK OF ALGEBRA. by the sum of the negative numbers : so that the sum ob- tained by adding the numbers seriatim, or one after another, is the same as that obtained by adding the sum of the posi- tive numbers, and the sum of the negative. In other words, tke order in which numhej^s are added in algebra is indiffer- ent. This is called the commutative law in addition. Thus, in examples (1 and (2 above (1. 5 + 15 + 1 = 21 ; - 4 - 9 = - 13 ; 21 - 13 = 8, answer as before. (2. 15; - 11 - 3 - 6 - 14 = - 34; 15 - 34 = - 19, answer as before. Remark. — The student has no doubt observed that when two opposite numbers are added the sum is their numerical difference. It is positive if the positive number is the greater, and negative if the negative number is the greater. The student may verify the truth of the commutative law for exercises (3 and (4. (5. Add -6,-7, 11, 15, - 3, - 21, + 4. (6. Add 19, - 3, - 6, 0, - 4, -f- 6, - 31, and 50. 29. Rule for Adding Positive and Negative Numbers. 1. If the numbers to be added have like signs, add them as in arithmetic, and prefix the common sign. 2. If the numbers to be added have unlike signs, add the positive numbers and the negative numbers separately, take the difference, and prefix the sign of the greater. 30. Exercise in the Addition of Positive and Negative Num- bers. Problems. 1. Add 51, 96, - 37, - 72, 101, - 49, -f 237. 2. Add - 79, - 106, - 304, + 40, - 9, + 1, and 362. 3. 142 _ 16 - 50 - 31 - 199 + 777 = ? AL(;KBUAir NOTA'IMOX. 37 4. 18 X 47 - 39 X 47 - 21 X 47 + 47 = ? 5. Addoi, -3H,8i, -193, _2Ql. 6. 29.4 - 3.3 + .079 - 1.001 - 20 + 43.05 + 11 = ? 7. A boy has 15 cts., his father owes him 12 cts., and one of his playmates 7 cts. If credits be marked +, how much has the boy, and how shall the amount be marked ? 8. William has no money, and he owes a grocer 8 cts., a l)ieman 4 cts., his sister 3 cts., and a playmate 10 cts. What is he worth, and how shall we mark the amount? 9. A ship starting from 17® north latitude goes one day 2° north, the next 3° north, and a third 4° south ; what lati- tude is she now in, north latitudes being marked -j- and south latitudes — ? 10. A vine that was 30 inches long grew in one month 9 in., in the next 12 in., and in the next 15 in. It was pruned back at one time 11 in., and at another time 5 in. What is its length ? 11. A boy has 25 cts. and owes 10; what is he worth? How shall 25 be marked, how 10, and how the result ? 12. George has 16 cts. in his money box, 10 in his pocket, and William owes him 10 cts. He owes his brother John 7 cts., and a confectioner 14. What is he worth ? 13. A thermometer which stood at 4 p.m. at 70°, and which by 8 o'clock had fallen 14°, and by midnight 8° more, and by 4 a.m. 3° more, rose from 4 o'clock to 8 o'clock 12°, and from 8 o'clock to 12 o'clock noon 25°. Where did the mercury stand at the last-named hour ? 14. A man who has $300 in the bank and $125 in his lK)cket owes A $550 and B $1000. He has due him from one party $400, and from another $640. But he owes as surety on a note $190. What is he worth ? 38 TEXT-BOOK OF ALGEBRA. 15. A rope which was 40 ft. long had at one time 8 ft. cut off, at another 6 ft. ; it was spliced with a piece 15 ft. long, after which 9 ft. was first broken off, then 11 ft., then 3 ft., when it was again spliced with a 20 ft. piece. How long was the rope then ? Query. — If a question like this were propounded, in which the parts broken off taken together exceeded the original length of the rope added to the sum of the splices, what would the student judge concerning the nature of the problem ? What then will a negative result sometimes indicate ? II. - SUBTRACTION. 31. Subtraction of Positive and Negative Numbers. — See 17. 1. 18 2. 15 ExPLANATiox. — To subtract — 8, 7 — 8 since it is a negative number, count 11 23 from 15, 8 units up to + 23. ExPLANATiox. — To Subtract -J ^v Z 'i - + l^j count 10 units down from the minuend. To subtract — 17, count -24 - 1() 17 units up from the minuend. 5. T) 6. 61 7. - 3 8. - 1 11 -89 + 7 - 12 - 5 150 - 10 +11 By Arts. 16 and 17 we perceive that subtracting a num- ber of one series is always the same as adding the corre- sponding number of the other series. Thus : +18 - +10 = +8, as also, +18 + "10 = +8 ; +21 - - 12 = +33, as also, +21 + +12 = +33 ; and so in every case. Now, since this is true, if we choose to do so, we can sub- tract by first changing the series to which the subtrahend ATX5EKUAIC NOTATION. 39 belongs, and then nddinfj it to the minuend. But to change the series of a nunibL*r, we merely change its sign. Coiis^*- quently we may use this rule : To Subtract, rhanf/e the sign of the subtrahend and add. a. Instead of actually chan£jln<; the sign of the subtrahend, it is more convenient to imagine it to be done. Thus, in Ex. 3, imagine the sign of 10 to become — ; and then, if — 10 is added to — 14, the sum is — 24. In Ex. «, conceive — 8i) to become + 8i), and adding + 89 to 01, we have + l.'iO ; and so in the other examples. 9. From IG take 23. 10. From — 15 take — 38. 11. 44 — 10() = ? 12. From - 399 take 183. 13. From — 1000 take 1001. 14. From 436 take :m. 32. Rule for Subtracting Positive and Negative Numbers. — Conceive the sign of tlic subtrahend nuniln'i- to be cluinged, and add it to the minuend. 33. Exercise in Subtracting Positive and Negative Num- b(r8. — 1. From 1428 take — 794. 2. From - 37 take — 50. 3. From the sum of 16, 32, - 79, - 37, - 109. 2, and 9, take the sum of - 19, - 108, - 42, 83, 2i), - (;o. aii.l 41. 4. From 19 X 99 take 73 X 99. 5. From 16 - 31 -f- 172 - 200 less 1(;9 - Ty^^^^ - 263 take 3<)() - 243 + 27 less 100 - 6 + 23. 6. On a certain day the mercury in a thermometer stood at 60°, and on the next it stood at 77°. Wliat was the dif- ference of temperature of the two days if -j- denote u})ward movement and — downward ? 7. A man who has property valued at $3000 owes, in various amounts, to the extent of S3355. What is he worth, and how shall we mark the number ? 40 TEXT-BOOK OF ALGEBRA. 8. What is the difference in longitude between New York and Berlin if New York is 74° 0' 24" west and Berlin is 13° 23' east? How shall we mark the result if east be + and west be — , the answer thus showing how far and in which direction New York is from Berlin ? 9. If north latitude be marked + ^-nd south — , what is the difference in latitude between two ships, one at 19° south and the other at 61° north ? 10. What is the difference between the average tempera- tures of January and December if that of December be 15° above zero and chat of January 4° below ? III. -MULTIPLICATION. 34. Multiplication in Algebra is the process of taking one number as many times as there are units in another, giving the product the same sign as the multiplicand's, or the opposite, according as the multiplier is a direct or reverse number. To illustrate (we suppose the first factor to be the multi- plier) : + 5 X + 6 = 30 , _ 3 X + T = - 21 ; + 4 X - 9 = - 36 ; _ 2 X - 8 = + 16. a. The definition of the multiplication of algebraic numbers given above is based on the nature of such numbers. One series (10) attaches a direct meaning to all its numbers, and is the first and nat- ural series. The other is joined to the first, and obtains its mean- ing from the first by always using its numbers in a sense contrary to that which is applied to those of the first. Looking at the two series as written in 11, the two parts are the same, or, rather, they are symmetrical, like the two arms of a balance. Nevertheless, the meanings attached to them place them on a different footing. So we find, while the product of two positive numbers is a positive num- ber, that, owing to the difference in their significations, the product of two negative numbers is not a negative, but a positive number. This last is not imlike the grammatical rule which says two nega- AI.(;KI!I:.\I(" Nr>TATloV. 41 lives make an atlirniutive. Thus, I said "not unlike," when I meant " alike,'" at U*ast, in some respects. The problems in 41 illustrate the definition. 35. Investigation of the Rule of Signs in Multiplication. — It can be showu Iroiu the definitions of addition and sub- traction that if a positive multiplier gives to the product the same sign as the multiplicand's, a negative multiplier will give opposite results. 1. When the sign of the niultiidier is -f- • (1. To multiply + 15 by + 7. Once -{- W is -\-lo; twice + !•"> is + 30; three times -j- 15 is + 45, and so on. 7 x + 15 is + 105. And so in general. (2. To multiply — 9 by + 8. Once — 9 is — 9 ; twice — 9 is — 18 ; three times — 9 is — 27, and so on. 8 X — 9 is — 72. And so in general. 2. When the sign of the multiplier is — : In the expression 6 X 12 — 13 x 12 it is evidently in- tended that 13 times 12 sliall be subtracted from G times 12, and the difference is therefore — 7 times 12. Now, 6 X 12 - 13 X 12 = 72 - 156 = - 84. (Art. 32.) Hence, - 7 x 12 = - 84. Again, let us evaluate the expression 3 x — 14 — 8 x — 14, which plainly means that 8 times — 14 is to be taken from 3 times — 14, and the remamder is, therefore, — 5 times — 14. But, .S X - 14 - S X - 14 = - 42 minus — 112 = + 70. (Art. 32.) Hence, — 5 x — 14 = -f- 70. 42 TEXT-BOOK OF ALGEBRA. This reasoning is plainly applicable to any other numbers. If, now, + refer to the direct series and — to the re- verse, these four products agree with the definition (34). But the last two have been obtained without any further reference to that article. Consequently, we are assured that the definitions of multiplication and addition and subtrac- tion are consistent. 36. Rule for Signs in Multiplication. — Like signs give +, and unlike, — . Or, more specifically, -|- by + and — by — give -1- ; while + by — and — by -|- give — . 37. Exercise in Multiplication of Positive and Negative Numbers. 1. 8 X - 2 = ? 2. 6 X -\-36 = ? 3. - 19 X 3 = ? 4. 42 X - 1 = ? 5. _ 72 X - 4 = ? 6. 2.V X - :n = ? 7. 14| X -^ = ? 8. 4.6 X - .23 = ? 38. The Commutative and Associative Laws in Multiplica- tion. — We knew in Arithmetic that it made no difference in what order numbers were multiplied; for example, 7X9 = 9X7, and 3x4x5 = 3x5x4 = 4x3x5, and so on. We will show that this is true in ordinary Algebra as well. 1. The Commutative Law. — That the numerical value of the product of two factors is the same, whichever be the multiplier, is known from Arithmetic. We are to show that interchanging the factors will not change the sign of the product. (1. If the Factors are both Positive or both Negative. — In either case the product would be +? both before and after changing the factors (36). AL(;khi:ai<' notation. 43 (2. If one Factor is Positive and the other Negative. — Here the pn^duct would be nejjative. both before and aftei- ehaiigiiig (36). Hence, the multiplier and iiiulti})licand may change places in Algebra, and the product remains the same. 2. The Associative Law. — To see whether the same law- will hold for three or more factors, let us first take three factors. Now, bv lUiiting any two of them into one factor, we have this prodiu^t nudtiplied by the third, in which multiplication, as well as in the first partial one, the order may be disregarded by the commutative law. And so also for a larger number of factors. Hence, it may be inferred that the product of any number of factors is the same in whatever order they are associated. 39. The Sign of the Product for Three or more Factors. The rule for signs, where two fa('tors are multiplied, has just been given. The rule for a greater number of factors is derived from that for two in the following manner: 1. The product of any number of j>ositive fju'tors is l)ositive. For, any two multiplied together give a positive product (36), and this product by a third factor is ])ositive; and so on for any number of factors, -f- by -(- always giving a lK)sitive product. Thus, -h r. X -f '•> X 4- - X 4- 1 = -f (I X 9 X L' X I = -f U)K For, -|-r»x-|-l) = -fr>4; +r>4 x +2 = +108; + lOS x + 1 = + KKS. 2. The product of any creji nund)er of negative factors is always positive. If there he positive factors, change the order so jus to bring them all together (38). Then, by taking tlie negative 44 TEXT-BOOK OF ALGEBRA. factors in pairs, since two negative factors give a positive product, these positive products with the positive factors, if any, give a positive product by 1 above. Thus, -6x-3x-2x -5x +4x +l=+18x -f 10 X + 4 = + 720. 3. The product of an odd number of negative factors is negative. For, if one negative factor be withdrawn (since one less than an odd number is an even number), the product then obtained from the others will be positive by the preceding case. This positive product multiplied by the negative factor withdrawn gives a negative result (36). Thus, +8x +2x -7 = 16x -1 = - 102. Also, +5x— 3x-9x— 4x — 2x-4 = 4-5x + 27x+8x-4=+ 1080 x - 4 = - 4320. 40. Rule. — If the number of negative factors in a product be odd, the product is negative. Otherwise, it is positive. 41. Exercise in the Multiplication of Opposite Numbers. 1. 6 X 11 X - 25. 2. 19 X - 1 X - 1 X - 2. 3. 72 X 2 X - 1 X - 40. 4. 11 X - 4 X -? X --. 3 3 5. _ .6 X - .G X - .3 X -.02. 2 4 3 6. -X-^X-'^X-^. 3 9 8 5 7. 11 X - 3 X - 4 X - 7 X - 21 X 2. 8. -Ix— Ix— Ix— Ix— 2x2x2x-3x - 3 X 3. 9. -6=-5x-4x-3x-2x-lxlx2 X 3. 10. 6 X 7 X 2 X 3 X 4 X 17 X 2 X - 1. ALGEBRAIC NOTATION. 45 11. A boy engaged in catching fish secures each day 50 tish for 10 successive days. How many did he catch in all ? 12. A hunlt'i- ustHJ 24 charges of ammunilion without securing any game, each charge costing him 2^ cents. If waste be marked — , how shall we write the factors and product indicating the expense connected with his sport ? 13. A manufacturer of telescopic object glasses, the casting of which costs $150 each, out of a certain number cast breaks 11 in the grinding. If the number of those broken be marked — , how shall we indicate his loss in the factors and product ? 14. * * The ten hindmost cars of a train which is going directly across a valley are still directed down hill, and besides overcoming friction are exerting a forward tendency equal to two horse-power each. How^ shall we represent their combined effect on the train by using numbers from the double series ? Solution. — On the level the force exerted by the loco- motive in a forward direction would ai)propriately be marked -f , while the resistance of each car would be marked—. On this supposition we mark every car's effect — , and in the case of the ten hindmost each — 2. But the ten hindmost cars act contrary to the original sup- position. Therefore we write — 10 X — 2 horse-power = -|- 20 horse-[K)wer ; i.e., 20 hoi*se-iX)wer in the jiositive direction. 15. ♦ * A l><»)k-agent selling a book at $(J.50 which costs him $4 finds him.self at the end of a certain year in debt to his company for 24 copies, the pay for which he never expects to be able to collect. The next year he sells 150 copies of a book at $5 a co])y, the cost of which is $.*>. r>ut at the end of this vear he Hnds he has been able to 46 TEXT-BOOK OF ALGEBRA. collect for only 148 copies. During the year, however, 9 copies of the first book have been paid for leaving only 15 still unpaid. It is required to represent his gains and losses by factors and products from the double series. Solution. — In his accounts the agent would naturally keep the old and the new items separate. The former as supposed loss would appropriately be marked — , and the latter as supposed gain +. Again the number of books paid for and the number of books not paid for i7i accoi^dance with his expectations, would be direct series numbers, while the number of books not paid for and the number of books paid for contrary to his expectations would be reverse series numbers + 15 X - $4 = - $60 loss; + 148 X $2 = + $296 gain; - 12 X $3 = - $36 loss ; - 9 X - $2.50 = + $22.50 gain. Adding, - $60 + $296 - $36 + $22.50 = $222.50 net profit. 42. Powers of Algebraic Numbers. — The term power of a number is used in the same sense in algebra as in arith- metic. Thus, 62 = 6X6 = 36; (-2)* = -2x-2x-2x - 2 = + 16. a. In algebra it is often convenient to place the exponent (i.e., the small figure written to the right and above the number) outside a parenthesis, as in the second example just given. This means that the number inside with its proper sign is to be taken as a factor as many times as the exponent has units. 1. The Signs of Powers. (1. If we use a positive number continuously as a factor, we always get a positive product (39). Hence any power of a positive number is also positive. (2. Even powers of negative numbers are positive (39,2). AICKfiKAir NOFATIOX. 47 (3. Odd powuiij ot negative numbers are negative ^39. S). 2. Exercise in raising nunihi'is to powers. (1. Square 16. (2. Raise -|- 4 to the fourth power. (3. Cube 16. (4. Raise — 3 to the fifth power. (5. Cul)e — 14. (6. Square — 11.1. (7. (-O)'*^? (8. (-2)«. (9. (Shy. (10. (-5X -3)*. (11. (3 X - 2'^y. IV. - DIVISION. 43. Division in Algebra is the process of finding one factor when the })roduct and tlie other factor are given. The product is the dividend, the given factor is the divisor, and the required factor is the quotient. Since division is thus seen to be the converse of multipli- cation, the law of signs in division may be inferred from that in niuUii>1i<-;ition. By thf (l»Miiiii»oii «)f division, it ^- 6 X 4-5=4- 30, then 4- 30 -^ 4- (> = 4- 5; i.e., 4- divided by 4- gives 4-. 4- <) X — 5 = — 30, then — 30 -i- 4- 6 = — 5 ; i.e., — divided by 4- gives — . _ (j X 4- 5 = — 30, then — 30 ^ 6=4-5; i.e., — divided by — gives 4-. — Ox - 5 = 4- «^, then 4- 30 ^ - 6 = - 5; i.e., 4- divided by — gives — . Thus the rule of signs in multiplication holds in division also; viz.. like signs give plus, and unlik(*. minus. 48 TEXT-BOOK OF ALGEBRA. 44. Exercise in the Division of Algebraic Numbers. I. Divide + 10 by - 2. 2. Divide — 15 by 3. 3. Divide - 50 by 5. 4. ^ ^ — 6 = ? 5. _3.i ^4? 6. 1.06 -^ -9? 7. _ 256 -J- - 8000 ? 8. ~ ^^'^^ — 2^ 9. ^ X -^ X - ?? X - .007 -f. - -? = ? — 3 — 13 2 .3 10. If in 11 days the mercury falls from 15° above zero to 18° below, how much is that a day, and how shall we mark the terms of the division ? II. A man losing 5 cents on every bushel of wheat he sold found he had fallen short $400. How many bushels did he sell ? Mark the terms. 45. Roots of Opposite Numbers. — The term root of a number is used in the same sense in algebra as in arith- metic. Thus, V^ = 0; ^-64 = - 4, etc. a. The radical sign ( y/ ) is used to show that a root is to be taken. What root is desired is indicated by the small figure above and to the left. When no figure is written the square root is under- stood. It will be convenient to first take up square roots, then cube roots, and lastly other roots. As in Powers, the si^ns of the roots is the main point considered. 1. 8(piare Hoots. (1. The s(piare root of a positive number maij be cither pins or minus. Thus, V 4 = + 2 or - 2. For, +2x+2=+4; _also -2x-2=+4. V 28 = -h 5.29+, or - 5.29+. For, + 5.29 X + 5.29 = + 28 ; also - 5.29 x - 5.29 = + 28 ; and so for any other number. ALGEBKAIC NOTATION. 40 b. To mark the two roots without rewriting the number the sign ± (read "plus or minus") is used. It indicates two different numbers. (2. The square root of a negative number is impos- sible. For, the square root of any negative number, say of — 36, to be an algebraic number must be either -|- 6 or — 6. Xow it is 7ieither, since (-)- 6)- = + 36, and (— 6y = -{- 36 also. Hence — 36 has no algebraic square root. 2. Cube Roots. (1. The cube root of a positive number is i)ositive. Thus, ^/Tt = + 3, since (-}- 3)» = + 27. (2. The cube root of a negative number is negative. Thus, V-~Gi = - 4, since (- 4f = - 64. 3. Other Roots. (1. Roots whose index is a prime number. — Of these, positive numbers have a positive root, and negative numbere a negative root, as in the cube root. (2. Roots whose index is a composite number. — Such roots may be found by extracting the roots denoted by the factors of the index, the one after the other. Thus, the fourth root can be derived by extracting the square root twice, the sixth root by extracting the s(juare root first and then the cul)e root of the result. The fifteenth, by ex- tracting the cube root and the fifth root, and so on. 4. Rules for signs of rcxjts. — We saw above that the square root of negative numbers is impossible. Hence we have (1. Even roots of negative numbers are im[)ossible. (2. Even roots of jiositive numbers are to be writ- ten ^. (3. Odd roots of numbers have the same sign as the powers themselves. Note. — These rules have reference only to wliat are called reed roots. This subject will be treated more fully in Chai)(er XVIH. 50 ' TEXT-BOOK OF ALGEBRA. 46. Exercise in extracting roots of algebraic numbers. 1. Extract the square root of 36, of 109, of 49, of 225,^ and of 121. 2. Extract the cube root of 8, of — 27, of — 64, of 125, and of - 343. 3. Extract the fourth root (i.e., the square root twice) of 16, 81, and of 256. 4. Extract the sixth root of 64, of 729. 5. What algebraic numbers are those whose squares are each + 144 ? 6. What algebraic number is it whose cube is — 512 ? 7. W^hat is the length of one side of a cubical box whose content is 216 cubic inches ? 1 If the student is not farniliiir with tlie rules for finding square and cube roots, the numerical results will have to be found by trial multiplications. ALdKIJKAIC NorATloN. 51 CHAPTER III. LETTERS USED TO KEPKESENT NUMBERS, DEFINITIONS, A N L) K X P L A N AT 1 ON S. 47. The Use of Letters to Stand for Numbers constitutes the seciond distinguishing chanu^teristic of algebra. We pro- ceed to give some ilhistrations of this use l)efore taking up the next topic. The solutions of the following questions show how a letter (x) may represent different numbers in different problems. 1. What number Is that which being added to twice itself the sum is 42 ? Solution. — Let x = the number, then 2x = twice the number, (2a; means 2 times x) and X + 2 x = the sum. But the sum is 42. Tlierefore x + 2 x = 42, or, 8 X = 42 (for x + 2 x = 3 x). Now if 3 X = 42, 1 X or X = i of 42 = 14. Therefore x = 14, which is the number. 2. What number is that to which if we add its half and ten more the sum is 43 ? Let X = the number ; then - x = one-half the number, and -x = 2 2 the number + - of the number. 2 Hut the iuiml>er + - the number -H 10 = 43. •> Therefore -x+ 10 = 43. 52 TEXT-BOOK OF ALGER 11 A. Now, if 10 added to ^ ic = 43, | x = 43 less 10 = 33; If - X- 33, then - x = i of 33 = 11: 2 ' 2 3 and —X, i.e., x = 2 x 11 = 22. 2 ' 3. In a store-room containing 40 barrels the number of those that are filled exceeds the number that are empty by 16. How many are there of each ? Let X equal the number filled, then x — 16 equals the number empty. Hence, x + x — 16 = 40, or, 2 X - 16 = 40. Now, if 16 has to be subtracted from 2 x to give 40, then 2 X = 40 + 16 = 56 X = 28 barrels filled, X — 16 = 12 barrels empty. 4. Three pieces of lead together weigh 47 lbs. ; the second is twice the weight of the first, and the third weighs 7 lbs. more than the second ; what is the weight of each piece ? Let X = the number of lbs. the first piece Aveighs ; then 2 X = the number of lbs. the second piece weighs, and 2x4-7 = the number of lbs. the third piece weighs. But the sum of these weights is 47 lbs. ; therefore, x + 2x-f-2x-|-7 = 47, or, 5 X + 7 = 47. Now, if 7 added to 5 x is 47, 5 x = 40. Hence, x = 8 2x = 16 2 X + 7 = 23 5. A boy bought a certain number of lemons and twice as many oranges for 10 cts., the lemons costing 2, and the oranges 3 cts. apiece; how many were there of each ? Let X = the number of lemons, AI.r.KrULMC NOTATION. 58 Now, X lemons at 2 cts. apiece amount to 2 x rts., and 2 x oranges at 3 cts. apiece amount to (> x cts. Then, to find the cost of both, 2x + 6a; = 40. 8 X = 40. X = 5 2 X = 10 6. A father gave his boy three times as many cents as he had, his uncle then gave him 40 cts., when he found he had nine times as many as he had at first ? How many had he at first ? Let X = the number of cents he had at first, then 3 x = tlie num- ber of cents his father gave him, so that 4x = the number he now had. His uncle gave him 40 cts. more, when the statement is made that 9x = his total amount. Hence, 4 x + 40 = 1) x. It is plain now, that if 40 added to 4 x makes 9«, then 40 = 5 x. If 5 X = 40, X = 8 the number he had at first. Exercises like the precetling will be found in Art. 85. SECTIOX I. Lki 1 hi;> u mi ihk Si<.n>. 48. An Algebraic Expression is anytliing written in the algebraic notation. Thus, a, -oO, «2 ^ ah, and ^^^-^^^ are all alge- braic expressions. 49. As suggested by the expressions just written, and by others already given, the signs appearing in arithmetic are used in algebra also, and retain the same meaning. Of them the most impoi-tant are -f-,— , x.-T-,( ),=. and the signs of powers and roots. 54 TEXT-BOOK OF algp:bra. 50. The Sign + (read " plus " ) denotes addition, and is used to show that the numbers ^ between which it is placed are to be added. Thus, a -{- b means the sum of the numbers denoted by a and b. 51. The Sign — (read " minus " ) denotes subtraction, and is used to show that the number following the sign is to be subtracted from the other. Thus, c — d means the number denoted by c less the number denoted by d. 52. The Sign X (read " multiplied by " ) is used to show that the numbers between which it is placed are to be multiplied. Thus, a X c means the product of the numbers denoted by a and r. a. The sign " • " is sometimes used instead of X to denote multi- plication. Thus, 6 • b ' c means that the numbers denoted by 5, 6, and c are to be multiplied together. b. In algebra no sign indicates multiplication. Thus, loabc denotes the product of the numbers 15, a, 6, and c. 1 Letters and Figures Representing Numbers. — A word Is the sign of :in idea. Now, a word (or figure or letter) that stands for a number is as different from the number itself as the written word wagon is different from a real wagon. A number as defined in arithmetic is a unit or collection of units, and 5, or five, for example, is nothing but a symbol which refers to this particular number. In ordinary language we drop this distinction. To illustrate : an older person pointing to the picture of a horse in a book asks " What is this ? " And the child immediately replies " It is a horse." Neither of the two refers to its being only the picture of a horse. So to the class in arithmetic the teacher says, " Tut down the number fifty," when a more correct statement would be " I'ut down the figures which denote fifty. Such ellipses are very frequent in all discourse. Since much is gained in clearness and brevity and nothing is lost if this explana- tion be remembered, it seems best to use the abridged expressions. Thus in the above a correct statement would be "the sign + denotes addition, and is used to show that the numbers between whose symbols it is placed are to be added. 'J'he difference between numbers and symbols of numbers, it is hoped, is here plainly emphasized, but circuitous language in definitions to insist on this distinction is avoided. ALGKHKAIC NOTATION 65 However, it must be understood that tliis does not apply to the arable symbols. Thus, 2524 means two thousand live hundred twenty-four in algebra as well as in arithmetic. 53. The Sign -r- (read '• divided by") denotes that the iiiunber preceding the sign is to be divided by the number following it. Thus, Gab -i- -i c means that 6 ab is to be divided by 4 c. a. The fractional form ^ is also used to denote a divided by b. b 54. The Parentheses, ( ), are used to show that all inside is looked on as one number. Thus, ( --"J j would be .34 treated just as if it were a simple fraction ttt. . In the same way f '^ ~ '^^ ~^ o) would also be regarded as one number. 55. The Sign = (read " equals to ") is used to indicate that the algebraic expression on its left has the same numer- ical value as the expression on its right. a. To explain further: The sign = denotes that the numbers or combinations of nrmbers on its two sides reduce, when simplified, to the same number. We proceed to illustrate this: — «-r2 + 4 11-2 X. 5 7 X :j X 4 ~ 3 + 18 -r 2 • For, -— — — — reduces to , which reduces to A; 7X3X4 84 * "' and -^ '- reduces to — — , which reduces to ^; 3+18^2 12 ^^ that is, when reduced, both have the same value, and therefore they are equal. Likewise, ab = He -\- a when a = 8, 6 = C, and c = ,5. For, ab, or 8 X 6 is 48; and 8 c + a = 8 x .5 + 8 is 48 also. Hence, ah = S c -h a, when a = S, h = 0, r = 5. Ef6 TEXT-BOOK OF ALGEBRA. 56. An Exponent ^ is a small symbol of number written to the right and above another number. When a whole num- ber, it shows how many times the other is used as a factor. Thus, 5 X 5 — ()-', a X a X a X a = a* ', 2 a X 2 a X 2 a = (2 ay. ¥ means c ^'s multiplied together. If c == 6, then ¥ = h X h X h X h X h X h. a. When no exponent is written, 1 is understood : — Thus, a, i. e., a used once equals a'. 6. The meaning and treatment of fractional exponents it is pre- ferable to give further on, in Chapter XIX. 57. The Product arising from taking a number as a factor a certain number of times is called a Power of the Number. Thus, a X a = a^ is the second power, or square of a. 5x5x5 = 5^ is the third power,, or cube of 5. (a -i-b) X (a -\- b) X (a -{- b) X (a -\- b) = (a -\- by is the fourth power of a -\- b. a. The exponent is called the index of the power. b. The second power of a number is usually styled its square ; . and the third power of a number, its cube. 58. The Sign ^ (called a radical sign) is used to indi- cate a factor which multiplied by itself some number of times will produce the number under the sign. 59. An Index of a Radical Sign is a small figure or letter placed to the left and above it, to show into how many equal factors the number is to be separated. Thus^ V25 = i 5 ; ^/64 = 4 ; a/81 = i 3 ; -^32 = 2 ; Vi = 1, in which a can have any value j for any root of 1 is 1. a. When no index is written, 2 is understood. Thus VlO = ± 4; Va^ + b^ means a number, which, multiplied by itself, equals a- + //-. 1 From Latin ex, " out of," and po7w, " to place; " i. e., it is placed out from the number to which it belongs. ALGEBRAIC NOTATION. 57 60. One of its equal factors is called a Root of a Number. Thus, 30 = 6 X 6, therefore is a root of 36 ; 243 = 3 X 3 X 3 X 3 X 3, therefore 3 is the fifth root of 243. a. One of two equal factors is called a square root^ and one of three equal factors is called a cube root. Thus, 4 a- = 2 a X 2a, and 2 a is tne square root of 4 a^. — 6 is one of the square roots of oO; y/Wl^ — ±Jx, read, "the square root of 49 X- equals plus or minus 7x." \/27 = 3, read, "the cube root of 27 equals 3." SECTION II. Classification of Symijols. 61. A Symbol, as used in Algebra, is a letter or sign with a distinct meaning. Thus, a letter, as a, stands for a numl)er ; -|- stands for the operation of addition; and "( )" indicates that an expression inside is to be looked u])()n as one number. a. The Symbols used in Algebra are (1) S>nnbols of Number, (2) Symbols of Operation, (3) Symbols of Relation, (4) Symbols of Aggregation, (5) Symbols of Omission, (6) Logical Symbols. 62. Symbols of Number. — Besides the use of letters and figures to stand for numbers, two other characters are used, (see 19), and oo . a. The sign (read "nought" or "zero") either denotes zero, as in Arithmetic, or an infinitesimally small number. It marks the bejxinning point for the two series. h. The sign x (read "infinity") stands for an exceedingly large number, greater than any that can be named. 63. Symbols of Operation. — Tlie.se are -}-, —^ X, 'y -h, the exponent, and the radical sign. Division is also indi- cated by writing the dividend over the divisor. 58 TEXT-UOOlv OF ALGEBRA. 64. Symbols of Relation. — These are =, > , or << is used to show that the numbers between which it is placed are unequal. The opening is always turned toward the larger number. > is read "greater than;" -< is read " less than." Thus, 8<7; 8 + 4>3 X 3; G<10. h. The symbol cr means " varies as." As the number on its right increases, the number on its left increases, and as the number on its right decreases, the number on its left decreases in the same ratio. Thus, a man's monthly wages a the number of days he works. c. A line drawn across any of these signs, thus, 7^, or -^ is used to deny that which the symbol unmarked expresses ; e. g., ^i 5, 11 5^110. 65. Symbols of Aggregation. — These are ( ), \ I, [ ], , I . Named in order, they are parenthesis, brace, bracket, vinculum, and bar. They are all used for the same purpose; viz., to show that an included expression is to be looked upon as one number. See 54. 66. Symbols of Omission. — Dots or dashes, usually called symbols of continuation, are used to show that certain ex- pressions have not been written, and are to be filled out from those that are given. Thus, a, a"^, a^ . . . a^^ ; or, x -{- x^ -■{- x^ -\- x* . . . x-'^' They may be read " and so on to." 67. Logical Symbols. — The logical symbols are those of reason and conclusion The sign •.• (read "since" or ''be- cause") is used to mark a reason. The sign .-. (read "there- fore," " hence ") is used to mark an inference or conclusion. AH.KI'.KAIC .N(»lAri(»N. 59 SEf'TloN III. Classification <>i- Ai.«.kiu: ak Kxthessions. 68. A Term is iiii algebriiic, expression the parts of which iiif nut separated by plus and minus signs; Thus, 3 ab, 9 a%e, 10 x are terms. Also Aah -\-2 ac — ,3 be is an ex])ressiou consisting of three terms : the first 4 abj the second 2 ac, and the third 3 be a. A simple term is a single expression, as 2 ax, 5 w, ^p- q. b. A compound tenu is any combination of simple terms looked upon as one expression. Tliis is usually indicated by a symbol of (cc c 69. An algebraic, expression consisting of but one term is called a monomial. Thus, Cyabr, 11 ?;/-, 16 ax are each monomials. a. It is plain that both wortls monomial and term can ])e applied to the same expression and with the same intent. To illustrate: 3 a6c can be called either a term or a monomial. Tenu is the more general word of the two, and may cover compound expressions. Monomial is more explicit in referring to a single tcrni. 70. A Binomial is nn algebraic expression consisting of two terms. Thus, II -f- b. 'J 11 — '•. '/-// — f'-. 71. A Trinomial is an algebraic expression consisting of thre.' terms. As. " -\- b -{- (', (lb -\- be -j- tn\ 'J ;/2-j- 10 m*;/^ 60 TEXT-BOOK OF ALGEI'.RA. SECTION lY. On the Term. 73. A simple term in algebra, such as 5 a^ x, usually con- sists of a coefficient and a literal part, each letter having an exponent expressed or understood. 74. The Coefficient ^ shows how many times the number denoted by the product of the other factors is taken. In 5 a^x, (i. e., a'^x -f a^x + a'^x + a^x -}- a^x), 5 is the coefficient, and shows how many times the number denoted by a:^x is taken. In ax, 6 a may be regarded as the coef- ficient, showing how many times x is taken. Of course, the a is a literal factor ; but we may, if we choose, regard some of the literal factors as part of the coefficient. In 3 {a + h), [i. e., {a + ^) + (« + h) + (^ + ^)], 3 is the coefficient of {(( -\- h). a. The coefficient standing in front of its literal part shows how many times that part is to be added; while the exponent of an expression, written to the right and above it, shows how many times it is to be used as a factor in a nudtiplication. To illustrate-: 5 ah means ah + ah + ah + ah + ah \ while {ah)^ = ah X ah X ah X ah X ah. h. The coefficient 1 is not written, although, of course, it may be if desired. Thus, he and 1 he are the same thing. 75. The Literal Part of a term is the part containing the letters. In 10 a%, a% is the literal part. a. One letter, as a in a%, is sometimes called a dimension of the term. 1 From Latin co or con, "with," and efficio, "to effect, or make;" i.e., the coefficient ii tliat which with the literal part makes up the term. ALGEBRAIC NOTATION. 61 76. Similar Terms or Like Terms are those whieli have the same letters affected by the same exponents. 7 aU^ and 9 cU/^ are similar terms. So also ^re 11 x^y^ and 14 ^y. Again, ax'^z and mx'^z are similar with respect to the literal i)art x'z. a. In arithmetic we could add and subtract like numbers, as $7 — $5 = $2 ; 19 qts. + 10 qts. = 35 qts. Likewise we can add 8 times a certain number to 5 times the same number and get 13 times the number. In the same maimer we can add and subtract what have been defined as similar terms. Thus, 1 ab^ -\- \) ab'- = Uiab-. For, if a = 4 and 6 = 3, ab- = 4 X 3- = 30. And 7 X 30 + 9 X 30 = 10 X 30. Likewise, 11 x^ y^ - 8 x*» y^ = 3 r' y^. b. For like and unlike signs of tenns, see 26. 77. The Degree of a Term is the sum of the exponents of its literal factors. (Remember 56, a.) The expression a%c is of the fourth degree. 6 mhi is of the third degree. 9 x^i/*z* is of the ninth degree. 78. Homogeneous Terms are those which are of the same degree. The following sets are homogeneous : 6 a%c^f 7 ab^c% 9 a%^c all of the fifth degree. 9 ary, 11 a:«, 4 x*x^, yz^ all of the sixth degree. a. A polynomial is said to Ik* liomogeneous when all its tonus arc of the same degree. To illustrate: (> ni^n^ — 7) m^n ;/' — ;>-'y» — 12 m'^-q- is homogeneous. So also is cC'b — ab- — abc — a:-c + ac- + be- 02 TEXT-BOOK OF ALGEBRA. CHAPTER IV. EXERCISES IN THE NOTATION. SECTION I. Exercise in Reading Algei3ijaic Expressions. 79. Read the following expressions : 1. a -^ b — c. 2. X -\- X 1/ — I/. Remark. — If the first were to be read as suggested in 50 and 51 we should say " the sum of the numbers denoted by a and 6, dimin- ished by tlie number denoted bye.'" The second should read "the sum of the numbers denoted by x, and the product of the numbers denoted by x and y, less the number denoted by y." It is assumed, however, that by this time the student is familiar with the idea of a letter standing for a number, and it will be sufficient to read the first "a plus 6 minus c," and the second "x plusx times y minus ?/." 3. V2 a - be + d (52, b). 5. a'' + 3 «. ( 57 ). 4. 2 a +3 b - c. 6. 2 d' -\-Ab^- 3 cl 7. tt + Z/ -f r + (/. Kk.makk. — This may be read "« plus 6, plus r, plus d ;" but it is regarded as more elegant to say "the sum of «, 6, c, and fZ." Also to use this mode of expression in reading binomials particularly. Thus, ex. 5 would be read, " the sum of a squared, and three times a."' Likewise, a residual binomial, such as a- — 4 6- would be read " the difference of a squared and four times h squared. 8. (; ab + 9 ar + 14 br. 10. 4 .r- — \) //-. 9. 44 abc — 11 bed. 11. 24 a - 5 A + 14 c - 7 d. AL(;i:iiKAIC NOTATK^N. 63 12. 8 abc — bed -I- i) cde — def. 13. V" H-^ (59, a). 14. c* H- rf2, or ^ (53, a). 15. 7 «' - 3 «'-^/» + c». SECTION II. K\KK( isK IN AVkiiim; Al(jkj;j:ak Kximjkssions. 80. Kxercise in writing expressions : — 1. Express the sum of a, 6, and c. 2. Express the double of h. 3. Express a, plus h divided by c. 4. By how much is a greater than 5 ? 5. Write the sum of a cubed, thn»e times c siiuaicd, and the product of />, r, and d. 6. Write cd over i, i)lus four times b divided by three times a^ minus cd divided by 24. 7. Write the sum of b .scpuire^l and c scpiared, divided by the (UfTerenee of two times e and three times a. 8. Express (• to the fourtli ])Ower. less four times /•(•u!»ed. plus three times c less G. 9. Write tlie sum of the sixtli jiowers of — o r -f- 4 r/) j-'V = (5+3-6 + 16) X :>- X 7 - IS X <;;; = 1134. J,is. 1. 8 a:- 4- (> if- — r> J-- = ? when ./• — 1. 2. y x-//•^' — ^' *^'V' + ^^ *^'"//' = "•' ^^I'^-ii ^* = -> y = ^^) •"^♦i ;i' = 5. 3. a^x-z 4- 3 ^xv; — U (//>^^^' = ? "vvhen a = o, b = 1\ and :r = 3, 5; = -. 4. 2 tn'tfz- — 'J n'-if-'S- -|- '*'intt/-.'r = '.' \n hrn v/i =4, 7t = 3, and y = 1, .^ = -. 84. Numerical Values introduced to verify Equations. Show that the condition of an (Mjuality is I'ullilhMl in llie following: 1. ^_~_^_ = a — x, when a = it, x = 3; wlicn (t = (), a -\- x a; = L 2. ^^ —^ — '^^ = a; — 0, when ^ = S ; also when x = 1). a; + 5 3. ?_ZL^ = x^ -|- J-// -f y-, taking x = any nund)er, y = a; - y any luimber. 85. To find the Numerical Value of a letter in an equation. Take the solutions in Article 47 as models. 1. a: + 6 = 20, a; = ? If 6 added to x equals 20, x must equal 14. 2. 2x4-6 = 20. 5. 3 ir — 6 = 24. 3. 2a: - 4 = 20. 6. .3x + I'O = 5x (.-.20; = 10). 4. 3 X + 6 = 24. 7. 9 r - 16 = o x. 8. Hx — 35 = x. QQ TEXT-BOOK OF ALGEBRA. 9. John is three times as old as James and the sum of their ages is 36 ; how old is each ? Let X = James' age, then 3x = John's age, and X ■-}- 3x = the sum of their ages. But the sum of their ages is 3(j. .'. X -\- 3 X = 3Q 4.x =36 - =^ \Aas. 3x = 21\ 10. There are four times as many girls as boys in a party of 60. How many were boys and how many were girls ? Let X — the number of boys, then cc + 4 cc = 60 is the equation. 11. Divide a line 21 inches long into two parts, such that one part may be | of the other. SuGGESTiox. X + 3 X = 21; I X = 21, J ic = I of 21 = 3; | x, or x = 4 X 3= 12: 21 -12 = 9. 12. A stick of timber 40 feet long is sawed in two, so that one part is f as long as the other. Eequired the length of each. 13. A farmer sold a sheep, a horse, and a cow for ^105. Eor the horse he received 10 times, and for the cow 4 times, as much as for the sheep. How much did he get for each ? 14. A man being asked how much he gave for his watch replied : If you multiply the price by 4, to the product add 70, and from the sum subtract 50, the remainder will be ^20, What did his watch cost ? ALCF.IilJAlC NOT A I ION. 67 15. Three boys together spend 50 ets. : the second spends 5 cts. more than the first, and the tliird three times as much as the first. How many ( t nis did each spend '! 16. Says A to B, " Good-mornmg, master, with your hun- dred geese." Says B, " I have not one hundred, but if I had twice as many as I now have and ten more I should have one hundred." How many geese had he ? 17. An apple, a ]>ear, and a peach cost 19 cts. : a pear costs 2 cents more than an apple and a peach 3 cents more than a pear. How much did each cost ? 18. Distribute $3.50 among Thomas, Richard, and Henry, so that Richard shall have twice as much iis Thomas, and Henry twice as much as Richard. 19. Divide the number 60 into three parts, so that the second may be three times the first, and the third double the second. 20. \Vhat number is that from which we obtain the same result whether we multiply it by four or subtract it from 400? 21. A boy bought the same number of tops, of marbles, and of balls for 65 cts. : for the marbles he gave 1 ct. each, for the tops 2 cts. each, and for the balls 10 cts. each. How many of each did he buy? 22. < )ut of 77 lK)oks a certain number were sold, when tlnif remained three more than were sold? How many wnv >nl,r/ 23. Jolin bought a certain number of tops and fifteen times as many marbles. After losing 14 of the marbles and giving away 30, he had only 16 left. How many tops did he buy ? 24. A man bought 3 liorses and 4 cows for $600. Each horse cost twice as much as a cow. How much did lie give for eiich? 68 TEXT-BOOK OF ALGEBRA. SiJGGESTiox. — Let X represent the price of a cow, and 2 x that of a horse. How much did all the cows cost, and all the horses? 25. Seven men and three boys were hired for a week (6 days), a man receiving three times as much as a boy. Altogether their wages amounted to $72. What did each receive ? Additional problems may be found in Tower's Intellectual Alarebra. AI.ci.r.llAIC KOTATION. 69 CHAPTER V. ADDITION. 86. Addition ^ is the process of uniting two or more ex- pressions into one called their sum. The sum is usually obtained in its simplest form. a. Meaning of Algebraic Addition. — An algebraic expression was defined (48) as anything written in algebraic language. In Chapter II. the rules for adding algebraic numbers were given. We now proceed to apply the same rules for signs to literal expressions. 87. Examples in Addition. — It is convenient to classify examples in addition into four cases : 1. When the terras added are similar and have like signs. (1. Monomials. (See 28.) (1) 2 a (2) - b (3) Sxy (4) -ISaV 8 a -10b 6xy - 2 ah/ 9 a — Ab Oa-y -20a^i/ 12a — _'?i 15 xY — a V ;n a (See 76, a.) - 18 b 38 xY - 41 a^ (2. I Polynomials. (72.) (1) Add the three trinomials given below. 4a^x- 5 a^x - 2a^x- 6 14 ah/ -\- 9 //2 ah/-\-n)r' ah/ -\-10if n a^x - 21 ahj-\- 36 if ' The etisential deflnition waa given in Art. 16. 70 TEXT-BOOlv OF ALGEBRA. This solution evidently proceeds on the assumption that the three trinomials can be added part at a time. Thus, we first add the quantities in the left column, next those in the middle column, and lastly those in the right column. Is this justifiable by the nature of addition ? See Commutative Law, 28. 2. When the terms added are similar, but have unlike signs. (1. Monomials. To add similar terms, add their coef- ficients (76, a). Consequently, in the examples now to be given the coefficients are added by the rules of 29, and the resulting coefficient is prefixed to the common literal part. (1) — 6 a'^ Explanation. — The sum of the posi- + Q 2 f % ^^^'^ coefficients is 11; the sum of the nega- T 4 ^^2 tive coefficients is 9 ; the difference is 2, _ 3 fi' ^iifl the sign of the greater is +• Hence, 2^1^ + 2 is the coefficient of «^ in the sum. . (2) bxy (4) Snm^ — Wdcy — 2 71171^ — 14 xy +21 m7i^ -{- 7 xy — 72 )fi7i^ — xy — 11 7)171^ — 14 xy (3) ■ 9xz'^ (5) 100 abc — xz' 4 abe — 7 xz' — 92 abc + 3xz^ + 42 abe + 10 xz"" - 9 abc -\-Uxz'- (2. Polynomials. The columns are added separately by the rule for iiKmomials. (1) a-2b-\-3c-4:d -2a -{-3b -4c -i- 5 d Remark. — The third — Aa -{- 5 b — 4-c -{- 7 d column cancels out of the 3 a — 4 i' + 5 c — 6 (/ sum. _ 2 2 a + 2 /; + 2 ^ ALGEBIIAIC NOTATION. 71 3. When the terms to be added are not similar, or some similar and others dissimilar. (1. Add 8 a»x- - li ax, 7 n.r - r» .n,\ \) .,•>,* - 5 ax + 2 x', and 2 a^x ^ -\^ xtj* -\- H if. Operation. 8 a^x- — 3 ax 7 ax — 5 xy^ - o -f- '.) a, 5 a, — 2 a, 7 a, and — 4 ((. (2. 10 am, — () (iin, 4 f and 18 dy^. (4. — 6 a'^, -\- 2 a-, — 5 a'^, 4 a"-, — 3 (r and a'-. (5. 13 vi^n, — 10 ^n^n, — mhi, 5 mhi, and — 4 mhi. (6. axy, —7 axij, -\- 8 axy, — axy, — 8 axy, and + 9 axy. 2. Add 3 «^ + o a%, 6 ah — 8 a'^h, 8 ah — a.%, and 3 ah 4- 2 an>. 3. Add T) ,r - 3 a +h ^ 7, and - 4 x - 3 a + 2 Z* - 9. 4. Add 3 a + 7 /> — + r — e, and — a. _ h^ _ c, - d. 5. Add 7 .r- — 2 x — 5, 2 .^-^ — 3 a- + 8, and — 9 if^ + 5 .7^ + 3. 6. Add 3 a%^ — 7 «^/^ + T) axy, — 7 a.%^ — 2 ah* — axy, aJA - 7 axy + 8 aH}% - 10 ah'' + crb^ + 3 rtit-y, and - 5 a-^»^ + 18 ah\ ALdKBKAlC NOTATION. 73 Note. — Here it will be necessary to change the order of the tenns as in 87, o. In the review, the student may solve these prob- lems mentally ; i. e., by simply running the eye along, picking out similar terms, and writing down th«' sums thus obtained with their proper signs. By this plan it will not be necessary to write the problem or to arrange the terms. 7. Add 2 (lb — lidx- -|- 2 a-x, Vllifv .S xij — o a -\- (S c — 3 in -f- ."» -f- xij -\- V2 -2 m. 10. Simplify a -{- b -\- c — d -- 2 e — 2 a -\- ?y b — r> c — r> d -\-a —\b -\-i\> (0 17 x'^ to + (), and we then add it to 11. So in the subsequent examples. (3. (4- 2. Polynomials. (Will it be correct to subtract yw?'?^ fft a time just as we added ])art at a time iu addition ?) - 13 .ry^ 10 Xff (5. (7. — 5 xy^ 9 a;// — 2.'! xff' — (*. (1 - ir>// (6. - 20 x' (8. - 12 ;/ - ?>y' AL(JKlil:Ai< N"iAili»N. 7") (1. From (> a7>V — 7 nhr^ -f- 1) or — :\ hr* take 4 it^b^r i )j>e ration. 6 n-O^r — 7 (i/tt-^ -f- *.) ae — 3 be* minuend. 4 a'-lr^c -\- 8 dbc'^ — S ac — 10 ^6'* subtrahend. 2 a-w* c.r — 4 .r- take r^.T^ — 2 r.r _^_ .S ,^x - 6 ('\ (-jr- -f- .) /'./- — .» r.r — 1 ./ - 3ear2-Trac - 4^^ - ^^a^jqp^ j7« Explanation. — Conceiving c'-^x^ to become — cV^ and adding we have 0, or simply c'^z'^ from cAc^ leaves nothing. Next we bring down 3 ex- since there is nothing to be subtracted from it ; and so in th«' fourth colunm. Conceiving the sign of '^ c'^x to become — , and add- ing — :i c-x to 0, we have — 4 c'^x. Lastly, changing — c^ to + (\ r-i and adding to (which amounts to bringing it down with its sit/n changed), we get 4- <')r'', 92. Rule for Algebraic Subtraction. 1. Write the suhtraliend under the minnend s(» that simi- lar terms shall stand in the same eolumn. It none of the terms are similar, unite the expressions after elianging the signs of all the terms of the subtrahend. 2. Conceive, in turn, the sign of e;wh term of the subtra- hend to he changed, and ad (a -\-b — c). Explanation. — As in 63, 6, the multiplication sign between 7 and the quantity (a + 6 — c) is dropped. Adding the coefficients as in simple terms, 7 + 4 — 2 — 6 = 3; .*. the answer is 3 (a + 6 — c). 2. Add 2 (« + ^»), 3 (a + b), and {a + b). 3. Simplify 4 (a: + z) + 5 (.r + z) — 11 (x -\- z). 4. Simplify 9 {a + bf + 10 {a + bf - (a + bf -2(a + by. (57). 5. Add 2 (.x- + I/) + S (x — y) and 'd {x + y) — 2 {x — tj). 6. Add 4 {ti + 2 by - 4 (« - m), - 20 {a + 2 by + 5 {a - m), 12 (a + 2 by -U(a- m), and - 5 (a -\-2by-\- 5 (a — 77i) 7. Keduce 17 abc (a* + a% + crb'' + «^>"^ -\- b*) - 6 abc {a' + a»^> + a-b^ + «i»3 + b*) to its simplest form. 97. The Formation of Compound Coefficients for a Common Literal Part. 1. Add ax, bx, and ex with respect to x. Explanation, a times x plus h times x plus c times cc equals the sum of a, 6, and c times x; or (a + 6 + c) x. Otherwise, ax + 6a: -}■ ex = (d + h + c) X. ALfiEBRAIC NOTATION. 79 2. Add axy and 3 jri/ with respect to xy. 3. Add ax -f- %? ''•^' + (f'J' ^^ -^ fih ^^^^^ i/-^ 4- ^^'J with respect to x and y. 4. Add aXy 2 car, and 4 dx. 5. Add a7^i + hn, and i;«. -|- «w. 6. Add ay -\- ex, 3 ay -\- 2 rx, and 4 y + (> a*. 7. Simplify aar + ay + rt;^ with respect to a. Explanation, x times n plus y times a phis 2 times « equals the sum of x, y, and z times (*. Or, ax + «y 4- (fz = f» (x + y + z). 8. Add 3 7/12!, 7 7/i//, — 7 777.^, and 4 m with respei^t to m. Sue. 3 771X = 3 X . 7w. 98. The Formation of Compound Coefficients for Similar Com- pound Terms. 1. Add a (x -\- y) and h (x -\- y). Answer (a -}- b) (x -\- y). Note. — As in other places the multiplication sign between par- entheses is omitted. 2. From a (x -\- y -\- z) take b (x -\- y -^ x). 3. Add a (x -\- y) -\- b {x — y), and 77i (x -^ y) — n (x — y). SECTION II. Principles and Rules connected with the Symbols of AormKOATioN. 99. Explanation of the Use of the Symbols of Aggregation. 1. By using parentheses the addition or subtraction of polynomials can be indicated. Thus (-h 5 a^cx'^ -f 4 aHKr^ -f mxhf) -(- (40 ahx^ - 2 a-^ bx' 4- 6 mx^if-) indicates the sum of the ^wo enclosed poly- nomials. 80 TEXT-JiOOK OF AL(iEBltA. So, also, (crb'^ + 12 aba — i) ax'^) — (4 ab'^ — 6 acx -\- 3 a^x) indicates the difference required in ex. 9, Art. 93. By the remark in 87 the two (quantities given first above can be added i)y simply uniting their terms. AVe have (5 (c^cx^ j^ 4 a'^bx^ 4- mx'Y) +. (^^> ((^^-f^^ — 2 (ri^x^ + mxhf) = 5 a'^cx^ + ^ (I'^bx^ -f- mx~ij^ + 40 a'cx'^ — 2 (i^^it"" + 6 mx^y^. It is plain from this that the only difference between the two modes of Avriting the sum is that the first indicates that it is made up of two parts, while the second represents it as one quantity. Consequently if no sign or a plus sign precedes a parenthesis, or other symbol of aggregation, the marks of parenthesis may be removed without altering the value of the quantity. Again, the rule for subtraction directs to proceed as in addition after liai'ing changed the sign of the subtrahend. Hence, after changing the signs of the subtrahend, it may be united to the minuend. Thus (a^b-^ + 12 abc — 9 ax) — ( + -A ab'^ — (3 acx + 3 a\r) = a^b'^ -\- 12 abc — 9 ax'^ — 4 ab'^ + (5 acx — 3 a'\r. This time when the parenthesis preceded by a minus sign was taken away, all the signs within were changed. Of course the reasons given apply to any other polynomials as well as to the trinomials used in illustration. Restating the foregoing, it is briefly this : to add a quantity (indicated by a plus sign preceding the parenthesis enclos- ing it) its terms are joined just as they stand; but to sub- tract a quantity the sign of each of its terms must be changed according to the rule for the subtrahend in sub- traction. It follows from this that any terms in a sum can be enclosed in a parenthesis with a + sign before it just as they stand, but must have their signs changed if enclosed in a parenthesis preceded by a — sign. 2. The parentheses are also sometimes used to enclose the series sign of a number. AI.CiKHKAIC NO'IArioN. 81 Thus, a — ( — b) nieaiis that b tioiii the negative series is to be subtracted from n. To remove such parentheses, we follow the same course as in the preceding case. a -\- {-\-b) — a -{- b\ a -\- {— b) = (I — b\ a — (-^b) = a — b; a — (— b) = a -\- b. 100. Rules for the Symbols of Aggregation — The previous article shows : 1. Any terms can be placed within a symbol of aggregation having a -f sign before it just as they stand. 2. Any terms can be placed within a symbol of aggregar tion hiiving a — sign before it, if the sign of each term be changed from -)- to — , or from — to -)-. 3. A syml)ol of aggregation having a -|- sign before it c^n Ije removed without changing the (piantity. 4. A symbol of aggregation with a — sign before it can be removed if the sign of each term within it be changed. Note. — The terms inside a symbol of aggregation are generally arranged so that a. positive term comes first, when, following the usual custom, no sign is written. The sign before the parenthesis be- longs to the whole quantity inside, and not to its first term. For example, — {a — b) = — (+ a — 6); or, w riting the same expression with a vinculum, — a — b = — ■¥ a — h. The expression — (7 — 3 -I- 4 — 12) = — (— 4) = + 4. Here the first term, 7, is positive The beginner should not forget this. SECTION 111. ExEiU'isE IX Kem()Vi.\<; thi: Symbols. 101. Exercise in Removing the Symbols of Aggregation from Simple Expressions. — The results are to be simplitied (88. 4). 1. 3 x' — 'J If -h CI ,r — 1). 2. ;■; .r + L> y - (.;• - //). 82 TEXT-BOOK OF ALGEBRA. 4. 2h -\-ih -2c) -\h-^2c\. 5. —{a — h'l — lb- ^] - [c — a]. 6. — S 7n -\- 2 71 — Sm — 2 n -{- 9 m. 7. ab — (m — 3 ab -{- 2 ax) — 7 ab. 8. 3a — b^lc — 2a-\-3b — 5b-^c-\-Zc — a. 9. 1 — (1 - a) + ( 1 - a + ^2) — (1 — tt + rt^ - a»). 10. («, _ .X _ 2/) + (x - 2/ _ ^,) + (c + 2 2/). 102. Exercise in removing the Symbols from Complex Ex- pressions. — The results are to be simplified. 1. 3a -[la -^2 + J4 6t - 5 - (- 2 a + 9 - a - 6) j] -10. Let us tirst remove the vinculum by changing the signs of a and 6, rule 4. 3 a - [7 a + 2 + §4 a - 5 — (- 2 a + 9 — a + 6) J ] — 10 ; Next we remove the parenthesis, rule 4. 3a-[7aH-2-h54a-5 + 2a-9 + a-6j]-10; Next, the brace, rule 3. 3a — [7a + 2 + 4a — 5 + 2a — 94-ft — 6] — 10; Finally, the bracket, rule 4. 3o^_7a — 2 — 4a + 5 — 2a + 9 — a + 6 — 10. By simplifying, 88, 4, we have 8 — 11 a. Ans. 2. 2a - \2b - {3c^2b) - a\. 3. 3a +0^- [a + 5a; — (3tt — 2a;)]. 4. \-\2-{l-x +^')S. Note. — This method of removing the symbols, viz., the inner- most first, then the next, and so on, is the best at first. The expert algebraist writes the result directly. Ai>(;i:i'.i:Ai(' notation. 83 6. 5 « — [a -f i) ji — >a — x — 3 « — 2 x}]. 6. a -\-2b — \(Sa — [3 /> -f- (8 j- — L> -f- // — x) -f- 4 a] — 3b\. 8. ax -\- ^cx — ( mx -\- ex — y) -\- [inx — {ex + y)]- 9. vi^\x-l-4.y + 2x^{ay-x) ^p-]. 10. {u'bc + 3 r^) -f 3 - 2 r) - J5 a — (10 6 + 5 r) 5 + |2 6 -(2/--L'.o ;. 12. J,, -[A + (.. + !>..)-( //-..)](. SECTION IV. Uses of thk Symbols. 103. General Exercise. The Uses of the Symbols of Aggre- gation. 1. AVrite the polynomial n -\- b -\- c -\- d as a binomial so that the sums of a and b and c and d may be regarded as single quantities. 2. Write a^ -{. 2 ab ^ b- — r^ — 2 cd — d^ as a residual binomial, regarding the first three and last three terms as single quantities. 3. Write a -^ b -\- e na a. binomial so as to show the sum of a and b as one quantity. 4. Write fl2 + 2 ab -\- b- — 2ae — e^-^2 bc — a-\-b—c as a trinomial of which the sum of the squares forms one term, the sum of the })roducts a second, and the sum of the factors a third. 5. Write 12 nx + 12 ay -f 4 by — 12 bz — 15 ex -{- 6 ey -f- 3 ez so that the terms which contain x may appear as one quantity, the terms which contain y as one quantity, and 84 TEXT-BOOK OF ALGEBRA. the terms which contain ^ as one quantity, but subtracted from the other two. 6. A man who owns a mill woi-th $a, has personal prop- erty worth $^ and has %d in the bank. But he owes %e for his engine and owes his millwright $/. However, the mill- wright is indebted to him for $^ worth of flour. How shall we represent what he is worth, indicating the worth of the mill as one part and all the rest as another ; also, keeping his personal property separate from his accounts, and indi- cating the difference he owes the millwright as one sum ? 7. Remove the parentheses from the preceding result, obtaining the man's fortune indicated as assets and liabili- ties. ALGEBRAIC NOTATION. 85 CHAPTER \ II. MULTIPLICATION. 104. The definition of multiplication was given in Art. 34. Consult again that article. We have, 4-3 X -1-4 = -f- 11'; .S X - 9= -27; -4xS = -.32; -12X -15= +1S(). -f J- X -\- ij = -\-xi/] X X — If = — xf/ ; — X X 1/ = — x}/ ; - X X — ij = -\- xt/. a. In Art. 36 there was given an investigation of the rule of signs in nmUiplication. It will be helpful to give similar proofs at this point, using letters, wliich may stand for any numbers, instead of figures. 1. To multiply + bby + a. Once 4- 6 is + 6; twice + /> is + 2 6; three times + h is+ :i /.: and so on. Then a times + Ms + ah. 2. To nuUtiply — d by c. Once — ri is — (/; twice — d is — '2 d; three times — d is — :i d. Then c times — dis — cd. 3. To nuiltlply ;j by — in. From what was learne 1 in Chapter II., this means that u is i,o be taken 0— m times; i.e., uo times less m times. Hut, x « = 0, and ut X n = wn. Subtracting the latter protiuct from tlie former, we have, — inn = — inn. 4. To multii)ly — n by — m. Hy tlie same reasoning as before, multiplying — n by — />/, we would have, — (— nin) = -\- nin. 86 TEXT-BOOK OF ALGKBliA. 105. Rule for Signs in Multiplication Like signs give plus, and unlike minus. This rule applies to all other expressions as well as to single numbers. 106. Multiplication of Monomials In the multiplication of monomials, as in the product — 5 d^x X + 8 cuj, there are three things to be considered ; viz., the signs, the numer- ical coefficients, and the literal part. 1. The Signs. — The fundamental rule of signs is that just given in 105. Extended, it includes the rules of 39 and 40. 2. The Numerical Coefficient. — The numerical coeffi- cients can be multiplied as in arithmetic, giving the numer- ical coefficient of the product. Thus T) aX&b = r^XC^Xah=: 'M) ah. 3. The Literal Fart. — We learned in 56 and 57 that a* is the shorthand way of writing a X aX aXa\ a^ of a X a X a-, a^ oi a X a; or, dropping the multiplication sign, a^ = aa, a^ = aaa, a'* = aaaa, etc. (1. Let us seek the product of, say, ah, a%, and ex". Writing out the factors, we get, aXaXcXaX a XaXbXcXxXx. Or,. by 38, '2, changing the order so as to bring the same letters together, we have, aX a X aX a X a XbX cX c Xx Xx = a^b(fx-. For, a X a X a X a X (^ = (^^^ (" X c = c^, and x Xx = x^. And just so it will be in every example of such multi})li- cation. A letter that appears in two or more factors is repeated as many times as indicated by the sum of its exponents. A letter that appears in only one term is repeated in the product merely as many times as there are units in its exponent. algebhak; notation. 87 (2. In the same manner let us multiply a;*" by jc". x^ =^ X- X X X X . . . X, m factors being multiplied together. x" = X- X X- X . . . X, 11 factors being multiplied together. x'^x^=xxxxxx . . . X, m -\-n factors being multiplied together. .-. x"* x" = x'" + " ; i.e., the exponents of the factors are added for the exponent of the same letter in the product. (3. Multiply together the following compound expres- sions. (a -J- by mx and (« + ^) (j*^ + '0 V' Writing the product in full, we have («-}-/>) (rt -\-h) {a-\- h) {in -\- n) mxy = (a -\- by (m -{■ n) mxy. 4. Examples of the Multiplication of Monomials. (1. 4 a- X abr = 24 a'^bc. Reasons. + by + gives -f ; 4 X 6 = 24 ; a'^ X a = a«, or abiding the exj)onents 2 -f- 1 = 3 ; b and c appearing in only one factor are brought down. (2. ~Sxj/Xxt/= -Sxhf. Reasons. — by -f gives — ; 3X1=3; x X x = x' ; .'/ X / = i/. (3. 7 m-n'p^ X - mn" X - 2 nY X - 3 jJ"^^ = - 42 Reasons. — An odd number of — signs, viz., three, gives — in the product ; 7 X 1 X 2 X 3 = 42 ; m^ X m = m«, or, 2 + 1=3; n' X n" X ii^ = 7i'+«+^ for s + « + 2 is the sum of the exponents of n: p' X p^ X p*~^ = /)'"•"% for the sum of the exponents is ^ 4- 2 -1- r — 2 = y? + ?'. 88 TEXT-BOOK OF ALGEBRA. 107. Rule for the Multiplication of Monomials. 1. An odd number of minus signs gives minus in the pro- duct ; otherwise the product is positive (39). 2. Multiply together the coefficients which are expressed in figures. 3. In the literal part, add the exponents in the factors for the exponent of the corresponding letter in the product, and bring down single factors. a. For convenience it is customary to arrange the literal part in the order in which the letters come in the alphahet. Thus, ax would be written, and not xa, unless for some special reason; x'kim'^y would be arranged into am^x^y; zhjx:^ into x~yz-; and so on. 108. Exercise in the Multiplication of Monomials. 1- (+ 8) (- 2). 8. _ 2 a^ X - '^ a%. 2. (+6«)(-2a). 9. a%<^XaH>'. 3. (+5wm) (+9??^). 10. 2xi/X2x'^!/. 4. (+ 8 ah) (- 3 c). 11. - \ ab-.v X T) ,tx'ii. 5. (2 ah'^x) (3 aVhr''). 12. - .vhj'- X - j'Yz-. 6. — xhj X2 ax. 13. 2 ./•'" X x. 7. 7 a^c X2ah. 14. //" X — )/. 15. - 3 (a + by X - 9 {a + bf. 16. 2 (:r + y) X 4 a'' {x + y). 17. 2mXnX — a X —2 b. 18. 8 rhj X — exy Xif. 26. \ , m times; — <\ vi times ; and -|- + d -f .) (o + ^. + y + r). Now, remembering 3, above, it follows tliat in the final product every term of the multi])licand woidd be niulti])lied by each term of the multi})lier. 5. The Distributive Law. — The multiplication of every term of tlie multiplicand by each term of the multiplier is termed tlie 2. (3 i/' + // - 2) j-i/ = ? 3a -':b Sf-\-i/-2 4a xji 12a^ - 8 ab 3xf -{- xf - 2 xi/ 3.-7 a^cx^ (4ai/ -3x -\-2 r) = ? 4 ay — .*) ./• -|- - r — 7 a*ex'^ - 28 a* cxhj -h 21 a^cx» -Ua^ ch [. (3< /^ _ 4 //^ _ 2 ^.2) X _ „-2^. ^ •> 3 a*- 4 A* - 2 t- — «'c . - 3 a*« -h 4 aVt^e -f 2 « V 2 ./ - -3.1-4 2x-' - 3 .i- — 4 4x* — (yx* — H x^ m\ilti[)l yinj,' by 2 — (•) ./•=' 4- 1» y- -\- 12 ./• multiply iiig by — .'! — H jr'^ -h 12 X -f 1() iniiltiplyiii<,^ by — 4 4 .r* — 12 ar« — 7 r' -f- 24 r + H5 J//.v. 6. ( 3/rV; - 2 <^// -f- //*) (2 (fh -\- h') = •: :\„v, - 2nh-'-\-h^ 2ab-\-b-' » ./■ U/ (5 aV>2 —4 aV»* -\- 2 ab^ multiplying,' bv 2 _}_ c) a^ — ah — 4 a" _ + a'¥r -/>« + A^ a^ + aHi' + a^ //• y^ 111. General Rule for Multiplication c 1. Monomials. See 107. 2. Polynomial and Monomial aia;i:iikaic notation. 93 Multiply every term of the polynomial l>y the monomial, according to the rule for monomials, and unite the products. 3. Tolynomials. (1. Two Polynomials. — Multiply every term of the multii)licand by each term of the multiplier and add the partial products. (li. Three or more Polynomials. — Multiply any two together and their product by a third, and so on. ). 4. (13 in^n- — 1 1 nV>'^ -f- 5 x*if^) X aha% 5. Multiply — /// — n — (I — c by — m. 6. {a-2b-'Sc) ( - 14 r). 7. -{a -h -r) {- 2abr). 8. 4 ah (a^ - :\ ah - 5 //^) (- 2 ah% 9. Multiply r -h 3 by J- 4- -• 10. Multiply .r + 8 by x - 2. 11. Multij)lyy/- -|- 7- by j- — >/'. 12. Multiply X' -f f/'^ by x^ - y'K 13. (x^- -\x -\- 16) (ar + 5). 14. {X — 9) {x — 5). 15. X -{- a and x -\- n. 16. m -|- <' «iii + ^• 29. a^ — a^-^a — 1 by ^//^ — ^* + 1. 30. (x*-{-2x^-\-3x'--}-2x-{-l) (-3x^) (-2x). 31. {«(^ _ (a - ^>) (^> + c)} -b{b-(a-c)} = ? SuGGKSTioN. — First perform the multiplication indicated in (a — b) {b + c). The product enclosed in a parenthesis takes the place of the factors. Then remove the two inner parentheses. Lastly b is multiplied into each term of the second bracket quantity, according to the distributive law, and this product is subtracted from the first brace quantity. 32. (x - 1) (x - 2) - 3 .^ (a: + 3) + 2 {(x + 2) (;r + l) - 3} . 33. 4 [4 (t - (4 ti + 1) (a -3 ) - 9] - a {(12 a - 2) 2 - (a -3) (a -4.)}. 34. a"'-\-b" by ff"'-\-b". 35. (l + ,y + y-^ + y^ + _,/)(l_y). 36. a'-\-a'^-\-a' by ^/ + 1. 37. (x -5)(x- ()) (./• - 7). 113. Use of one Letter to stand for a Quantity expressed by any number of letters. In 109, 4, we had x, y, m, and 7i. each representing sums of other letters. Likewise, in any example, one letter may ALGEBKAIC NUTATION. 95 take the place of any coiiibination of letters. For, when the combination is reduced to its numerical value, it gives one number, and the letter used to take the place of the com- bination has this number for its value. Thus, A may stand for ^!—t m or, d may replace (3 c^ -\- 7 xyf. For, if a = 5, i = 4, c = 3, m = 11, a^-h3^>c^ _ 5'^-h3-4-3 _61 m ~ 11 ~11* Thus A must have the value — In like manner, if c = 2, a- = 3, and y = 1. c/ = (3 •2»-f 7 •3- 1)2 = 45-^ = 2025. '' " second quantity. 96 TEXT-BOOK OF ALGEBRA. The square is found by simple niulti- ^J- ' :? plication. The result, since it is true for \ i^ A T^ ^^^ quantities, may be translated into a T ^1^ , T.2 general theorem as follows : — A^ + 2 AB 4- B-^ G^- Theorem I. I'/te square of the sum of two quantities is equal to the square of the first, ])lus twice the product of thf first hij the second, plus the square of the second. (4. ExEKcisK. — The result is to be written down without multiplying. (1) (3 + 4)'-^ = 3--^ + 2 3X4 + 4^ = 9 + 24 + 1(3 = 49 = 7^. (2) (4 ax + 9 ijf = 16 a-x- + 72 )\ (2. Generalization. (-t) (4 «(2 - - 6 »')-. (5) (fi x" - -If. («) (ix;,- -3i '■)•-■ A -B A-B Let A denote any quantity, and B de- note any second quantity. Then the ^' I AB+ B-^ ^'1''^'"^ of A - B is A-^ - 2 AB + Bl A^_2AB+B^ ^'''''' (3. Thkokem II. The square of the difference of two quantities is equal to the square of the first minus twice the, product of the first by the second, plus the square of the second. (4. Exercise. — The answers must be written down by the theorem. ALGEBKAIC NoiATioN. 97 (1) {a-'6aby. (4) ((>y-7«)«. (2) C^x-Syy. ip) (3/-^')'. (3) (4y'-3y. (6) (10-D//m)*. 3. The product of the sum mid difference of two quantities. (1. ExAMi'LEs. — Set down and multiply as in 112. (1) (pa-\-b)ipa-b). (2) (3//r-i-r)/>)(3//i-'-5yv). (3) (11 m' - 6 w*) (11 m- + 6 n'). (4) (15«*-l)(loa^-f-l). (2. Generalization. A +H A — B Let A denote any ). (2) (6«6-f 9)(Gr/^-9). (3) (lxyz-\){lxi,z + l). (4) (a" -I- b") {a" — b"). 115. Exercise in the Use of the Theorems of Multiplication. 1. (a +6) («+*). 6. {x-\-y)\ 2. (a-f 4)(a-f4). 7. (3a + 26)». 3. (7 + 5)(7 4-r>). 8. {n<*d^Ae(Py. 4. (a-.r)(n-x). 9. (\ m^ -^ \ n^. 6. (m -\-7i) (m — v). 10. (4r/7;* - \)\ 98 TEXT-BOOK OF ALGEBKA. 11. (6 + 3) (6 -3). 14. (imhi-iy. 12. (4 a-' 4- 3^-) (4 a- -3 6'). 15. (a\-f,')\ 13. (8 abc — 9 ab-cy. 16. (a-"' — by)\ 17. {a"'-^b"y-. 116. Other Noteworthy Cases in Multiplication. 1. Prove by Theorem I. that (a-\-b + cy = a'^b'^ c- -\-2ab-\-2ac-\-2 be. We have, \_{a + ^') + cj = {a^by^2 {a -\-b)c-\- &= a" -^2ab-\-b'''^2ac^2bc^c\ 2. Prove by the same theorem that {a^b ■^c-\-d)- = (r + b'^ + c" + ^- + 2 ab -\- 2 ac -{-2adi-2bc-i-2bd-\-2 cd. lia^b) + {c + d)J={a^b-f^2{a^b){e^d) J^(c^df = a''-]-2 ab + b' + 2 ac^2ad -^2 be -\.2bd-\-c- +2cd + d\ The formulse of this and the preceding example general- ized give the following theorem. 3. Theorem IV. The square of the sum of any number of quantities is equal to the sum of their squares, plus twice the product of each quant it tj by all those that follow it. 4. Prove by aetnal multiplication that (a + by = {a + b) (a+ b) (a + b) = a^ + 3 a% + 3 ah^ + b\ From this formula we have, 5. Theorem V. The cnhe of the sum, of two quantities is equal to the cube of the first, pdus three times the square of the first times the second, plus three times the first times the square of the second, pdus the cube of the second. 6. {a 4- by = a^ + 4 aH) + Ga'"' li' + 4 a// -\- b\ Prove this equality by actual multiplication of the former result {(I -j- by by tt + ^. State a theorem for this case. ALGKBliAlC NOTATION. 99 7. Perform the multiplications indicated in the following: (1. (x + 3)(x-h4). (7. (x-h2)(a:-6). (2. (x-f6)(u:-hy> (8. (x-3)(x-4). (3. (X -f 13) (x -h 17). (9. (X + 9) (x - 12). (4. (X -hi) (or -5). (10. (x - 15) (a- + 16). (5. (X + 18) (x - 4). (11. (x — 12) (x - 8). (6. (x -h 20) (X - 4). (12. (X - 25) (x - 30). 8. TiiKMKKM \'I. 77n' prodKct of two binomials Jnirlnfj a term common is equal to the square of the common term, plus the SUM of the other terms times the common term, plus the PRODUCT of the other teiins. 9. Exercise in the use of the theorems of this article. (1. (x -h 11) (x -h 13). (6. {a-b-\- S)\ (2. (x-6)(x-l). (7. (Aax-\-5b + 6cy. (3. (x-20)(x + 8). (8. (15x-9y- 12«)'. (4. (x -h 100) (x - 2). (9. {Sm-\-n-{-2p-\- 1)'. (5. (« + // + !)'. (10. (a + 2^»-(3c-"b\ (4) Develop the foUowiug: 3 (4 xy, (-2a.V)^ ( - ^rxf, ( -y../)«, 2 ( - by, \(ay}\ (5) Develop and multiply, 7 ( - 2 .r//)' X ( - 2 am) ^; ( - 1) ( - l)"^ ( - 1)M " Y' Kkmahk. — Since 1 = 1-= 1^ = 1+ = 1^, etc., we may regard las liaviiig any exponent desired. 2. Powers of Polynomials. (1. Square of Polynomials. — See Theorem IV. of the last article. Being true for any number of terms, it includes Theorems I. and 11. of Art. 114. ^^ G^''^^^ ALGKHHAK; NoTATIoyV \^\ (2. Cul)e and Fourth Power of Polynomials. — See 116, 5 and 6 for cube and fourth power of a binomhd. The further investig^ation of the cases already given and of the development of hi«]:her powers is remanded to a subsequent cha]>ter on Involution. ^rr fVid 102 TEXT-BOOK OF ALGEBRA. CHAPTER IX. DIVISION. 118. The definition of division was given in Art. 43. All the rules for signs were explained there. 119. Division of Monomials. 1. The signs. — See 43. 2. The Coefficients. — Divide the coefficient of the divi- dend by the coefficient of the divisor to obtain the coefficient of the quotient. 3. The Literal Part. — In multiplication we added the exponents of the same letter in the factors. Thus, a^ X a^ = a'. Hence, by the definition of division (43), we must subtract the exponent in the divisor from that in the divi- dend for the exponent in the quotient. Thus, a' -^a^ =a^', x"^ -^ x"" = x"*"". Three special cases deserve attention. (1. JVhen a letter is found in the dividend and not in the divisor, it must appear in the quotient, else when the divisor and quotient are multiplied they will not give the dividend. (2. When a letter contains the same exponent in both dividend and divisor it gives rise to the factor 1 in the quotient, and is not written ; for, any factor is contained in itself once. Thus, Qa^b-^2a^=Zl-b = 3 b. ALGEBRAIC NOTATION. 103 (3. When a letter is found Imving a greater exponent in the divisor than in the dividend, or appearing in the divisor and not in the dividend, the quotient is usually written as a fraction. A factor in the dividend cancels an equal factor in the divisor, as in the previous case. Hence, we subtract the exponent in the dividend from that in the divisor, and write the letter with this exponent in the denominator. When a letter is found in the divisor and not in the dividend, it is written in the denornirmtor of the quotient. A fraction as here used is to be regarded as an expressed division, the numerator being the dividend and the denom- inator the divisor. 4. Examples in the Division of Monomials. (1. (\ a*b' -^ 2 a'b = :UrlK Reasons. -|- -t- + giv<'« -j-, (j ^ '2 = 3', 4-2 = 2; 2-1 = 1. (2. 9 m,'' Ji^j, ^ 3 n'^ = 3 vi'^ p. ^ 4 s (4. - 45 ? — 7iy 122. Leading Letter in a Polynomial. In most polynomial expressions in division there is what is called a leadhig letter. Thus, in ax^ + hx} + ^'•^'^ + ^^^ + ^•^" +/^ + (h ^ is the leading letter. Usually it is quite easy to distinguish it. Sometimes, however, one letter seems as ])roniinent as an- other, as in a^ -|- 2 ah + ^"^ + 2^/^ + 2 he ~\- o-. In such cases it is convenient to regard that l^tt^r which comes first in the alphabet as the leading one. and to let the others come ALGEBRAIC Nn lATlON. * lOo after it as nearly as may be iii their alphabetical order. Thus, a would l>e the leadinj^ letter in the above expression. In the terms in whieh a does not appear would l)e the leading letter, and so on. 123. A Polynomial is said to be arranged with reference to the powers oi the leading letter when the exponents of this letter either increase or decrease in regular order. For example, 2 a* -I- 2 a-'h -h r» a-h' - (> al/' -f- 4 //* is anunged with reference to the descending powers of a. Also, 1 -|-2.i-r,. 1-3 + 8.1-5 is arranged with reference to the ascending powers of x. 124. Division of Polynomials. — Explanations. There are two ca.ses corresponding To short and long division in arithmetic. 1. Division of a Polynomial by a Monomial. As in multiplication we multiply every term of the poly- nomial by the monomial; so in division, for similar rea.sons, we divide every term of the i)olynomial by the monomial. Such examples may be set down as in short division in arithmetic. 3 ,,.'//) () ,,7,2 -f 15 a-b^'c - 3 a'-be' 2. Division of a Polynomial by a Polynomial. (1. Investigation of tlie Process of Division. — Since division is the ccmverse of multiidication, let us nndtiply two iK)lynomials ; then, using the product as a dividend, and one factor as the divisor, let us try to obtain the other factor as quotient by retracing our steps in multiplication. 106 TEXT-BOOK OF ALGEBRA. 2 a;2 - 3 ax + 62 Sx^ + 4bx- a2 Qx^-9ax^ + S h^x^ + 8 &a;3 - 12 ahx"- + 4 Wx - 2 a2a;2 -f 3 gSa; - ^252 6 a;4 - 9 ax^ + 3 ¥x'^ + ^hx^ — 12 a6a;2 + 4 6% - 2 a^a:^ + Z a^x - u^b'^ For reasons given farther on, we arrange the dividend with refer- ence to X. Divisor. Quotient. 2 a;2 — 3 ax + b^) Dividend. (3 x^ ■^4bx — a^ 6x*-9 ax^ + 8bx^-2 cfix^ — 12 abx- + 3 b'^x:^ + 3 a^ic + 4 b^x - ofil>^ 6 X* - 9 ax3 + 3 b^x^ 8 6x3 — 2 a^x^ — 12 a6x2 + 3 a% + 4 bH — d^U^ 8 6x3 - 12 a6 x2 + 4 68x - 2 d^x^ + 3 a^x - a-^62 - 2 a2x2 +3a3x - d-b^ Explanation. — We know in advance that the quotient is the former multiplier, 3 x^ -f- 4 6x — «-. In order to obtain the first term of the quotient, 3 x-, the first term of the dividend^ 6x\ is divided by 2x2, the first term of the divisor. Then, as in arithmetic, the divisor is multiplied by the first term of the quotient, giving 6x^ — 9 ax^ + 3 6^x2, and the product is subtracted from the dividend. In order to obtain the second term of the quotient, 4 6x, we divide the first term of the remainder, 8 6x3, by the first term of the divisor, 2 x2. Then we multiply the whole divisor by this quotient term, and subtract the product from the first remainder. To get the third term of the quotient the first term of the last remainder, — 2 d^x!^ is divided by the first term of the divisor. After multiplying the whole divisor by the quotient term just found, and subtracting, there is no remainder, and the division is completed. a. Remark on the Process of Long Division. — If we look closely at the example given above, we see that the process of division suc- ceeds in separating the dividend into three trinomial parts, into each of which the divisor is contained exactly. The same separation of the dividend is seen in long division in arithmetic. 25) 675 (27 i.e. 25 ) 500 + 175 = 675 ^^ (Q) 20 + 7 = 27. 175 175 ALGEBRAIC NOTATION. 107 (2. Divide 4x^-x*-\-4xhy2x* + '3x-\-2. DiVIDKND. 4 a-6 — ar» -I- 4 ar | 2g'-f 3g +2 Divisor. 4x*-|-Ga-* + 4ar' 2x^—3x*-\-2x Quotient. — 6x* — 9x^ — (jx^ 4ar»H-6ar=' + 4x Here the dividend is separated into (4a:» + 6ar* + 4a-«) + (-(>x*-9a-«-(>x') + (4a:» + 6x'+4a:). By adding these we may verify that their sum is the dividend. Writing the dividend in this form and dividing as in short division 2x» - Sj-- + 2x b. Necessity of dividing the first term of the remainder by the first term of the
  • as a tenn of the quotient! 125. Examples in the Division of Polynomials. 1. Divide :i aH* -f (> a*b" — 3 ab^ by — .'i ab\ — 3 ab' ) 3 a"'b^ -\- 6 a*b'' — 3 ab^ — a'"-^b — 2a^b»-^-\-b Ans. 2. Divide 4 y* — 5 y* + 1 by 1 — if— y. Ari-anging lK)th dividend and divisor afCH)rding to the ascending jK)wers of y, we liave 108 TEXT-BOOK OF ALGEBKA. 1 — 5 7/2 -I- 4 ?/4 I 1-f-?/ — 2//- DIVISOK. 1 + // — 2 v/^ 1 _ y _ 2 if- Quotient. — y — 3 y'^ «. Had both dividend and divisor been arranged according to the descending powers of ?/, it would merely /«rtre reversed the order of the terms in the quotient. Sometimes one arrangement is preferable, sometimes another. h. In writing the partial products it is often difficult to have terms come under those to which they are similar. When dissimilar tenns are written in the same column, care is necessary on the part of the beginner to keep from subtracting their coefficients. c. The divisor and quotient of the first example of the preceding article were written respectively at the left and right of the dividend as in arithmetic. For convenience of multiplying in Algebra both are written at the right. d. When there is no term in the minuend similar to a term in the subtrahend, the latter is subtracted from zero, and according to tbe rule for subtraction, is brought down with its mjn changed. 3. Divide d^ + 1 by d + 1. d'^1 \d±l d' d^ - d' + d' + d d-\-l d + 1 d-1 As in arithmetic, the remainder is written over tlie divisor and added to the other terms of the quotient. In this addition a -|- sign must be used. See 52 b. ALGKHKAIC Nnl.vrioN. lUU 126. Rule for Division. 1. Moiioiiiiiils. See 120. 2. Polynomial by a Monomial. — Divide each term of the Polynomial by the Monomial. 3. Polynomial by a Polynomial. (1. Arrange both diviiU'nd and divisor according to the descending or a-seending powers of the same leading letter. ('J. Divide the tirst term of the dividend by the lirst term of tlie divisor and place the resnlt in the quotient ; multiply the divisor by this quotient term and subtract the product from the dividend, and arrange the remainder with reference to the leading letter. (li. Divide the first term of the remainder by the first term of the divisor for the sec(md term of the quotient ; multiply the divisor by this term and subtract as before ; and so continue until all tlie terms are brought down. (4. If there is a remainder write it over the divisor, and add this fraction to tlie other terms of the quotient. 127. Exercise in Division. 1. :U)a*h -. 5«-/i = ? 2. 27 m'n* (a -\- i-y -4- 9 mu* (a -|- rf. 3. 3x"*^" -^ .f'"-". 4. a' -J- <('. 6. (So mhj -\- 28 vrtf — 14 /////') -. 7 my. 6. (12 ay — 10 a^if^) -h 4 ct'if': 7. Divide 'Myp^q'h'K^ — 7'2 jrf/'-rf^ — S-ij^f/'-rx- by — 12 /»(/-. g 4fr6c-f 8 tf&r^-f 12«^V 2abc 9 27 (« - by - 18 (g - by + ^^ja ~ bf 110 TEXT-BOOK OF ALGEBRA. 10. Divide 2 x^ -\- 7 xi/ -^ 6 y"^ hj x -\- 2 f/. 11. Divide ac -\- be — ad — bd by a -\- h. 12. (12 - 4 « - 3 a^ + iv") -V- (4 - a^). 13. (4 a^ - 5 aH>' + ^^4) -^ (2 a- — 3 ah + Z-/^^. 14. Divide 5 .r-y -f y' -j- x^ + '"> ^y^ lL)y A:xy-\-i/-\- x^. 15. Divide 1 a:*^ + II ic'^ - :^ X + 5 by - ic + 3. 36 4^4*^3^ 16. Find the multiplicand when the product is 3x^ -\- lA: x^ -\- Si x -\- 2 and the multiplier is o;^ -f 5 ic + 1. 17 (a-* — ^'') -=-(«- if). 18. (a^ - 243) -^ (^/. - 3). 19. Divide a^ — 3 if'* by a -f a:. 20. Divide G a V — 14 «5.r« + 12 a^x^ — a'' by 2 aV _ ^s 21. Divide ax^ — air- + IP-x — x^' by {x -\- b) (a — x). 22. (a^ 4- ^»3 + c^ — 3 a/>c) -~ (a'^ + Z*'^ + c^ — Z»c — ac — aZ»). /l 1 7 3 > 23. K' ^'' + 2 ^"^ "^ ^''' ~ '^' "^ 2 '^'' ~ 2 '"^ "^ V2 2^2 24. Divide x^" + x-"i/-" + //« l)y ic-« - x"f/" + y-". 25. Divide x"^ — 2 x' + 1 by x'^ — 2 a? + 1. 26. Divide x"" - a'^ by x- -f- 2 ^/o-^ + 2 a.^x -\- a\ 27. Divide a' + aV/' + aV/ + a-b'' + ^»« by a' + a'Z/ + a-Z*^ -^ab''-\-b\ 128. Zero and Negative Exponents. — Theorems of Nota- tion. Zero exponents and negative exponents arise when the exponent in the divisor is equal to or greater than that in the dividend. Thus, a^-ira' = a'-^ = «<>, and b^ -j- b^'^ ^ ?/-i<> = b-". ALGEBKAIC NOTATION. Ill Their use was obviated in 119, .'i, as explained in (2 and (3 of tliat Article. 1. Zero Exponents. Let A be any (juantity and n \w any exponei Then, Al = A"-" = A«> ; but — = 1, A" A" .-. A** = 1. Thus, 50 = 1, VI'' = 1, 1(X)0<> = 1. Thkorkm T. Avij qiuintity irhose exj)o?ie7it is is equal to unit I/. Note. — There is a sliarp distinction between zero used as a co(V'-" BC^ 1 In like manner, any other factor could be transferred from tlie denominator to the numerator or from the numer- ator to the denominator, by changing the sign of its exponent. Theorem II. Any factor ukii/ he transferred from one term of a division to the other, providimj the sign of its exponent is changed. Kemakk. — Here we s^e the opposite nature of the signs when prefixed to exponents. If a poa'itixe exponent denotes that its quan- tity is to be multiplied into another quantity, a negative shows that its quantity is to be used as a divisor of the other quantity. The student is asked to note that we have come quite naturally to tlie meaning of a negative exponent. In subtracting exponents we have simply followed the rule for division (which in turn came from the rule for multiplication), and have merely subtracted a greater exponent from a less. ALGKBKAIC NoiATloN. 113 a. The following (U'tiiiition may prove helpful to the student. a) . integral power of a number is the continued ) I negative j quotient of 1 by that number. h. The Reciprocal of a quantity is 1 divided by that quantity. The 1 1 a~^ reciprocal of « is . But by this article —•= -- — = a~^ . Hence a a 1 the exponent — 1 denotes the reciprocal of a quantity. 3. Exercise in the use of Zero and negative Exponents. (1. What will x"~'^ equal when w = 2 ? (2. AVhat does a^ X «"* X a"* equal ? (3. Reduce the expression — - — -— to an equivalent 6 arm~'U one having positive exponents. By cancelling the a^s and transposing those letters which have negative exponents U) the opposite term of the division, we derive Gahn-^d (ybe*d (4. 3+* = ? 3-^ 3<», 3-\ etc. Write all the values in a series from the exponent + 4 to — 4. (5. Reduce to positive exjwnents 6 a~^^V h- 2 abc~^; M-i^-n _j. ,^-1. (,3„,;,-4„ X 2--nni) -4- ^rOin-ni 4 (6. Reduce to positive exponents, -, _, 1 a-^b'c^ (7. Simplify 129. Simple Roots. — Dividing a number by one of its roots two or more times will give the quotient unity, since a root of a number is one of its t^puil fa<*tors (60). 114 TEXT-BOOK OF ALGEBRA. 1. Roots of Monomials. (1. Example. — Having a root of a quantity, say the fourth root of 81 a^h^'^c'^, to find, we see that if we extract the fourth root of the coefficient and then divide each exponent by 4, we get a quantity (?>a%^c), which being divided into 81 a^h^-c^ and then into the quotient, and so on four times, gives a quotient 1. Therefore by the defini- tion, 3 a^hh is the fourth root of 81 a^b^'^c'^. This process, it is plain, applies generally. (2. Rule for extracting the root of a monomial. (1) Extract the required root of the coefficient as in arithmetic. (2) Divide each exponent in tlie literal part by the index of the root required to obtain the exponents of the respec- tive letters in the root. (3) Find the sign of the root by the rules in 45. (3. Exercise in the extraction of the roots of mono- mials. (1) -y/W^I^. (5) ^-^WW. (2) ^W^iW. (6) ^-S2a^^ (3) ^WxY- (7) ^6'2^pY'- (4) ^'256. (8) ^8T^i«. (4. Different ways of regarding the same ])Ower. (1) Tlius S ,f^ = (a')' = (ay ; a'' = (a")- = {a^ = (a^. (2) rAriM\ 115 130. Theorems in Division : Divisibility of Binomials. — 1. TilKoHK.M i. Th<' difffrenn' of the slus and minus. 3. Theorem III. The sum of the same odd powers of two quantities is always divisible by the sum of the quan- tities. (1. Verify the following divisions. (A + B) -J- (A + B) = 1 ; (A^ + B'^) -- (A+ B) = A^ - AB + B-^. (A^ + B^) -f- (A + B) = A^ - A^B + A^B-^ - AB^ + B^ (A^ 4- B^) -^ (A + B) = A« - A^B + A^B^ - A^B^ + A^B* - AB^ + B«. These examples suggest the theorem. (2. For the quotient use the rule given for theorem II. Remark. — When the second term of the divisor is minus, the process of division in these examples makes every successive divi- dend plus, so that eveiy term of the quotient is plus. When the sec- ond term of the divisor is plus, only every other successive dividend is plus, and the signs of the quotient are thus made to alternate. The process of division shows also the reason for the rule given above for writing down the quotient. In actual division we first multiply a quotient term by the second term of the divisor and afterwards divide by the first term of the divisor. By the rule we divide first and multiply afterwards, which is easier. 4. Theorem IV. The sum of the same even powers. of two quantities is not exactly divisible by either the sum or the difference of the quantities. (1. Perform the following indicated operations : (A2 + B2) -J- (A + B) = A - B + /^' i.e., is not ex- A 4" jd actly divisible. ALGEBRAIC NOTATION. 117 ( A^ + B^) -V- (A + B) = ? (A« 4- B«) ^ (A + B) = ? ^A« 4- K«) ^ (A + B) = ? (A« -f- B«) ^ (A^ -f B'^) = ? (A-^ 4- B-^) ^ (A - B) = ? (A^ + B^) -^ (A - B) = ? (A« + B«) -f- (A - B) = ? (A« 4- B«) -r- (A-^ 4- B-) = ? The last is exactly divisible, because it really comes iiider Theorem III., although it seems to belong here. (2. The quotients may be written by the previous rules, -f the proper remainders are also found and added. 5. P^xamples in the theorems of Divisibility. — (With l>articular reference to powers which are themselves simple powers.) (1. What is x^ — 21)f divisible by? Ans. a: — 3y; tor it is the difference of the third powers of x and ^y, Theorem I. The quotient is x^ 4- 3 a; 4- 9. For x* -i- x = x"^; then to find the second term of the quotient (x"^ -^ x) X S = '^x', to find the third term, (3 a; -4- ic) X 3 = 9. The signs of the quotient are all plus. (2. 16 1/* — 81 is divisible by what binomial ? By theorem 1. it is divisible by 2y — 3. For, -v/lfTy^ = 2 y and ^Hl=3. To find the quotient: 16y* -5- 2y = 8y«; (Sy» - 2y) X 3 = 12./; (12 y^ -j- 2y) X 3 = 18y; (18y - 2 y) X 3 = 27. The signs are all 4-- Hence the quotient is8y»4-12y*4-18y + 27. (3. By what is x* 4- 32 y* divisible ? Am. a- 4- 2 y. Theorem III., since V^=a: and -N/32p = 2y. To find the quotient : — x^ ^x = :r'; (./•* ^ x) X 2 y = 2 x*i/; (2 x^y -i- x) X2 y = 4 x'Y; (4 x'Y -^x) X2y = Sxy*; (Sxf/*-i-x)X2y= 16 1/. Quotient, x* — 2 x^y-i- 4: x'Y — Hxy^-\- 16 1/. (4. What is m? - 7i« divisible by ? (Of. 129, 1, (4, for this and the following examples.) 118 TEXT-BOOK OF ALGEBRA. This quantity is to be looked upon as the difference of two cubes, viz., of ni^ and tv^, and by theorem I. is divisible by m^ — v?-. (m^ — n^) -7- (m^ — v?-) = m^ -\- m^n'^ -\- 7i^ by the rule. (5. What is x^ — j/^ divisible by ? By Theorem II (x^ — j/^) -j- (x'^ + y) = ^''^ — -^V + ^^l/^ ~ 'f (fourth roots.) {:z^ — if) -^ {x^ + if) = x"^ — f (square roots.) Ix' _ y^-) ^ (x-^ - y) == '' (Theorem I) (6. What is x^ — tf divisible by ? By Theorem I {x^ _ ^9) ^ {x --y)=x^^ xhj + ./•«,//' + ^V' + a^y + -^'V + x'if -h x,/ + //^ {x'' — y^) -H (cc^ - 7/) = x" + x'l/^ + ?/«. (7. What factors has x^ — y^ ? By Theorem I., (.x« - ;/) ^ (.x - y) = ? (x^ - ?/) -^ C;^^ - r) = ? (x« - //«) - (..^ - :/) = ? By Theorem IL, (x^ — ?/) -^ (x + //) = ? (x« - y') -r- (x' + //^) = ? (x' - f) -^ (x' + y') ■■= ? (8. What are the factors of a^ — 1 ? (See Remark in 117.) By Theorem I., ((/« -!«)-- (a - 1) = a' + a' + rt^ _|. ^^2 + a + l. (^6 _ 16^ ^ (,,2 _ 12-) _ 9 (,,6 _ 1) ^ (,,3 - 1) = ? By Theorem II., (a^ - 1) ^ (a + 1) = ? (^^' - 1) -^ («' + 1)=? (9. AVhat are the factors of x^ + x^' ? One is x*^ + x^ ; what is the other ? 131. Exercise in the Use of the Theorems of Divisibility. — The quotient factors are to be written down directly, by reference to the appropriate rule, and not to be found by division. AI.(il.I'.i:.\l»' NOTATION. 119 1. {a^- fr) -- I" -A) = ? 2. I • > -^ ( /// ^ /' ' — '^ 3. - ' - -^ I ^ - - - ' 4. (1 — //) ^ (1 — //> = ^' 5. '(S^r'-fj;//'). 8. {a^'-h'). 9. (/'-^^ + /''')• 10. i»''~-/:>-2 4-^-^. (4. ory-xYi-yy. (5. 14 bc^j- - 21 ^VV + 7 //VV. (G. G hr-.r - IT) Ar"^ - :; //-V-^. )i. In factoring; tli«' Hrst thinji to Im^ done with any jjiv«Mi poly- nomial is to scok for a monomial far-tor in it, and if on<' is found to removr it. SECTION 11. BlN<).MIAI>. 134. Tin* Faitorinsj^ of Binomials. 1. r/tr (Ufference of iim st/mnws is fdefnnil info fJir sum II nd difference of their roots. This is the converse operation of tin* third th«M»r»'iii nf mhitiplication By the first theorem of division 9 7n^ — U) n- is divisilih* by 3 711 — 4 », and by the second it is divisible by 3 m -\- 4n. Or, more simply, reversing tlie theorem of multijjli- cation the differenc^e of the two sqnares is factored into tlu' prodnct of the snm and difference of the sqnare roots. 9 7n'^ — IG )r = (.*i ;// 4- 4 >/ ) ('^ )n — 4 //). 122 TEXT-BOOK OF ALGEBRA. (2. a"" - h\ (3. a^U^-c'd:'. (4. a^- —. ^ 49 (5. 2. 4. The Sinn of the Sff/tn' ftm odd /joirers. Tlieorciii III., 130. (1. 3-8 + 27. I>y the theorem t^ + 27 is divisible by x -{■ 3, and tlie (juotient .s .v'^ — 3 a- -f- 9. See 130, 3, (2. (2. ./••-[- 27//. (9. .r''4-04.-». (3. x' H- .t;^ (10. :r« + 128 xz'. (4. 27 m^ + 8 w^ (11. x^ + //«. (5. ;s' -h 1. (12. a^*^ + f,'\ (6. 1+04 2 ^ (^2 ^ o ah + 2 P) (a' - 2 ^/> Exercise in factoring such binomials will be given in 136, 2. SECTIOX 111. Thixo.mi AI,S. 135. Trinomials. 1. Trinoinidl Square,^.. — See thecu'ems T. and II. in mul- tiplication. Reversing the process of squaring in theorems 1. and II., 114, we derive tliis rule for factoring trinomial squares : Extract the square I'oot of the first term and of tlie last; if the middle term is twice the product of these two, the trinomial is a perfect square, and the sum or dif- ferenc^e of tlie square roots is one of the two equal factors according as the middle term is plus or minus. -s/x^ = .r : V(>4 = 8, 2X8- x = 16 .r, .-. .r^ + 16 x + ()4 = (x + S) (x + z\ (;>>. .r-2+ 4 ic + 4. (5. \) X' + :^() X + 25. (C. y^,f + 22.r//- + 121 .V-. ^S. 4./-' -f 12. /••-// + *.>.'/'. (1). ;^{;./-^4-S4.'-v--f H> V AlAiKHIlAlC .\t>i.\IH)N. 1 ll."> (10. 9/ + 30/ + 25. (11. a:«— (3^ + 9. \x'^ = a-. V9 = .S, L' X .<• X .S = () .1', .-. ;f- — () ./ -^[) = U-:\)(^r-3) = (x-'3y. (VJ. a--^ — I'C, // + 1(;9. (1^. a-^_a- + i. 14. 4x-^--a-// + ^. 3 -^ ^ 9 9 15 (15. — t/'^ // + 25. (K). y' — '2x+l. (17. r><.^— I()a«^; + 5i2 (18. 24 ic-7> — 72 x'b^ + 54 «?.». (19. «2» — 2^j + 40 c\ (18. a-2 4- 9 if — 22. Here the sum must be + 9, and the PRODUCT — 22. x^ _|_ 9 a; — 22 = (a- + 11) {x — 2), since 11 -2 = 9, and 11 X - 2 = -22. 3 m 1 "4 4* (19. a^ J^2a- 15. (22. m-^ + (20. x^ + x^- 30. (21. ,f + \y ^ 16- (23. chl^^-cd6- — e\ ^ ^3 9 (24. a-y + 7 ic^^ — 8. (25. ./■- — 4. T — 21. Here the sum must be — 4, and the PRoiH'CT — 21. y^ _ 4 .r - 21 = (x, - 7) {x + 3), a.p - 7 + 3 = - 4, and - 7 X 3 = - 21. (26. x' -X - 30. (29. a' (27. 7/-^ - 4 //.^ - 45 -^•^. (30. m'P - 3 v/i"^^' - 70. (28. cWc" ^ Q, abc — m, (31. x^' - ^yx^'i/' - 104. yK AUJKIiKAlC NOT.vrioN. 1 -J 1 3. Trinomials the jiriKhnf of ninj tint himnuials. (1. Let us first take any two binomials which will give a trinomial product, say 2x -\-'^ and .S ^ -}- 4, and form their product. '2 X -\- '^ '^' i'<»^^'> ^> '''' + 1^ •'■ + 1- h ' 3x -j- 4 given to find its factors, we know 6 ic^ -}- y ^ jiist three things concerning them •^ %x -\-V2 as the multiplication in the margin i\x^-\-\'l X -{-V2 shows. First, that the product of the two coefficients is (5. Second, that the product of the last two numbers is 12. Third, that the sum of the cross products is 17 x. Now, if the product is given to find the factors, we do not know whether the (> is })roduced by multiplying 2 by 3, or 6 by 1; or, whether the 12 results from the multiplica- tion of 3 by 4, or 2 by 0, or 12 by 1 ; for, we do not know in advance what arrangement of the coefficients will, upon cross multiplication and addition, give the 17. Wliat has been said suggests the following rule. (2. Kule. (1). Take a set of four coefficients, the product of the first terms l)eing ecpial to the first coefficient, and the prod- uct of the last terms l)eing erpial to the last coeffijrient, and examine whether upon (rross multijilicitioii iIkv will give the middle coefficient of the trinomial. (2). If the first set chosen docs not give the iniildh' coefficient, try another arrangement, and so on until a set is found .satisfying the last condition, or it is shown that none will answer, which would indicate that the trinomial is a prime quantity. 128 TEXT- BOOK OF ALGEBRA. (3. Exercise. (1). Required the factors of 6 x^ — 25x?/ -\- 4: y^. Using 2 and 3 as the factors of 6, we write the following sets, 2 — 2 2 — 4 2 — 1 3 — 2 3 — 1 3 — 4 middle coefficient = — 10' middle coefficient =■ — 14 middle coefficient = — 11 Next using 6 and 1 as the factors of 6 6 — 2 6-4 6 — 1 1 — 2 1 — 1 1 —4 middle cocfficent = — 14 middle coefficient = — 10 middle coefficient ^=- — 2o The last arrangement gives the middle term as desired. Hence, ^ x^ — 'liS xy -\- ^ if = (^ x — y) {x — A: y). Ans. (2) 9 a;2 + 9 .^ + 2. (11) 12 .X'- - 31 x - 15. (3) 3 ic^ + 13 x + 14. (12) 15 z' - 224 z - 15. (4) 4a;2_|_llif-.3. (13) 2-ix'-2^)xy-\7/, (5) 9 ^.2 _^ 04 ^ + 7. (14) 'S-\-l\x^-4.x\ (6) 3ir'- + 10.T//-V - 37 ahc - 72. (9) 2 6--^ - 13 rcZ + 6 f/-^. (18) 6 + 32 x- - 21 .^-. (10) 2 tn' - 3 my - 2 y\ (19) 5 x'- - 1 .r.v — fV x\ 4. Trbiomials of the form 9 a^ — 4 tt'^^^ -j- 4 ^^. By adding 16 a'^h"^ to the second term and subtracting it again in a fourth term, as was done in 134, 5, the trinomial becomes a quadrinomial which is the difference of two squares. Exercise in factoring such trinomials will be given in 2 of the next article. 5. Trinomials the Product of a Binomial and Trinotnial. If we multiply 3 .r^ -j- 4 ./• + 5 by 3 a^ — 4, the coefficient of x^' in the product is zero, and the product reduces to a ALGKIillAlc NolAriON. 1«3 trinomial. In the suniu way o x- -\- li x — 4: ami 3a*-f 4 multiplied together give zero as the eoetticient of the first IKJvver of X in the product. Such trinomials, having their coefficients related in one or other of these two ways may be factored. (1. 4r'*-43x-21. To find the factors assume that 2 is the first coefficient in eacdi factor. The second term of the binomial is one or other of the factors of 21. Trying 7 as the second coefficient in each and 3 as the third term of the trinomial, the factors assumed are 2ic- -|- 7ir + 3 and 2 a: — 7, which, upon cross- multiplication, give —43 a;. Hence these quantities are the two factors required. (2. 2oa-»-fila--12. (4. 21 a-^ + 2« a-' + 25. (3. 8«» — 24ar'-|-2o. (o. \) x^ -\- o x -^ TAK SECTION IV. QUADIMNOMIAI.S. 136. Quadrinomials. 1. I'he cube of a binomial. (1. The product obtained in 116, 4, is the form of a binomial cube, as «* -|- 2 nb -\- h'^ is the form of a binomial square. From the cul)e i)roduct there given, viz., {a -f by = a'-f- ^a% -\- '.U/b'^ -f- b^y we derive the rule for ol)taining the cube root. (2. Rule. (1) Extract the cube root of the two leading terms, the first and last as usually arranged. (2) With these roots see if the middle terms are respec- tively three times the square of the first into the second, and three times the first into the square of the second. 130 TEXT-BOOK OF ALGEBllA. (3) The sum of the roots or their difference (depending on whether all the terms of the quadrinomial are positive, or the second and fourth minus) is one of the three equal factors of the quantity. (3. Exercise. (1) 8 a^ - 36 a'b + 54 ab- - 27 b\ -v/8^ = 2a; -\/27l« = 3^*; and 3 (2 a) ^ (3 5) = 36 a^^*, the second term ; 3 (2 a) (3 ^)"^ = 54 ab'^. Since the second and fourth terms are negative the cube root is 2 a — 3 6. (2) a3_|_3^^2_^o^_^l^ (3) Ub^ + A^b^^l2b^ ^ 1. (4) 8 ic^ - 60 x^ij + 150 xf - 125 y\ (5) 2\Q>x^ -\mx^y^\^xy^-y\ 2. Qiiadrino7)iials the difference of two Squares. (1. JFactor x^ -\- 2 xy -\- y^ — a^. Writing the quadrinomial as the difference of two squares, we have {x + y)'^ — a^ which by 134, 1, is factor- able into the sum and difference of its square roots. {x + yY _ a'^ = ( (x + y) + a) {{x^y) - a) = {x^y + a) (x f // - a). (2. 4 a^ - (9 h"- -^bc^ c"). (3. 4a2_(9^2_ 12 be -^ic^). (4. 4:a'--9b''-\-c''-\-4:ac. In this and some of the following examples, the order of the terms is disarranged, and to unite the terms properly is the first thing to be done. Here the first, third, and last terms go together to make the square of 2a-\-c. Hence we have (2a-\- ey — 9 b'^, the factors of which are (2a-\- c ■i-Sb) (2a-^G-3b). (5. P — m'^ — n^ -\- 2 m.n (6. 2ab-\-a'' -x'' + b.^ (7. b^-l-2ab + a\ AIXiEBUAIC NOTATION. 131 (8. (Sa^-^by- (a2-;UV (9. 10 - O^a - 2/0^. (10. x*-y' i-vjj- .;<; (11. -4ir-\-x'-\-'J//r:~\iz\ (12. 42 aO -1-1-41) a- - \) //-. (13. -V2af,--\a--\-i)y'-'Jb\ (14. 2ab-a'-b'-\- 1. (15. a^ -f 2 be - ^-^ - 6=^. (l(>. y"" -(i/- z)-^". (17. »' + .,-y/^-f/A (S(H' 135, 4.) Suggestion. — Add and subtract a-lr. Thus we get {n* + 2 a-'^^^i + 6*) - a^/^ = {n^ + '/- + ch) {,r^ + /,^ - rr/;). (18. 25x<-36a-«y*-h4//'. SuOGESTlo.N. — To have ;i 1 1 iiiuinial square, the inichUe term ought to be 4- 20 i-^//- or — 20 J-?/"- If — 10 xV" *>e taken out of — 30jr'V" a"tJ a fourth term b«' made of it, tlie expression can be factored by this case. (25 I* - 20 xV + 4 y^) - 10 x-^y-' ^ (5 x^ - 2 //-^ + 4 xy) (5 x-^ - 2 y2 - 4 xy). (19. ir>x^-i7ry + //. (20. 9ir* + 38j:y + 49./. (21. 9a* + 21aV + 25c*. (22. 25 X* - 41 a;2 1/ + 16 /. ,'_';;. .,•" -|-r,4. (See 134. ."».") SicjGESTioN. — A(hl and subtract 10 x-^. Thus, x^+ 10x^ + 04 -16x-^. (24. 4a;* + 81«^ (25. 64xy + 81«\ (26. 4 m* -h ()25. 3. Quadrinomials the product of binomials. (1. Examples. (1) Factor (tr -f od -\- he 4- hd. lo2 TEXT-BOOK OF ALGEBRA. Here a can be taken out of the first two terms and h out of the last two, giving which is plainly divisible by the binomial c-\-d^ c + d) a{c-\-d)-\- h (c + d) a -{-b Therefore the factors of ac -\- ad -\- he -{- bd are c -\- d and a-\-h. (2. Factor Qax — 2 by -{■ ?,hx — ^ aij. Taking 3 x out of the first and third terms, and 2 y out of the second and fourth Q ax - 2by + ^bx - 4.ay = ^x {2 a ^h) -2y (b +2 a) 2 a + b) :ix(2a + h)-2y{b^2a ) Zx-2y .'. 6ax-2by-\-3bx — 4:ay = (2a-\-b)(3x-2y), (2. Rule. (1) Take a monomial factor out of two terms, and a second monomial factor (if necessary) out of the two remaining terms. If this is done properly, a binomial fac- tor is seen directly (i.e., if the quadrinomial can be factored at all). (2) The other factor is found by division. (3. Exercise. (1) ac — bd -\- be — ad. (2) a^-{-a^-\-a-{-l. (3) jcfj -3y^2x-(j. (4) Q ax - I ay- 21 xy-\-Uy'. (5) 9 am — 4: bm — 27 an -\-\2bn. (6) m,^ — jnn -\- m^n^ — n^. (7) x^ — xy^ — x^y -{- 7f. (8) cdx'^ — cxy -\- dxy — y\ (9) x^ — xhj — xjf + y^. ALGEIJUAIC NOTATION. 133 (10) abcy — h^dy — acilx -\- hdhc. (11) 12x»-8j-// — 9a;y + Gy«. (12) 15^^«-J-12«-H-10«-|-8. 4. Qiiadrinomials tiie product of binomials and trinomials. (1) 4 a;2 _ 9 y2 _2 a-« -f 3 yz. Factoring the first two terms and taking z out of the last t\v.> 2 a;- 3 y) (2 a; -f- 3//) (2x - 3//) - ;g (2 a; - 3y) 2x-f3y-« (2) a;» - 6 a;2 -f G a; - 1. a;-l)a;»-l-6a;(x-l) x«-|-a; + l-6a; = a;2_5x+l. .*. the factors are x — 1 and x^ — 5 a; 4- 1. (3) m» + 5/»2 + 5 7/?. -h 1. (4) 4 a%' - 1(>1) c' 4- 6 aZ»(/ + 39 cd. (5) *♦ 10j;» + aj*^-:z;-28. Assuming that this can be factored into a binomial and trinomial, let us set down trial coefficients, leaving the middle term of the trinomial blank. ^ .J _ / V _ - Choosing as coeflficients, 5, — 7, 2, and 2 - 4-4 4, as indicated, and multiplying ox^ by 4 i Q j^ J- / w^ — 14 3; *^^ product is 20 x^. Since the coefficient 20x* — 28 of x^ in the product is 1, the product of 2 10x8 — 28 t)y the blank coefficient of x must be — 19. But this value makes the coefficient of ar, — 52 instead of — 1. Hence this ar- K^ _7 rangement of the coefficients fails. ■xci'T* 4. t\'\ -2 4- *>n '1^\\\» arrangement gives — 14 j*"^, which 1 \^ ^2 -^\)x - 28 ^q"''^» + 0^">) J^'^ I.e., the coefficient + 3 10 x"* + x'^ — — X ^^^^28 ^^ ^^^^ trinomial. Now + 3 in the tri- nomial gives — dC in the product as desired. Hence, 2 x* + 3 ac 4- 4, and 5 x — 7 are the factors sought. (6) 4x«-21a-2 + 44a- — 30. (7) 6a;»-3a-2-33a-- 6. 134 TEXT-BOOK OF ALGKBllA. SECTION V. Polynomials of more than Four Terms. 137. Factoring Polynomials of more than Four Terms. 1. Expressions which are powers. (1. A^ + 4 A«B + 6 A'-^IV^ + 4 AB^ + V>\ If an expression of five terms have two fourth powers and three other terms formed out of the two roots as the middle terms above are formed from A and B, it is a per- fect fourth power, and can be factored into an expression of the form, (A + B)^ or (A — B)^ (2. 16 x^ — 96 x^y + 216 x^'if- - 216 x?/ + 81 7/. VT6x' = 2x', ^/Sly' =-^37/; now, 4 X (2 x)^ X — 3 y = — 96 x^ ; 6 X (2 xy X (- 3 j/Y = 216 xY ; 4 X (2 x) (- 3 yY = - 216 xi/. Therefore the factors are (2 x — 3 i/y. Note. — Higher powers than the fourth contain more terms, and involve a greater number of conditions, but are solved in the same way as the third and fourth powers. 2. Polynomials the difference of two squares. (1. Factor a^ _^ 2 ah -\- U' — c'' -\- 2 cd — d} = (a + hy — {c — dy. The factors are the sum and difference of the roots a -\-h and e — d. (Cf. 134, 1, and 136, 2). a -{-h -{- e — d and a -\-h — c — d =^ or, a -\-h -\- c — d and a -\-h — e -\- d. (2. (« + ^> + cy -d\ (3. 4 ^2 _ 12 m/i + 9 71^ — ^)2 _|_ 4^^ _ 4 ^2^ (4 {^a^h-^cy-ie^f^gy. (5. {a^h^c-\-dy-{e-fy. (6. (3 x' — 4 X — 2)2 -f (3 x-^ + 4 X — 2)2. (7. 4 {ah + edy - (a^- 4- IP- — (^ ~ d'^Y. AL(;i;i!i;.\ic N(>'r.\'rM»v. 135 3. Polynomial the square of a trinomial. (1. a^ + 4 b' 4- 9 6-=^ -h 4 ai + 6 af 4- 12 be. Fixtnicting the roots of the square terms, we have a, 2 6, and 3c. The other terms are double the products of these, and therefore (See 116, 3), the factors are {a -\-2b ■\- i^ r).'^ (2. a.^ -h 9 1/' + 25 c2 — 6 ab -f- 10 ac - SObc. 4. Othef Polijnomiah. (1. x^ + L> x>j -f ,/ + C. ./• -h ('. // .»!•, {..' + ///- -f <> (.^ + y) wliicli is evidently divisible by (jc -}- //). ^ + y) («_+j^)' + 6 fe+_y) . (2. * 6 x*-^ — 11 a^y -h 3 //-^ - u-^ — 7 //,i; - 2 ;?2 Factoring ^Ae ,/zVs^ three terms as in 135, 3, and setting down tlie factors in the customary way, we have, 2a.-3y, ^x — y. We see now that if -|- 5; be annexed to the first factor and — 2 « to the second, we sliall obtain the additional terms of the expression. Tlierefore 2 x — 3// -|- ,~ and .» x — // — 2 z arc the fac- tors sought. (3. x'i -2xy-\-y'^^ 5x — 5y. ^ (4. 2x^-xy-3y^-5yz-2z*. (5. 2a^'\-(jax — 18 « 4. 4 x* — 24 x 4- 36. (6. 2a^ — 4ab-4 ae -f 2 b' -|- 4 /yr + 2 e\ _j_ lei) _^ (; ,^/,,. This is the form of the cube of a trinomial, as may be verified by forming and arranging the iircxhict (a 4- ^^ + ^) (a + ^» 4- c) (a 4- ^ + ^). (8. «» — ^» — r« — 3 {a% — ab'' -\- a^c — ac' -\- b'^c -\- be') 4- 6 a^c. 186 TEXT-BOOK OF ALGEBRA. General Remark. — When the number of terms in a poly- nomial exceeds four, it is usually difficult to factor it. And, as a rule, the greater the number of terms the greater the difficulty of factoring. In the foregoing classification of algebraic expressions with respect to the number of their terms it is intended that the expressions are to he in their eximnded forms. Other- wise there will be more or less confusion. Thus {a + ^)^ — c^ is a quadrinomial, not a trinomial or binomial. 138. Promiscuous Exercise in Factoring. 1. a^ _|_ 2 aH)'' + h\ (2. 1) y' - 49 a""}/. 3. 18 x'' + 33 axij + 14 ah/\ 4. 16 ay'h^c^ + 24 alA-^ + 9 h^n\ 5. 8 c" -i\cd-:) d\ 6. (1^ — .r«. 7. .,3/>3 _|_ 512. 8. a' — if'^ — 2yz — .-2. 9. (a + hy - r\ 10. 8 a^ -\- 6 ah/ — 9 n ,/' - L>7 /A 11. 54 a^inx + 12 ahn^ -f- IS aui. 12. (7 X + 4 7/)2 - (2 X 4- 3//)2. 13. x^ — 6 x'^t/ 4- 12 x>r — 8 /A 14. 6 a?^ — 5 .T?/ — 6 ?/"^. 15. 8 «2 + 2 rj - 3. 16. ?/i^x' + m^i/ — ?iV — 7?^//. 17. 1 — a^x^ — //y + 2 r//>x;/. 18. ah/ - />V' - ^''^^//' + hMx\. 19. »-^ + hx^ -\- ax -{- ah. 20. .//^/^ - r^'^ - Z;2 + 1. 21. rr* — 7 a-- ~ 18. AI <;iJii;AI(' NOTATION. 187 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. x(x-{-z) - 1/ (// -h z). ^' + //^ + '^ ^if (-^ + y)' a%c — acSi — (ifnl -\- hcj-. in^-^ n'^ -\- p'^ + li mn -\- 2 mp -\- '1 np. c^d* _ c^ _ a^»d^ -I- a'\ (a + t,y' -h (a + b). a'-' - /A aVr 1 18 x^t/z + y a-=^//-^,t + () x\f/z^ H- 3 .r}/'z\ () am -\- 4 an -\- \) bm -\- i\bn. 1 _ (r _ yY- a^ - (b - r)\ (a _3:r)2-16//2. S(r + //)«- (2 .>•-//)». a^ - /A«. „4 _|_ 0.%-^ - //V - r*. :r^ — 7 a*V* + ?/• 31 J- _ 3;-) - 6 x\ ax Of + A") + />// (^''•^^+ ^W 1 4- (A _ ,,.2) j.-J _ ,thx\ 3 «« _ 6 ^/> -j- 3 //^ + 6 ^r — <•) A/'. (a-^-f //- + 2 <^^»)- - (^'^ -h //-^ - 2 ///>)'^. (^' + y' + v' - -' ^// + 2 .r.tr - 2 //-) - 0/ 4- ^y z^+{x-,jy-2z{x-y). m* — w* — m (m^ — n^) + n (m — w)*-^. (a-^-^a -4)- -4. J.- _ (J. _ 6)2. 138 TEXT-BOOK OF ALGEBKA. CHAPTER XL COMMON FACTORS. 139. Common Factors. — Highest Common Factor. 1. A Coiunioii Factor of two or more quantities is a fac- tor that appears in each of them. 2. The Highest Common Factor of two or more quanti- ties is the product of all the prime factors common to each. a. A common factor is the same as a common divisor. The high- est common factor (abbreviated into h. c. f.) in algebra corresponds to the greatest common divisor in aritlimetic. Indeed, by many authors the latter is the term used in algebra. SECTION I. First Method. Th?: Highest Common Factor by FACTORiNa. 140. Principle involved in finding the h. c. f . by Factoring. — If we have a quantity expressed as the product of its fac- tors, each factor is a divisor of it, and furthermore, the product of (iny set of its lyrime facfors is also a divisor. Thus, the prime factors of 2?A0 are 2, 8, 5, 7, and 11. Now any one of these, of course, is a divisor of 2310 ; the product of any two of them, as .33 (= 11 X 3), is likewise a divisor ; so, also, the product of any three of them as 70 (=2X5X7); and so on. And finally the product of all of the factors is a divisor, being the number itself. AL(Ji:i;i;ai<' notation. 1B9 Similarly the product of any set ot t\w i)riine fiictors of a^ — i^, (ly X — I/, X -^ //, ./•■- -f- //-, and x* -f t/*) is a divisor ofa:* — /. It is evident from what has just hceu said that the prod- uct of all factors common to two or more quantities, (i.e., found in each of them) is a divisor of each of them. More- over, this is the h. c. f., since if an additional factor were introduced all of the quantities would not contain the result. If, then, we have the different quantities exi)ressed as the ])roduct of their factors, we can pick out, merely by inspection, all those which are common, and so have the factors of the h. c. f. To illustrate. Suppose it is required to find the h. c. f. of a* — 2ab -\- b'^, a'^ — P, and ar — he Factoring each quantity, «2 _ 2ab + b'^ = {a — b) (a - b) , a^-b^ = (a + b) (a - b). ac — be =: c(a — b). Here a — ^ is a divisor of each product, and therefore by definition the common factor. To illustrate still further, let us find the h. c. f. of 12 nV> (a ~ xy (a + xf, 3 a% {a - xf (a + ./•)«, and 9 a^b' (a - xf {a -I- x). 12 a% (a - xy (a -f x)* = 12 a% (a - x) (a - x) (a + x) (a^x)(a-\-x). 3 a^b (a - xy (a -f xy = 8 a% (a - x) (a -^ x) (a + x) (a -{- X) (a -^ x). 9 a*b'' (a - xy (a-{.x)= 9 a'b'' (a - x) (a - x) (a - x) (a+x). By inspection, we see that 3a^b is the product of all the monomial factors common ; (a — xy is the highest power of a — ar, and (a -{- x) the highest power of a -j-o;, wliich are contained in each of the quantities. Therefore, by defi- nition, the h. c. f. is 3 a^b (a — x)'^ (a -\- x) 140 TEXT-BOOK OF ALGEBRA. 141. Rule to find the h. c. f . of two or more Quantities by- Factoring. 1. Separate each quantity into its prime factors. 2. Choose out the greatest coefficient and the highest power of every other factor that will still be contained in each quantity. 3. The product of these will be the h. c. f. 142. Exercise. — Represent the h. c. f . as the product of all the prime common factors. 1. 6 a%, 9 a%, 24 a^'xij. 2. 284 a, 126 a, 210 a. 3. 12 ab\ and 25 Z/V. 4. 48.ry;^^ 12 xhfz% 24:xyz\ 20x'ifz\ 5. 2a^-2 ab% and U {a + bf. 6. 35 a^2)xy, and 42 b\xy. 7. 12 a^xhj - 4 a^xif, and 30 a^xhf - 10 a^xhf. 8. .T^ — ?/2, x^ — ?/^, and x'^ — 1 xy -\-^ y'^. 9. a-^ _ 2 J! - 3, and 0^2 _ ^ _l_ ^2. 10. 3a^8_^6iz;2_24ic, and6cc^-96:r. 11. 3:^2-6^ + 3, 6a:2 + 6aj-12, andl2a'2_l2. 12. ^6 _ ^8 _ 30 .p6 _ ;i3 ^3 _^ 42^ and ir« + x^ - 42. 13. ic^'" + o!'" — 30, and ic^'" _ x'^ — 42. 14. «c (a — b) (a — c), and ^c (^ — a) (b — c). Note, b — a = — (a — b), and a — ^ is contained in (b — a), — 1 times. 15. a^b^ - 4 a'b', and a«^2 _ iq ^2^6^ 16. «« + 3 a^^ + 2 ab% and a^ + 6 a»6 + 8 a^^,^. 17. a^ — a^o;, a^ — ax^, a^ — a^r^ 18. x^ — xy^, and x^ -\- xSj + o-y + t/^. 19. a:;3 4- 3 ic^y _|_ 2 ic?/2, ic^ + G .t2?/ + 5 x?/^. AI.(;i;i5KAIC NOIA'PION. 141 20. 3a^ — 'iab-\-0',^a'-6a''0-{-a'0\ 21. 2h' - 5h -^2, rjb"^ -Sb-' -3b -\-2. 22. x' - 1, x' - 2x - 3, iSx^ -X - 20. 23. (ja^^lax — 3 x\ (S n^ + 11 au- + 3 x\ 24. 3x2-|-16ar-35, 5x^4-33.r-14. 25. ch:^ — d^f acx"^ — bcx -\- adx — bd. 26. 2x2 H- 9a; + 4, 2x2 + liar + 5^ gx^ - 3x - 2. 27. 3x^ + 8x« + 4x2, 3x« -h 11 x^ -I- x», 3./-^ - lGx« -12x2 28. 8a» + l, 16a^4-4a2_|.l. 29. 8x«- 27, 16x^ + 36x2 + 81. Kf:m AKK. — The method by factoring is not adapted to the sohition of problems whose expressions are difticult to resolve into their fac- tors; and recourse is then had to tlir method by continiu'd divisioij. SECTION 11. FlM»IN<. TIIK H. V. V. \\\ CONTIM KI> DIVISION. 143. Principles involved in finding the h. c. f. by Continued Division. 1. The definitions and |ninciples of the first method. 2. A divisor of a nnniber (or quantity) is a divisor of any nund)er of times that number (or quantity.) If, for example, 5 is contained in 20, it is contained in {'^) (=z '.\ X 20) three times as often a,s in 20. It is contained in 20 four times, and in (>() tliree times four times, or twelve times. If ft is contained in J. b times exactly, it is containe(l in 2 J. 2b times exactly, and in ///.I. /ttb times exactly. 3. A common divis(»r of two numhers (^or (quantities; is a divisor of tlnMr sum or difference. 142 TEXT-BOOK OF ALGEBRA. This follows from 124, 1. Thus, ^) 24 + ^JL6 ^_9^ . 4 times -[- .11 times =15 times 11 times 2 times = 9 times 144. Application of the Principles to justify the Method of finding the h.c. f. by Continued Division. 1. Application to an arithmetical example. — To find the g. c. d. of 258 and 731. Explanation. — 731 is divided by 258 and contains it twice with a re- 258") 731 {9 mainder 215. Then 258 is divided 515 by this remainder, and so on. 43 is 215) 258 (1 contained in 215 exactly and is the 215 g. c. d. It is required to prove that 43)215(5 this process gives the g. c. d. of the 215 two numbers. (1. Is any divisor of 258 a divisor of 516 ? By which principle ? Is any common divisor of 516 and 731 con- tained in 215 ? By which principle ? Does it follow that any common divisor of 258 and 731 is a divisor of 215 ? Why? Can it be inferred from this reasoning that the g. c. d. of 258 and 731 cannot he greater than 215 ? (2. Furthermore, is any common divisor of 215 and 258 (and, of course, of twice 258) also contained in 731 ? State the principle. Would it be allowable then to drop 731 entirely and proceed as before with 215 aad 258 ? Also, after a second similar operation with 43 and 215 ? (3. Is 43 the g. c. d. of itself and 215? By the fore- going reasoning is 43 contained in 258 and 731, and can their g. c. d. be greater than 43 ? Does it follow then that 43 is the g. c. d. sought ? ai.(;i:i;i:ai<' \< m a i m -n. 1 j;*, By tlie preceding lUftliod we always nave two num- bers before us from which the h. c. f. required is to Ih^ fouiul. First, the two numbers themselves are taken ; then tlie divisor and a third number derived from the first two; then the last derived and a fourtli, and so on. Now, every derived number used was obtiiined by principles 2 and />'. But these j)rinciples will give other numl)ers. Thus, a divisor of l^'iS and I'M is a divisor of their sum *.)89, or their difference, 473; and 25, may 1) • Uoed instead of 258 and 731. 2. Algebraic demonstration of the metliod of finding the li. c. f. by continued division.^ To find the h. c. f. of A and B. Dividing as in the margin, just as in the arithmetical example, letting the Q's and R's stand for the successive (juotients and remainders, sup^)ose R2 is contained in Rj, Qa times exactly ; then Rg is the h. c. f. sought. The i)roof follows the lines of that for the arithmetical problem. A) B (Q QA • B - QA = H) A (Q, Q.R A - Q,R = R,) R (Q, R - Q2H1 = R2) Ri (Qg Ri— Q8K2=o (1. Every divisor of A is also a divisor of QA (prin. 2) ; every common divisor of I^ and QA is a divisor of iXw'n difference, R (prin. 3) ; hence, any common divisor of B and A is a divisor of R, and so, of course, cannot he higher than a. Furthermore, any common divisor of R and A is also a divisor of B Cprin. 3); therefore B can be dropped, and < Tilts drnionstratioii .«liutild be- omitted by beg^innera. 144 TEXT-BOOK OF ALGEBRA. the process can be continued with A and R ; in the next operation, the common divisor of A and R cannot be higher than Bi, and so on, as long as it is necessary to continue the operation. Hence, the h. c. f. cannot be of a higher degree than the last divisor. (2. The last divisor is contained in the given quanti- ties, as also in each of the remainders, as is made evident by the factored values, (140). By hypothesis Rj contains K2, Qs times. Therefore II = Q2H1 + R2 = Q2 (QsRs) + 1^2 = (Q2Q3 + 1)^2 (43) A = QiR + R, = Q, (Q2Q3 + 1) R2 + Q3R2 = (Q1Q2Q3 + Qi + Q3)K2 B = QA + R = Q (Q1Q2Q3 + Qi + Qs) B2 + (Q2Q3 + 1) B2 = (QQ1Q2Q3 ^ QQi + QQ3 + Q2Q3 + 1) K2 145. Method and Principles in the Solution of Algebraic Examples. — Modifications. The argument by which we have proved that the process of continued division gives the g. c. d. of two arithmetical numbers applies equally well (as we have just seen) in algebra. But there are a number of very important modifications in algebra. To show how these arise, and how they are dealt with, an algebraic ex- ample is given. It is re(pured to find the li. c. f. of S./.5 _ K)./.^ __ ;5(l./-'5 -^ lS.r-. and IS./-* — m,,-'^ — (k- -f 12x. Taking out tlie monomial factors 8^.5 _ \{\jA _ ;5(|.;.J _^ is.r-' = '2x- {A.r' - S./-'^ - ITu- + 9) 18.r^ — (\{)y^ - {\y' + 72j- = iSx (}\x^ — \i)x' - X + 12). In this form we see by inspection that 2x is the monomial factor common, and therefore a factor of the h. c. f. Since we have no read}' nunins of factoring the polynomials to ALGEBRAIC NOTATION. 14o see if they have a cominon la* t i. we resort to division as in 144 to tind their h. c. i". IJut we are confronted at the outset with the dihMimia that whichever polynomial is made the dividend, the first term of the other will not be contained in it exactly. Thus, 4x^ is not contiiined in 3x*, nor vice versa. To obviate this dithculty, and others of a similar character, certain coursf^s are pursued in algebra which were not necessary in arithmetic;. They depend upon the principles already given. 1. IVe maf/j if we choosej multiply one of the two quanti- ties in hand by a factor not found in the other. 2. We may disrard a factor found ifi one quantity and not in the other. For 710 cowjnon factor is either introdured into, or taken aivay from the h. c. f. However, we mast not multiply one quantity by a factor found in the other ; for in so doing we make this factor common to the two, and introduce an irrelevant factor into the h. c. f. 3. It is immaterial at any time whether a quantity itself is used, or its opposite. The h. c. f. when found may have one value or its opposite. For, finding a h. c. f. is simply a question of divisibility. To give a simple illustration, if 6 is a divisor of -f- 30, it is containeil in — .*)() also; and if — 3 is contained in 18, so ;.l>.Ms + :;. a. In order to change the sign of a quantity, we must change the sign of each term (100, 2, 4). To change the sign of a polynomial in the present case. It is customary to divide through hy a negative factor if any is to come out, otherwise by — 1. It is preferable to retain the first term of polynomial quantities positive, changing signs wlien necessary. Let us i)roceed now with the example begun above by first multiplying ^or* — li)r^ — ar + 12 by 4 so as to make it divisible by 4x* — rotiucts (prin. li), and use this result with either of the given polynomials, or with a second result similarly obtained to continue the operation. This plan may b;* followed when the regidar process is about to give large numerical coefficients. f/. When a remainder is obtained which does not contain the letter of arrangement, there is no common divisor. 148 TEXT-BOOK OF ALGEBIIA. 147. Examples in finding the h. c. f. by Continued Division. 1. Given x' — 6x-\-S, and 4 x"" — 21 x' -f 1 5 .x + 20. 4.r^-21.r=^ + 15ic + 2() 4:X^ — 24.x'-\-32x 4:X-\-H Sx^ — lSx - \-24: h. c. f. x — 4.j x'' — (yx-\-H( x--2 x'^ — 4 X ~^Yx + 8 — 2a; + 8 2. Given 7 x^-\- 2ox — 12, and 17 x^ + 67 ic - 4. The solution of the problem in this form promises to give large numerical coefficients. AVe prefer therefore to derive two other expressions (146, c). 7 x^ -{-'2ox — 12 Ux^-\-50x — 24: 7ic2 + 2oic-12 17 X- + 67 X — 4 14 a^-^ 4-00 a- — 24 3 a:;^ -j- 17 a? -j~ -^^ l**' derived expression. 17:/-^+ 67 »• — 4 9 3oa;"^ + 125a; — 60 34 ic^ _|. 134 a; _ 8 34a:^ + 134a- — 8 X'^ ^ X — 52 2d derived expression. We proceed with the two derived expressions as if they were the quantities given. The following arrangement of the work of tinding the h. c. f. avoids the necessity of re- writing divisors. 2d quotient. ic — 13 1st d visor and 2d dividend. £c2_9^_52 x^ -\- Ax -13X — 52 -13.x -52 1st dividend. 3.^2 + 17 a; + 20 3a-2 — 27 a; — 156 44( 44a; + 176 h. c. f . a; + 4 Ist quotient. 3 ALGEBRAIC NOTATION. 149 148. Exercise in finding Highest Common Factors. — The choict' of method in left to the student. (See reiiiiirk in 142.; 1. 21)45 and 3441. 2. (;;i(>. IKii, KWO. 3. ;J094, 4420, 2G52, 4(>G2. 4. ;i y- -i-Wx — So, and 5 r^ + 3.'? ./• - 1 4. 5. 2x'^-xt/-6i/%andSx^-Sxi/-\-^l/-. 6. 6a^-{-7ax — 3x^, and Sa-^ + 11 aa- + Ga^. 7. « V — 4 a^rm -{- 3 ncm'\ and a*c^ — (> a^c'^ni -j- 5 r-m^. 8. 6 a-« — 7<^a-2 — 20 <^ V, ;ind :\ ./•- + 7 o.r -\- 4 //-. 9. 2x* — 7x* 4- oa-^ and ar* -j- Sj- — 4.#'. 10. 12x* — 51a-« + 12a-, and 2a-* — 4a-^ — 2a'» + 4ar2. 11. 3y«-13y2 + 23y-21, and 6y» + / — 44y + 21. 12. ^/ V» — a-/>r-^y + . 16. J-* — 2 a'\r^ -\- a\ and r* 4- 2 r/.r« 4- ./// 4- h\ 3 ^/« - 3 a^h 4- ^///- — /A and o' — h\ 18. a'«4-7a;2 — a- — 7. .r^ 4- 5 .r-^ — ./• — T), and (.r-^ — 2x 4- l)'^ (136, 3). 19. Ga-^ — 2r);r 4- 14. 4 r' - 20.r + 21. 2 a-"^ - 1 r> .r 4- 2S. and 2x'» 4- 5a- -42. 20. 2 a-* — 4 .r» 4- S ./••- — 1 L\/' 4- (1. and ."> .r< — 3 ,r^ — (J .r- 4-9.r — 3. 21. a-^ 4- ^/a-" — 1)^/ V- 4- 11 r/«a- — 4a^ and ar^ — aa-« — 3^/ V 4- 5 «V — 2 a*. 150 TEXT-BOOK OF ALGEBUa. 22. 3x^-^17x-i-20,3indx^—9x — 52. The following example and exercises should be passed over until the subject is reviewed. 23. Given 6 a^ + 13 a'd — 73 ah' + 60 b^, and 8 a^ — 26 a% -^21ah-' — 9h\ We shall use the first, and the difference of the first and second instead of the two quantities themselves. Isl Divisor. 1st D.vidend. 2a3-39rt26 + 100«62-r)9 63 6«3+ iSa^ft- 73a62+ 60^3 3, 05 'ZA. Dividend 6 a^ — 11? asj, + .'^(jo ab^ — 207 b^ 1st a 2d quo. 130 rt^ - 130 «i- -23 6jj- 2535 ii-b + 0500^62 — 4485 b^ '6T.Ufib+ 20~rtf>2 2 162 «26 + C233 ab'- - 4485 6^' 94^2 - 271 «6 + 195 62 3 l^) 1 30 (CH) - .373 a62 + 267 63 130 a2 -373«6 +207 62 quo. 260 a2 -740a6 +53462 282 ar- 200 a2 — 813 «6 + 585 62 746 a6 + 534 62 22 a2 — 67rt6 + 6162 132 a2 130 a2 3 402 a6 + 30662 373^6 + 267 62 a- 13 4tli 4th Dividend . 2 rt2 2a2 — 29 «6 + 39 62 3«6 22a2- 22 a2- - 67rt6+ 5162 -319a6 + 429 62 3d quo. quo. ~ 20 a6 + 39 62 26 a6 + 3962 120 6^ 252 a6- 378 62 2a — 3 6 4cl, D.Tisor. ll. c. f. Explanation. — Having found the second remainder, — 2162a-6 + C233 ah'' — 4485 />'', and the numbers being very large, by tlie method of g. c. d. in arithmetic their greatest common divisor was found to be 23. Using 94 a^ - 271 ah + 195 />^ and 180 a' - 373 ah + 267 h\ we derive 22 or — (57 ah + 51 &"^ by 146, c. Next, using 22 a' — 67 ah + 51?/- in a similar manner 2 a^ — 29 rt6 + 39 //- is derived. Finally, using 2 rt- — 29 ah + 39 6"^ and 22 or — 67 ah + 51 //- the h. c. f. is obtained in the usual manner. 24. 6:r» + 13a-2 + 15£c — 25, and 2x» + 4:^^ -f 4a: — 10. 25. Q>x^ — x^ — 3x''-Ax — 4, and ^x^ — 2x^ _ ig^c^ + 3x + 10. 26. 7/ -57/2 + 117/ -15, 7/^-7/2 + 3^ + 5, 2l/-7 7/ + 16// -15. 27. 2 if -\-2i/ — 3 if — 5if-Uf — 7 ?/, and 2if-\-2f — r)f-ryf-7y-'7. 28. .T« + 4.T^-3.x'^- 16.t3 + 11.t2 + 12.t -9 and 6x^ + 20£c4 - 12 ic^ _ 48x2 + 22x + 12. ALGKBltAlC NUTATION. lol CHAPTER XII. COMMON MULTirLES. 149. Common Multiples — Lowest Common Multiple. 1. A multiple of a qiuintity is literally the (juautity mul- tiplied by something. 2. A common multiple of two or more quantities is a quantity that is some numl)er of times each one. 3. The lowest common multiple of two or more quanti- ties is that one which is of the lowest degree. a. \ multiple of a quantity is often defined as one that will contain it; and a common multiple of two or more quantities as a (luantity that will contain each. The lowest common multiple (ahhrevlated into 1. c. m.) contains no factors except those necessary to make it a multi])lc of the several quantities. Of conunon nniltiples there can be any number, but there can he only tnir lowest common multiple. To illustrate this, if we take any quantity as a + m, then '.\ {a + m), ({ n 4- V, or any higher exponents. (3. aWc^ still contains each. (4. aV/r* the 1. c. m. 2. Find the 1. c. in. of '.) n (x^ — 1), (yab (x^ — x), and 4 fc^b (x — \). Factoring the (|uantities and writing the re- sults side by side, ;j a {X- — 1) = 3 a (x 4- 1) (j- — 1) 6 ab (x^ — x) = 2-3 abx (x — \) 4 .rV; (j-^ - 1) = 2-2 a% (x --i) (x' -\- x + 1). .-. the I.e. ni,l)y rule 1, is 12rrYy.r (x + 1) (x - 1) (x' + ;r -f- 1). 3. Find the 1. c. m. of x'' — irx^ -|- U j. — 6, x"' - 9 x^ -|- •jC, .r _ L'4, and ar* - 8ar* + 19 ar - 1 2. 3^ _ 9 j-2 -f 2r> X - 24 (ar'^ -Gx^-hlla;— 6 7^-(Sx^-\-nx-iS (1 - ^ (-33-^ -f ir)x- 18 xjj-_5x -h 6)a:^ -(\x^ -\-\\ x - (S {x- 1 ar' — o.r'- -j- ^> ^ — x^ -\- r» .r — (5 — x^-^r^x ~() ... ar- — 5 a- -f- 6 (= (a- — 2) (a- — 3)) is the 1. p. m. of these two quantities. Dividinj? the third quantity, .r*" — S a-* -|- 19 a- — 12 by J--— 5 a- 4- r». tlicir h. <•. f. is soon found to l)c 154 TEXT-BOOK OF ALGEBRA. X — 3, We are now in position to factor the three quanti- ties, since we know two of the factors of the first two quantities, and one factor of the third. We find x^-6x^-{-llx~ 6 = (x-l)(x-2)(x -3) x^-9x'-\-2(Jx-24: = (x- 2) (x - 3) (x - 4) x^ - S x^ -\-19 X - 12 = (x - ^) (x - 3) (x - 4). By inspection, we have for the 1. c. ni. (x — 1) (x — 2) (X - 3) (X - 4). 153. Exercise in finding the 1. c. m. of Sets of Quantities. 1. 54, 81, 24, 27. 2. 60, 12, 120, 48, 3G. 3. 432, 270. 4. 18ax2, Taif, 12xij. 5. x^ and ax -\- x'^. • 6. x'^ — 1 and x'^ — x. 7. 3a% 4.b% 2cH, 9ad^ 8. x^ — y^, x -{- y, and x — y. 9. ah'^c^x^, a%G^x^, and cv'U^cx. 10. 3x-yz^, ir)xy^z% 10 xhj'^z^. 11. 3 a'hc, 27 a^h'', and (>. 12. 9 a'b% 12 b\ a^c^, 36, 8. 13. 14a2^,2^ 7^»V, 3, 2, 5. 14. 2x(x —y),4. xy (x^ — y^), 6 xy^ (x + ?/). 15. 4 (1 + x), 8 (ic - 1), 1 - ;r2. 16. a'^ + «Z», ai» + b\ 17. 3x2, 4ic2_|_g^ 18. Ax^y — y, 2x^ -\-x. 19. x*^ (x - yY, if {x 4- yy, xy {xP- - y^) 20. Sa''b-\-%ab% Qa-Q>b. Ai.(ii:ni:Aic noia ri<>N, loo 21. a (x — b) (x — c), b (/• — jc) {x — a), ami c (a — x) (b — X). (See .Sug. Ex. 14, 142.) 22. X' 4- 2 X, x'^ + 3x + L\ 23. '^ {X -it y\'i {X - y)rS{x' ■\- f). 24. a -f b, a^ -\- 2 ab + b'^, and «* - b\ 25. a-- -f 5 X -j- 0, and x'^ •\- (Sx -\- 8. 26. a--^ -f- 1 1 J- 4- 30, and x'^ -\.\2x-\- 35. 27. a'^ — X', a^ — 2 ax -\- x'-, a^ -\- 2 ax + x^. 28. x' -l,x^-\-x-\-l. 29. Ox^-x- 1, 2a:'^-3ic-2. 30. {x - x^, x'^ — 1, and 4 x (1 + x). 31. x2 - 4 «''^, (x + 2 ay, and (x - 2 af. 32. x2 — 1, x3 + 1, x» - 1, and x« + 1. 33. 3 («« - ^«), 4 (« - b)\ 5 («<- ^^), <; (^/- - ^2)^ and 34. x^ - 2 x^ + 1, X* H- 4 ar» + 6 x2 + 4 X + 1. 36. 3 x^ + 2G x« -f- 35 x*, 6 x-^ + 3S x - 2S, and 27 x» H-27x'^ — 30x. 36. 12 X- - 23 xy + 10 if, 4 x'-^ _ 9 xy + 5 y-, and 3 x^ — 2n. X* + ffX"' 4- a^x -f- O^ X^ + «2^2 _|_ ^^4^ 38. 15x3- 14x'V-f 24xy2-7y'', j^,^j 27x» + 33x*V- 20x^=^ + 2^. 39. x2--3x?/- lU7/-,x2-|-2xy-35//^x- - Sx// + 15//-, and x^ -\- A xy — 2\ tf, 40. .f'^-f- 7x + 9, x*-^ - 3x -h 7, and x^ - 2x + 11. 156 TfiXT-BOOK OF ALGEBRA. CHAPTER XIII. FRACTIONS. 154. A Fraction in Algebra is any expression written in the fractional form, that is, with a numerator and denom- inator, and used to indicate a division. «. Thus, if a and h represent whole numbers | represents a simple fraction in algebra. The unit is divided into h parts and a parts are taken. Or, if we look upon the fraction as an indicated division, a is divided by />, just as 3 is divided by 5 in %. But letters are supposed to have any values, integral or fractional. Now, if, for example, « = ^ and b = f, we cannot any longer say that the unit is divided into f parts and i part is taken, for such language has no meaning; but we can say that {> is divided by | giv- ing an arithmetical complex fraction. For this reason it is custom- ary to look upon all algebraic fractions as indicated divisions (Cf. 63, r/), the numerator being the dividend, the denominator the divisor, and the value of the fraction the quotient.' As in arithmetic, the numerator and denominator are called the terms of the fraction. Besides such expressions as the one given, I , the fraction may have any complex quantity for either or both of its terms, e.g., ^> «^ + 9 qc + lo 6^c ^^^ ^ trinomial for its numerator 16 ahc + 9 c2 and a binomial for its denominator. h. It is the form of an indicated division, and not any nufherical value it may have, which makes an expression fractional. Indeed, what we should call an integral quantity, as 3 a&, may have a frac- tional value ; thus, putting a = |, ?> = 4, 3 ab = |. While, on the other hand ~ which is an algebraic fraction becomes equal to 2, an integer. J For a thoroughgoing treatment of all the fundamental questions of algebr.a the teacher should consult treatises on the subject like Peacock's or Chrystal's. ALGEBRAIC NOTATION. 157 SECTION I. ('I.A>SIFI(ATI<)\ AM) 1'IM\« IIM.K<. 155. Classification of Fractions. 1. With respect to their origin. (1. A Simple Fraction, the original form of the frac- tion, contains entire (quantities for its numerator and denominator. Thus, p^, J^±^''+'\. 5 nrn 2 nif/ -{- 3 nr: — f- (2. A Complex Fraction arises upon dividing one frac- tion by another. A complex fraction has a fractional ex- pression in one or both of its terms. nf — 71' ") „/, in III -f 7 7'S Thus, (3. A Mixed Quantity is one which is partly integral and partly fractional. E.g., r,„+£ a 2. With res[>ect to their capability of reduction to a mixed quantity. (1. When the numerator does not contain the denomi- nator an entire numl)er of times, the expression may be called by analogy a proper fraction. ,, iSnhi' 3r/ 4-2// mn \-\- i c (2. When the numerator does contain tiie denominator an entire number of times, it may be called an improper fraction. E. fr, A«_+ -i ^' =3 4- ^"^ (126;. ^' 04.64-c ^ a-l-6-f-c ^ 158 TEXT-BOOK OF ALGEBRA 156. Fundamental Principle in Fractions. — If a fraction is regarded as an expressed division, then multiplying or dividing both terms by the same number will not change the quotient, i.e., the value of the fraction. A proof of this principle may be given as follows : — Let — denote any fraction, and x its value ; then x = — . b ^ h Whenc^e. a = bx (def., 43) Let in be any number; then from the equation just writ- ten, it follows self-evidently that ma = mhx (207, o) or, ma = inh ' x (38, 2) Therefore, -'^ = x. ' (def. 43) 7nb i.e, ^' = « Q.Ji.n mb b It follows, conversely, that ])oth terms of a fraction may be divided by tlie same quantity without altering its value. 157. The Three Signs connected with every fraction. There are three signs expressed or understood belonging to every fraction, viz., those of the numerator, denominator, and fraction itself. The same is true of a polynomial numerator or denom- inator. To illustrate. . <^ - ''' - '\ =+ +/"' 7 '" ~ "'] , — ab -\- ac — be -\- (— ab -{- ac — be) ^ , + 0^-f,'-e^ — (ab — ac -\- be) parentheses being used to constitute the polynomial one quantity. ALGEBRAIC NOTATION. loO The sign before the fraction applies to the valtie of the frnctlon, i.e., tlie quotient of the numerator divided by the (h^noniinator. 1. p]ffect upon the fraction produced by changing these signs. (1. Evidently, changing the sign l)efore a fraction changes its calue from plus to minus, or from minus to ])lus. (2. Changing the sign of eitli«*r numerator or denom- inator (dianges the sign of the quotient, (43) wliich is the value of the fraction. Thus, — = -h o, while — - — = — 5, and = — 5. 4 4 — 4 (3. If the sign of eitlier numerator or denominator and at the same time the sign of the fraction be changed, tlie fraction is changed back to its former sign and remains unaltered. (4. If the signs of both terms of a fraction be changed, the value of the fra(^tion remains unaltered. rp. 15 . , -15 K — 18 18 o Tims. — = ;). and = 5 ; = = — 6. :; -3 () -(> (.">. Finally, since changing the signs of the two terms leaves the fraction the same, changing all three signs changes the sign of the fraction. 2. Ivules for changing the signs of a fraction. (1. Changing one or all three, i.e., an odd numl)er of the signs of a fraction, changes the sign of the fraction. (2. Changing any two of the signs of a fraction does not alter its ^^alue. 160 TEXT-BOOK OF ALGEBRA. SECTION II. Keduction. 158. Reduction of Fractions. — Keduction in all mathe- matics is the process of changing the form of a quantity without altering its value. a. There are five cases of reduction of fractions commonly given: reduction to lowest terms; reduction of an improper fraction to a mixed quantity; reduction of a mixed quantity to an improper frac- tion; reduction of an entire quantity or a fraction to the form of a fraction having a given denominator; reduction of two or more fractions to equivalent fractions having a least common, or any connnon denominator. I. -TO LOWEST TERMS. 159. Reduction of a Fraction to its Lowest Terms. Prin- ciple and Kule. — A fraction is in its lowest terms when its numerator and denominator are })rime to each other. By 156, dividing both terms of a fraction by the same quantity changes its form, but does not alter its value. Therefore we may, 1. Factor the numerator and denominator into their prime factors, and then cancel out of both terms the factors common. 2. Or, find by continued division the h. c, f. of the terms, and divide both terms by it. The resulting fraction is in its lowest terms. 160. Exercise in reducing Fractions to their Lowest Terms. 1. . Dividing both terms by 3 ah^c we get — . 9 aire 3 h Ans. 12. ALGEBRA IC NOTATION. 161 15 jn 243 n^-n%-^ 45' 144' 1(52 n'-b' , 1274 18(>07 ,^ x^-b^ 2002' 24587 ' y' -\. 2 bx -{- b'^ 105 ^y 16 (^-^f^y+i^^-i^r 15 bi/' ' a*-b* 12da%'^x'^ a- -a -20 27 o'b\r* a' + a - 12 a'-x' — 16 u'^ 18. a"-^^ iix^-{-dax-\-20a 8 ''*''" 19 -^« + <^* _ nx"-^ ' 18tt-6a-^H-2tt» o{a^-U') ' x?^xif 10. ^^-^ 21. 26^-2^x 2xy-\-2ij 'lbx^-\-^:h^x-^2b^ ax -\-x:^ 22 ^* '*^ + 7 ajc — 3 a;'** rtc* 4- c=^ * 6a^ + llaa; + 3a:'^ aAr -h_^ 3 rf x< 4- 9 tf 6« 4- 6 M^ arx -f- r* * a^ .\^a% — 2 d^b'^ 24. 25. 6,. -3(/ Remark. — In most instances hereafUT it will be desirable to use fractions in their lowest terms, and in some eases it is necessary. II. REDUCTION OF FRACTIONS TO MIXED OR ENTIRE QUANTITIES. 161. To Reduce an Improper Fraction to an Entire or Mixed Quantity. PriiK-iple and Kule. Since a fraction is an expressed division, divide the numerator by the deuoniinator, and if there is a remainder 162 TEXT-BOOK OF ALGEBRA. place it over the denominator. This amounts to the same as dividing both terms by the denominator, which is justi- fiable by 156. 162. Exercise in reducing Fractions to Entire or Mixed Quan- tities. 1. "^ + ^^' operation. <^^ + '^fia+x Operation. a-\-x ^ ax-{-x X' rjf,^ ^ H ; — Ans. "^ a -\-x 25 147 75 1425 2 a' -2ab-\-4:b^ "? TTT' 8' 19 13' 111 a-b ab + b^ . 22 a^b' - S3 a'b' -\- a 11 a:'b X —y 5. tJlf X 4:X- — 2x 2x- — X -{-1 10. 11. 12. 13. i ax X a^ -1 -hx'- -x"" 1 ■ a -\-x — a — ab -f a% 10 ab-h a2 _ 13 ax - -Sx' „ a^x — 3 ax^ 7. : ;j— 14. w^ — ax- 2a — 3x ^ x^-^Sx-]-2 jg x'-^i/ x-\-3 ' x^y 12 c^ 4- 8 ac-x^ — 3 acx, — 2 a/x^ 3 c + 2 ax~ 163 Dissection of a Fraction. — Principle and Rule. — Divide eacli term of the numerator by the denominator (126, 2), writing the quotients in the fractional form con- nected by their proper signs. Reduce each fraction to its lowest terms. 164. Exercise in Dissecting Fractions. adn -{- ben — bd m _ adn ben bdm bdn bdn bdn bdn = « + ^_ _ » (159). bdn ALGEBRAIC NOTATION. lt)3 2 6 V2 ' 4 (r/ - bf 3 tlL 6 ^l^^'*^ III. -MIXED QUANTITIES TO IMPROPER FRACTIONS. 167. To Reduce a Mixod Quantity to the Form of an Im- proper Fraction. rrin('ii>le and Kule. This case is the reverse of the preceding. There the dividend and divisor were given to find the quotient. Here the quotient, remainder, and divisor are given to find the dividend, which being found is written over the divisor for the equivalent friution. Therefore, the entire quantity is 164 TEXT-BOOK OF ALGEBRA. to be multiplied by the denoiiiiiiator and the numerator added to the product (or subtracted if the sign before the fraction is minus), and the result placed over the denom- inator. 168. Exercise in reducing a Mixed Quantity to the Form of a Fraction. , o , ,2 ax — xif 1. 3a-fy + 1. Operation. (3 a + y) (x — y) = 3 aic — 3 ay -\.xij — }f Adding the Numerator. 2 ax — xif i) ((X — 3 at/ — y- Q , , 2aa? — xu oax — oaa — ?/"' i .'. 3 « + // -|- ^ = -y iL. Jus. x — y x — y 2. 2 + 3y/ y 4y Operation. (2 + 3y) 4y - (^ - 5) ^ 12yM^7,H:J ^ 4 y 4 // «. The minus sign before a fraction changes the sign of every term of the numerator according to the rule for sub- traction. (92, 1). Thus, - —^ is the same as ~ f "^ '^ . V ' / '4 2/ 4:y 3- m T^ 4^. Remahk. — In arithmetic we write, e.g., 4| and not 4 + |, while in algebra, « + ^ must be written so, and not a-^, since this last would mean a times K This illustrates a difference in one particu- lar between the arithmetical and algebraical notations. It may be said that when no sign is written in arithmetic + is understood, while in algebra the sign of multiplication is understood In arith- metic the numerator is always added to the product of the denom- inator and entire quantity; in algebra it is added or subtracted according as the sign before the fraction is + or — . (Cf. 62, b). AiJ.i:i'.l;Al( NolAl'lON. 165 4. 2a-'Jh-\-'' ~'^ 10. n-'-ax-\-x^ ~ 3 a -\- X 5. (f-\- 11 v^iAii"N. 107 but not so the denominators. Let the student follow this ru'e, never to imtUqdy fartor.s toi/et/nr unlil it is seen to ftc )U'C('! _ 1 - x' (l-|-a-)(l - x){l -{-x+x')' 1 - x' (l-^x')(l-\-x) (l-\-xXl-x)(l-^x-\-x^) 2 <• ".> 1- 7 ^ a c e \) 12' 20' 10 ' V d' f 6. -^, _!_, -^ 5 be 10 ac ti ah . ^^ ^^ - 3 a; 2x ?n 2 a oa^ n a b 1 X x^' ar» 3a 56 lb ' 21 r 168 TEXT-BOOR OF ALGKBllA. ^ 9a 7b_ lla 7^i -^ b) Hx 36ic' 28 ' Ax 9 a — b a — € b — 'lx^ — 2xy^' ()(ir + //)"-^ 16. 1 1 1 (^a + by a (a - by b (a' - P) SECTION III. The Fundamental Opehations in Fkactions. 173. Addition, Subtraction, Multiplication, and Division of Fractions. The Rules for these operations are the same as in Arithmetic. I. -ADDITION. 174. Addition of Fractions. Principle and Rule. Let — = a-, and - = y,he two fractions having a common e c denominator. a — xc ^ Then, ^ _ ^ ^ - (Def. of division, 43) Adding these equal quantities (Ax. 1, 207) a + b = (x + y)e (97) .-. ' —x-\-y (Def. of division, 43) ALGKlilJAlC NofArioX. 109 Thus, when two fractions have a conniion denominator, the numerator of the sum is tlie sum of their numerators. This demonstration may be extended to include any number (»f fractions united by lx)th -f- and — si<;ns. RuLK. Reduce the fnictions to equivalent fractions hav- ing a lowest common denominator, add their numerators and i)lace the sum over the conimon denominator. 175. £zercise in Adding Fractions. 1. —^ + — ^ +;i X — 1 X -f- 1 Operation. — We write 1 for the denominator of the integer. 1 (x — 1) (or -f- 1) is the 1. c. d. The equivalent fractions are, xjx-^V) X ( x - 1) 3(a;-l )(a;-f 1) {x^X){x-\y (x + l)(x-l)"^ (x + l)(a:-l) Expanding the numerators (171, ^/) and adding them, we have, x^ -\- X -\- x^ — X ->^ ?t x^ — .'> = 5 .r- — 3, which, placed over the 1. c. d., gives, 5x^-3 x-^-1 Ans. 4. 2 ^4-^ 4-11 7 ''^^-L^.l 3 " ~ "^ I ?.?_ZlA_ 8 ^^ ~ ^^ -I- ^ "^- 4- ^ ~ ^ 5 ' 2 (fO ac be 8 ^12^4 ^~2~^:> ban i \) ♦ 6. x-f- 1 — 11. Add3x,xH , 4x 2 3 4 o Suggestion. — A<1<1 th«' entire parts and the fractions separately. 170 TEXT-BOOK OF ALGEBHA. 12. 4:a^ -\ 1 j \- oa^ -{- 13. 3 ' 3 ' 7 1 . i .^ n , l-2n 1 = — 16. — ~ h ic + 2 ic + 3 n — ln'^ — l 14. ''' + i> 17. 1-^' _p 1+^' ?/t 4-i^ ^^ — p 14-^"^ 1 — x^ 15 'i]ia^^J2^ , 'nia — X 1 , 1 in -j- 71 m -\- 71 X (x — y) y (^ -\- ])) 19. Add^L-, _L_, ^_ a — h h — c c — d 20. Add « 1 _ a ' (1 - «)2 ' (1 - a)3 21. -^^i-^+ ' ic"^ — 7 X + 12 ic'^ — 5 ic + 6 22 ^ „™ 4- ^ 5 _^ ic _ 18ic' ^ 2 + 5ic + 2a;2 1 2 23. Add a;-2_7a; + 12' cc2_4x + 3' £C'^-5ic + 4 24. — -V + — V- + 7-;V^, + "^ 8-8ic 8 + 8ic 4+4 x'^ 4-4ic* II. - SUBTRACTION. 176. Subtraction of Fractions. Eule. Reduce the fractions to equivalent fractions having a lowest common denominator, and subtract the numerator of the subtrahend from that of the minuend for that of the difference. If several fractions are connected by + and — signs, reduce them to the lowest common denominator, and add the numerators, those whose fractions are preceded by a minus sign being taken negatively, i.e., with their signs changed. ■ See })riii('i|)le ex})lained in addition. ALGEBRAIC NOTATION. 171 177. Exercise in Subtracting Fractions 1 ^-_ ^* L'd~r7 2. --1 - + - 3. 1 _ 3 o X — a 4 a-f z; c-d a — b c-{.d 1 1 b. _1 a: — a ^ x-Jfa 6. l-\-x^ 4x^ x-\-a l-x^ 1-x' ,. =.,l£^«_( ., + --?) 8. 2a-3x + 9. X- ="' a —■ w ( ^ . X — 2 a2-62 a-\-b (a-bXb-c) ib-c)(a-b) (a-6)(6-c) b-c (160) 17. b — a a — 2b 3cr {a — b) X — b b -\- X b'^ — x'^ l-2«^l+2;r ^ 4rx^-l 19. ^ - + i! ,+ (^a-b)(n — e) (b - r) (b - a) (e, - a) {e - b) Suggestion. — For the sake of system, it is convenient to have regard to the cyclic order of the letters. To do this we think of them as placed in a circle so that they follow one another in regular order, the last being succeeded by the first. Thus, if there are four letters a, b, c, cZ, we go from a to b, from b to c, from c to d, then from d to a again. Of the six expressions in the example only three are written in cyclic order, viz., a — b, b — c, and c — a. Changing the others to this order as in the suggestion, Ex. 16, we get, a2 ?>2 c2 (a-b)(c-a) (b-cXa-b) (c-a)(b — c) the 1. c. d. of which is now easily seen. 20. r-jj-q ^ q -^r-p ^ r -\- p - q {P - q) (r -p) (q - r) (q - p) (r - p) (r - q) 21. yH-'^ _ ^ + ^ _^ x ^y (^x-y){z-x) {y-z){y-x) (x - z) (z - ij) Suggestion. — In the third expression both factors of the denom- inator will have to be changed, which will not change its sign (107, 1). 22. - -f 1 4- ^ (b — g){c — a) (b — a){a —c) (a — b)(b — c) 23. i ^ 1 _ 1 X (x - y) {x - z) y (y - x)(y - z) xyz AlJJKIUiAlC NOTATION. 173 III. - MULTIPLICATION. 178. Multiplication of Fractions. Rule. (Ex. 21, Art. 255.) Factor the terms of the fractions, cancel common factors, and then multiply the numerators together for the numera- tor of the product and the denominators for the denomi- nator of the product. a. Integral factors are to be regarded as numerators over tin* denominator 1 undei-stood. h. Quantities to be multiplied (or divided) must lirst be reduced to either integral <}uantities or simi)le fractions. 179. Exercise in the Multiplication of Fractions. 1. -_L' X () a' Operation. _ X ^^— = 4 ah. «5 a 2. ,. X -^ X •'/ 3. -'^'X"* cd 4. 4j-yX 4 jrt/ 8. !"l^J': X (nr' 4- 2 mn -h //-) ft/ -\- n 3^. 1 5. loa'V 24 rV 6. -<') 7. 4.1a-7/V'' 24;^r//V 180 a'^6c« 27a:y;. 9. 17.^^-4 *2abx-\-b Xab LO. x*-a -X3y LI. •t' + -x 5x 3 2a- -1-1 d(x-\-f/)(x-ij) ^ ^^^ 13. \^ a J X 174 TEXT-BOOK OF ALGEBRA. 5 a ^ 2 - a I/ — 2 -„ a -\- X a — x ^0^'+^0' 25. (l_., + .,-2)^ 1_^1 y ^ a ' a^ 26. Multiply together ic^ — X — 20 ic^ — X — 2 ic^ + oa? 27. Multiply 1 + 1 + -Uy 1 _ 1 + -1 a G a G ALGEBRAIC NOTATION. 175 IV. - DIVISION. 180. Division of Fractions. Kule. — Invert the divisor ml proceed as in niultiplit'ution. See Ex. 22, Art. 255. 181. Exercise in Division of Fractions. 1. Divide ^''' by ^' "*. Inverting tl 9m 3 2a 3 2. , 3d — ah -^ -- 2 (lb 3. 4. ''^ 3b 6. ' ^oh'' 27 ^2 n ahe 40c ^ 81rfJ • 14rZ« 6. _8w * 21n^xhj \ '^ 81m7i 7. 8. [b^> — 2 ^'^ l ^ aM;^^> [_ ^2 _ ^-2 — • - 3 J ^ 7t^ry 182. Complex Fractions. — When the division of one frac- tion by another is indicated in the fractional form by writing the dividend over the divisor, there resnlts what is called a complex fraction. (155, 1.) To simplify such fractions write the numerator as the dividend, and the denominator as the divisor in a problem in division of fractions, and proceed according to the rule. (180.) Or, what comes to the same thing, multiply the extremes for the numerator and the means for the denom- inator, and then simplify the resulting fraction. 183. Exercise in the Reduction of Complex Fractions to Simple Fractions. 1. 7- 1/ a 5 — c 7 — y 5 — c a 5 a — ac 1 (» _^ tL __ ^ s/ = ' *' X ' a X 1 — ij 1 X — xy ALGKin:Aic NcnwriMX. 177 a -f h »> iU // 7 axy ^; 14. 9 Uo^ + X 4 ^ 2x-\ 15. ^ + 1 ^ ^_1 ^ +2 2 1 1_ 16 1 + ^^- 1 — <^ 1 1 + x^ 1 — a ^ 1 -\- a i + 'i^ i + "^-l ^ a — x -j a^ — x^ a -\- X a'^ + x^ 2x 3 17. x-'d 12. TT-r 18. 1 + 1 + x-?y "^1_^ 184. General Exercise in Fractions. — Siiiii)lify the quan- tities of tliis article. ^ ax 27 a -\- a^ (fx- — ax 18 a - 6 a^ -{- 2 a^ 5 a^l, + 10 a%^ ^ .7-2 _ 5 x-\-6 :Ui'b-^ -\- 6 ab^ x^-1 x+12 ^ 2x — ?yy,x^2ii :^ .T — 2 ?/ 5. -\- — "^ — _ '- alokbi:ai(' Nor a'iion. 179 6. 7. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. { 1 r> + a * 1 ,^-2 + Sx-l 21. 22. />-7 _j. ^--/^ 4. 7_ir y^Y y>;- qr a^ — ar* ^ (a -f xy a^ -\- xr^ {a — x)' 180 TEXT-BOOK OF ALGEBKA. 23. 3V_ 5y/- + 7 _ (>.r-^-ll xy^ xy^ x^y 24. ^ + ±_ 25. ^^' f^lytlffl 2g 4a^-16^>^ ^ 5 a a -2^ 20<^-^ + 8a^ + 8^2 27. ^xi-^ rtQ ma^ — 7ix <60. (a -j- ^) (m — w) .x-8 + 1 1-x^ 1 -y^ ( X \ ' ~ 1 + 7/ ^ x^ + x^ -y^ + i-xj 32 ^ ' 3 (1 _ ^) "^ 3 (^ _ 1) "^ i 34. _^ .. + ^ 7i x(x^ — y^) y{x^-\-y^) 35. X 3 % c^ — x'^ d^ — ax X -\- a x-\-3 a a — x x — 3a 2-bx _ 3H-a; , 2a;(2a; — 11) ALGEBUAIC NOTATION. 181 38. l+« ^ 1+* _L 1-f^ (,, _ A) (a _ r) {b — c){b-a) {c — a) {r — b) ar-hl (^.4-l)(x + 2) (a; + l)(a;-f2)(x + 3) * ^-^f ^'^y' y'-x' {x + y)ix^^y^) ni^ -\- n^ n rti^ — n^ n m A 42. 3a-[/>+S2«-(^-.)n + |+ 2-eTT- 1 43. a- 4- ^ ■+3-^; '-■1+47 '-^tf a; -f 1 x^ — 1 X x--^^-- ^{x-l){x-2) 45. If 46 x- 1 47. f-+l4-lU(-4.iL + ^) \x y z I \x y z J 48. p +^ 4- ^ ~^1 -^ r* J- ^ _ '' ~ ^1 49 1 r ^ I ^+M •«(T+~)(i-4)*(:--)(7-l) SECOND GENERAL SUBJECT.^ — SIMPLE EQUATIONS. CHAPTER XIV. SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 185. Definition. — An Equation, is an Expression of the Equality of two Quantities. By this is meant that the numerical values of the two quantities will be found upon reduction to be the same number, integral or fractional. (See 55, where the mean- ing and use of the equality sign, =, was explained.) Equations considered in a general way may be regarded as expressing the relations of numbers. SECTION I. General Definitions. 186. Equations are of two kinds, Identical and Conditional. 187. An equation in which one side is but the develop- ment of the operations indicated in the other, or is exactly the same on both sides, is called an Identical Equation. Thus, (x + 3) (x-9) =x^-6x- 27, and ax -\- b = ax -\- b, illustrate the two kinds of identical equations. 1 A discussion of powers and roots as forming tlie remaining subjects in the "Algebraic Notation" ought theoretically to be given before taking up equa- tions. Other reasons exist, however, for changing this order. 182 siMi'i.i: i:(,n AiioNs. 183 The examples given in addition, subtraction, multiplica- tion, division, factoring, and t'rac;tions set equal to their answers would be equations of this sort. a. A prime characteristic of identical equations is the property that any letter may have any value, provided it has the same value on both sides of the equation. This is true because the two sides of the equation are either actually or virtually the same quantity, and conset (juaii- titirs jn-eseiit in a conditional ecjuation, the known and the unknown. Tiie latter are commonly represented by the last letters of the alphabet, (x, y, or z), while the former are represented by figures, and by the first letters of the al- phabet. This, of w)urse, is purely conventional. 195. The Degree of an Equation is the same as the highest |)o\ver of the unknown (juantity; or, in equations contain- ing more than one unknown, it is the same as the degree of the term (see 77) containing the greatest exponent, or the greatest sum of exponents of the unknown quantities. Thus, ox + 6 = 91 is of the first degree. ax' -\-hx = c is of the second degree. 11 X -\- {) j-^ -\- 7 y- = 4li is of the third degree. ^z"* -{-3z'"-^ — 2z '» - - = 7 is of the m"" degree. G x» + 9 xY + 14 //* = 21 J-// is of the fifth degree. 196. Equations of the First Degree are also called Simple Equations. Sec licading of this cha})tcr. ". Of all equations those of the first degree ;irc thr easiest to solve. The treatnient of sinijde equations only, and not of equations in general, will l)e taken uj> at this time. SECTION II. Aloebrak'.vi. Mktuoi) of Thk.\tmk\t. 197. Nature of the Treatment. Tlie metliod by wliich the problems of 47 and 85 were solved was of a special nature, similar to what is called analysis in arithmetic. Scarcely any two were reasoned out in the same way. The treatment to which the equaticm is now to be subjected is of a very different character, more comprehensive and more scientific. 186 TEXT-BOOK OF ALGEBRA. I. -LOGICAL TERMS. 198. A Proposition is a statement presented for considera- tion. A proposition may be true or false, and may therefore be proved or refuted. 199. Propositions are of two kinds, theorems and problems. 200. A Problem in algebra is a question proposed for solution. 201. A Theorem is a proposition to be proved by a demonstration. 202. A Demonstration is a chain of reasoning by which the truth or falsity of a proposition is made evident. 203. A Corollary (cor.) is a truth inferred from a prop- ositicjn or from something in its demonstration. 204. A Scholium (sch.) is a remark concerning a proposi- tion or its proof. 205. An Hypothesis (hyp.) is a supposition made at the beginning or in the course of a proof or solution. 206. An Axiom (ax.) is a self-evident truth. II. -AXIOMS. 207. The following are the Axioms used in the Solution of Equations. <(. Addition and Subtraction. 1. If the same quantity or equal quantities be added to equal quantities, the sums will be equal. 2. If the same quantity or equal quantities be sub- tracted from equal quantities, the remainders will be equal. b. Multiplication and Division. 3. If equal quantities be multiplied by the same quan- tity or equal quantities, the products will be equal. 4. If equal quantities be divided by the same quan- tity or equal quantities, the quotients will be equal. SIMPLE lOQl'ATIONS. 187 c. Powers and Roots. 5. If e(iual (luaiitities, or the same quantity be raised to the same power the products are equal. * 6. If equal quantities, or the same quantity have tlie same root extracted, the results (there being more than one; are resjiectively equal. 7. Quantities which are equal to the same quantity are equal to each other. 8. General Axiom. — If the same operation be per- formed on two e(pial quantities, the results will be equal. XoTK. — It will often be convenient to refer to the aildition and subtraction axioms tiKjcther, as axiom «, and to the multiplication and division axioms as axiom h. When the addition axiom alone is referred to, it will be by the number, 1, and so for the others. III. - SOLUTION OP EQUATIONS. 208. Solution of Equations of the Form ((jc = b. To solve this equation x must In* made to stand alone on one side of the equation. (193.) We have ax=:h (Hypothesis, 205) Dividing both sides of the equation, that is the two equals by a, a; = - (Axiom 4) a Notice that this solution is genera, in its nature. Kor, a may stand for any co-efficient of x, and h for any quantity whatever on the right side of the equation (113). This element is characteristic of all algebraic solutions, and must l)e remembered to understand them. Notice further the use of the axiom to accom})lish the end desired. 188 TEXT-BOOK OF ALGEBRA. 209. Exercise. 1. 9.r=36. Dividing the two equal quantities by 9, by axiom 4, the quotients are equal. .-. ir = 4, answer. 2. H)x = 64. 8 m = nx. 3. 9:^=144. 9. 45?/ = 120. 4. \\x = 29. 10. 17 a;^ = 3 «, ^ = ? 5. «^x = 16. 11. 6 a- + 5 X = 33. 6. 3r^/' = 7c. 12. 7x — 3a; = 33. 7. 35 = 5 J'. 13. 2ic + 3.x + 9^ = 28. 14. 2j'-hl4.T-13:r = 29-14 + 12 — 3. (87,4.) 15. 9 // + 7 y - 2 // + S // — // - 2 ^ = 36 + 10-8. 16. 25 + S- 3 + 20 -9+1 =- + 2;v + llrv — 7;s. 17. {a + /> + <') -r =(l + (' +/■ 18. m.r + //.'• + px + y./- = /•. (97.) 19. (i.r + />■/' = ly through by the lowest romino)i denominator, thereby obtaining simpler expressions, and that more readily. ScH. IF. — Tliis proposition takes any equation ami ex- plains how to clear it of fractions. It is then ready for tran8iX)sition, and ultimately for solutiou. 192 TEXT-BOOK OF ALGEBRA. 213. Exercise. 5 — ox _Sx — 9 2 '~3~~' ISoLiTioN. — The 1. c. d. is 0, and we multiply through by it. o—'3x 8 ./ — 9 2 — :} lo — \)x = Ujx — lS _().,._ 10.x- = -15 — 18 — 25.r = —33 2. X XX x-\-2 2 "^ 3 4 " 2 ' 3. x-1 x—3 x—3 2 4 ~ 2 ' 4. 5 = 8-^. X 5. •: - I-:,' = 42,-v ./■ ./• ./• X X X , ^^ ^ ;5 4 "" 4 7. ^■_:!l:i_^. •'■ + •' - -'+:=i(^-E)-:;+3(^^-i} 9. '+'s = '~^- 10. ^ .. _ .;}(*, A)x — .18 .2 .9 11. ax+l, = l + l. 12. X — a X — b b — X ~ a — X 1Q ,. , 4 , (Hyp-) (Ax. 3.) (Ax. 2) (86) (Ax. 4) X + 2 ' X + (i SIMI'LK K(,H ArioNS. IIK) 214 Examples of the Complete Solution of Equations. 11 ^ 13 1' Mui»KL Sun HON. — 1. c. (I. =L'X 11X13. 182 .c - 208 H- 330 j: -f IKi = s:>8 j? - 4433 + 143 or (Ax. 3) 182a- -f 330a: - Ho>^ j- — 143 j' = 2(»8 - 17G - 44;i3 (Ax. a) (86) (Ax. 4) -489a: =-44 101 a:=y. rKRIFH ATIOX. 7x9-8 ir)X04-8_ .. ^, 11 ' 13 .-■■"-•i-^ i.e., r, 4- 11 - 27 - 11, or, K; = IG. o ^ + 4 _ a:-h5 (55, «) 3. ;^.r_8 3a' -7 3(> 4r> a K "^ ^("-^^-i^("-'^=5^"-'^ + 48 Caitioxs. — The student should be careful not to make mistakes in addition, fnultiplication, fractions. sii;ns. etc., and then attribute the failure to get correct results lo som. tiling unknown about equa- tions. We will refer to common .sounts of error by articles which the student would do well to look up and fix in mind. Reduce com- plex fractions to simple ones (182), and i)erform all indicated opera- tions either before or after clearing as may be found convenient. See 100, 4 ; 168, a \ and 176. Study carefully 171 «, and its applica- tion to e(juation8. Reduce answers to their lowest tenns (169). An answer may have to be changed to make it identical with that given (see 167). For factors in parentheses, see 111, 2. In clearing of frac- tions remember to multiply tlie integral (piantities by the 1. c. d. Other references might be given, but these will suggest such mistakes as are liable to hv made. 194 TEXT-BOOK OF ALGEBRA. 215. General Rules for Solving Simple Equations with One Unknown. 1. Normal process. (1. First clear the equation of fractions. Theorem 212* (2. Next transpose all the terms containing x to the left member, and all the known terms to the right member of the equation. Theorem, 210. (3. Then by 87, 4, collect the terms on both sides, and when necessary form the coefficient of x as in 97. (4. Finally, use axiom 4 to find the value of the unknown. 2. General process. (1. Proceed in any way that seems advantageous by use of the axioms (207) to reduce the equation to the form ax = h. (2. Finish the solution as in 208. Rf:mark. — The first rule is straightforward in its method, and so best for the beginner. The second is superior to the other for the expert student. It may be emphasized here that any process in accordance with the axioms will always lead to correct results ; and any jrrocess or operation not in accordance with them will lead inevi- tably to incorrect results. 216. Find the Value of the Unknown Quantity in the fol- lowing, and verify the answers by substituting them in the original equation (189, a). 1. l|.=8^. 3.|x + 12 = |x + 6. 2. ^1+23^9 4 l-"_4 = 5. X — 1 . ^' 7x-\-U 17 -3a; 4.^ + 2 5. 5 - 6x + ^3— = —^ ^-. 8IMPLK EQUATIONS. 195 9. T) X - (8 X - 3 [K; _ (i ^ • + (4 - 5 xj]} = 6. 10. « (x - 2) + 2 .r = G + «. "■K"-"0-.'('-y+; lA. , = 0. Solution. — The 1. o. d. of cocni.i.'iiis .■ juils 12. 3rt = (111, 2) 5 X = 2 a — a + — f/ = — a 5 5 (Ax. a, 174) ICATION. (Ax. 4) ^ f ?^ _ 1!^ _ ^ f ?^ _ -^ 4. 1 f — _ -^ = 2\2r> 3/ aV^o •*/ 4\25 5/ l/_jfir_\_l/7^\ l/3a\^2^_7i^ 9ja ^ 2\ 75/ ;H100/"'' 4\2r, / 300 300 "^ 300 "• NoTK. — In some literal eciuations verification is easy, in others it is more or less difticult. In the following examples the verification Is rather complex, though of course it can always be perfonned. In numerical problems verification is always comparatively easy. 12. {it H- a-) (/> -f ^) = (w + x) {ii + X). 13. ^ - - = ^ _ p r 7 196 TEXT-BOOK OF ALGEBRA. 3a; , 7 1 "^^- 4+0 7x = 5 "6x ' 4 7 7 1 _ 5 4 6 7x 6x 7 49 a; - 6 = 35 - 24 X 49 a; + 24 X = 35 + 6. 41 x = — . Ans. . 3x + -J- • (163, 159) (Ax. 2) (Ax. 3) (Ax. 1) (Ax. 4) 15. 8 (a; - 1) + 17 (x - 3) = 4 (4a; - 9) + 4. a c 16. a — x X — 1 4x - ^ 7 X- 6 ^ X — 2 3 a;— 9 3 + 5 ~ 8^ ^"^ ~Y~ + 10 18. Q>hx -\-A.a'' -\-2ax = % a%x - 3 ax. ,n 3a; + l x-2 3(a;— 2) ~ a; - 1 X — 7 2 a; — 15 1 x-\-l ~ 2a; - 6 2 (a; + 7) 4 7 37 21. 4- o; + 2 ^ a; + 3 ~ ar' + 5 a- + 6 22. 7 _ 6a^ + l _ 3(l + 2^^) 'a; — 1 a;H-l a;- — 1 a!-3 _ x-^ 1^ ^^' 4 (a; - 1) - 6 (a; - 1) + 9 ' 1 a — h 1 a -\- b a — ^ X a -\-b ^ x a — b a + b 1 1 Solution. X X a + b a — b a — b — a — b a — b — a — b X cfi — 62 2 6 — 2 6 X a^ 1 1 (Ax. '^) (176)> (Ax. 4) (Ax. 3) X «2 _ ^2 X = a^ — /A Ans. This solution is shorter and easier than if the equation had first been cleared of fractions (215, 2). SIMPLK KQITATIOXS. 197 25. ax-\-bx _ __ 1 _ 1 b a — b a SuooESTiox. — Add the nu'iiibers of the equation as they stand before clearing. 26. {X - b) (X -2) -(x-^o) (2x - r>) + (0- -h 7) (x- 2) = 0. 27. > (L'x - 10) - ^ Qix — 40) = lb - ^ (57 - x). 28. 3i-;L>8-g + l>4);-=;H{2i+f}- 29. (a ■i-x)(b-\-x)- a (^ + c) = -- + x\ b 30. (j'-h 1)' = S^) - (I -x)\x- 2. 31. .5 a: — .3 ar = .25 X — 1. 32. 4.8^- — .72 a- — .05 = 1.6ar + 8.9. .5 «3 1 3 ^1-x 34. ''•7' + -^^ = .ooia. + .(;---2 5 .5 .05 36. 1 ■ 1 1 36. «i — ay be — b)/ ac — ay 1111 X — 2 a* — 4 .r — () .r — 8 SuooKSTiON. — Add the members (176). „ ax^ -\- bx -\- c ax -{' b 01 , — J = . px^ -\- qx -\-r px -}- q 38. ax- — hx = hx^ — ex. Suggestion. — Divide through by x. 39 0a?4-3 , 3x-6 ^2 3a' + 22 27 ■^2x-5 3"^ 9 198 TEXT-BOOK OF ALGEBIIA. Solution. — Clearing of monomial denominators only (215, 2), 81 x - 102 9aj + 3-f -^ . - . .- 2 _ r " — Ao T y X -r oo li^x. .3; 81 X - 102 _ ^^ 2x-5 (Ax. 2, 86) 81 X - 102 = 102 X - 405 (Ax. 3) - 81 X = - 243 (Ax, a) x = 3. (Ax. 4) 40 ^ (^ ^ - 3) 11 .T - 1 _ 9 g; + 11 14 + 3 0.' + 1 ~ ~~7 .^7x — 6 cc — 5 .T 85 6;r-l()l~5 4^ 6-5x 7-2x-^ 1 + 3.T 10:r 15 14(.r-l) 21 30 ^105 ^2 cc — l X — 2 _x — ic — 6 a'-2 ic — 3 a; — 6 x-T Solution. l+,_, 1 ,_3-l+^_, 1- ^_, (161) 1111 X — 2 X — 3~x — X — 7 -1 -1 (176) (x_2)(x-3)-(x-e)(x-7) -x2+13x-42 = -x2 + 5x-6 (Ax. 3) 8 X = 30 (Ax. 0, 86) x = 4i (Ax. 4) 44. Solve br = p regarding, first h, then r, and lastly /> as the unknown quantity. First, Second, Third, hr=^p (Hyp.) hr = p (Hyp.) ^'r = p (Hyp.) Z* = 1^ (Ax. 4) r =1 (Ax. 4) 'or 7^ = hr. si.Mi'ij-: K(^(Arn »Ns. 199 This equation hr =jt> is the fuiuhiinental formula in Pei- eentage in arithmetic, b standing lor kise, r for rate ex- }»resse(l (h'cimally, and j) for the percentage. The two equations obtained, b = J—^ and /• = — , are tiie formulae r p for the cases, given the percentage and rate to lind the base, and given the base and percentage to find the rate. 45. Solve a = b (1 -\- r) for ^, and r. (a = " amount ''). 46. Solve d — b (1 ~ /•) for b, and r. {(l = '• difference "). 47. Solve I = jjrn for jo, r, and n in succession. By the definition of interest, (/ = interest, p = principal, r = rate, and 7i = time in years), i = pm ; we are to find the principal when the rate, time, and interest are given ; next, to find the rate, when the principal, time, and interest are given ; lastly, to find the time when the principal, rate, and interest are given. 48. (1) Solve f = i for « and c. (2) Solve t == 1 for e. c c 49. Solve — — -|- = for rt, i, c and d in succession. b — c b^ d 50. Solve ^ -f 1 = in the same manner. cd 217. General Remarks on the Solution of Equations. — By Kule '2, the student i.s at liberty to pursue aiiij course in con- sonance with axiomatic principles. Certain of the exam- jdes of the preceding article have exhibited an advantage gained by deviating from the normal i)rocess. The follow- ing precepts have already been exemplified in one or another of the exercises. 1. Mark at the outset and after every reduction whether the equation can be divided tlirough by any factor, mono- mial or other, and remove it. (Axiom 4.) 200 TEXT-BOOK OF ALGEBRA. 2. Study the original equation and each derived equation to see if by uniting certain terms as they stand, or after transposition, quantities can be made to disappear, and the equation be thereby simplified. 3. Examine whether any simple reduction, as that of an improper fraction to a mixed number, will pave the Avay for the simplification suggested in 2. 4. Consider, before applying axiom 3, whether it would not be better to clear the equation of only a part of its de- nominators, and then simplify, clearing of the remaining denominators afterwards. SECTION III. • Problems Involving Slmple Equations containing One Unknown. 218. The Problems of Algebra (see 47 and 85) are quite like those of arithmetic, except that the former are usually much more difficult ; but the algebraic treatment of them is different. In arithmetic we were accustomed to start with what was given and work towards the result ; while in algebra we conceive the problem to be solved, and pro- ceed as if we were verifying our answer, using meanwhile a letter to stand for it, since we do not know its value. Going through the form of a verification, the letter used everywhere taking the place of the unknown number, an equation results, the solution of which gives the value of the letter, and solves the problem. This is called the analytic method. 219. The Solution of a Problem involves two operations. 1. The Statement of tho Prohlem. — The statement of a problem is its expression in algebraic symbols in the form of an equation. siMi'r>K K(.)r.\Ti(>\s. -201 2. The Solution of the Kqiidtion. — Section II. of tins clijipter dealt with the systematic solution of equations, and the student is supix)sed to be familiar with this part of the in'ooess. 220. In the Enunciation of every problem we notice two things. (See those in 47 and 85.) 1. The description of an unknown number (or quantity) together with one or more others dependent upon the first for their values. 2. The a.ssertion that two numl)ers or quantities, obtained in different ways, are equal. The rule to be given explains the translation of the prob- lem into algebraic language. It will be convenient before giving the rule to define the term ^'function of a quantity." 221. A Function of a Quantity is one which depends upon it for its value. Thus, '^x -\- i) is a function oi x. If a- = 6, then the function equals 3 X (> + 5 = 2.S ; if a; = 9, the function equals 3X94-'^ = 32, and so on. From this we can see that any quantity which contains x is. a function of x, and any quantity which contains a is a function of r^ and so on. 222. General Rule for Solving Problems. — Compare with the solutions of Arts. 47 and 85. 1. Let a?, or some convenient function of x, represent the unknown cpiantity. 2. Express the different functions of x to be used in forming the equation in terms of x. 3. Construct the two expressions described as equal, and make them the two members of an equation. 4. Solve the equation. 5. Obtain the values of such functions of the root as the [)roblem asks for. a. Illustkatioxs. — In the examples as set down in 47 and 85 we first wrote x for the unknown quantity; then underneatli iU func- 202 TEXT-BOOK OF ALGEBRA. tions. Afterwards the equation was formed, which was solved ac- cording to the best light we then had. To particularize, let us take Ex. 5, Art. 47. The functions of x are 2 x and 2 x + 7. The equation is then formed, after which it is solved, giving x = 8. But the prob- lem requires two other answers, viz., 2 x and 2 x + 7, which are found equal to 16 and 23 respectively. h. The two sides of an equation must be expressed in terms of the same unit. Of course, one side cannot be expressed as a certain number of dollars, and the other as an equal number of cents, neither can one side denote metres and the other centimetres, and so on. Careless thinking sometimes leads a student even into such absurdities as these. 223. Proportions are Transformed into Equations by means of the fundamental property tliat the product of the means equals the product of the extremes. 224. Exercise in the Solution of Problems in One Unknown Quantity. 1. A gentleman divides two dollars among twelve chil- dren, giving to some 18 cts. each, and to the others 14 cts. "each. How many were there of each class ? Let X = the number receiving 18 cts. each, (222, 1)^ Then 12 — a? = the number receiving 14 cts. each, (222, 2) 18 0? = the number of cents the first class received, (222, 2) (1^ — ^) 14 = the number of cents the second class re- ceived, (222, 2) 18 X + (12 - x) 14 = 200, (222, 3,. and b) 18 a^ + 168 -Ux = 200 (110) 4:X = 200 - 168 = 32 (Ax. 2) x = S I (Ax. 4) 12-^ = 4 \ (222,5) Verification. 8 X 18 + 4 X 14 = 144 + 56 = 200. 2. There are three consecutive numbers such that if they be divided by 10, 17, and 26, respectively, the sum of the quotients will be 10; find them. Sl.MI'J.h i.i,'( A i loN.s. 20^ Here the student may liave trouble in expressing the three num- bers (220, 1). The U\v _|_ o _ ia 3. All estate is divided among four children in sutdi a manner that the first has $200, more than ^ of the whole, " second 340, '' " ^ " " ** third 300, " '* ^ " " fourtli 400, " '' i " what is the value of the estate ? Here the beginner may not see how the statement of 220, 2 is true of tliis problem. It must be taken rather by implication than by the direct statement of the problem. Let X = the value of the estate (222, 1) ^+ 200 = the number of dollars of the first child's port ion, (222, 2) 4 ^-1-340=" " " " 5 ^-h:J0O = " " " " - + 400=" 8 |-+ 200 -I- 1+340 +1+300+1 + 4-+!+l+^-^=-^-^^ 30x + 24x + 20x + 1.5 X - 120x = - 148800 _ :31 X = - 148800 x = 4800. second " " (222, 2) lliird " " (222, 2) fourth " - (222. 2) = x (222, :J) 204 TEXT-BOOK OF ALGEBRA. Verification. 4800 . 4800 4800 4800 -^— + 200 + -^ + 340 + — ^ + 300 + -^ + 400 = 4800. 4 D O O 4. A boy bought an equal number of apples, lemons, and oranges for 56 cts. ; for the apples he gave 1 ct. apiece, for the lemons, 2 cts. apiece, and for the oranges 5 cts. apiece. How many of each did he purchase ? 5. Divide the number 181 into two such parts that 4 times the greater may exceed 5 times the less by 49. 6. A boy ate I of his plums and gave away ^ of them. The difference between the number he ate and the number he gave away was 3. How many had he ? 7. What number is that whose half, third, and fourth parts together equal 63 ? Verify the answer. 8. A father aged 54 years has a son aged 9 years : in how many years will the age of the father be just 4 times that of the son ? 9. A father is now 40 years of age and his daughter 13. How many years ago was the father's age 10 times that of the daughter's ? 10. A man's age and his wife's age now bear to each other the ratio of 6:5. But 15 years hence they will be to each other as 9 : 8. How old are they now ? 5 X In this problem we might let x = the man's age, and — phis wife's age ; however, we can avoid fractions in the folloVing preferable way : Let Gx = the man's age, (222, 1) then 5 x = the wife's age, (222, 1) a X + V) = the man's age 15 years later, (222, 2) 6x + 15 = the wife's age 15 years later, (232, 2) 6x + 15 :5x+ 15 ::9 :8 48 cc + 120 = 45 X + 135 (223) 15:25 + 15 ::9:8. X = 5 633 = 30 5x = 25 1^ ERIFICATIOX. 30- SIMTLK KgiATlONS. 205 11. A gentleman is now twenty-live yeai*s old, and his yoi^ngest brother is 15. How many years must elapse before their ages will be as 5 : 4 / 12. The ditferenc;.^ between two numbers is V2, and the greater is to the less as 11:5. What are the num- bers ? 13. What two numbers are as 3 : 4, to each of which if 4 be added the sums will be to each other as 5 : ? 14. A farm contains 26 acres. Three times A's part is 6 acres less than 4 times IVs part. How many acres had each? 15. Find two numbers differing by 10 whose sum is equal to twice their difference. 16. A man has four children the sum of whose ages is 48 years, and the common difference of their ages is ecjual to twice that of the youngest. Find their ages. 17. Three boys are talking of their money. A says to B, '' I have three times as many cents as you have." Says C to A and 1^, *' I have as much as the difference between your sums.'- Now, A's money achled to twic(^ IVs and twice C's makes 03 cents. How much had each ? 18. A grocer has two sorts of sugar, one wortli *.) cents. and the other 13 cents a i)ound. How many jHuuids of eacli sort must l>e taken to make a mixture of a liundrcd ])()unds worth VJ cents j)er jxjund *.' 19. \ ]>ayniastcr wisliing to draw for use on pay day the sum (jf .il»113<)() recpiested tlic teller to m.ike up the sun* with l)ank-l)il's of different denominations as f(jllows : a certain number of lOO's, twice as many 5()'s, twice as many -O's as 5(Vs. twice as many tlTs as 2()*s. twice as many 5';s a,s lO's, twice as numy l''s as 5's, and twice as many I's as -■- How nianv of each denomination were needed V 206 TEXT-BOOK OF ALGEBRA. 20. A grocer bought two casks full of oil, one of which held twice as much as the other. I'roui one cask he drew 10 gal. and from the other 1 gal., and then drew from the larger i of the remaining oil, and found that the two casks now contained equal quantities of oil. How much did each hold when full ? 21. The garrison of a certain town consisted of 125 men, partly cavalry and partly infantry. The monthly pay of a cavalryman is $20, and that of an infantryman is $15, and the whole garrison receives $2050 a month. What is the number of cavalry and what of infantry ? 22. Divide 88 dollars between A, B, and C, giving to B f , and to C f as much as to A. 23. A man having completed f of his journey, finds that after traveling 30 miles farther only f of the journey remains, liequired the length of the journey. 24. Two-fifths of a pole is in the water, one-tenth in the mud, and the remainder out of the water. There are 6 ft. less of it in the water than in the mud and above water. How long is the pole ? 25. There are two silver cui)s and one cover for both. The first weighs 12 oz., and with the cover twice as much as the other without it; but the second with the cover weighs 1 more than the first without it. Find the weight of the cover. 26. I have a certain number in mind. I multiply it by 7, add 3 to the product, and divide the sum by 2 ; I then find tliat if I subtract 4 from the quotient I get 15. What num- ber am I thinking of ? 27. A certain number multiplied by 5, 24 subtracted from the product, the remainder divided by 6, and this quo- tient increased by 13, results in the number itself. What is the number ? simimj: I ..•! \ I loNs. '201 28. A |)erson spent \ of his moiu'y, alter which he earned $'.i. Then he lost .\ of wliat he then had, receiving after- wards $2 back. Lastly he gave away } of what he now had, when he found he had only $12 remaining. "What had he at first ? Lot .r = the number of dollars lie Imd at first, (222, 1) tlun - J" = remainder after spending one fourth, (222, 2) 4 2 /3 \ .y I 7- X + ;H = the second remainder, (222, 2) !! r| r^ J- 4- :J j + 21 = the third remainder =^ 12. (222, :i) 2l(^^^)+7-^^ ao9,.) 7 + 7 + 7 " ^^ 3x+ 12 + 12 = 84 a: = 20 Ana. Vkrificatiox. i of 20 = 15; 15 + 3 = 18; | of 18 = 12; 12 + 2 = 14; ? of 14 = 12= 12. 29. A basket of oranges is emptied by one i)erson taking half of them and one more, a second person taking half of the remainder and one more, and a third j)erson taking half of this remainder and <*> moi-c. How in;niy did the basket contain at first. 30. In making a journey a traveler went on the first day ^ of the distance and 8 miles more ; on the second day he went ^ of the distance that remained and 15 miles more : and tlie tJiird day he went ^ of the distance that remained and 12 miles more; on the fourth. day he went .S5 miles ;nid finished his journey. What was the distance traveled . 31. A man who has $2(X)0 invested in a mill from whicii he receives a certtiin per cent, and $10(K) in real estate from which he derives oidy J of the previous rate, has an income from l)oth of $330. What rate per cent does he receive ? 208 TEXT-BOOK OF ALGEBRA. Let X = rate % on the $2000 investment, '^ X = rate on the $1000 investment ; .-. 2000 X ^ = 20x = the interest on the first, and ivA' 1000 x-i^ - 100 30 X ■ 4 = interest on rate per cent. 20 X 30 ic "*" 4 " = 330 80 X + 30x = : 1320 110 x = 1320 . X = ix = 12% 1 9%) 32. What is the property of a person whose income is $430, when he has § of it invested atAfc, i SitSfc, and the remainder at 2^. 33. A person who is worth $12000 invests a certain amount in railroad stock from which he receives 6^. One- third of the diiference pays 4^ interest, and the other two- thirds 5fo interest, and from all his funds he clears $896. What amount was invested in railroad stock ? On what amount had he realized but $600 ? (See 252, 7.) 34. A man has $5000 invested at a certain rate, $2000 at double the former rate, and $1000 at triple the first rate. He received from a property an income of $1900, but pays out $2000 for improvements, personal expenses, insurance, etc. He finds he has $500 remaining at the end of the ;[ear. What does he receive on the $2000 ? 35. A workman engaged for 48 days at the rate of $2 l)er day and his board. But for every day he might be idle he was to pay $1 for his board. At the end of the time he received only $42. How many days did he work '.' 36. A man hired a laborer for one year at the wages of $90 and a suit of clothes. At the end of 7 months the laborer quit his service and received $83.75. At what price were the clothes estimated ? S1M1MJ-: i:(,>i A rioNs. 209 37. A boy engaged to carry «'50 glass vessels to a cer- tain place on condition of receiving o cents for every one he delivered safe, and f< rh itini; 12 cents for every one he should break. On settlement he received 99 cents. How many did he break ? 38. A tree standing vertically on level ground is 60 feet high. Upon being broken over in a storm tlie upper part reached from the top of the trunk to the ground just 30 feet from the foot of the trunk. What was the length of the part broken off ? Note. — This problem tlei>ends upon the theorem proved in geometry and stated in most arithmetics, that the square on the longest side of a right-angled triangle is equal to the sum of the squares on the other two sides. 39. A can perform a i)iece of work in () days, B can per- form the same work in 8 days, and C in 24 days. In how many days can they finish it if all work together? Let j; = the number of days when all work together. Now, A does j^ of the work in one day, and in x days he will do x times jl or ^ of the work. For a like reason B will do g, and C, ^4 of the work in one day. But they do one times the work in x days, therefore 1+1+ 2T=^ (219,1) .-. x = 3.* Ans. (219,1) Note. — The whole work, as is customary in intellectual arith- metic, is called one; then the fractions of the work which each one does added together make the whole work, or one. TIk' equation i H" i "I" 5V '^^ x (virtually the same as the one just given, being obtained from it by dividing through by x, Ax. 4) is found by saying that if it takes all of them x days, in one day they will do ^ of the work. Solving this equation x = 3 as before. 40. A and ?> together can do a piece of work in 12 days, A and C in 15 days, and B and C in 20 days. In what time can they do it all working together ? > Hereafter, when it Is cuiiveiiUnt to leave the solution of an equation to the student, the answer will be given witli a refi r< iic< to 219, >'• 210 TEXT-BOOK OF ALGEBKA. Let X = the number of days in which all can do the work. Then X — tV ^^ the part C can do in one day, etc. 41. A cistern can be filled by one pipe in 15 minutes, by another in 12 minutes, and by a third in 10 minutes. In what time can it be filled if all are left open ? 42. A tank can be filled by two pipes in 24 and 30 min- utes respectively, and emptied by a third in 20 minutes. What time will it be in filling if all are left open ? • 43. A man who can perform a piece of work in 14 days, works 4 days when he calls in a boy, and together they fin- ish the job in 6 days. In how many days could the boy do the work alone ? 44. A cistern can be filled by two pipes in 9 minutes and 12 minutes, and emptied by two others in 15 and 18 minutes respectively. The first pipe is left open one minute, then the third is opened. At the end of the second minute the second pipe is opened, and at the end of the third minute the fourth is opened. How soon will the cistern be in filling counting from the time the first pipe was opened ? 45. A boy buys apples at the rate of 5 for 2 cts. He sells half of them at the rate of 2 for 1 ct., and the rest at the rate of 3 for 1 ct., and clears 1 ct. by the transaction. How many did he buy ? 46. A market woman bought some eggs at the rate of 2 for 1 ct., and as many more at the rate of 3 for 1 ct. She sold them all at the rate of 5 for 2 cts., and found she had lost 4 cts. How many of each sort did she buy ? 47. A train on the Northwestern line passes from Lon- don to Birmingham in 3 hours ; a train on the Great Western line, which is 15 miles longer, traveling at a speed which is less by 1 mile per hour, passes from one place to the other in 3^ hours. Find the length of each line. SIMl'LK Kijl AlloNS. 211 Hemauk. — Problems of distance, time, and rate occur so fre- quently, that it may be helpful to the student to be familiar with the answers to the following gemial (jih stions : — Given the time and rate, how is the distance found ? Given the distance and rate, how is the time found? Given the distance and time, how is the rate found? Similarly in problems involving price, number of articles, and total cost. Given the price and the number of things, how is the total cost found ? Given the price and the total cost, how is the number of things found ? Given the number of things and the total cost, how is the price found ? So also we might discuss problems involving length, breadth, and thickness, area and solid contents. The student must decide in every problem what kind of imits the two sides of the equation are to contain, and prepare the functions accordingly. This will tend to make the process of solution clear. 48. A person walks to the top of a mountain at the rate of 2]\ miles an hour, and clown the same way at the rate of .'U miles an hour, and is out 5 hours. How far is it to the top of the mountain ? 49. A boy who runs at the rate of 12 feet per second, starts 20 yards behind another whose rate is 10^ feet. How socii will the first boy be 10 yards ahead of the second ? 50. The distance from M. to L. is 31^ miles. The express down train leaves M. at 11.30 a.m., and arrives in L. at 1 - ;< > P.M. The up train leaves L. at 11.45 a.m., and arrives ui M. at 12.35 p.m. Supposing the speed of each train to be uniform, when will they meet ? 51. A bought eggs at 18 cts. a doz. Had he bought 5 more for the same money, they would have cost him 2^ cts. a dozen less. How many eggs did he buy ? 52. A rows 4 miles and B 3 miles an hour. A is 14 miles farther up stream than B, and they row towards each other 212 TEXT-BOOK OF ALGEBRA. till they meet 4 miles above B's position. How rapid is the current ? Suggestion. — Let x = the rate of the current, then x + 4 equals A's rate down, 3 — ic B's rate up. 53. A boatman who can row 6 miles an hour in still water rows a certain distance down stream and back again in 3 hours. How far did he row, supposing the stream to flow 3 miles in two hours ? 54. There are two numbers in the ratio of ^ to f , which being increased respectively by 6 and 5, are in the ratio of f to ^. What are the numbers ? 55. A farmer makes a mixture of rye, oats, and barley, using 3 bushels of rye as often as 4 of oats, and 5 of barley. The whole amount of grain used was 66 bushels. How many bushels were there of each ? 56. The estate of a bankrupt, valued at $21,000, is to be divided among 4 creditors proportionately to what is due them. The debts due A and B are as 2 and 3 ; B's and C's claims are in the ratio of 6:7; and D's claim is equal to A's. What sum must each receive ? SuGGESTiox. — Let 4 X = A's and 6x= B's money. 57. A general arranging his men in the form of a solid square finds he has 21 men over, but attempting to add one man to each side of the square finds he wants 200 men to fill up the square. How many men had he ? 58. The length of a room exceeds its breadth by 8 feet. If each had been increased by 2 feet, the area would have been increased by 60 square feet. Find the original dimen- sions of the room. SIMPLK K(^l ATIONS. 21 ^J ( HAPTKK XV. SIMPLi: KQUATIONS CONTAINING TWO OH MOIJK I'N- KNOWN QUANTITIES. ELIMINATION. 225. Simultaneous Equations are sets of equations which are to l)e satisfied by the same values of the unknown quantities contained in tlieni. a. All proper conditional problems (188) give rise either to a single equation containing one unknown quantity, or to two or more equations containing as many unknown (juantities. In the latter case some of the funrtlous (222, 2) of the unknown are regarded as other unknown quantitten. Consequently we may solve problems in one of two ways : either regard the problem as containing but one unknown (x) and express everything in terms of it, as we did in the last chapter ; or regard the problem as containing two or more unknowns (x, y, z, etc.), and form two or more equations. To make this plain let us take Ex. 18, 224- 1st. With one unknown. Let X = the niunber of lbs. of 1» ct. sugar, (222, 1) then 100 — ar the number of lbs. of 1:] ct. sugar, (222, 2) J- + 13 (100 -x) = 1200, (219, 1) x = 2r> ) (219,2) 100 - .r = 7.") S 2d. With two unknowns. Let X = the numl>or of \hf. of ct. sugar, and y = the number of lbs. of 13 ct. sugar, Equation (1) x -\- y = 100 ) /g^g jv Equation (2) j + 13 // = 1200 S // = 100- X (From Eq. (1) by Ax. 2) 9 J' -H 13 (100 - J-) = 1200 (Substituting 100 - x for y in the second equation) X = 2r> (219, 2) y = 75 (By substituting 25 for x in (2)) 214 TEXT-BOOK OF ALGEBRA. In tlie second solution, a new unknown quantity, ?/, is first intro- duced and then removed from the second equation by substituting its value obtained from the first. Tlie resulting equation is the same as the equation of the first solution. In this problem it makes little dif- ference whether we solve it with two or with one unknown quantity; but if the second quantity is a very complex function of the first, the second method has an advantage over the first. h. A single equation containing two unknown quantities is " ?n- determinate,^'' since the value of either of the unknowns cannot be found until that of the other is known. To illustrate this, suppose 5 a; - 3 2/ = 18 DX = lS + Sy (Ax. 1) _18 + 32/ o (Ax. 4) li y — 1, then x = ■ — ;: — = 4^ lfy = 2, then x = ^^ '^ ^ "" ^ = 4* If ?/ = 3, then x = — ^ = 5§ ^^^^^ ^^ ^^^ If, however, two equations are given, such as (1) Sx + 4y = 9 (2) 2x + 9y = 11 the value of y can be found from equation (1) (1) i^x + 4y = 9,4y = 9-^;\x,y = '^-^^, and then placed in Eq. (2), obtaining 2x + 9 5^^ =17, and we have resulting one equation containing but one unknown, which solved, gives x — — ^ . Similarly, the value of x might have been found from (1) and substituted in (2), whence the value of y might have been found. c Any proper axiomatic operation performed on an equation will not change the value of any letter, and so will not change the " iden- tity " of the equation. It will still be the same equation, only under a new guise. (Consult Art. 360.) SIMPLE EQUATIONS. 215 It will be convenient to use a system of marking equations by which it can be indicated that a certain equation appears in a new fonn. The figures written at the left in parentheses will be used to designate the equations, while the same figures with subscripts will distinguish the several equations in their new forms. Thu3, (1) Sx + 4y = 2'l (2) 2j--3iy = 21 (1,) 9 X + 12 y = 81 [Eq. (1) multiplied by 3, Ax. 3] (2,) 8 x - 13 2/ = 84 [Eq. (2) multiplied by 4, Ax. 3] (1.) a; + 1 2/ = [Eq. (1) divided by 3, Ax. 4] and so on. 226. Elimination is the process of deducing from two or more e(|Uiitions containing two or more unknown quantities a single equation with but one unknown quantity. rt. There are three kinds of elimination usually given : elimination by substitution ; by comparison ; and by addition and suhtrartion. We shall take up in section I. the case in which there are but two unknowns. Then in the next section that in which there are three or more unknowns. SECTION I. Solution of Simple Equations containing Two Unknown QlANTITIES. 227. Equations Containing Two Unknown Quantities are solved by the methods of elimination. We give first the methods of elimination with exercises, and then problems involving such simultaneous equations. I. -ELIMINATION BY SUBSTITUTION. 228. Elimination by Substitution is performed by finding the value of one of the unknowns from either of the given equations and substitutin<^ it in the other. 229. After the value of one of the unknowns has been found, it is substituted in one of the foregoing equations, thus giv- ing an equation conttiining only the other unknown whose value can then be found. 216 TEXT-BOOK OF ALGEBRA. (1,) 8.T = 68 — 5y 68 -5?/ 8 230. Exercise in Elimination by Substitution. 1. Given (1) 8.r + o// = 68, and (2) 12a^ + 7?/ = 100 (Ax. 2) (Ax. 4) (228) (178) (Ax. 3) (229j (1.) X -. (2) Vl(^^^^^\ + ^y = im 204 -15^,^ _,.^ 2 — + < y = 100 ^ 204- 15ij + Uy = 2()() y = 4 J^is. (1) 8;r + 5x4 = 68 S ^ = 48 X = (J. Ans. 2. Given (1) 1; 3^ _ 3j/^- 4 :r + 7 5 To ~ r> ~^ ^"^ 3 ■+■ 2 ~ 20 "" -^ - 15 +6+10 (li) 4 v/ - 9 ci" = 18 y - 24 X + 42 - 25 (Ax. 3) (1.) 15 ic — 14 y = 17 (Ax. a, and 86) 15 X — 17 (la) // 14 (Ax's n, and 4) (2i) 20 y - 20 + 30 .T - 9 y - 60 = 4 ?/ — 4 .r + 10 .i- -|- (> (Ax. 3) (2^) i>4 .r -[- 7 y = 86 15 .T -17' (2.) 24 .r + 7 14 86 48.'r + 15.r — 17 = 172 (',3 X = 189 X = 3 A71S. (1,) y = lA><^|::il^ = 2. ^,«. (228) (229) siMi'Li-: i:()r.\'ri(>Ns. 217 XoTE. — As either unknown may Ix- eliminated the student is left to discover which one it will be advisable to eliminate first. As a rule that unknown should beelimiiiated which has the bust coefficient. In substituting in one of the preceding eciuations (229), choose the simplest form. Complex equations, like Ex. 2, must be simplified before proceeding with the elimination. 3. Given (1) 3 x + 4 // = 10, and 4 x -\- // = <). SuooESTiON. — Find the value of ?/ in (2) and substitute it in (1). 4 j(l) X + 2//=:13 • } (2) 2a: + 3// = 21. 5 (1) 40^-3// = 26 " ~ -4,/ = U. // = 10 6. J (2) 3x [(1)0.- (2) 5^3 10. 7. 8. (1) I ••^ = 1 (2) ox — 3 // = 10. i (1) Si)x-\-17 t/=H6 I (2) 50 .r- 13// = 17. (2) 1,1 4^ + r,//=«- 10. < 0) r^^ + .y = i<5 11. (2) ^ + l=U. I '* 1(2) ^ = ^-1. I ^ 3 r> r(i) 3x-7// = o 1(2) U^P^' 218 . TEXT-BOOK OF ALGEBRA. 13. Given { 1) ax -\- by = c and (2) mx -\- ny =p. (IJ ax = c — by c — by (1.) ^- a (2) m r-~^\ + ny^p (328) mc — mby -h any = aj) (Ax. 3) (an — bm) y = ap — cm (97, Ax. 2) ap — cm . an — bm The value of x may be found by substituting this value of ?/, but in literal exercises it is often simpler to go back and eliminate the other unknown. (I3) y '- c — a2 b (2) mx + > •{*- — ax\ wl lience x p (228) ^^E^Z^. Ans. (219,2) 6m — an Or, substituting the value of y in (1) we have ax + b f^^P-^^A = c (229) \an— bm I whence x = ^^ ~ ^^^ as before. (219, 2) bm — an a. Wlien equations are symmetrical (i.e., such that when corre- sponding letters are inter changed^ the equation is not altered) one answer may be obtained from the other by inspection. Thus, to get the value of x from y, change y to x, a to ft, b to a, m to n, and n to ?yi, which, can be quickly done. 14. (1) cx-\-dij = 1 and (2) ex ^ fy = 1. 15. (1) ax = by and (2) hx -\- ay = c. II.— ELIMINATION BY COMPARISON. 231. Elimination by Comparison is performed by solving for the same unknown in both equations, and setting the two values thus obtained equal to each other (Ax. 7). SIMPLE EQUATIONS. 219 232. Exercise in Elimination by Comparison. 7 + 5 11 (1,) r,-ir)r + 21 7/-7 = 70 (2i) 3a;4-//-hlly = 99 (Ax. .S) (lo) - 15a! = 72 - 21 // (22) 3ar = 99 - 12y lo 5 (Ax. 4) 7?/ -24 (2.) as- 4y 7y- 24 = 165 -20// 27y = 189 y = 7 A71S. a;= 3.^-4x7 = 5. J;?.s. (Ax. 7) , (1) 2x + 7y = 65 • • (2) 6a:-2y = 34. ' I x ^ I (2) 3 - '/ = - 1. ( (I) .Sx +.-,,/ = 29 g J (2) ;•. X - r. y = 19. 6. 3 a- // ('>) — — ' = 2 (I) 8- x+3 4 ~ 7 — 3 a- -2?/ 5 (2) 42- 8-y 3 ~ = 24A 2ar-|-l 8. (1) 21// + 20 a- = 105 (2) 77y — 3()r = 295. 2 _ 3 (2) 5(x+3)=3(y-2)+2. 220 TEXT-BOOK OF ALGEBRA. (1) lhx = UtJ-\-4j%. ^' ' (2) Ux = ^y-21^,. 10. (1) x-2?j = a. (2) 2x^H?/ = 0. X y ,,.. X If 2 (1,) x = a + a\ (2J x = 2« + || (Ax. 7) 26-^ (Ax. 2) y = 2}). Ans. (Ax. h) (Ij) a; =^ a + — r— = :J «. ^ns. (229) 12. (1) ? + -^ = TO (2) ? + '^ = ». ^ a ' cd III. -ELIMINATION BY ADDITION AND SUBTRACTION. 233. Elimination by Addition or Subtraction is performed by making the coefficients of one of the unknowns the same in both equations, and then adding or subtracting the equations member by member according as these coefficients have un- like or like signs. 1. In order to make coefficients the same in both equa- tions, proceed as follows : — (1. Keduce each equation to the form ax •\-hy ^= c, i.e. collect into the first term all the terms containing x, into the second term all the terms containing y, and into the right member all the known terms. (2. Find the 1. c. m. of the two coefficients of the un- known to be eliminated. (3. Multiply both members of each equation by the quotient obtained from dividing the 1. c. m. by the coeffi- cient in that equation of the unknown to be eliminated. SIMPLE EglATloNS. 221 2. Upon adding or subtracting the new equations mem- ber by meml>er, the terms with equal coefficients cancel and one unknown is thus eliminated. 234. Exercise in Elimination by Addition and Subtraction. 1. Given (1) ^i:^ + 3 = i^ ^^ 4 2^3 (1,) Ioj:— l>r)//-|-30 = 4.c + 2// (U) 11 « - 27 // = - 30 (233. 1, (1) (2.) 9(>-.3x + 6z/ = Ca^ + 4y (2,) -<)a: + 2v/ = -9() (233,1,(1) (I3) 99 a; - 243 y = - 270 (233, 1, (3) (2,) -99a;H -22 y = — 1056 (Ax. 1) -221y = -1326 ?/ = 6. .i;i.*f. (lij _ \)x -I- i> X (; = - 90 .-. X = 12. .l;/.s-. (229) 2. (1) 8 a: - 21 // = 33 (2) 6 x -f 35 y = 177 (li) 2Ax - 03 y= 99 (2,) 24 X + 140 y = 708 203 y = 609 y = 3 Ans. (1) 8 x - 21 X 3 = 33 .-. X = 12. Ans. (229) 3. (1) 10x4-17// = 500 (2) 17x — 3y=110. (li) 48x-f 51y = 1500 (2i) 28 9 X - 51 y = 1870 3;f7x =3370 (Ax. 1) X = 10. Ans. y = 20. Ans, (229) 222 TEXT-BOOK OF ALGEBRA. 4. 7. 8. 10. 11. 12. 13. ( (1) ;^x + 2,/ = = 19 U2) 2x-3i/ = = 4. (1) < M=« (2) M=«- (1) (2) 5 ';'=3 x-y X 2 "^ -J-^=«- (1) 2x y_ 3 ^4 " 16 (2) x^l = U. |(1) 2 1 5^-12^ = 3 (2) 4a: — ?/ = 20. (1) f + l = ^^ r (2) . + 1 = 4. !• ^■(1) 1(2) 4^_2j/^ 5 5 6 ;r = 9 y. = 4 (1) x + 37/ _ 2 7i < (2) 4^ + 5y 4 = 8. (1) 3 2 'lO^ (2) 8^ + 6^ = = 7. ((1) 4ic -fSy = = 2.4 1(2) 10.2 05 -6 ^ = 3.48, SiMl'LK tAii A 1 iuN.> 23 14 Ul) ax-\-b,j=p ((2) ax^dy=^q. ((1) ma:-h«// = i(2) // = «a-+/>. 15 16. 17. 18. fa b J (1) iT^j = 3M^ l (2) rtJ- + 2 />// = ^/. (1) x + i(3 x - (2) §(4 0. + '^l/) (1) 11 15 (2) 5 + 1 = x y 9 5 ■)=l+f(y-l) Solution. — Inequations containing reciprocals it is usually best not to clear of fractions^ but to solve directly for ^ and * . 19. 20. 11 (i) X "^ y ~ 5 (2) 3 4 9 x^ y~ 6 2 2 y"5 10 = 2y y = 5 ^n«. (10 1 11 2 1 x~15 5"~3*- (1) 5 + '' = .s (-0 15 + 5 = 4. (I) 2li+ 12^ =5 0? // (2) 1 _ 1 _ 1 // X 42* x = 3 (233, 1) (Ax. 2) (Ax. 3) (229) 224 TEXT-BOOK OF ALGEBKA. 21. 22. 23. 24. (1) 4-! a; + 27-1 = 2/ 42 (2) 14 1_15 U a- y = 1. f(l) 1(2) 4 a?- ' + 3y-' = ' + 82/-' = f(l) 1 -\'/ — a- = 4 icy ~ (-') 4_ !=»■ 3 x^ 5 8 -y-15 I (2) .),;_22.<-=-^. IV. -SPECIAL METHODS OF ELIMINATION. 235. Special Methods of Elimination. In these the elim- ination is attained by one or other of the three processes already explained. The difference lies in the preparation of the equations for the elimination. A, Elimination by Undetermined Multipliers. 236. The General Process of Elimination by Undetermined Multipliers for equations of any degrees is called Bezout's Method. 1. Given {1) hx-ly = 2Q (2) 9 ic - 11 2/ = 44. (1 J 5 mx — 7 my = 20 ni (Ax. 3) (2) 9x-ny = U (3 ) (5 m — 9) X — (1 m — ll)y = 20 m — 44 (Ax. 2) Now, if 5 m —9 = 0, then x will be eliminated from the equation. If 5 m — 9 = 0, m = ~. Substituting this value in (3), we have .-( 9 \ 9 7 X V - 11 ) 2/ = 20 X - - 44 r. 8 40 Or, —^„z=z- _ o 7/ = .") .r ^ 11. (229) SIMPLi: I'Xa'ATlONS. 225 If 7 '^ 'Hi'l .; + -^ = '39.2. 5. (liven X -[• aij ^ b and ax -)- Z»// = 1. B Elimination by Means of one or more Derived Equations. 237. When One of the Regular Processes of Elimination gives ri.se to large products, the work can often be materi- ally shortened by means of derived equations. If either or both of the given equations be multiplied by some numl)er and the resulting equations added or sul>- tracted member by member, we get a new equation whicli will he satisfied by the same values of x and y as the given equations (Ax. a.) This new equation, it is true, has both X and y still in it, but with smaller coefficients. Now, this equation used with either one of the given equations, or with another similarly derived, since two equations suffice, will give the values of x and y by ordinary methods of elimination. x + y = 16 9x + 13?/ = 184 9x + 9y = 144 226 TEXT-BOOK OF ALGEBRA. 1. (1 ) 9 cc + 13 ?/ = 184 (2) 13 £c + 19 ?/ = 268 (li) 27 X + 39 ?/ = 552 (2i) 26 X + 38 y = 536 (3 ) ~^+ y= 16 (Ax. 2) (1) (3i) 4 */ = 40 (Ax. 2) 2/ =10 .•.a-= 6. . (229) 2. (1) 14 X - 17 ?/ = 159 (2) 29 x - 37 // = 324. Suggestion. — Multiply (1) by 2 and subtract from (2). 3. (1) 139 X + 152 ?/ = 1377 (2) 35 x + 37 ?/ = 348. 4. (1) 755^-564 ?/ = 2074 (2) 1133 x - 847?/ = 3113. 5. (1) x + 50?/ = 557 (2) 50x + ?/ = 361. Sr(;(;i-:sTiON. — Add the two equations and divide through by 51 ; then subtract the one from the other and divide through by 49. 6. (I) 59x + 73y = 390| 7. (1) 23x — 41?/= — 8. (1) 28x — 35?/=56 9. (1) {a-\-2h)x— (a- (2) (a + 30i/-(a- C. Elimination by Method of finding H. C. F. 238. This Method depends upon the Same Principles as the last and is really only a special way of applying them. 1. Given (1) .5 ^- - 7 ?/ = - 8 (2) .5 y + 82 = 7 u- (1,) 5 .r. - 7 ?/ + 8 = (2j) 7 x - 5 ?/ - 32 = 7 X - .5 ?/ - 82) 5 :/• - 7 ?/ + 8 5 X — 7 ?y + 8 (J_ 2 )2y/+2?/ — 40 X + ?/ - 20) .-> .r - 7 ?/ + 8 (5 r> X + 5 ?/ - 100 - 12 ?/ + 108 = - 12 ?/ = - 108 7/ = 9 .-. x = ll. (229) •f (2) 73x- ■f59?/ = = 379}. 2^ (2) 39x -25y = = 63. (2) 29 a:- -13y = = 151. -2b}!,= = C«c - 3 c) X == . 4 ab. simpm: i:(^uation's. liiii Explanation. — Kach etiuation is so transposoii that one nu'iiibcr is zero. Then the divisor is multiplied through by some factor, and the result subtracted from the other given equation, giving a new equation, which is used with the divisor (or dividend), and so on. The last remainder does not contain j; i.e., it has been elhnhiated. This remainder, e(|ual to zero, gives the value of y. 2. (1) Sx - 21 t/ = 33 (2) (> J- + 35 // = 177. 3. (1) l(j x + \7 !/ = 500 (2) 17 .r - 3 // = 110. 239. Exercise in Elimination. The choice of the method is left to the student. Addition and Subti-action usually gives a more elegant form to the solution than the other two methods. 1. 4 or -h 9 // = 51 and 8 a- — 13 // = 9. 2. 7 .r - 9 // = 7 and 3 x + 10 // = 100. 3. 4 ./• -f S y = 2.4 an. I 1 0.2 ./• — (*) // = 3.4S. 4. X -\- If = 24 and x = 5 y. 5. 2 // -\- 79 =^iyx and 3 ./■ — 7 = 4 -f ./• + //. 6. :£ + -•" = (•, ami ^^+'^ = 0. 3 .) 3 i> 7. ^+^ = :^ + 2and^4--^ = -^ + 4. ^ ^ 2 4^3 10 4 5 8. = and 2 ./• 4- 5 // = 35. 5 + y 12+ J- ^ •' 9. 10 J- = 2 -f 2 // and 4 // ^ 20 - 1 r. 10. •'* t ^' + '"^ V - •■'! ;"'<' -^ ' '' + 10. *• = ly- • > 1 11. ox -\- />// = r — (/ and mx = n//. 12. jj^±J; = '!amlS.,-4 = !»y. 13. '+-=l'aM,liLL±iii^ = /,. ./• — // X -{- a 14. .-) ./• -h 7 // : 3 r + 1 1 : : 13 : 7 and 1 1 r + 27 : 7 ./' -|- oy::19:ll. 228 TEXT-BOOK OF ALGEBRA. 16. 2 (2 X + ;^ I/) = 3 (2 ./• — o y) + 10 and 4 a; - 3 // = 4 (() y - 2 .r) + 8. 17. + -^^^ = 2 and '-^ -[- y/ = 1). 7 5 11 18. l(x-\- 2) + ^(y - a-) = --^ - ^ ^11^^^ ^ (- // - ^^) + / 4 o 1 (8 ^ + // — 4) = 3 X + 4. o 19. 2 a; : y : : 29 : 4 and y-^lx^^S ^ 4 //-^ + 13 .r// - 12.x--^ 4 y - 3 ./• - 1 on ^ 3 1 18 ,2 1 1 /I , 1\ , 1 20. - I - _|_ _ and- _- = ^_ + - + ic ' y ' 4 2/ ^ 2/ ^ \^ y J J-^ (See Ex. 18 in 234.) 21. _^- + _J^ = 2«. and^-:^l = l. a -\- b a — h 4: ah 22. §^i±iidy^ - ^-+^-V = 5 + -^^^ and 10 15 5 9?/ + 5a7 — 8 X -\- y 1 X -\-Q> I2 4'^ ^ ~Tl~" ' 23. 3 «x — 2hy ^ c and c/'-^ic + ^V = ^ ^^'• 10 9 2a; + 3y- 29 _ 7a; — 8 y ,^ 2 3 ~^ Suggestion. — Put 2 ic + 3 ?/ — 29 = ?t and 1 x — 8y +24 = tj; then the two equations become i5_^^8and!! = ^, u V 2 3 from which (228) u = 2, and u == 3. Substituting these values, we now have 2 ic + 3 ?/ - 29 = 2, and 7 X - 8 ?/ + 24 = 3, which, being solved, give the values of x and i/. SIMPLE EQUATIONS. '2'2y + 10//= 10 J- -fssi. SUGOEHTiox. — Put = M and 5 X — 8 w + 44 = r, transposing lU ij. 240. Problems containing Two Unknown Quantities. 1. When the greater of two numbers is divided by the less tlje quotient is 4 and the remainder 3 ; and when the sum of the two numl)ers is increased by 38 and the result divided by the greater of the two numl)ers the quotient is 2 and the remainder 2. Ijet X equal the greater number and 1/ equal the less. (225, o.) Then (1) ^^ = 4 and (2) """^-^^^^"^ = 2 (222, :\) (l,)r-4//= :) C2,)x - y = 36 3 y = .'« (Ax, 2) ?/=ll x = 4- (229) 2. A farmer paid four men and six boys 72 dimes for Liboring one day, and afterwards at the same rate he i;aid three men and nine boys 81 dimes for one day. What were the wages of each ? 3. Find two nunil»ears were bad, but she sold the good apples at 2 cts. each and the good jx'ars at 3 cts. each, realizing $2.00. How many of each fruit did she buy ? 16. If a certain numl^er be divided by the sum of its two digits, the quotient is G and the remainder 3; if the digits be interchanged and the resulting number be divided by the sum of the digits, the (piotient is 4 and the remainder 1). What is the number ? SoiATioN. — In order to represent numbers in the Arabic scale of 10 by letter H^ we must say, Let X = the number denoted by the figure in unit's place, and y = the number denoted by the figure in ten's plaro, and so on, usinj; other letters for hijjher orders. Then, lOy 4- j represents a number of two figures, 100 2 + 10// + X represents a number of three figures, and so on. Wlien tlie order of the digits is reversed, 10 J -f y = the number of two places. The equations of the above problem are lOy + x-3 lOx + y-9 (1) \^^^ =6 and (2) -_^^ = 4. 16. Find a number which is greater by 2 than 5 times the sum of its digits, and if 9 be added to it the digits will be reversed. 232 TEXT-BOOK OF ALGEniJA. 17. Find tliat number of two digits, to wldcli if the num- ber found by changing its digits be added, tlie sum is 121 ; and if the less of tlie same two be taken from the greater the remainder is 9. 18. In 11 hours C walks 12^ miles less than D walks in 12 hours ; and in 5 hours D walks 3| miles less than C does in 7 hours. How many miles does each walk per hour ? 19. A man has two measures. Nine of the first or fif- teen of the second will fill a certain vessel. Using both measures 13 times in all, how many times is each used ? 20. A banker has two kinds of money. It takes a pieces of the one or b pieces of the other to make a dollar. If c pieces be given for a dollar, how many of each will be used ? 21. A pound of tea and 3 pounds of sugar cost $1.20. But if tea were to rise 50% and sugar 10% they would cost $1.56. Find the price per pound of eacli. 22. A grocer knows neither the weight nor the first cost of a box of tea he had purchased. He only recollects that if he had sold the whole at 30 cts. per pound he would have gained $1. But if he had sold it at 22 cts. per pound he would have lost $3. Eequired the number of pounds in the box and the first cost per pound. 23. Two persons 27 miles apart setting out at the same time are together in 9 hours if they walk in the same direc- tion; but if they walk in opposite directions towards each other, in 3 hours. Find their rates. 24. Find two numbers in the ratio of 5:7 to which two other required numbers in the ratio 3 : 5 being respectively added, the sums shall be in the ratio of 9 : 13, and the differ- ence of whose sums equals 16. 25. Two trains set out at the same moment, the one to go from Boston to Springfield, the other from Si)ringfield to Boston. The distance between the two cities' is 98 miles. SIMl'I.i: KC^TATIONS. !28o They meet eacli other at the end of one honr and 24 min- utes, and the train from lioston travtds as far in 4 liours as the other does in .'1 \\ hat was the speed of each train '/ 26. If the sides of a rectangular liehl were each increased by 2 yards, the area wouhl be increased by 220 square yards. If the length were increased and the breadth diminished each by 5 yards, the area would be diminished by 185 square yards. What is the area ? SECTIOX II. Simultaneous Equations coxtaining Three or More Ux K xowx Qr a xtiti es. 241. When Three or More Unknown Quantities appear in as Many Equations the process of eliminating, properly con- ducted, will cause the unknowns to successively disai)pear, and ultimately lead to the soluticm of the problem. a. When there are three equations containing three unknown quantities, we first select that one of the unknowns which can be most easily made to disappear, and proceed to eliminate it by usinf, say, the first and second e(|uations, obtaining a new (fourth) equa- tion ; then tlu» same unknown is eliminated by using the first and third (or, if preferable, the second and third) equations, giving an- other new (fifth) equation. These two equations, the fourth and fifth, contain but two unknowns, and can be solved as in Section I. of this chapter. The wliole process becomes more intelligible by studying an example. 242. Examples of Equations containing Three or More Un- known Quantities. 1. (Ij :r-2i/+:]z = () ] (2) 2x-f .3//-4« = 20 I (3) 3a--2// + r>5: = 26 J On inspection we j>erceive that of the three unknowns x can most readily be eliminated, having the smallest co- efficients. 234 TEXT-BOOK OF ALGEBRA. (li) 2x-4.y-\- 6z = 12 (2) 2x -\-3i/ - 4.z==20 (4) ' 7 // - 10 ;^ -= 8 (lo) 3x ~(j?/ -\- 9.^ = 18 (3 ) 3x - 2?j + 5 z = 26 (5 ) 4y/- 4^^"T (5,) y/- ^=2 (52) 7 ?/ - 7z = U (4 ) 7 y - 10 ^ = 8 3 ;v = 6 ^ = 2. Jws. (5i) ij - 2 = 2 .: y = 4. Atis. (1) X -2X4 + 3X2 = 6.-. it- (Ax. 3) (Ax. 2) (Ax. 2) (Ax. 4) (Ax. 3) (Ax. 2) (229) = 8.J7^s.(229) Note. — In deriving equations (4) and (5) aiiij f/ro j)airs of the three equations may be used. 2. (1) ^ + 2y/-3y^.+ z = 4. ^ (2) 2 .?• - y + 2 7t - 3 ,~ = 1 (3) 5.:^ -3.//- 7^-2,- = 11 (4) ;j ic -p 4 // - r> y^^ + (>- = _ 9 . (li) 3 X + G !i - 9 n + 3 - = 12 (2 ) 2 £c - y -f 2 v^ - 3 ;^ = 1 (5) hx^hy-lu =13' (I2) 2a; + 4y/-6«+2;>;= 8 (3) 5x-37/- 1^-2;^ = 11 (Ax. 1 1 (6) 7ir + y -1 n =19' (2i) 4a:-27/ + 4y^, -G.^= 2 (4 ) 3 :r + 4 ?/ - 5 ?^ + C) ,t = - 9 (Ax. 1) (7) 7.r + 2 7/-7^ = -7'. (Ax. 1) Equations (5), (G), and (7) constitute a new set. (5 ) 5 a? + 5 ?/ - (00 35a. + 5 v/- r)5 y/ 13 95 (8 ) 30 X (81) 15 a; - 28 u = 82 -Uu-= 41 (Ax. 2) (Ax. 4) SI.MI'LK KQrATlONS. 235 62) 14a; +2y- Un = 38 (7) 7a; + 2y~ ». == - 7 (9 ) 7 x - 13 M = 45 ■ (Ax. 2) Equations (8) and (9) constitute the third set. ( 9 ) 7 j; — 13 ?« = 45 ( 8i) 15 a; -14 7/ =41 (10 ) 8 ar - u = -4 (Ax. L>. 237) (lOi) 104ar-13w= - r>2. ( 9 ) 7 a; — 13 ^ = 45 97 a; = - 97 (Ax. 2) .T = — 1. Ans. (10) H X - 1 - u = - 4 .'. >f = - 4. Ans. (229) (6) 7X - 1 +//- 7 X -4 = 11) . // = — 2 .i/w. (229) ( 1) - 1 + 2 X - 2 - 3 X - ^ +5; = 4 .•.;s = -3. yi«..s-. (229) 243. Rule.' — The Normal Process of Elimination, wliere tliere arc three or more unknowns, may be described as follows : — 1. Select the unknown to be first eliminated. Then com- bine the equations in pairs in the most desirable manner to each time eliminate this unknown. However, the derived equations must be hidependent, i.e., such that no one of them can be derived from one or more of the others. (Cf. 246, a.) The number of new e(iuations is one less than the given ecjuations. 2. Proc(»e(l in the same manner with the derived equa- tions to eliminate one of the remaining unknowns. Then, in like manner, with the set thus obtained, and so on. until the value of the hist remaining unknown is found. 3. Substitute the value of the last unknown in one of the l)receding eijuations contiiining but two unknown ([uantities > A far more abridged anil elegant method of solving these problems Is by means of Determinants. 236 TEXT-BOOK OF ALGEBRA. and the value of the other is immediately known. Substi- tute these values in an equation which contains a third un- known, and so on. a. If in any case one or more of the equations do not contain one of the unknowns, it will usually be better to regard them as belong- ing to the second set. b. Special metho^te of elimination in particular problems fre- quently far surpass the normal process in elegance and brevity. 244. Exercise in Equations containing Three or More Un- known Quantities. [(1) :r+y + ^ = 6 1. -j (2) 5 ic + 4 ?/ + 3 ,^ = 22 [ (3) 15 X -h 10 ?/ + 6 ^ = 53. [(1) x^2y = 2'S 2. <^ (2) 3 :^ + 4 ^ == 57 [(3) 57/ + C^ = 94. Suggestion. — Eliminate z between (2) and (3), and com- bine the new equation with (1). r(l) a: + .y-.^ = 132 3. -^(2) x-y-^z = (\5A [(3) -x^y^x=-\.2. (1) ^+J/_^2,v = 21 3 (2)?/_±-_3x=-65 . (3) ■S.r + y -.-_3g_ f (1) i {.^ + ~-- '-.) = // - .t 5. (2) =2.»--ll 1(3) =!)-(.r + 2.t) f (1) and (2) y + ^_= •1±5 = ^_.±J/ &■ \ ' ^ 3 2 1(3) a^ + y + .t = 27. siMi'Li: i:(,M-.\ rioNs. 237 8. 9. 10. 11. 12. (1), (2) .r: //:.. = 5:1-: 13 (i.e., .r : // :: 5 : 12 and jc : z :: ii : lo) (IV) ,- + // -f - = 27. -. (2) i (.r +;•) = <>-// [(•^) 1 (.'•-::) = 2// -7. [(1) .i--h<< ^ !/ + .-: ^ (2) // + ''= 2./- + 2;. [(;i) x + a = :\j'-^:\^. [(1) «.r + /y// = r ^ (2) rj- -h through B and C, I nnist pay $6.10 hack fare. If I go from A to B through D and C, I must pay $5.50. Going from A to C through B, I pay the same as from A to C through D. On the other hand, from B to I) through A costs 40 cts. less than from B to J) through ('. What are the distances A B, B C, C D, and D A, if the fare is 10 cts. i)er mile ? 12. A, B, and C in a hunting excursion killed 90 birds, which they wish to share equally ; in order to do this A, who has most, gives to B and C as many as they already have ; next, B gives to A and C as many as they had after the first division ; and lastly, C gives to A and B as many as they each had after the second division. It was then found that each had the same nund)er. How many had each at first ? 13. A piece of work can be completed by A, B, and (' in 10 days; by A and B togetluM- in 11 days, su])p<)sing (' to have worked 5 days and left ntt ; liy \\ an<] (' it !*> works 15 days and C works .30 days. liow long will it take each alone to do the work ? 240 TEXT-BOOK OF ALGEBRA. CHAPTER X\ 1. INDETERMINATE AND REDUNDANT EQUATIONS. — PROBLEMS IN ALGEBRA. SECTION I. Indeterminate Et^uATioNS. 246. Indeterminate Equations occur wlieii the number of unknown quantities exceeds the number of Independent Equations. (See 225, b.) a. Equations seemingly different sometimes reduce to the same equation. Hence two given simultaneous equations might be mde- terminate. Thus, (1) 2 (8 x - 17i) = - H y. (2) 10x + ly = m- Upon clearing of fractions and transposing, both become 60 ic + 42 ?/ = 350. 247. Examples of Indeterminate Equations. These may be classified into two kinds: those on which are imposed no restrictions ; and those in which only positive integral values are permissible. The latter are often called Dio- phantine Equations. a. An example of the former kind was given in 225, b. It is there made plain that any number of results can be obtained from such an equation, every one of which is a solution. Also two or more equations may be reduced by elimination to a single equation containing two or more unknowns, whose solution would be like that in 225, 5. All that can be done with these equations, then, is to assign arbitrary values to one or more of the unknowns, and de- rive the corresponding values of the remaining one. These equa- tions find an appropriate use in analytic geometry. SIMPLE EQUATIONS. 241 b. If, however, the answers be conhned to integral positive num- bers, the problem becomes definite, and may be solved, although the method of solution is very unlike that of other equations. 1. Given r}x -f- 11 y = 47 ; required to find the pairs of i)ositive integral values of x and y which satisfy the equation. 5x = 47-lly (Ax. 2) x = ilz;lij^ = 0-2y + ^ (162) x-\) + 'J!/ = '-y^ (Ax. 2) o Since x and y are to be whole numbers, the integral quantity, aj — 9 -|- 2 y, must be a whole number ; and if the left side of the equation is a whole number, the 7'lf/ht side must be a whole number also. Consequently values of y must be selected which will make — —^ an integer (posi- o tive, negative, or zero). Such values are y = 2, making ^7^ = (for H- T) = 0) 5 y = ~. making --7 "'^ = — 1 5 y z= 12, making '1 s — 2 5 Finding the corresponding values of x by substitution (229), we have y = 2, a; = 5 ; y = 7, x = — 6. But negative values are excluded. Furthermore, larger values of y would give still larger negative values of x. Hence x = T), y = 2, are the only pair of integral roots. 2. Given 41 or -|- 11 y = 79() !/ = ^-^ = < 1 - 3 J- + — ^— . (162) 242 TEXT-BOOK OF ALGEBRA. As before ~~. — must be an integer, and the value of x must be found from it by trial. We may, however, derive a shti2)ler expression from which to find the value of x. For, if — — is an integer, any whole number of times it must also be an integer. Let us multiply this fraction by 7 (since 7 times 8 is one greater than a multiple of 11), and reduce the new improper fraction to a mixed quantity. Now, if 5 — 5x-\- is integral, since the part 5 — ox is integral, the fraction is likewise integral. From this fraction, then, values of x may be found which must make it integral. Putting . = 8, (1=^ = O), , = ^p ' = ''■ ''"'■ X = 19, (^^ = - 1^ .y ^ 790-41 X19 ^^ ^,^ No other values of x will give y positive. 248. Rule for tiie Solution of Indeterminate Equations. 1. When there are more than one of the given equations, reduce them by elimination to a single equation containing two or more unknown quantities. 2. If more than two unknowns remain in the single equation thus left, arbitrary values must be assigned to all but two of them. 3. Solve for that one of the unknowns most readily found in terms of the other, and reduce its value (when SIMPLE EQUATIONS. 243 (livisil)le) to the lorin ot a inixiHl quantity. The iniutjon thus ubtuined can Impiently he transtoinied into a simpler fraction by multiplyiuL; it 1)\ ^oiue number (always less than its denominator) wliich will make the coefficient of the unknown one greater than some multiide of the denom- inator, and then reducing this new fraction to a mixed quantity. The fractional part of this mixed (|n:intity may be used instead of the original fraction. 4. Supply successively such values to the unknown in the numerator of the fraction as will make it zero, or some multiple of the denominator, and each time find the cor- responding value of the other. Such pairs of values will be answers so long as they are both positive. Remark. — The method here given is very elementary. Other letters used as auxlHaries in the solution are often employed, but their use is not essential, and they have not been introduced in the explanations given above. 249. Exercise in the Solution of Indeterminate Equations. 1. 2 a; -}- 3 // = L^5. 2. 5 ic + 7 y + 4 = 50. 3. x4-3y-f-2«= 10. Suggestion. — Here it will be necessary to assign arbitrary values to one of the letters, say z. Zero values being excluded, z may equal 1, or 2, or 3. These values give three different indeter- minate equations to be solved for x and y. 4. 2;;7:{= Vix-\-24y. 5. 17.r +5;5y— 123 = 411 - 19 .r -f- 15 y. «■ 1S3^^^^;I51. ' ^«-=^«"'> I (2) 4 a; — 5 // — G ,t; = — GG. J (1) 29// = 8a^-4 ■ / (2) 4o2^17x — z. \(\) 2x + 3i/^7z = V.) 9. i (2) 5ar + 8y + ll.v-24 [ (3) 7 X -h 11 y -h -1 - = 43. ^^.M. 246, '') 244 TEXT-BOOK OF ALGEBRA. 250. Problems involving Diophantine Equations. 1. Separate 71 into parts of which the one is divisible by 5 and the other by 8 without remainders. 2. " Had I two times as many eggs as I now have " said one peasant girl to another, " and you seven times as many as you now have, and I were then to give you one egg, we should each have the same number." How many eggs had each ? 3. In how many ways may 100 be divided into two parts one of which shall be a multiple of 7 and the other of 9 ? 4. What is the least number which when divided by 5 and 3 leaves remainders 3 and 2 respectively. How many such numbers are less than 100 ? Suggestion. — Let z be the nmnber and x and y the quotients. 5. A person bought 40 animals consisting of pigs, geese, and chickens for f 40. The pigs cost $5 a piece, the geese $1, and the chickens 25 cts. each. Find the number he bought of each. 6. Divide 17 into three such parts that if the first be multiplied by 5, the second by 4, and the third by 7, the sum of these products is 80. ' 7. A number consisting of three digits, of which the middle one is 2, has its digits inverted by adding 198. What is the number ? 8. A farmer buys oxen, sheep, and hens. The whole number bought is 100 and the total price $100. If the oxen cost $35, the sheep $3, and the hens 2o cts. each, how many of each did he buy ? 9. A boy sees that he can buy oranges at 2 cts., 3 cts., 4 cts., 5 cts., or 6 cts. apiece and spend all his money ; but if he buys at 7 cts. apiece he will have 5 cts. remaining. How much money has he if he has the least sum possible ? Suggestion, — The 1, c, m, of 2, 3, 4, 5, and 6 is 00. Let 60 x = Ills money, SIMPLK KQIATIONS. 245 10. Find four integral numbers such that the sum of the first three shall be 18 ; the sum of the first, second, and fourth 16 ; and the sum of the first, third and fourth 14. SECTION II. Redundant Equations. 251. Redundant Equations occur when the number of equar tions exceeds the number of Unknown Quantities. These equations may be chissed into two kinds : 1. Compatible Redmidant Ef/uatlojis are such as are satis- fied by values of the unknowns derived from the other equations of the set to which they beh)ng. a. Redundant equations were used in 237. The new equations formed out of the old were satisfied by the same values of the un- knowns, i.e., were compatible. When new equations were con- structed an equal number of old ones were dropped. We can also have independent redundant equations. Thus, (1) 7 X -H 2 7/ = .>i, (2) 12 X + 5 y = 94, (3) 14 x - 11 y = 70, are all sat- isfied by X = 7, y = 2. All that can be done with such equations is to test for comi)atibiIity, or incompatibility. 2. Incompatible Redundant Equations are such as are not satisfied by the values of the unknowns obtained from the other equations of the set to which they belong. Thus, in (l)2a; + 3y = 23(2)4x-5y = -9(3) llx + 17y = 6, (1) and (2) give r = 4, y = 5 ; (2) and (3) give a; = — 1, // = 1 ; (1) and (3) give x = 373, y = - 241. SECTION III. \koativk and Inconsistent Solutions in Pkoblkms ixvolv- ixo SiMiM.K Equations. 252. Problems involving Arithmetical Inconsistencies. 1. If from f of a certain number 1 be subtracted, the re- mainder equals the sum of twice the number divided by 7, 246 TEXT-BOOK OF ALGEBRA. five times the number divided by 14 and 3. What is the number ? |c.-l = ^'^ + ff + 3 (219,1) a; = - 224 (219, 2) Here the negative answer points to some absurdity in the problem viewed as an arithmetical one. On examination we find that — -\ is greater than — ;-, and should be di- minished rather than increased to produce — . 2. A father's age is 40 years ; his son's age is 13 ; in how many years will the age of the father be four times that of the son ? 40-f ic = 4(13 + a;) (219,1) X = — 4 (219, 2) In this question the negative answer shows that the tacit assumption that the epoch named would be in the future was wrong. The question should have read " how long ago," and the equation have been 40 — a? = 4 (13 — x), whence cc = + 4. 3. A and B went into business agreeing to divide the profits in a certain way. Twice B's money diminished by $4250 indicated A's financial standing in the company when they began. A made $4000 and B made $750 when it was found that A had $500 more than B. How much had each when they began ? (1) X + 4250 = 2 2/ (2) X + 4000 = 7/ + 750 + 500 (219, 1) a; = — 1250 = A's, and y = 1500 = B's (219, 2) The result indicates that A was in debt when he first be^jan business. SIMPLE EQUATIONS. 247 4. A man worked 7 days and had his son with him 3 days, and received as jjay $2.20. He afterwards worked 5 days and hiKl his sou with him one day, and received for wages $1.80. What was the father's daily wages, and what was tlie effect of the son's presence ? (1) 7a; + 3y = 220 (2) 5x + f/= ISO a; = 40 cts. father's wages, y = 20 cts. son's expense, i.e., the father paid out 20 cts. a day for the sou. 5. A and B travel in the same direction at the rate of 6 mi. and 4 mi. respectively per hour. A arrives at a certain place P at a certain time, and at tlie end of 8 hours from that time B arrives at a certain place Q. Find when A and B meet. P Q R H 1 1 1. Su[)pose the distance P Q 50 miles. Let X = the number of hours from the time A is at P, till they meet at R. Then since A travels at the rate of miles per hour, the distance P R is 6 x miles. Also B goes ov(»r the distance Q R in x — 8 hours, so that Q R equals 4 (x - 8) miles. Now P R = P Q + Q R. Hence Gx = r)() + 4(a; — 8) x = 9 (219, 2) 2. But if the distance P Q is 20 miles the equation would be 6j- = 2() + 4(x-8) Wlience a: = — G. The negative value of x indicates that they met to the left of P instead of at the riglit. It is plain that in 8 hours B would walk 32 miles, a number greater than 20. Con- sequently they met before A reached P. 248 " TEXT-BOOK OF ALGEBRA. 3. N^ext suppose A travels 4 miles, and B G miles per hour, and suppose P Q = 50 again. Then 4 X = 50 + 6 (x — 8) In 8 hours B travels 48 miles. He would therefore be just 2 miles beyond P at the time A arrived there. They liad met one hour before A arrived at P. 4. Lastly suppose A and B travel in as 3, but that P Q is 20 miles. 4.x = 20 + G(:r — 8) .-. :r = 14 Cases (1 and (4 are in accord witli the idea of meeting in the future as implied in the statement, while (2 and (3 con- tradict this supposition. 6. A grocer has twO barrels of molasses, one of whicli contains twice as much as the other. From the larger cask he draws 16 gals, and from the smaller 10 gals. Then after a fourth of what remained in the larger cask had been witli- drawn, the two casks were found to contain an equal num- ber of gallons. How much did each cask hold ? Let 2 X = tlie number of gals, in the larger and X = '• '' " " " " smaller. Then f (2 x — 10) = x — 10 x = 4 2a; = 8 Here the answers are positive, but that does not save us from the absurdity of drawing 16 gals, from an 8 gal. cask, and 10 gals, from a 4 gal. cask ! Verifying the equation, 3 x — 8 = — 6. Thus we learn that the existence of the minus quantities in the process of solution vitiates the result for the arithmetical problem. 7. Ex. 33, Art. 224, illustrates the point brought out in Ex. 6 still further. By the statement of that problem we SIMPLK FX^UATrONS. '240 are led to suppose that the amount invested at G ^ is only a part of the $12000. The solution shows that the man borrowed $132(K), on part of which he paid 4 ^ and the other pai-t r> '/i . He got 6 % on the whole $25200. Six per cent on $\'J(M)() amounts to only $720'. 253. The Problems of the preceding Article illustrate, 1. That a iiegative result indicates either some arithmeti- cal incongruity in the statement of the ])roblem, or at least that the number obtained as the answer should be tiiken in a sense conti-ary to that implied in the statement of the l)roblem. 2. That positive results do not iipr'«>ssarily ])rove that l)roblems ai*e satisfactorily soK.d. Ilcuce all ])roblenis which do not admit of algebraic, interpretation (i.e., tliose which do not admit of both positive and negative values for their quantities), when solved by algebraic methods should have their answers tested for arithmetical ccmsistency. SKCTION IV. LiTKKAi, I'lMuu.KM"^. — Generalization. 254. The Generalization of a Problem is attained by repla- cing the known iuinil)ers by lrtiri> and then deriving its solution. (I. Problems are ronimonly first stiiditd liy usin^ particular num- bers; afterwards, on account of the freciucncy witli wlil<'h they occur, they are generalized. After such a sohition has been worked out, all that is necessary afterwards is to substitute the numbers of any particular case of the problem in the answer of the generalized solution, and reduce as in 81. 255. Exercise in the Solution of Literal Problems and Gen- eralization. ii. In some of the first problems, and also those difficult to state, reference to a particular form of tli<- umxcii jnohlem, or one similar 250 TEXT-BOOK OF ALGEBRA. to it, will be made. The student will find these references very helpful if aid is needed in the statement. 6. The following set of problems includes such as contain some- times one, at other times more unknown quantities. The student is left to decide to which class any given problem ought to be assigned. 1. (1. Particular form. — Divide |>183 between two men so that ^ of what the first receives shall equal ^\ of what the second receives. i^~x = j% (183 - x) (219, 1) cc == 63 ; 183 — x = 120 (219, 2) (2. Generalized form. — Divide $a between two men so that - of what the first receives shall be equal to ^ of what the second receives. --x = L(a-x) (219,1) np -f- mq ^ ' ^ (3. Special case. Solution by substitution. — Divide 168 into two parts so that -§ of one shall equal {^ of the other. Here a = 168, m = 5, n = 9, p = 5, q = 12. Hence ^ = o"^v!^rc^^r^^io = ^^- ^ns. 168 - 72 = 96. Ans. 9 X 5 -f 5 X 12 Verification. § of 72 = f>^ of 96. 2. A boy bought an equal number of apples, lemons, and oranges for c cts. ; for the apples he gave I cts., for the lemons m cts., and for the oranges n cts. apiece. How many of each did he purchase ? See Ex. 4, 224. 3. A father is now a years of age, and his daughter b years of age. How many years ago was the father's age n times that of the daughter ? See ex. 9, 224. Obtain the answer in the following cases by substitution. SIMl'LE EQUATIONS. 251 a = 51, /> = 24, n = 2\ ; ti = 75, h = 19, 7i = 5 ; a = ;]7, h = 12, Ji = ;i. 4. The sum of two miiuUt'is is .v, ami their (lirt'crt'iicc • and X — 1/ ^ d X — t/ = d then 2 ar =s-\-d 2 // = .s- — rf (Ax. a) S -f- f / s — ' ? 17. Two bodies move towards each other from points I meters apart. The one p meters a minute, the other q meters. In how many minutes will they be r meters from each other. I*ut / = 500, j) = 30, fj = 2r>, ;• = 00. 18. The fore wheel of a wagon is a feet and the hind wheel b feet in circumference. Through what distance must the wagon pass in order that the fore wheel shall have made n more revolutions than the hind wheel ? 19. A person distributed a cents among n beggars, giving b cents to some and c cents to the others. How many were there of each ? 1.) A father divides $8500 among 7 chihlren, giving to each son $1750, and to each daughter $500. How many of his children were sons and how many daughters? 20. A, B, and C hold a ])asture in common for which they l)ay q dollars a year. A puts in a cows for m months, B, b cows for n months, C, c cows for p months. Required each one's share of the rent. 1.) q = $181.20, a = 6, ^ = 5, c = 8, m = 30, 7j = 40, JO = 28. 20'. A boy who had three studies, Latin, Greek, and mathematics, received at the end ot a certain term 95 in 254 TEXT-BOOK OF ALGEBRA. mathematics, 90 in Latin, and 85 in Greek. The mathe- matics recited 5 times a week, the Latin 4 times, and the Greek 3 times. What was his average for the term in all his studies ? To make a convenient formula for calculating an average grade a*, let a be an approximate value of the average grade (as 90, or 80, or 60), and let Z, w, n be the differences (taken with proper signs) between the given grades and the assumed average grade. Also let ^, c, and d be the number of hours per week the respective studies recite. Then X = (^-^ + ^ + ((f' + fii^) e -{- (a-\- n) d __ ^^ . lb -^mc -\- nd h -^c-{- d '^ h j^c^ d which expression is more convenient for calculating the value of X. 21. To derive the rule for multi[)lication of fractions. a c Let y and -. be the factors and x their product. Then x = ~fX-j h d a e hdx = j^-bX~^-d (Ax. .3 and 38, 2) = a X c (by definition of division, 43) (Ax. 4) a c ac ^^^' h^d = M Hence, to multiply fractions multiply their numerators for a new numerator, and their denominators for a new de- nominator. The result is then to be reduced to its lowest terms, by striking out equal factors in the numerator and denominator. To save unneQessary writing this cancellation is done before and not after the terms are multiplied. S!.MIMJ<: KQUATIONS. 255 22. 'l"u UfiiM' ihf nilt^ i'or divisiuu oi iiiuLion>. [a-I __ ami — be the tractions and x their quotient. Then b (I (I c c ^ d f a c\ e d But l)v tlic (l(»tiuiti()n of division ( - -^ -]X , = , , and \0 a J d c d by the preceding ^ X 7 = 1 . Then a d , V (t r " ^d ad Hence, to divide one fraction by another, iiivtTl tlui divisor and multiply. Kkm AKK. — It was thought that the student would be able to understand these proofs better after studying equations. They might have been given just as they stand in articles 178 and 180. They include as particular cases the principles of division often given : ,,.... , S dividend ,. . ,. , S divisor Multiplying the J ^^^^^^ator «^ ^^'^'^'^^ the ^ ,,,.„,,,„i„ .... , S quotient multiplies the J ^,^,^^^ ^^ ^,^^ ^^^^.^^^ . , S dividend , . , . , S <^i Dividing the ^ n„,„erator ^' "»"ItM>ly.ni: tlu- J ^,^, itor divisor 'nominator divides the fraction. THIRD GENERAL SUBJECT. (THE NOTATION CONCLUDED.) POWERS, ROOTS, AND RADICALS. CHAPTER XVII. OF POWEU8. 256. Involution as a term in algebra signifies raising qnantities to powers. a. Involution is merely one case in multiplication (56, 67) where the numbers multiplied are equal. And so some exercises in involu- tion have been given under multiplication (117). For present pur- poses it will be convenient to treat this subject under three heads : monomial powers, binomial powers, and polynomial powers of three or more terms. SECTION I. —Monomial Powers. 257. Exercise in Involving Monomials to Powers. — The student should familiarize himself anew with 117 and 128. a. To develop a fraction both terms nuist be raised to the proper power. (178.) 4 7 a"* 3^V' 2~J 1. Square «'V, 11 Ire', — 4 a%^x', — | trx'^, — o— g- ""'h'% (^l~~j , on-\ 2-^ • 2-^ (3|)-^ (- a'xy, (- af (- by 93 9 (- C)\ ^5, % a«j"«+-; yP-\ yU-X ym^l 256 POWEKS, HOOTS. AND RADICALS. 257 2. CiiIk', — «•, a'\ h^if, n^^ X a- -- e;:j a) NoTK. — 111 tlio last (!Xi>ressiou it is not known whether lu is oven or odd, and, consequently, wliether the result is + or — . In such cases (— 1) "• may be written as the si(jn coefficient. 4. Kaise — ((ibr)'^ to the Hftli i)o\\t'r: niise — crb^c^ to the With power, when m is even. 5. Find the cube of (— >i/*\-'-} (— ^/•^^V") ; the square of 4a*b*c (— xy.-v*) ; the fifth \n)\\(tv of — "^ x" , the cube of — -! ; the seventli power of abc^ ; the sixth power of i> b~^ (?)'{f)" 6. Simplify 2 a (—3 ft2,/iwy; (4a^")'; o (W)" ; a{5nf\ 7 (7 pqh-y ; 3 ( - 2 < (" 1)"" ' "8 •" '^ "^ 2rV/Y .•fa/icj SECTION II. — Binomial Powers. 258. Investigation of Binomial Powers. — xV study of the development of binomials leads to Xewton's theorem. The development of the square of a binomial was studied in Thr'nnMns [. and IT. in nmlti]>lifation ; then the other 258 TEXT-BOOK OF ALGEBRA. simple cases of the cube and the fourth power of a binomial in 116, 4 and 6. But each of these led to a special theorem. We are now to seek for a general law applicable alike to all powers. Let A and B represent any two quantities. Then A -|- B or A — B is a binomial. We proceed to form the powers by multiplication. A +B A +B A^ + AB + AB + B^ A2 + 2 AB + B2 A + B A^ + 2 A^B + AB-^ + A^B + 2 AB^^ + B^ A« -h 3 A43T3 AB^ + B8 A + B A^ + 3 A^B + 3 A^B^ + AW + A^B + 3 A^B^ + 3 A B« -f B ^ A* + 4 A^B + 6 A^B^ + 4 AB^ + B* A + B A« + 4 A*B + 6 A«B2 + 4 A^B« + AB^ + A^B + 4 A^B'^ + 6 A^B^ + 4 Al^^ + B^ A*5 + 5A^B + 10A3B2 + 10A=^B« + 5AB* + B5 = (A + B)s A -B A -B A^^^^TB - AB + B^ A^ — 2 AB 4- B2 A — B A3 — 2 A'^B + AB^ - A^B + 2 AB^ - B« A3 — 3 A^B + 3 AB^ — B^ = (A — B^^ A - B A* — 3 A«B + 3 A'^B' — AB^ - A^B + 3 A^B^ - 3 AB« + B^ A^ — 4 A^BTTA^B^ — 4 AB^ + B^ POWERS, ROOTS. AM) KADK \LS. 259 A careful sciutiny of tliese iiiiiltiplicatioiis leads us to the following eonclusions. whicli if true in all the above cases we may guess to Ix- tun d all powers. The (h'nn^n- sf ration of Newton's theorem which is rather too difficult to be inserted here does justify the conclusions about to be given. In tlie development of a binomial, as (a-\-b)"y we learn 1. The number of terms is always one greater than the exponent of the power to which the binomial is raised. Thus there are five terms in the fourth jwwer. 2. The exj)onent of the leading letter in the first term of any jK)wer is the same as the exponent of the })ower of the binomial. In the second term it is one less, in the third term one less than in the second term, and so on, the lead- ing letter not appearing in the last term. On the other hand the other letter begins in the second term with the exi)onent 1 which increases by unity each time, until in the last term it is the same as that of the power of the binomial. Thus the product is symmetrical with respect to the letters. 3. When the binomial is a sum the signs of all the terms are ]X)sitive. "But when the binomial is a residual, every term which contains an odd power of the second letter is minus. 4. llu' coefficient of the iirst t* mi is 1 (understood), as also that of the last term. The coefficient of the second term is the same as the exi)onent of the power of the bi- nomial, as also that of the next to the last term. Further- more, the coefficients increase to the middle term, or terms, and then diminish to the last, those equally distant from the extreme terms l>eing ec^ual to each other. 5. The third Coefficient can always be derived from the second by multiplying the second by 1 less than itself and 260 TEXT-BOOK OF ALGEBRA. dividing the product by 2. The fourth coefficient may be derived from the third by multiplying it by 1 less than the previous multiplier, and dividing by 1 greater than the previous divisor (i.e. by 3), and so on, the multiplier be- coming 1 less and the divisor 1 greater than the last for every new term. Thus the coefficient 6 in the third term of (A -f- ^y is derived from the 4 by multiplying 4 by 4 — 1 = 3j and dividing by 2 ; the coefficient 4 following the 6, by multiplying 6 by 2 (1 less than 3, the previous multiplier) and dividing by 3 (1 greater than 2, the previous divisor). The last coefficient, 1, from the preceding coefficient, 4, by multiplying it by 1, and dividing the product by 4. Unlike the preceding laws, the fifth is probably too com- plex for the student to have perceived it unaided. a. There is a memoriter rule which can be more easily followed when A and B stand for two first powers : — Multiply each coeffi- cient by the exponent of the leading letter in that term and divide the product by the number of that term from the left. The quo- tient will be the coefficient of the next term. Thus in the fifth term of (A + B)5, 5 AB*, the 5 is found from 10 A^B^ by multiplying 10 by 2, the exponent of A, and dividing the product by 4, the number of the term 10 A-^B=^ from the beginning. It may be observed that the divisor is always 1 greater than the exponent of the second letter. b. The student will find it advantageous to remember the bino- mial coefficients up to the fifth or sixth powers. 259. Examples of the Use of the Binomial Theorem. 1. Develop (A -|- B)^ by the theorem. In this expansion there will be 6 terms, and hence when three coefficients are found the others can be written down directly, being those already obtained in reverse order. The reasons in parentheses refer to sections in the previous article. POWKHS. KOOTS, AM) i;\l>l("AI>S. 261 (4,-. (4,2) ^ ,... ura, .; (4,2) (A -h ])f = \' -h r> An5 -f '^"-^ A«B'^ + 10 A'^B» (4, -'J l4. I.') -f 5 AB^+ li^ 2. Develop (x — y)®. Here there will be 9 tei-ms in all, and it will be necessary to determine 5 coefficients. (.1,4) (5, or a) (3,4) (^ _ ,/). = ..* _ 8 ^'y + ^ (= 28) ^y - ?5^ (3.j)ro) (4) (= 56) xY + '^^^ (= 70) j'Y - 5(> :ry + 28 xY - ' 8a-y + / 3. Develop (2 x'^ — 3 (///)*. Here A = 2a;^, and !> .'I'/y. Fo preserve their iden- tity they are written in parentheses, treating (2 x^) as A, and (.3 «y) as H. 4X3 (2 «* - 3 at/y = (2 x'Y - 4 (2 j--^)« (3 «y) + -^— (2 x^y (3 ay)^ - ^' 3- (2 o-'^ (3 «y)« + GJ ^^V)* The single terms must now be simplified. (2 xY = 10 .rs . 4 (2 X'') » (3 .^y) = 1)0 «.rV ; -^ (2 a;^« (3ay)''^=216«*V/ 5A? (2 ««) (3 fly)" = 21 (; ^/ 1/ -//' : ( .; ^/y)* = 81 ay. Hence (2 J-- - 3 «y)*~16 x« - m ux''i/ + 216 a^y - 216 a»«y -f 81 aY- Note. — The student must remember to keep distinct the bino- * mini coefficients, and any numbers that may come into the result from numeriral coefficients in tlie quantities developetl. We could never, for instance, obtain the third coefficient 210 in the expansion just given by multiplying 90 (!) by «.') and dividing by 2. For the binomial coefficient of that term is not 00 but 4. 262 TEXT-BOOK OF ALGEBRA. 260. Exercise in Raising Binomials to Powers. 1. (J99. - ^'Y- 10. (2 r — 6 my. 2. {2 m— pf. 11. (la-?, by. 3. (x'-^4.,fy- 12. (5 ic5 _ 4 ^4^); 4. {5 a — bey. 13. (a - ly. ■ 5. (2 ^2 + axy. 14. (§ ^ - fl yy- 6. (1-2 by. 15. ix'- - 2 xi/y. 7. (a + iy. 16. 8. ( 2^V 17. {x^ - 2y. 9. io-^xy. 18. {x-^ - Iff. 19. Find the first four terms of (2 am -\- ly. 20. Verify 99^ = (100 — 1)^ 21. Cube by the theorem 999 = (1000 — 1). 22. Raise 9999 to the fourth power. 23. Raise 12^ = 12 + i to the fourth power. 24. Raise 1892 = (1900 — 8) to the third jxnver. SECTION III. — Polynomial Powers. 261. Development of Polynomial Powers. — rolynomials are most readily developed by regarding them as Binomials, and using Newton's theorem. a. Polynomial squares are easily written out by the rule given in 116, 3. 1. Required to cube a-\-b-\-c. To make a-j-b -\- c a binomial it is written {a-{-b) -]- c. Next the binomial thus formed is developed, and afterwards the different powers oi a-\-b are expanded. Last of all the resulting indicated opei-ations are performed. POWEHS, tcfM.i>. A Mi KADICAL8. 263 By section II. we have [(« + b) + cf = (a + by + :; i '/ + h)-r + o yn + h)c-^ -f c* = a"* + 3 aV, + 3 ab'^ -{■ h^ + :'» -.' ' + i' ab + b'^)r. + 3 (a + 6)c2 + c» = ^3 + 3 (I'b + 3 «6- + /;•» + ;i r/-r + r. f/^f + 3 b-^r + 3 ar-2 + 3 bc^ + c« ^1 tlH. = a-^ + /*■» + 6-84- 3 (a^/) 4- a-r + ab'^ + ac- + b-r + ^f-^) + 6 rr^r. State this result in the form of a theorem. 2. Required the fourth power of 'An -|- 2 A — r -f . [(3a + '2b) - /<• - ^\ J* ^ (3a + 2 />)* - 4 (3 a + 'J b)^ U- |\ + (J (3 a + 2 />)-• ^- - '-'"j '- 4 (3 a -\-2b) (''- tY = (3 a)* + 4 (3 a)3 (2 6) + 6 (3 a )2 (2 by + 4 (3 a) (2 by 4- (2 b)* - 4 [(3 ay + 3 (3 ay (2 6) + :; (:5 (0 (-^ ^)*-^ + (-' '^)' ] [^ " ^ ] + C. fna-i + 12 rt/> + 4 />-'"] fc-^ - C(I 4- ^1 -4[3a + 2/>] [c^-3r:^ 2 + -^ ^" T " S J rpoii developing tli«' iii<»n<>iiii;il txpi. -^i(.ll■^ and ix'rforniinii the imdtiplications. [:; a f 2 /» - /• -f I) - 81 a^ + 2l»; a'/, 4- 2ir, a-7/-! 4- •.»«; afri + ir. M - 108 a-^r - 2 Hi a-^r - 144 at^r - 32 />^r + 54 a^d + 108 a-bd + 72 a^tZ 4- K^ Ifld + 54 rt-^f'-J 4- 72 abr^ + 24 fi^r^ - 54 a^rvi _ -JO ,,}„',! - 24 //-i^-f? 4- — «2'''- + 1>< "/"'-' + <•' ''■-'?■- - 12 ac» 4- ]>«> (/'-■'/ — 1* ard^ + -^ ar/i — .s /k-- -t- i:.' hr-d — tj ftccZ^ 4- biV' 4- r* - 2r«/Z 4- :; ^-'?- - :; rfZ» 4- -^ ■ 264 TEXT-BOOK OF ALGEBRA. 3. Square xy + yz -\- zx. 4. Develop (1 + x + x' + x^. 5. Cube a -\- b — r. 6. Develop (1 — a — a^)^ 7. Develop (1 — 2 x -\- x^. 8. (2 rt»* - b'' -h cr^y = ? 9. Cube ax + />// + 1. 10. Develop (a -^ 2 h — c)*. 11. Develop (1 -{- x + x^. 12. Develop (ax + /;// + 1)^ 13. (x-{-y-{-iy = ? 14. Cube 1799 = 1000 -f- 800 — 1„ and verify the answei I'oWl.KS. iMtoTS, AN1> 1! A 1 >I< "A I.S. 265 CHAPTER XVIII. OF EXACT ROOTS. 262. Evolution iis ;i tt'ini in algebra signifies extracting the roots of quantities. e root of polynomials; other roots of polyiiomials. SECTION I. — Monomial Roots. 263. Roots of Monomials. — As in simple multiplication and division, there are three thinj^s to be considered : the coefticient. the sign, and the literal ])art. Consult articles 45. 46. and 129. See also 58, 59, 60. 264. Signs of Roots. (45.) The reason for the existence of two scpiare roots so different in character may at first puzzle the beginner. It is a consequence, however, of the rule of multiplication that minus by minus as well as plus by plus gives ])lus. I*n)p('r1y s])eaking, the sign -j- slionld l)e read " jdus ami minus." inst-ad of " plus or minus.'" 1 1, now, it is known that a power has been produced by multi- plying two positive numbers, then its root is positive; or, if by two negative factors, it is negative. In tlie solution of arithmetical iiroblems. the nature of the i)roblem may admit of only positive aiisuiMs: tln'ii only the positive root is taken for the answer. 265. Imaginary Quantities. — It can be shown that a real meaning may be assigned to what have been called '' imagi- 266 TEXT-BOOK OF AUilCBllA. nary," or "impossi})le " quantities (45). But it is only by going out of the realm of algebra proper that it can be done. Whenever an imaginary value is found for the un- known in a problem, it shows that the problem is algebra- ically impossible, just as a negative result shows its problem to be arithmetically impossible. 266. Exercise in Extracting the Real Roots of Monomials. 1- ^a%\ -, AND KADK ALS. 207 mial theorem would serve to develop the powers of any polynomial, so here the investigation of hinomial roots will include the general case. 1. Required to extract iIk Sipiare root of A^ J:; 2 AB + B'^. Referring to 258 we see that when a power Is arranged the root of its first term is the tirst term in the root. Hence, in extracting roots in order to make a heginning, the quantity is arranged with refer- ence to the exponents of its leading letter, and the root of its first term is then extracted. Explanation. — The first term of A2 + 2 AB + IP (A + B ^j^g ^^^^ ^ having been found, as just „ 1 -- explained, it remains to find the otlier 2 A + B I 2 AB + B-^ ^^^^^^ ^^ ^^^^ ^.^^j^. ^^^^^j ^j^^^^ j^., j^ ^j^^ ' square of A alone, A is squared and subtracted from the polynomial, leaving 2 AB + B^ = (2 A + B)B. Of course, in this example, w know perfectly well that the square root of vV^ ± 2 AB + B-^ is A ± li. (See the tlieorems in mul- tiplication.) Hence B is the second term of the root sought. Man- ifestly, now, if 2 AB is divided by 2 A (double the root term alrea«ly found), B is the quotient. Similarly — 2 AB ^ 2 A = — B, which is the second term of the root in this case. Lastly we observe that if B be now annexed to 2 A and the sum multiplied by B, the product, 2 AB + B^^^ subtracted from the previ- ous remainder leaves zero, and the process is complete. To show how this method applies wlien there are more than two terms in the root, let us take the s(|uart' of a + h + c = a^ + 2ab + //- + 2 ac +2 he + cr which can \n- w i itten (a + 6)2 + 2 (« + 6) r + '-. Thus, when the first two t. iiii>. d + 6, are given, the third can 1m' found l)y dividing tin- lir>t tnni of the remainder, 2 (r/ + />) r, by twice the root already found. Now the root terms, a + h, must be found from the first terms of the polynomial in a preliminary ojieration similar to that given above for A + B, and its S(iuare sub- tracted leaves the remainder 2 (« + 6) r + c'-. If there are still other tenns in the root, t6 find each succeeding term all of those already found are considered as one quantity. To better illustrate this, we give another example: 2. Required to extract the square root of Ax"^ + n'^if -*- 9 .t;* — 4 axil + 1 2 -rz^ — i\ oi/x^. 268 TEXT-BOOK OF ALGEBRA. Arranging this with reference to the powers of x (and y and 2), we have 4: x^ — 4c axy + 12 xz'^ + a^y- — ayz- + 9 z^ ((2 x — ay) + 3 z'^ 4x^ 4x — ay — 4 axy + 12 xz^ + ahj^ — 4 axy + ary'^ 4x — 2ay + Sz^ 12 xz^ — 6 ayz'^ + 92* 12 xz^ - 6 ayz^ + 9 2* In this problem after two terms of the root are secured, they are considered as one quantity and doubled for tlie trial divisor. Divid- ing we get the third term of the root, which, when found, is an- nexed to the others with its proper sign. And so in all cases, the root already found is treated as one quantity. 268. Rule for Extracting the Square Root of Polynomials. 1. Arrange the polynomial with reference to its leading letter, extract the square root of the first term, and subtract its square from the polynomial. 2. Divide the first term of the remainder by double the root quantity, and the quotient will be the second term of the root, which is written at the right of the previous or trial divisor, as well as in the root. This complete divisor is then multiplied by the root term just found and the pro- duct taken from the remainder. 3. Double the two terms of the root for the next trial divisor and annex to it the new root term, when found, be- fore multiplying. Continue the operation until there is no remainder, or as far as desired. 269. Exercise in Extracting the Square Root of Polynomials. 1. W - 2 AB + A2 ( = A^ - 2 AB + B"^). The root is B — A, which differs in sign only from A — B, the root previously obtained. Thus, B — A = — (A — B). (See 45, 4, (2.) Moreover, this is true of all polynomial square roots. If a polynomial is a square root of a quantity, the same polynomial with all of its signs changed is also a root. POWKItS, i:()(/rs, AND RADICALS. JOU 2. /* + r)/V 4- \)j\ 4 ^9 4. fi-^ - 2 + .) '■('+:)'-('+'-' 10. 1 -2,v + -^^-^«-|-^'. 11. ./•- - 2 ./• 4- 1 + 2 .,•// - 2 // -f //■-. 12. .r(.r-f-l)(.r-h 2 )(./• + ;;) + 1. 13. (13 r-^)- -f- (4 .r')- + (7 .!•)- + 2H> .r' - 12(Lr^ 14. Obtain four terms of ^/„i _j_ y\ n^^' bn^ b'-n'^~'' n"^'- hit'' tr ' \?~ "•" G^ + ~T44' + ./ + 12 + 4' 270. Extraction of the Square Root of Arithmetical Numbers as Polynomials. — Tlic lulc for tlu* extraction of iiuiiil)'is ('written in the Arabic notation) is a sjiecialized form of tlmt for i)olynomial.s. We jiroceed to illustrate this by e.xamples. 1. Required to extriu-t the square loot of :{44.")(;<.). We write the number as a ]>olynomial whose terms li;i\ <• resj)ectivrly an even number of ciphers. 270 TEXT-BOOK OF ALGEBRA, ^40000 + 4500 + 69 |500 + 80 + 7 250000 1000 + 80 (1080) 90000 + 4500 =- 94500 8()400 1160 + 7 (1167) 8100 + 69 = 8169 8169 Explanation. — Starting at the decimal point the numher is separated into terms of two tigures each, annexing an even number of ciphers to fill out the vacant orders. Evidently two ciphers in the power will give one in the root, four in the power will give tw^o in the root, and so on. This enables us to get the first figure in the root. For the first term of this number polynomial Mill give the first term in the root. In the present example, the first figure of the root cannot be (> for 600 squared is 360,000, which is greater than the given number 344569. Taking 500 as the greatest number of hundreds and subtracting its square from the first term gives 9000, which added to the next term makes 94500. Doubling the root found, 500, the trial divisor is found to be 1000. This is contained in the remainder 90 times; but upon add- ing 90 to the trial divisor and multiplying by 90 the product is 98100, which is too great. Hence eighty is the second term of the root instead of 90. Writing 80 in the root and adding it to the trial divisor, we have 1080 for the complete divisor. Multiplying the complete divisor by the second term of the root and subtracting the product from 94500 the 'remainder is 8100, to which the last term of the polynomial, 69, is added. Doubling the root already found we have for a new trial divisor 1160, from which 7 is found, and the process is completed by adding 7 to the trial divisor and multiplying. The 500, of course, corre- sponds to A, and 80 to B in the first operation ; then 580 to A and 7 to B in the second. The process here explained succeeds in separating the number into the terms of a polynomial, which is a perfect square. Thus, 344509 = 250000 + 86400 + 8169 = (500 + 87)-^. 2. We next solve the same problem, abrldyirig the work as much as j^ossible. roWKKS, ROOTS, AND llADKALS. liT 1 Explanation. — The miiuber is separated •A 4o o9( o87 juj^j periods of two figures each for the reason "*^ .. given above. Having found the first figure of j^jjlj the root from tlie first periocl, it is square. I ml 11<57 )SUi{) subtracted, and tlie next iH»riod of two Hgures 8101) is annexed to the remainder. Then the first figure of the root is doubled for a trial divisor, anil divided into 045, or rather 1)4. It is contained, as we saw above, not 9 but 8 times. This figure 8 is written afti-r 5 in the root, and also annexed to the trial divisor. After multiplying by 8 and sub- tracting, the next i>eriod is brought down, and so on. A careful comparison of this solution with the precetling will make the whole process plain. The difference is that in the second solution all the ciphers are omitted. 2. Extract the square mnf ti\' tin- (Icciniiil fi-action l-.\ ri.A.N A I lo.N. — Here as bet'ore .00W02'70( .01904 + ,i„. „,„„her is divided off into periods ^« r^A/^ of two figures each commencing at the 29 ).000202 , . , . , ,„, -. , -. , ,, ^ 261 decimal point. Ihe first figure of the 3804 UTCPOO ^^^^^ having been found, it is s<^ »*^xt period is brought down. The first figure 1 is then doubled for a trial divisor and the process is continued. A cipher is annexed to 7 to make its period full. Thereafter two new ciphers would be used for each new ])eriod. The second trial divisor :J8 not being contained in IT. ;i cipher is i)laced in the root and also at the right of the trial divisnr. after which a new period of two ciphers is annexed to tlic rrniainder. The trial divisor is then contained 4 times. 271. Rule for Extracting the Square Root of Numbers written in the Arabic Notation. 1. ('onnueiH'in*; at tlw^ dccinuil iM)iiit divide tlie niimher otf into i>eriods of two figures eacdi. 2 Find the greatest square in tlic left hand peri -|- l'()\\i;i;s. i;n..r<. wi. i; \ i »i('.\ i.s. :i73 A' -h 3 A-li + '6 A13- -h J3 'C A + li Koot A8 3A- + :; AH + B^ )3 A2B + 3 AB-' 4- B" :; A^B + ;} AB-' + B=» Exi'LANATioN. —The first term of the root having been obtained by extract in ji; the cube root of tlie first term of the polynomial, it is eubed and the resnlt subtracted. The remainder is '4 \'^ B + 3 A B- + B'. By reference to 258 we know tliat A + B is the cui)e root of the polynomial, and that the second term of the root is B. In'ow we see that the second term of the root can be gotten by dividing the first term of the remainder by three times the square of the first term of the root already found, or, 3 A'-^ B by 3 A'^. If to the 3 A-, which becomes a trial divisor for the next term of the root, we now annex 3 A B + B=^, i.e., three times the first by the second, plus the s<|uare of the second, and multiply the sum by the second term B, ui)on subtracting, nothing remains, and tlie operation is complete. It <'an Im' show n, as in 267, that the binomial solution is of general ai>plication, and that when the root is to consist of three or more terms, we first find two and then proceed in the next operation as if they were one quantity, and so on. 2. Extract the cube root of S ./• - :)(> ,/• ' + 1 14 .r« — 207 x=' + L'ST) .r^ -225j + 12.-). t>jc-- ;w X" ■ + 11-1 .c -*.. ., , / - rjr) x + 125 \'Zx--iix + i 8*» 12x< — 18x3 + «x» |-»J«3 + IHx* -'J07x I -'• Wa;-'4-54x<-2r a<' 12x»-36x« + 27x» 30ar»-46x -»-25 36x1 + 57 x«-45x + 26 60 X* - 180x3 + 285 x2 - 226 x + 125 60 x< - 180 x» + 286 x» - 22ft X + 126 Explanation. — The 12 x« = 3 (2 x^) 2 ; the - 18 x» = 3 x -3j; X 2x2; 9^2 ^ (lix)^ In tlie next operation, 12 x « — 30 x« + 27 x2 :::; 3 ( 2 x2 - 3 x )2 ; 30 X- - 45 X = 3 X 5 (2 x2 - 3 X) ; 25 = 52. 274. Rule for Extracting the Cube Root of Polynomials. 1. Arrange tlic polyiioiniul witli ivterence to the exixnuMits of the leading letters, extract the cube root of the first term, and subtract its cube from the polynomial. 274 TEXT-BOOK OF ALGEBKA. 2. Divide the first term of the remainder by 3 times the square of the root quantity, and the quotient will be the second term of the root. ISlow to the trial divisor is added 3 times the product of the first term of the root by the sec- ond, plus the square of the second (3 AB + B^), thus form- ing the complete divisor, which is multiplied by the second root term, and the product subtracted from the first remainder. 3. If this does not finish the operation, square the root quantity (of two terms) already found, and multiply by 3 for a trial divisor, and continue as before. 275. Exercise in Extracting the Cube Root of Polynomials. 1. ^a^ — maH^ 54 ab'' — 27 P. 2. Sa^ — S4.a^x-\r 294 ax' — 343 x\ 3. x^-{-3x^-{-6x^-i-7x^-{-6x-'-{-3x-\-l. 4. a« _ 3 a'h + 6 aW — 7 a^b^ + 6 a'~ b^ — 3 ab'' + b\ — 6 ahc. ^3 Q ^2 6. -,-'^+^xy-y\ y^ y 7. 8 ic^ — 60 x^ z" + 150 x^ %" — 125 z^. 8. 1 — 6 a;'" + 12 £C 2;« _ 8 ^ 3m_ 9. 1 — a? to three terms in the answer. 10. 60 6'2 x" + 48 cx^ — 27 c^ + 108 c^ a; — 90 c* iz;^ + 8 x'' — 80^3^:^ 276. Extraction of the Cube Root of Numbers in the Arabic Notation as Polynomials. 1. Required to extract the cube root of 158252632.929. 1.58000000 4- 2.52000 + 632. 4-. 929 | 500 + 40 -|- + .9 125000000 750000 60000 1600 33000000 + 252000 = 33252000 811600 ;:{:i4G4U0U 3(540)2 == 874800.00 1458.00 .81 876258.81 788000 632 788632 -f- .929 = 788632.929 788632.929 r<)\\Kl;>. IKMds. AM) KADICAI.S. Z<0 Explanation'. — Starting from the deciinal point the niunher is separated into parts of three figures eaeli annexing ciphers to fill out omitted orders. It is plain that one cipher in the root will give three in the power, or conversely, three in the power will give one in the root, six in the ix>wer will give two in the root, and so on. Seeking the greatest cube root in the first term, oOO is found, which is cubtnl and subtracted from that term. Squaring the root found and multi- plying by :j we get 750000 as a trial divisor. Dividing, the quotient 40 is set down as the second term of the root. As in the polynomial rule, taking three times the product of the first term of the root by the second, plus the square of the second, the results are (>00(X) an l: A I HCA I.S. Zii 280. Prime Roots other than the Square and Cube Roots. These may be iouiid by means of rules derived ironi inninda* in the same way that the rub's inrs(juare and eub( r^dt wiir found . For instance, the filtli root. (A + B)*= A^+(5A^-+ 1()A»1J-H l()A-r.--hr)Ar>»-f B^)B Here 5 A*, i.e., 5 times the fourth i>ower of the first term of the root, is the ti-ial divisor ; and to it must be added, for the complete divisor, 10 A»B + 10 A^B'-' + 5 AB» -f B* ; i.e., 10 times the cube of the first times the second, etc., etc. 281 Exercise in Extracting Roots other than the Square and Cube Roots. 1. Find the fourtli root of li'ya* — \Hi c^x -\- 'J\i\ ^•' -6X-I-1. 6. Find the eighth root of x« -|- H-70x*-h56x»-f 28x^-|-Sx-h 1. 7. Find the seventh root of 128 x' - 448 x« -(- (nli a* _ r>m r* -f 2s(> r^ _ 84 y- -f 14 r — 1. 278 TEXT-BOOK OF ALGEBRA. CHAPTER XIX. OF FRACTIONAL EXPONENTS. 282. Fractional Exponents result naturally from the law of exponents in the extraction of roots. By 129 each ex- ponent IS divided by the index of the root to be extracted. When therefore this index is not contained exactly m any exponent, a fractional power results. a. To illustrate. (See Ex. 266.) y/7iFb^ = a%^ = a^h"^', \/^ = '«" = c''; etc. If now instead of v^aW we had Va'^^, according to the same rule of division by the index of tlie root we should get Va'b^ = a^b^. 283. Meaning of the Terms in a Fractional Exponent The numerator of any fractional exponent evidently denotes the p)otver to which the quantity is to be raised while the denominator indicates the root to be taken. a. The expression ^a« means the ui'^ root of the n'* power of a. Hence we see that a number affected with a fractional exponent has a perfectly definite meaning, and performing both operations gives rise to a resulting number, the value of the expression. 284. Fundamental Principle Governing the use of Fractional Exponents, Any quantity affected with a fractional exponent may be separated into its factors, each factor taking the original fractional exponent. This principle holds for integral exponents, thus, (abe)"* = aH^'c"". The question arises does it hold for fractional exponents as well. To show this let us prove that the expression m m m (1) (ab)" =a^' X b\ row Kits, uooTS. \\i> i; A I lie A I, s. 279 Kaising botli ineiubers ot (1^ to the n"' power, since by definition the denominator n means the ?i'* root in each case, (2) {ah)'" = a'" X h'" = aH"" (Ax. o). The resulting equation (L'; is ])hiinly true; and conse- quently the members of equation (1) are equal. Moreover, it is easy to see that the same reasoning would apply gener- ally to any fractional exponents and to any number of factors. Hence the theorem : — Any root of a product is efjunl to the product of the like roots of the factors : and con- versely, the product of like roots of two or inore factors is t'ljuiil to tJie same root of thru- /nod net. (I. If a ninnlwr affected with a fractional exponent is to ]n\ evaluated it is immaterial whttlit r it )><■ t'nsi r .|niil»l.(l ili.- (|uantily is squared; but when the denominator is doubh-d a s(juare root must he taken in addition to the root denoted by the old index, and the two operations cancel each other. We may reasqn in like manner for any other factor by which both terms of the fractional index may be multiplied. 285. Fractional Exponents and Radical Signs — Kadical signs are used for the same jjurpose as fractional exponents, and the two notations are employed interchangeably. (See next chapter.) 280 TEXT-BOOK OF ALGEBIii^ It would he for in'oforahlr [f the radical sign notation were entirehj displaced by fractional exponents. As both are in constant use, both liave to be taught. 286. Object of Treatment of fractional exponents in the present cliapter. — Tliis object is twofold : 1. To show that the same rules apply to fractional ex- ponents as held for integral exponents. 2. To furnish the student exercise in the use of fractional exponents in all the simple operations. 287. Fractional Exponents and the Fundamental Operations. 1. Addition and subtra(;tion of quantities involving frac- tional exponents. Here it is plain that we can always add or subtract sitnilar quantities whether affected by integral or fractional exponents. Tluis. n a¥ -f () a¥ - 4 a¥ = 7 aW. 2. Multiplication and division of (piantities involving fraction:d ex})onents. Let us seek, e. g., the i)ro(hu't of x'' by .T^ x^ X x^ = x^ X .r« (284, b) x^ X x^ = (.T^)« X (x')^' (284, a) (.r^ X x')^ = (.r«)« ^ x^ (284) Now §, the new exponent, is the sum of the old expo- nents -^ and ^, the very process of determining it giving their sum. In other words, the old rule of adding the ex- ponents \w multi])lication holds for fractional as Avell as for integral ex})onents. Since division is the reverse of niulti])licati()n, the ex- ponent of tlic divisor is stdifracfcd from tliat of tlie dividend. I'oWKKS. lands, ANh l:.\l)I("ALS. 281 3. Fractional powers and roots of (piantities affected with fractional exponents. Let us take, e. g., («')*. («')' - {[(«')']'}* = {(«')'}♦= {«•)» = {(«')'}♦ = w^ = «.\(284,a) since the fifth root of tlie tliird root is, by definition, the fifteenth. Now ^ the new exponent is the product of the old exponents f^ and ^. Hence the old rule of multlphj'mg the exponent of the (quantity by the exi)onent of the power holds ioT fractional as well as, for integral exponents. Thus, generally, (a»)l = a'm -^ and conversely, flf, IW = {(I w ) « = (^< v) '» = (i, and 1>7^ 2. W n i hi- -f- 9 ^/ ^ hr' — 1 ;'> ii i hi-. 3. 15;//'y/* -f"'"^"'' — ni^ n^. 4. Find the sum of "1 a.v '"ij^ -\-\\ Uc — .\ " ~ V " ^ -|- *► /> ; \\as-'"ip^'lhr-\- " ., — I'A: ;in * — „\f,l^(\rifl. o^i,i _y,„hi,i_:>„\f,i. .,,,,1 _ 'j„l/,i^,.^,l 6. From tlH» sum of 5 ox^ — (.r -j- //)* -f- (/r — h)^ , — 7 ^m"^ + 2 (j- + y)i — ^ (a — h)^ , and 12 ././-^ — ;i (.r + //)i -f 12 (a — //)* take the sum of .'W/x* -|- 4 (.r -}- y)i -f- ("^^ — A)i and ^'.r« -(•'• + . '/'^ :!-•»(" -^:)*- 7. T(» T) .^r■- — 7 f,r -f S ;//i —'Jr->'. add :'. />/• — I '• " + 2 (iw'^. and then sul)tract |() iir' — ."i ./•// -j- .'J ///• — 12 r "". 282 TEXT-BOOK OF AIJJKiiUA. 8. From 9 c^ d"^ -{- 10 rj d"^ — 17 chl^ t^ke rJd^ — 20 c^d^ - 11 c'dK 9. Taking the even roots positively, simplify 16^ + 8^ + 16^ + 125^ — 512^ + 100«-^ — 810-". 10. Calculate, 36 1% 49 ^^ 4"=^% 8-^^, 9-<>-5 11. Multiply — 3 ?/n and 6 y^ ; also 9 a^ h''' and 2a^h'^. 12. Multiply together 7^ , 7% 7l 13. a^'d'^-a'-a^^^l a%^ ' a'^h ' a-^]/^ = 2 14. a-^ • .T~^ = ? m^ • m^ = ? ?i X M^ X ti^' Xn ~^ = ? 15. Multiply a^ -\- a,-'^ -\- a hy a-'^ — (t^. 16. (^tH-'c^^-^) -^(a^Z'-"c-i?^'^)=? 17. Multiply 2a-^ — 7x-h/—llrJ by r/a!?/-^^^. 18. Multiply x'^-\-'if by ic^ — yK 19. Multiply m^ + m^ ^^ + ?i^ by m^ — nK 20. Multiply a^ — Z»-3 by a^ — b. 21. Simplify (x^ X a: ^)'*. 22. Find the product of (^ ]Y^) and ( -^^ 23. Multiply a?^ — xy^ -\-x^y — y^ by a* -|- x^ y^ -j- y. 24. Multiply a^ -\-l/^ -\- a~^h\)j ah~^ — a^ + h^. 25. Divide 11 a^ — 33 a^ by 11 (A 26. I )i vide (^f '5* — /> '^ by a^ — h~ ". 27. I )ivi(le x'^ -{- a-^y^ -\- y^ by x^ ~\- xhf -{- y^. 28. 1 )i vide 5 :^" — 1 ^ ^ '" + 15 x'^y by o x^. 29. 1 )i vide x"^ — a;^ — .x-^ + 6 ^r — 2 cc^ by a;^ — 4 ic^ + 2. 30. Divide a — h ])y i;aih('A1,>. '2> a;— 5ri— 14 * a-*H-2 * «/> — ^* a^—b 284 TEXT-BOOK OF ALGEBRA. CHAPTER XX. RADICALS. 289. Radical Quantities or simply Radicals are quantities of which some root is to be extracted. They are expressed sometimes with radical signs, sometimes with fractional ex- ponents. (285.) a. When a root of a quantity can be found exactly the radical sign or fractional exponent disappears in the simplified form (272). It is only when the root cannot be found exactly that these signs have to be retained. As a consequence the subject of radicals deals almost exclusively with the reduction and use of quantities whose roots can- not be exactly found. 290. Radicals whose Roots cannot be exactly extracted are termed Irrational Quantities, or Surds, (the English term). Thus, ■\% ^/7^ etc. a. A fraction always expresses the ratio of one number to another, viz., that of the numerator to the denominator. Upon reduction to the decimal form it gives rise to either a finite or circulating decimal {i.e., one in which, sooner or later, sets of figures are repeated.) Now, the process of extracting the roots of numbers never gives rise to a circulating or repeating decimal, as the student can easily see with a little reflection. Consequently the latter can never exactly equal a fraction, and thus express a rafio. Hence the propriety in the use of the word In-ational. 291. The Treatment of Radicals embraces: 1. Kediu-tion of radical quantities. 2. Addition, subtraction, multiplication, and division of radical quantities. I'OWKUS, KooTS. AND KADU'ALS. 285 3. Eciuations containing radicals. a. It will often be convenient to solve, /»«»/< /m.^.-m, iIh- f»aiii»- t-wi- eise in tlie two notations. To indicate this brackets will be written at tlj»' left. SECTION I. —Reduction of Radicals. 292. Kinds of Reduction. — The ti-catnient will U' iiinler three heads. 1. Sini])litieati()n. 2. liediiction to surd form. 3. Reduction to a coniinon index. 1. - SIMPLIFICATIONS. 293. Reduction of Radicals to equivalent ones having a Lower Index, or to Rational Quantities. 1. Simplify (4f/2)4 ^ ^4^,, (187. /.) { (4 ./•-)i = (( 4 a')^ )* ■= (2 «)i (287, 3, and 129) ( \^4 «-^ = y/ V4 a' = V2 a (^et- 59. // ) 2. Simplify (1) u*ir)^ = -^i) aV/^ ( (9 a*b'')^ = ((9 «V/2)* )* = Qi a%y (287. :;. and 129) ( y/^a*f^ = y/ V9 a'b^ = "v"?"^ 3. Simi)lify (— 8 m«7t«)* = ^—8 vihi'^ ^ (_ 8 ?nV)* = ((- 8 w«««)i )i = (— 2 //tn-')* ', -^'ZTHfn*?!^ = y/ ^irir/w»7i« = V- '2 nnr 294. Rule for reducing Radicals to Lower Indices. — Take an exiK't root of the (pumtity (whose index nmst he a factor of the given index). The result is still affected by a radical sign with the other factor as its index. 295. Exercise in the Simplification of Radicals liy nduction to a lower index. 1. Simplify -^Uya*b*c*y y/21 a*d^^ ^/25)^ = i 5 « O'^O^ 1 V25 a% = V25^- X -Vab = -j- 5 a V^//> 3. Simplify (24«>^3c^)^ = V2r^^V (24 a^bh'-y^ = (8 «,3^,3>>J X (3 ac')"^ = 2 .//. (3 .^r^)^ ^/24a*P? = ^8"^« X \/3^' = 2 ./A v:5 ./r- 4. Simplify 7 (625 a^b'c)^ = 7 -v^625 a'^b'c 7 -^6257?^ = 7 a/(;25 ^^Z/'^ X c = _[- 7 X 5 o^^' a/7= 35 (^Z* ^^ { 5. Simplify (3 ax^ = V27 a^x^ ^'WaFx^ = V9 d^x^ X V37t = i 3 ^/.r'^ V3 « 6. Simplify ^^_ d ^VTT ~ ^^ V "4" ^ Vr/ - ^ 2/. Vc 297. Rule. 1. Separate tlie radical quantity into two factors one of whicli is the greatest perfect power of which the desired root can be taken. 2. Exti'a;-t th(^ root of tlie fact(n' Avhicli is a ])erfect power and place the result in tlie coeiticient of tlie other factor affected as before. lN.\Vi;i;S, !;(.(•!>. AM) I:.\1)1(AL8. -^1 298. £xerci8e in removing Factors from Radicals. 1. V288, 2 V75^ V !-'"•'•. \ -T8T 2. V9a*a;, V^G a», 4 v-7t/^^S V50aZ»-'c-, V80a»;r» 3. S;i28 a' ^»^ V— 108 x* y», V25 (5 a» + 5 a* b)^ >^-512a«y», \^^iM-~^ 5 \/80 299. Simplification of Fractions under the Radical Sign. «. Tliere are good reasons for i)rt*ferring to liave (luantities under the radical sign intcf/ral. This object can always be attained by means of the well-known principle in 156. In order to remove the denominator from under the radical sign both tenns of tlu' fraction are nudti])lied by such a factor as will make the denominator a perfect power ; tlu'n extracting its root the result is placed in the d(>nominator outside, as in the previous case. 1. Siiuplifv (5|)* = V'^ \2x'l/J \2xhj \.r.r) \i^''y) = . ±.(20 abxi/')^ 2xy^ Hk.m.\uk. — The student may write this example in the radical notation. He should accustomhimself hereaft<'r to use either at pleasin-e. 3. Reduce tii f " J -^m\l" to its simplest form »Y^ = ,«,yV. X-_ = „,y ,^ = .,_^„,, '" V '' . r'' 288 TEXT-B(J()K OF AJ.GEBltA. 4. Reduce y — '— to its simplest form. 4 c- // 1(1^ X^ (IX (1 (IX jit^ 1/ (IX Y 4<'"^ // J r \ 1/ 2 (• \ 1/ 1/ 2 I'll ^ .. ,.,. 'Ill ( h' \\ 5. .Sim])liiv - ,-( , ] S(>LUTI<>X. 'Ill ( h^ 'lii\S '111 h 6. Reduce ( '- | = V ' to its simplest form. ff — ./' ! II — X . . II -\- X 1 \ II + X V II + X II 4- X a -\- X 7. Reduce y/v ^^^ ^^^ sinii)le8t forui. /TT 7 5 2"X3"X8 1 ,,— V 12 "" V 2 X 2 X 3 X 2 X 3 X 3 6 300. Rule for Simplifying Radical Quantities by removing the denominator from under the radical sign. 1. Simplify the radical by the previous cases. 2. Multiply both terms of the fraction under the radical sign by such factors as will make the denominator a per- fect power of the degree denoted by the index of the radi- cal, and in so doing use no unnecessary factors. 3. Extract the root of the denominator and place it as a factor of the denominator of the coefficient of the radical. 301. Exercise in Simplifying Radicals in the Fractional Form. , /I /2 ,, /3 1 /3 2 fn flS\h '- V/3'V5''V/8'3V7'5VT'(,25) 9 aJ^^' ^l2ll^' (n'b'\h p la ,e lis POWKKS, I:(M)'rs. AM> UADK'AI.S. '2^0 3a«ty/i^ II. -REDUCTION TO SURD FORM Converse Operation. 302. Reduction of Entire Quantities to the Form of Surds. 1. Kediice \\ax to the form ot a .s(nuire root. W or = \/9 rtV-*, or (9 a^^) , by squaring. 2. Change — \ (i%h to the form of a cube root. 3. Put — 3 a^x~'^ under a radical whose index is 4. 6 /o — «V5-;/¥=(^)' 303. Rule for Reducing Rational Quantities to the form of .surds, or ra(.lical.s to eciuivaU'iit one.s of higher degrees. In- volve the quantity to the pro{)er power and place it under the sign. 304. Exercise in Reduction to the Surd Form. 1. Reduce — 5 (I'b to the form of a cube root. 2. Place 6, 2a%, — ;/, — 7 m-ti under radicals whose in- dices are 2. 3. Reduce J, a — x,{j a^x'*, '— — to the form of cube roots. _ ^ny 4. Express Var*, a^, o , and a as sixth roots. 290 TEXT-BOOK OF ALGEBRA. 5. Express V-, 2v/(), as surds of the same order, viz., sixth roots. 6. Express Vo' Vll' vl^^ as surds having the same index. 7. Reduce — to the form of a cubic surd. a -{- b -\- X V2 8. Reduce ^ to the form of a fourth root. 5 305. Reduction of Radical Qualities to the Form of Entire Surds : In Other Words, Placing Coefficients within the Sign. 1. Reduce 3V5 to the form of an entire surd. 3V5 = V3 X 3 Vo = V45. * 2. Reduce 2 « V2 a^^ to the form of an entire surd. 2 a^/2^ = ^/J2~^' X V2^ = ^{2ayx2a' = V64^ 3. Put the coefficient of — ~--^ 1/ — ^ within the sign. -l^M^-_JA^ x^ = -v/^ ^y'V3a2 Vl25/ 3a'-^ V75/y^ 306. Rule for Inserting Coefficients under the Sign. 1. Raise the coefficient to the power denoted by the index of the radical. 2. Multiply this result into the quantity already affected by the sign, and write the product under the sign. 307. Exercise in Placing Coefficients under the Sign. 1- "' vn, \ V3, ~ y/^, 6 V4, 2 ^5, i (4)' • 3 ^4 /_f_ ^^ » A^ — ^"^ a -}- h f a — b\^ I'oWKI.s l;(K»|s. \\1> I;AI)1( ALS. 2'Jl 4. i'iace tliu - 111 Liiu coetticieiit oi '2x-\/'.\(ib within tli«' sign. 5. (a - h) y/a' + 0- + L' o/>. /,\ H a 3 x V liT III. - REDUCTION OF RADICALS TO THE SAME INDEX. 308. Reduction of Radicals to Equivalent Ones having a Common Index. (See 284, b.) 1. V3, V^, = (r. + g)i, (g - c) * 292 TEXT-BOOK OF ALGEBRA. 309. Rule for Reducing Radicals to a Common Index. 1. If not already written with fractional exponents place them in that form. 2. • Reduce these exponents to equivalent ones having a least common denominator. 3. Develop the radical quantities to the powers denoted by the numerators. 310. Exercise in Reducing Radicals to a Common Index. 1. ^/8, V3, ^6. 5. 3% 2% oK 2. ^/a^, Va. 6. Va% Va\ ^uK 3. a/5, V4. 7. — r> Vf^. 5. 2 V.3~- I VlT 4- 4 V27" - 2 V7^ 6. V48^2 + /> VWa + V3a(a — 9 6)«. 9- V2rtar« — 4 «ar -h 2a — V2 ax^ -{- 4 ax -\- 2 a- /a -f- a; 10. From (a — x) Va* — ar* take r/ (a - x) -\l- • 294 TEXT-BOOK OF ALGEBRA. 12. 3Vl47-^\/5-v/i- 13. i/W^b + ^/la% + a/o4 «^^ + V^. 14. ^3)^ — (16 a"' ' %')'' + (2 «*'" + ') ^+ (2 c^a^O^ • 17. 71 a/7^ + 4 ^0.21875 - 5 ^^0.0256. 18. c V^^*WV^ - a VoiW? + /> -\/(7P?. 314. Multiplication and Division of Radicals. (See 287, 2) a. By 284 two radical quantities can be multiplied together pro- vided they have the same index. We readily see that they cannot be multiplied until reduced to the same index. Thus V6 X ^5 7^ V3() i^ ^M. By 309 radicals can always be reduced to a common index, and therefore they can always be multiplied or divided as desired. 1. Multiply V6 by V3. A7is. Vi8 = 3 V2. 2. Divide V^/*^'Vy by ^/a%f/\ Ans. — Vat' -f" b Va3" +^> + V^ — Vac — V^c — c — Vac — V^c — c. 315. Rule for Multiplying and Dividing Radicals. 1. If the radicals have the same index, multiply and di- vide the quantities under the radical signs as desired, placing the result under the common sign. 2. If tlie radicals do not have a common index, reduce them as in 310. 3. When polynomials involving radicals are to be multi- plied or divided. ])r()ceed as witli rational (piantities, observ- iiiL: the rules tm the iinilti])lication and division of monomials. a. The roeflficirnts arc to ho lunhipliod or «livi(hMl as in V3^. 296 TEXT-BOO\' OF ALG1-.15::A. 3- \/^ X v/'l'"' ; ^2-a X ^47^ 5 V3 X 7 Vi X V2. 4. Divide V40 by V2 ; 6 V'54 -- 3 V2; 77 a/9 -- 7 -^. 5. Multiply ^/lO by 4 ^2; V2 X a/3 X V^. 6. a Vx X ^» V^ ; 2 -^I X 3 V|. 7. ^6ai^c-i X -v''3-i«-Hc'^; ^ "^S -^ ^ v"^. 8. a/2 • -s/l'--- ^3 ; (V5 + 2 V7 + 3 VlO) 2 V5. 9. Multiply 5 a* by 3*; AaH^ XoaH"^-, ^S^e X i/~2ac. 10. Multiply 3a^^?/^ by 2x'^i/^ and express the product without fractional exponents. 11. Divide y^x^ by ^/x^ ; -y/y^ -j- -\/?/^ 12. Multiply together j- ^ax, -y VZ>// and — -i/oz. 13. ( y/f X y/^) ^ y ^ ; ^64 - 2 ; V;^^^:^- (./ - a.). 14. (3H-V5)(3- V5); (|+tV|)a-7V^.) 15. (^5-2^G)(3^4-^3(3);(y/^ + y/|) 16. ( V2 + V3 - V5) (V2 + V3 + V5) ; (,,2 _ ^ V^ _ 5 ^,) _^ (,, _ 3 V^). 17. («2_|_2aH^-4a^^»^ — 8^*^) -f- («^ — 4?>^). 18. ( V.^ + Vv/4- V.^) ( V^ — Vy — V^) ( V^ — Vy + v^). 19. (a/6 + 4 Vis _ 3 - 8 V2) H- V3 ; (V7^ + V32 - 4) -- a/8. 20. (2a/8 + 3 a/5 — 7 a/2) (a/72 — 5 a/20 — 2 V2). 21. (2 V3 - V2) (2 + a/9) ; (o + -^4 + 2 v^) ( a/6 + V5). i»oWi":i:s, iiouis, ANi) i:ai>I( Ai.s. 29' 22. Divide 20 by V40; m \J ~^ ^ '* y ^TZl* 23. 4 V2 X (3 V8 -f- i Vli) X e V2; {a + ^ V^^) 317. Rationalization of Denominators. — There is one case of reduction of radicals (292) which it is desirable to treat at this point. It is that of rationalizing denominators ; i.e., transforming fractions with radical quantities in their de- nominators into equivalent ones whose denominators are rational. Many fractional expressions containing radicals in their denominators can have them removed by multii)ly- ing both terms of the fraction l)y the ]>roj)er factor (156). 2 1. Rationalize the denoiiiiiiaim of V3 2 2 ^ V3 2V3 . .,.^, = - X - - = -71— Am^. (156) ^/:^ v/3 V3 '^ * 2. Rationali/.*' tl»e (h^iioiiiiiiator (»f 2 + V3 4 _ 4 _ 2 — V3 4 (2 — V3) 2 -I- V3 2 -h S/3 ' 2 - V3 4-3 = S - 4 V3 Ans. 3. Rationalize tin dc nniuiiuitor of V3_4 V5 — 2 V7 2 V3 - V5 + V7 * I^y actual multiplication and addition, we have V3-4V5-2V7 __V 3-4V5-2V7 2 V3 — V5^-f V7 ^ (2 V3 — V5) + V7 2V3-V5~-V7 40 — 9 V15 - 5 V21 -f 6 V3g ^(2V3-V5)-V7~ 2(5-2Vl5) 298 TEXT-BOOK OF ALGEBllA. To rationalize tliis result botli terms are multiplied by 5 + 2 V 15, giving (the student should verily the result), 35 V~i5 + 35 V 1^— 70 2— vI5— V^ 2 X — 6o 2 K To rationalize a fraction of the ion a"' ± fj- luetjj be the 1. c. ni. of m and ?i. Then . . . =f (f,i>Y ' (?,y 130. 1 and 2) = Q (say) K If now both terms of \ v Ix' inultiplieil by Q, we have a'" + h K KQ 1 L_ — / ~i\p 7^'v' since the (quotient Q multi- ])lied by the divisor ^,*1 _[_^^^ ought to ei the equivalent fraction with a rational denominator. In Examples 2—4 the multiplier in each instance is termed the eomplemenfary radical. In the case where the radical or radicals are square roots, as in Ex's 2 and 3, wlien the denominator is a siirri the C()m])lementiiry factor is a dif- ference, and vice versa. 318. Rule for Rationalizing the Denominators of Fractions. 1. For Monomial Denonimators. — ^Multiply both terms of the fraction by such a factor as will rationalize the denominator. POWEUS, UOOTS, AND liADHALS. 299 2. For Biiioiiiiul Deiiuiiiinators. — Multii>ly both terms of the fraction by the factor coniplcim'iitary to the deiiomiiiiitor. 3. For Polyiioiniiil Dtii'iiiiiiiators. — Regard one pari ( t the denominator as one t»'i m and the others as the second term of a binomial, and pioc ccd as in 2. a. The advantage gained by rationalizing denominatoi-s may be shown by an example. Thus, ^ if calculated just as it stands 4 V? would rc(iuire the extraction of two roots antl the division of one long decimal by another. Whereas if the denominator be rational- ized, the extraction of only one root is necessary, and the division is a decimal divided by an intej^er (usually not a larirc nninbor). So, likewise, with binomial denominators. 319. Exercise in Rationalizing the Denominators of Ex- pressed Divisions. m 2 _v3 :5+V8 a -f \/o ^ ^* V^' y/V)' ^"- \ -• \-^^- 3 6-h8 :. -f Vir> 300 TEXT- -BOOK OF ALGEBRA. 8. 1 1 1 1 -\/x— -yjy\x + V.x^ — 1 X — Vx" - •1 9, V2 + A/3. V2 - ^3 10. 1 + V3 + V6 5 V2 1 -f V2 - V3 V2 + V3 - V5 320. Powers and Roots of Radicals. For Principles and Rules see 287, 3. a. Cases arise, however, in the extraction of roots for which the rules heretofore given do not apply. These problems will be treated separately in the next article. 1. Square 3 ^^3. (3 -^3)2 = (3 iZff = 9 (3)' = 9 (3'^)^ = 9 a/9. Or, (3 ^3)2 = 3 ^3 X 3 ^3 = 9 v'9. (315) Cube -s/ax^ ; raise ^ — 10 to the fourth power. 2. Raise .\yO to the fourth power; 2V;V to the fifth power. 3. Raise |V — ^xhj to the fifth power; — 3 Vf to the third power. ^•(-^-)^v/(i)V(iT■(v/^" 5. (^^'^yf, (v'2)2; raise V27^ to the nW\ power. 6. Square V3 + ic V3 ; cube 2 - V3. 7. Square a-' — ?/-§; raise V3 — V2 to the fourth power. 8. Raise V^c — 2/ to the third power ; cube -\/ a -{/a -f- x 9. Cube V^ + 3 Vy; cube — -^'^a — ^bc. POWERS, HOOTS, AND KAMI ALS. 801 10. Extract the square root of ^a*; ol (Va^)^- 11. Exti-act the fourth root of IG a^ ^/2 c j of y/y/a*b'^ ' 12. Extract the cube i-oot of 32 -J^27 a*x^y of (\/49~i5^)* • 13. ^6T««~S^^ Vy^, \/l25"^7^y/3 . 14. Required the cube vuot of 125 a: i, of 04 a^l>* V2 cd. 15. Extnict the fifth root of 486 a -^ia^. 16. Extract the scjuare root of a-J -f Ga:§//i -f <> //; tlie cube root of (a -{- x) y/a -\- x. 17. Exti*act the fifth root \/32 x" ; the eighteenth root of \/^»«^'-». 18. Extract the cube root of 8 Vx* -\- 3(> xf/ -f 54 i/^ Vx + 27y^. 19. Extract the square root of 9 x — 6 Vary -f // — ^> Vx + 2 V^ + 1. 321. Roots of Radical Quantities not in the Original Pro- duct form. — Binomial Surds. a. If X and y be replaced by 2 and 3 respectively m Ex. 18 of the last article, we have 16 V^ + '^^^ x 2 x :] + 54 x 9 V^ + ^7 X 27 = 945 + 502 V*^ which has only two terms, one being a surd, and may therefore be appropriately called a binomial surd. In order to extract the cube root of 945 -I- 502 V^ by the usual cube root rule, it be- comes necessary to arrange it in the form first given. This is gen- erally difficult to do, when one does not know how it was obtained. Such roots, when they exist, may often be discovered by special processes. Only the sim])lest case, viz., the square root of binomial surds, is presented in most treatises on elementary algebra. The method of solution for other cases is very similar. Two or three lemmas are a necessary preliminary to the deriva- tion of the root formula. ■\/'H = « + V m n = a^ + 2 a -\Jm -\- m _ 2 a ^vi ■■ = a/' -\- rn — n s/vi - = a- -\- hi — n 2 a 302 TEXT-BOOK OF ALGEBRA. 222. Theorems Relating to Equations containing Radicals. 1. The square root of a rational qua7itity cannot he partly rational and 'partly surd. Let n be any number, and suppose it has for its square root a + -yjni . Then (Ax. Thus a surd equals a, rational fraction or a whole number (dei)endent on whether 2 a is contained in the numerator exactly) which is impossible. (See 290, a.) Q. E. D. 2. In any equation containing radicals, the rational part on one side equals the rational part on the other, and the surd part on one side equals the surd part on the other. Thus in the equation a -f- V^ = x -\- Vy, if ci represent all the rational quantities on the left side, and x those on the right side, they must be equal ; also V^ = Vy. «+V^=.T + V// _ (Hyp.) V/> = a; — a + -yj'y (Ax. 2) Now, if X is not equal to a, -yj h is partly rational and partly surd, which is impossible by 1 of this article. Whence x. must equal a, and if x = a, V^ = V?/- "(?• J^- ^• If Vrt 4- V^ = V.r + V^/, then will ^ a -_^h = Vx — Vy V^+ V^ = V^ + y^ (Hyp.) aJ^^f,=x^2 Vxy + y (Ax. 5) ^.^ a = X -\- y, and V0 = 2 Vxy = V4 xy (2, above) .'. a — -yjb ^x — 2 ^xy -\- y = {^x — ^yf ... V(i- V^ ^ Vx -Wy. Q. E. D. POWERS. ROOTS, AND RADICALS. 803 323. The Square Root of Binomial Surds. — Demonstration, Rule and Examples. 1. Let A represent the rational part and VB the mdical pai-t of any binomial surd. ^Suppose Var + Vy to represent its square root. Then (1) Va -f VB = V a- -f Vy (2) A = X -f // and (3) B = 4 xy (322, 3) Now, {x — i/Y = (x ^-yY — ^xy^ A-^ — B (Identities) (4) .-. x-y = ^^:'-\^ (10 .r + y = A. 2x -A + VA'^-B (Ax. 6) (Ax. 1) ,-_i/A + VA«-B ^x \l 2 2//= A-VA^'-B (Ax. (!) (Ax. 2) ,__./A-VA-^-B Vy - V 2 Hence, (A.i. (i) Also, H + ^A- -VA^ 2 -B Vi - V^ = y/A + V A^-B _^A~VA-^- B 2 (322, 3) 2. RrLE. — Tlie rule is written from the foi-mula just found. (1. Sc^uare the rational quantity, subtract the quantity under the radical sign fnun the result, and extract the square root of the remainder. If this cannot be done there is no root of the form sought. (2. Add the root just obtained to the mtiontvl quantity and also subti'act it from the rational q\iantity, aiid divide 304 TEXT-BOOK OF ALGEBllA. the results by two. The sum or dittereiice (according as the binomial surd is a sum or difference) of the square roots of these quotients is the root sought. ((. Instead of following the rule, infipectioii will often enable one to determine the root. 3. Examples. (1. Extract the square root of 31 + 10 V6. Here A = 31, and VB = 10 VO = V600, or B = 600. A2 - B = 961 - 600 = 361 ; and VA^ - B = 19 /- 7- , / 31 + 19 , , /31 - 19 .-. V31 + V600 = y — |— H- y ■ — 2 — ^ _^_ (5 _|- V6) Ans. 2. Eeduce V '/ijo + 2 m^ — 2 m Vnjj + ^f^^ to its simplest form. Here VA^ — B = \^n'^p'^ = /ip. Hence, the answer is 3. Vll + 2 V30 = ? Vll -j- 2 V30 = V6 + 2 V30 -|- 5 = V6 -f V5 Ans. 324. Exercise in Extracting the Square Roots of Binomial Surds. 1. 3 - 2 V2, 49 - 20 V 6, 87 - 12 V42, 10 - V96; 42 + 3 Vi742. 2. 75 + 12 V2T, 4^ - ^ V3; I + V2: 3. £c — 2 Va;— 1, a; + ic?/ — 2 ic Vy, 2 — V4 — 4 a'^ aaj — 2 a V«^ic — a^, X -\- y -\- z -f-2 V^c^ + y^. 4. Solve by inspection 4 + 2 V3, 6 - 2 V5, 9 - 2 Vli, 23-8 V7, 11 + V72, 28 -~ 5 Vl2. roWEliS, KUUTS, AM> KADiCAL6. 305 5. Extract the fourth root ot 17 + 12 V-?, 50 -f- 24 V^) ^ V5 4- 3i, 248 + 32 VOO. 6. Va V5 + V40 == ? V V1573 - 4 V78, V2] + V5. y ^ -h [; Va' - cS V28 + 10 V3 + Vg7 - 1() V'S. 325. Fundamental Operations with Imaginary Radical Quan- tities. Any pure imaginary quantity, as v — a, can be reduced to -^ Va~^V— 1 (284), of which Va may l>e regarded as the coefficient of the imaginary V— 1. Thus, all pure im- aginaries can be expressed in terms of V— 1. Let us examine the powers of V— 1. ( V— 1 )^ = - 1 by dejinltion of a square root. ( v^^i y = ( V- 1 y V- 1 = - 1 v^=T ( V - 1 )* = ( v^^i y ( v=n )2 = - 1 X - 1 = + 1 ( V— 1 y" = (( V— 1 )*)" = + 1, where 71 is any integral number. ( V--T)*"+i = ( V- 1)*" V-^l = V- 1 ( V^^)*"-^* = ( V-'i)*" (V— T)2 = - 1 1. Add V— a'^ and V—b'K .-. V— 7a'' 4- V—b^= ( ± « i 6) V— 1. Or, taking posi- tive roots only, = (a -\- b) V— 1. 2. From 4 V--27 take 2 V— 12 4V--27 = 12>A-3; 2V— 12=4V-3; ... 4 V--27 - 2 V- 12 = 8 V— 3. 306 tp:xt-book of algebha. 3. Multiply c -\/ — ahy d ^ — b c V — a = c -\ a ^ — 1 ; d ^ — h = d -sjb V — 1 ; now since V— 1 X V— 1 = — 1, -l^y delinition, c V— (t X d V — b = cd ^/ab X — ^ = — c d ^ab. Note. — It should be remembered that the sign ± is only written in case of ambiguity (See 264). When it is known by what factors a product has been formed the appropriate sign must be prefixed. Thus, V+1 X ^/^^l = + 1 ; V^ x v'^ = - 1 Here c V— a X d V— b equals not ± cd V-\- ab but — rd v + ab. 4. Divide 3 V^^ — 2 V— 11^ + VO — 9 by 11 V -^. (3 X 2 V^^ — 4 V:> V^^ + V + 2. Squaring V— (> + 2, we get — () 4- 4 V^^ + 4=4 V^^^ — 2. 326. Exercise in the Application of the Fundamental Rules to Imaginaries. 1. Add 2 V— 48, ;{ V— 12, 5 V— .S, and — 7 V- o2. 2. Find the sum of 2 + V— 1 and 3 — V— 04. 3. Add -v/^=^ and ^^TTi ; -J/Hl and ^/^9. 4. From V — 18 subtract V — H; from o -\- V— ^ take a -f- V— c. 5. Multiply 4 V- 3 by 2 V- 2 ; 4 V^^ X 9 V-~12. 6. 2 V^l> X r> V-~4 X 3 V=^; 2 ^/::^T x :; v - To. 7. Find the third and fourth jn^wers of aV — 1. 8. I )i vide (> V^^^ by 2 V^=^; (4 + V^-2) -4- 2 — V— 2. 9. Divide 2 \/S— \^— 10 by — V— ~2"; — V^^^-^ — (> V- 3). . , 1 + V— 1 S 10. Simi.lilv - ; ' ' 1 _ V— 1 - 1 -f \ - u ;; 11. (7 \ -o) (10—;; \ -<;,: < .:■ _ T) v^" V — T) (7 _ I \ :; \ _T). 12. Of wliat number are 24 + 7 V— 1 and 21 — 7 V— 1 tht» factors ? 13. Find tin- c-Dntinucd ]>rndiict i^\' ./• -f ". ./• -f- " V — 1, « — a, and x — a V—l* 14. Multiply, (x —J, — y V— 1) (x —p + y V— T). 16. Kaise a -j- It \^— 1 to the third power. 308 TEXT-BOOK OF ALGEBRA. 16. Extract the square root of o -j- ^ V— 22. 17. Extract the square root ofSV— 1 (=0 + 8 V— 1. 18. Extract the square root of — 2 + 4 V— 6 ; of 31 + 42 19. ( v^nrr + V^==l9) (V-iT9 - V-T33) = ? 20. Prove that an imaginary quantity of the form V— a cannot be partly real and partly imaginary, as V — a= m _|. ^ZZVi, (See 322, 1.) 21. Prove that in any equation containing imaginaries of the form a-\-h V — 1 = m + ^ V — 1, the real part on one side equals the real part on the other, and the imaginary part on the one side, the imaginary part on the other. In particular, if a -\-h V — 1=0, then a = 0, and ^ = 0. (See 322, 2.) Remark on Imaginaries. — It was not till the beginning of this century that Argand invented his diagram, since which time great progress has been made in this and allied sub- jects. An extended presentation of imaginaries would be out of place in any except an advanced algebra ; but some little idea of the graphical representation of imaginaries will probably be very suggestive and instructive even here. It was early shown that multiplying a positive number by — 1 reverses its character and makes it negative, and mul- tiplying a negative number by — 1 reverses its character and makes it positive. Naw we may conceive of — 1 as an operator which revolves the whole positive series around into the negative, and the negative around into the positive, zero being the pivot. But we saw that V— 1 X V— 1 = — 1, or V— 1 X V— 1 X a = — a; i.e. the operator V — acting on a twice revolves it from the positive to the negative position. We are thus led to the idea that V— 1 acting on a once would revolve it POWERS, nOOTS, AND RADICALS. 309 half way, or into the position at right angles. Generalizing this result we have the iniaginaries as represented in the figure. .fiovri + »-/31 + »V=1 + Tv^ + tv=i -t.«vcn^.~ .^« ■*-*V=i ! 5 + *V=1 : a. + 2>/^ i i + ^ri 10,-». -»,-:, -6, - 3. - f. -a, - 1', - i ,-»-4, +B,+6,+r, +8. +». +10, f « I... A 1 — r 1 * — 4^—1 ..J4. In Chapter IT., we regarded tin* luimhors of the double series as fixing lengths and directions on the line containing it. Thus, fixed tlie origin, -f- 3, the distance to the point 3 units to the right from the origin, and so on. Similarly, here, -|- 2 V— 1 fixes the line from the origin to the point 2 units above it, and — 4 V— 1, the distance from the origin to the jwint 4 units below it. In Argand's diagram the dis- tance and direction of any point in the whole plane may l>e fixed by a complex number^ i. e., one partly real and partly imaginary, as ::i: a ^ ft V— 1, by letting a denote the per- 310 TEXT-BOOK OF ALGEliKA. pendicular distance to the vertical line (measured parallel to the real series), and h the perpendicular distance to the horizontal line (measured parallel to the imaginary series). Thus, the line joining and 1 in the figure would be repre- sented by 4 -|- 5 V— 1 ; the line joining and 2 by — 3 + 2 V— 1 ; the line joining and 3 by — 4— 4V-"1; the line joining and 4 by +6 — 5V— 1. The meaning of sums, differences, products, and quotients of complex numbers may now be interpreted on the diagram. But the examination of these operations would lead us too far astray from the subject of radicals. SECTION III. Equations Involving Radicals. 327. Equations containing Radicals. — Since there is no reason why such equations should be of one degree rather than another, this subject might be postponed until the solution of quadratic equations is learned. Still, because the preliminary transformations are much the same in all cases, they can be treated here. Only such as reduce to simple equations are given in the following set. 1. Solve Vic + 5 + ;3 = 8 — Va; V^-f5 + 3 = 8 — V^ ^x -^^=-5 — Wx (Ax. 2) x-\-5=2o — iO V^ + X (Ax. 5, 114, 2) 10 -Vx = 20 (Ax. a) Vx = 2 • (Ax. 4) x = 4. (Ax. 5) Solve 2i/5x-35 = 5^2x-7 S(5x-35) = 125 (2 0^ - 7) (Ax. 5) x = 2^ (219, 2) »(>WEKS, UOOTS, AND UADICALS. 311 3. Solve \/12x — o-l- V3ar - 1 = V27x — 2 12 r — 5 + 2 y/Mx^ -.21x + f> + .'5 x - 1 = 27 x--_2__ (Ax. 5, 114, 1) V'3(> X* — 21 X -\- iS = iSx -\- 2 (Axs. a and 4) 3(> X- — 27 a- -h 5 = 3() j--^ -|- 24 :r + 4 (Ax. T)) .-. ./• = ,V ^1/^s-. (219, 2) But upon substitutinj^ this value in the given equation we find it is not verified. Thus, V-^W + V- n 7^ i V- H i.e., 9 V- Vt -f 4 V- iV ^ i 5 V- ^ It is sufficient for our present purposes to know that values of the unknown obtained by such ])rocesses as these may not satisfy the original equatin, and are therefore not roots of it unless the equation is taken in a certain way. In the present case ^^ is a root if the sign of the first or second terms be taken negatively. To avoid mistakes, roots {)l)tained for radical ecpuitions should be verijied by substitution, in the original form of the equation. (See 360.) 4. Given a; + « = v ^^ _j_ x V^* + x^ to find x x2 -f- 2 flj; + a^ = a"" + x V^^T^'^ (Ax. 5) x^ -\- 2 ((X = X -s/li' -f- . /••- (Ax. 2) X -I- 2 tt = y/b' + X' (Ax. 4) a.2 ^ 4 ax H- 4 rt2 = ^2 4. x' (Ax. 5) //-' - 4 «2 .•. .r = - . 4 a Verification. 6* , / . , 6«-4a» /l6o2(>2+6«-8a''62'+T(J«' 4^ = V« + -TS-V 16tfi _ 1/ 10 a< -i- ft* - 16 fl* _ i!^ 312 TEXT-BOOK OF ALGEBllA. 5. Solv^ e V2 .7- - -8) = 2 , 2 -8 V2a;'^ ■-16x- V2 x' - 2x^- 2x''- Ux - Wx - 4 X = 3() () - 12. T + 36 which is a quadratic equation, whose solution we are not now in position to obtain. 328. Rules and Directions for solving Equations involving Radicals. No general rule can be given. The advantage of devia- tions from the normal process is here even more marked than in simple equations. (See 217.) 1. When but one radical appears in an equation, transpose all the other quantities to the opposite side of the equation, and then raise both sides to the power indicated by the index of the radical. 2. When two or more radicals occur, it is usually best to place one alone on one side and then raise both members to the desired power. 3. The process just described will often have to be re- peated one or more times in the same problem before all the radicals disappear. a. When fractions occur, it, is generally best to clear of fractions before squaring or cubing or raising to other powers as the case may be. Sometimes an advantage is gained by reducing a fraction to its lowest terms, or by rationalizing a denominator, or by transposing and uniting, etc. h. Care should be taken to combine all similar terms before rais- ing to a power. c. By example 3 the student is admonished that the value of the unknown quantity should be verified in tl\e original equation. iM>\vi:i:s, uoo'is. and kaiucals. 'US 329. Exercise in the Solution of Equations containing Rad- icals. 1. \ .,• - r> = :;. 2. r -f L' — V H) 4-2-^ = 3. J- = 9 — VxM^i). 4. 8-}- V3^~-f0 = 14. 6. V2a;-f-ll = Vs. 6. Vr -f~2r) = 1 -f- V.r. 7. v.r -fy^ — .1- — .\ = 0. 8. V;^-"4 4- '^ = Va--f 11. 9. V4 ic -|- 5 — V^ = Var + 3. 10. s/ax -\-2ab — n = l>. 11. V3 -^j^ — A/3+li^= 0. 12. 3 + ^ar»-9ar'^ = r. 13. V4 a + X = 2 VA + >• — V.r. 14. V4 — ViP*^*^'^ = .r - 2. 15. V9ar _4-|-().= S. 16. V4 .r« - 7 ^ H- 1 =2 ./• - 1 1 17. \ ./• -h v'4 -f .1- = ^ • 45 18. .. + V9 + .«^^. 19. V.'- 4- •' + v.'- + « - V4 rr'-fTM = 0. 20. \ ./• + \ " \ 'IX 4- ar* = V''. 21. = 22. \ y^/r -f ;5 ^ \ \ / — :; = V2\ /• 314 TEXT-BOOK OF ALGEBKA. 23. =5V^-8 + -^r- 1 , 1 ,/l / 4 -9 25. - + - = T- 3/7^ 26. V64 + ic^ - 8 a: = 4 + a; a/4'+^' 3 0- — 1 ' , V3 X — 1 V.r) X -\rl ^ FOURTH GENERAL SUBJECT. —QUAD- RATIC EQUATIONS. CHAPTER XXI. QUADRATIC EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 330. Quadratic Equations are equations of the nerrrmf ilc- (free. There are two kinds. Complete and Incomplete. Complete quadratic e(iu{itions (also called (idfected (piad- i-atic equations) contain both the second and the first powers of the unknown. Incomi)lete quadratic equations (also called inwe quadratic e(iuations) contain only the second jx>wer. Thus, itx' -\- bx = c is a complete quadratic equation, while inx^ = n is an incomplete quadratic equation. «. These are the only eases to l)e considered. For, the first equa- tion contains the s) ix + ^) = IS X- -72x- -2i<; ISx' - 144 ./• = 1 1 1. Given 1^ > = 4 + x 2 ./••- = 4 u- + (> (Ax. ,S) j-^ — 'Jx= :i (334) 2 Vx^= '^ ./• ; 2 .r -T- 2 u- = 1 ; 1- = 1 (335) x--i_2.r + 1 =;{ + 1 =4 j: — 1 = i 2 (Ax. u) i.e. J- — 1 = +2, or j—l = —2 . . ./ = ;^, or j:* = — 1. .///.s'. VkKIKH ATlo.N. 1st. X = »S ill the given ec^uatiou, 2 X .'i = 4 + ♦; i.e. 6 = 6. 2(1. ./• = — 1 in the given equation, 2X— 1 = 4 + -^ , i.e., _ 2 _. — 2. 2. Solve i)0 x'^ — 15 x = 27 100^2 _ 3() ^ ^ 54 (334, 2 VlOO r-« = 20 X ; 30 X -i- 20 -p = ^ ; (•;!)- = i( KM) X- _ 30 J- + I =r 54 + 24 = mi = 5(>.25 10 j: — I = -1- 7i i.e. + 7i an == - 5- '^"•^■• 3. Solve (x 4- 1) (-'^ + 3) = 4 j:'^ _ 22 2x2 -f. 5a: 4. 3 = 4^^-22 — 2«2-f-5a; = -25 a;2 _ ^ jc = _j_ :;^^. f-2^1; (t)-^ = fg «' - B^ -h f ^ = ¥ + e = Vc*'^ a: - ^ = i Y X = 5, or - 2.V. 4. Solve 3x^-\- 10 a: = 57 36^2 -f- 120 a- = 6S4 (334) 2 V36 3^2= 12 X ; 120 a- -^ 12 a- = 10 ; 10'^ = 100 36a;« + 120a; + l'^ = 4:nq -\- p^ 2nx -\- p = ^ V4 ^j.g' + pi^ — p^ V4 Tiy + p^ X = 7. (py-p' 2n Observatiox. — This example shows that if an equation be multi2:)lied through by 4 times the coefficient of x^, the square of the old coefficient of x, viz., p, is the quantity to be added to complete the square. See also Ex. 4. 337. Rules for Solving Complete Quadratic Equations. — One rule only is really needed, the first and second rules as given below being special cases of the general process ex- plained in the third rule. 1. First rule. (1. Reduce the given equation to the type form ax"^ -\- bx = c, and divide through by the coefficient of x^, if it is not already 1. (2. Add the square of one half the coefficient of x to both sides of the equation and extract the square root of the two members. (3. Solve the two resulting simple equations for the two values of x. (See Examples 1 and 3 of the last article.) 2. Second rule. (1. Reduce the equation to the type form (removing any monomial factors which may exist in it, Ax. 4) ; then multiply through by 4 times the coefficient of x'^, merely in- dicating the multiplications in the left member. (2. Add to both sides of the equation the square of the old coefficient of x, and complete the solution as in rule 1. (See Exs. 4 and 5.) QUADRATIC EQUATIONS. '>-•> 3. Tliinl rul(*. li. Reduce the e(|uaLi«»ii to llie t \ pt- Inini tiiid umltiplv or divide tlirough by such a (juautity as will make the coef- ficient of x'^ a iMii'fect square, and at the same time shorten to the greatest extent the subsequent reductions. (2. Add to both sides of the equation such a quantity as will make the left member a j)erfect trinomial scjuare. To do this ilivide the second term by twice the square root of the first term, and square the quotient for the third term. Complete the solution as in the first rule. rt. The first and second rules are easier for beginners. The rule just given is really the best because it is most flexible. However, considerable experience is needed to be able to use it to the best ad- vantage, .Suggestions will be made from time to time to show the student its sui)eriority over tlie others, and tlierefore the desirability of mastering it as well as the others. The second, which ensures the avoidance of fractions, is sometimes called the Hindoo Rule. 4. Fourth rule. (1. Reduce the equation to the type form. (2. Substitute the values of n, p, and q from the given j)roblem in the answer of example 5 of the last article. (See 255.) Thus, given y - ^ + ' ^ = 8 12 .r'^ — 8 .r = 15 . (to type form) nx'^ + jw = y (Eq. of Ex. 5) So here, « = 12 ; yy = — 8 ; and «/ = 15. Substituting these values in the value of x found in l^x. 5. ^ _ - (- 8) -i- V64 + 4 X 12 X 15 2X 12 . ^j. - 8 (- 8) - V64 + 4 X 12 X 15 2X12 « ^ + ^'8 ^ .3 ^^^. 8 — 28 ^ _ 5 "* 24 •_'* "' ' 24 (5 Ans. 324 TEXT-BOOK OF ALGEBRA. 338. Exercise in the Solution of Complete Quadratic Equa- tions. 1. x'^ —16 = 45 — 4 X. 7. X- — 341 = 20 x. 2. a;2 + 22 ic = 75. 8. 23 ic = 120 + x\ 3. a;2 = X + 72. 9, x^ — 6x = 6x-\- 28. 4. 3 ?/2 + 48 = 30 I/. 10. 4 o;-^ + 4 x- = - 1. 5. 5 ./•- 4- 20 X = 25. 11. ^2 — *^ ;;>• = 32. 6. x^ — () ./' = 0. 12. '" 4- 350 — 12 /■ = 0. 10 ^ ,_ .T+22 4 _ 9^-6 j_^, — — — , 3 X 2 14. (./• — 1) (X — 2) = 1. ,, 19 11 15. __ a- = — — ic^ 5 5 16. Sa;'-^ — 20x = 21. Suggestion. — Multiply through by 2. 17. ^' + i^x = 20 .r + 25J - ^^x\ Suggestion. — After clearing divide through by 6. 18. 252 x"- + 360 x= — 125. Suggestion. — 252 = 7 x 36. Therefore if an extra factor, 7, is introduced, the product is a perfect square. Multiplying the equation through by 7, we get 7 X 252 x2 4- 7 X 360 x = - 875 2 ^7'-^ X 36 x2 = 2 X 42 x; 7 X 360 X ^ 2 X 42 X = 30; (30)^ = 900. (42 x)2 + ( ) + (30)'-^ = 900 - 875 = 25 42a; + 30=-t5 x = - If, or, - f ( ^l^ — 15#^ = — 815 = — 125 Vek.k.cation. - \ j^. _ 3^,^ ^ _ j|._ i^UAUUATTC EQFATTOXS. •>-•> Kem AKK. — To see the advantage gained, the student should solve this exercise by the first and second rules. 19. 72 y- + 40S r = 1222. Si'OOESTiox. — Divide through by 2. 20. 84 a-'^ -f 45 = 124 J-. Suggestion. — 84 = 7 x :J x 4. Hencemultiply through by 7x3. 21. 96 x* = 4 jr -f 15. Suggestion. — tMi = (i X US. Hence multiply through l»y «*». 22. 622 J- = 15 x' -f- 6.3-)' SiGGESTioN. — Transpose and unite, without clearing of frac- )ns. 32. J ' — {ft -\- ^0 ^ -\- "^ = ^*- 33. X 34. (m — 7i) x^ — nx = VI. 35. 1 1 _ 3 -h rr* a — X a -f- ' ~~ a* — ^* 36. 1 .. Ill „ —h 4- .r n 1, ^ .,' 326 TEXT-BOOK OF ALGEBliA, 37. Vx -1 =x—l. 38. X — Vx = 20. 39. V 2 X + 2 + VT + 6 ;:c = V7 a? + 72. 40. 7 (a: + 7) + ^ J ^ = 0. 41. 9 a'b^x^ - 6 a%^x = b'\ 42. V :c + <^ — ^x — a = V2 X. 43. (^a?'-^ — 2 ex y/d = ax^ — cc^. SECTION III. PROBLEMS. 339. Problems Involving the Solution of Quadratic Equa- tions, Complete and Incomplete. a. Algebra is a formal science made to cover all cases, and with- out any reference to particular problems. Some problems by their nature admit of negative numbers: in such a negative answer has its proper significance; while in others negative values for the answer are inadmissible. Moreover, in algebra, imaginary values for the unknown denote that the problem is hupossible. 1. Find two numbers one of which is 5 times as great as the other, and the difference of whose squares is 90. Suggestion. — Let x = the less, then 5x = the greater. Then the equation is (5 xy — x^ = 96. 2. The square of a certain numljer diminished by 17 is equal to 130 diminished by twice the square of the number. Required the number. 3. A person bought a quantity of cloth for $120; and if he had bought 6 yards more for the same sum the price per yard w^ould have been $1 less. What was the price ? (ji AhK.vric i:(jUATioxs. 327 Let X = the price per yard, then the equation is l^ -£-\. Therefore X = 5 or - 4. (219,2) X But — 4 can have no meaning in this problem. (253). We can, however, so modify the statement of tlie problem as to be able to make use of the second answer. Thus, — The exchange account of a banker amounted in a certain number of days to $120, during which time exchange remained the same. Had the period differed from what it was by days, and in his favor (i.e. more if premi- ums and less if discounts) it would have made a difference of $1 per day. What was his daily premium or daily discount ? Let X = the daily premium or discount. Then referring to the equation above, we see that if x is positive the statement is satisfied; also if x is negative, ^ is negative and 6 added diminishes the number of days numericaUij. The two answers 5 and — 4 signify a daily premium of $5 or a daily discount of $4. 4. Find a number such that if you subtract it from 10 and multiply this luimber by the number itself the product shall l)e21. 5. Divide the number 346 into two such parts that the sum of their square roots shall be 26. Let X = the square root of one part, and 26 — x, that of the other. 6. A merchant bought a piece of cloth for $.324, and the number of dollars he paid for a yard was to the inunher of yards as 4 : 9. How many yards did he buy ? 7. If a certain number be added to 94 and then the same number be taken from 94, the product of these derived num- bers is 8512. What are the numbers ? 8. A man traveled 60 miles and if he had traveled 1 mile an hour more he would have required 3 hours less to per- form the journey. At what rate did he travel ? 9. Find three consecutive numbers whose sum is equal to the product of the first two. 328 TEXT-BOOK OF ALGEBRA. 10. A rectangular field, whose length is 3367 and whose breadth is 37 yards has by it another field of an equal number of acres whose length is to its breadth as 13 : 7. What are the dimensions of the latter ? 11. An individual bought a certain number of kilograms of salt, 4 times as many of sugar, and 8 times as many of coffee, and paid for each of them 40 times as many cents as there were kilos of that material. How many kilos of coffee did he buy if he paid altogether $32.40 ? 12. If a certain number is diminished by 3 and also in- creased by 3, then is the sum of the quotients which we get by dividing the greater by the less, and the less by the greater equal to 3^y. What is the number ? 13. A capital stands at 4^ interest. If the number of dollars of capital be multiplied by the number of dollars in 5 months' interest, the product is 10o3375. How many dollars are there in the capital ? 14. The hypothenuse of a certain right-angled triangle is 2 ft. greater than the base, and 9 ft. greater than the per- pendicular. Find the sides of the triangle. 15. A girl bought a number of oranges for 40 cts. Had she bought at another place she would have received 3 more oranges for the same money, each orange costing § of a ct. less. How many did she buy ? 16. Two peasant women together brought 260 eggs to market, and both lost the same amount. " Had I sold your eggs," said the first to the second, '' and had they brought my price, I should have lost on them 7.20 marks." '' That may be," responded the other, " but if I had sold your eggs at the price mine brought, I should have lost 9.ens if 2c is less than d^ ? Thus, e.g., take d = <) and 2r = SO. (See Art. 265.) 24. Two iK)ints start out together from the vertex of a right angle along its respective sides, the one moving m ft. l)er second, the other n ft. jmi s. ( ond. How long will they recpiire to be r ft. apart ? 26. It is required to tind three numbers such that the product of the first and second equals a, the product of the first and third ecjuals h, and the sum of the .squares of the .second and third equals r. 26. A set out from C towards I) and travelled 7 miles l>er day. After he had gone 32 miles, B set out from I) towards ( ' and went every day ^\i of the whole journey, and after he had tmveled as many days as he went miles in a day, he met A. Required the distance from C to D. 330 TEXT-BOOK OF ALGKBUA. 27. A reservoir can be filled by two pipes, by one 2 hours sooner than by the other. If both pipes are open at the same time the reservoir is filled in 1| hours. In how many hours can it be tilled by the smaller pipe alone ? 28. A farmer sowed one year a hektoliter of wheat ; the next year he sowed what he harvested the first year, less b hektoliters, and reaped c fold of what he sowed and d hek- toliters beside. Assuming a like fruitfulness both years how much did he reap the first year ? 29. It is required to divide each of the numbers 21 and 30 into two parts, so that the first part of 21 may be 3 times as great as the first part of 30 ; and that the sum of the squares of the remaining parts may be 585. 30. A looking glass, in size a X b inches, has a frame of uniform width and of the same area as the glass. What is the width of the frame ? Suppose a = 12, and b = 18, i.e., a glass 12 X 18, and substitute in the answer as by the fourth rule. 31. A square courtyard has a rectangular walk around it (on the inside). The side of the court is 2 yards less than six times the breadth of the walk ; and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 92. Find the area of the court. 32. The driving wheels of a locomotive are two feet greater in diameter than the running wheels ; the running wheels make 140 turns more than the driving wheels in a mile. What are the diameters ? (The circumference of a Avheel = 3f times the diameter, nearly.) 33. Generalize the preceding problem. 34. A grazier bought a certain number of oxen for f 240, and after losing 3 sold the remainder at $S per head more than they cost him, thus gaining $59. What number did he buy ? (,>r.\i>i;.\iic i.<,»i \ ri< "NS. 331 35. Generalize tlie iUtli iirobleiiL 36. A wall is ])iiilt l)y two masons of which the first ht'<,niis to work 1.^ days later than the other. It takes o^ (lavs from the time the first began. How long would it take to finish the wall should the first work li days less than the second ? SoMTioN. — Let X — the number of days the first requires in which to build the wall, and y = the number of days the second requires. Ihen — u =1 whence - = y^ X ^ ^»ence ^ ^^^ There is need for still another unknown which we may denote by z. Let z = the number of days n'(|iiin(l wluii the first works 3 days less than the second. Then, J- \ 14x / 14 z- 42 4- 2x2-11 2 = 14 a-. Thus it appears that we have to find the values of x y, z, k 8/«f//e"quadratic equation in two unknowns, which is indeterminate. The method of solution will be like that of 247. Solving for x, 32-42 , z-28 14-22 22-14 To make x positive the signs of the two terms of the fraction must be unlike. For this we can assume z = 8, and 10, both of which give positive integral values for x, but only 2 = 8 gives y pos- itive and integral. Thus we have 2 = 8, x = 9, y = 18. Of course in the present problem there is nothing to interfere with z having fractional values within the limits determined by the above equation. 37. A man has $130(), which he divides into two poi-tions and loans at different rates of interest. If the first portion had been loaned at the second rate, it would have produced $3G ; and if the second portion had been loaned at the first rate it would have produced ^49. Required the rates of interest. 832 TEXT-BOOK OF ALGEBRA. 38. Required the rates in the preceding if the two por- tions produce equal returns. How does this problem differ from the preceding ? 39. There is a certain fraction to each of whose terms if 1 be added the product of the resulting fraction and origi nal fraction is |. Required the fraction. QUADKATIC EQUATIONS. 333 CHAPTER XXII. SIMULTANEOUS QUADRATIC KQUATIONS. 340. Simultaneous Equations in Quadratics. See 225. Two cases may be distinguished. First, when two or more of the given equations are of the second degree ; and second, when there is but one quadratic equation given. Perhaps the latter ought to be called simultaneous quadratic and simple equations. See 195. Thus, -] ?^2 T * '^ Z ijj 1^ «i^^ example of the first case ; while, \ 11 -^ y + *^ -^ + 1*;,^ // = }•> iy an example of the second case. SECTION I. Simultaneous Equations, One of wnicu is of the Second Dkgkee, and tmk Others Simple Equations. 341. Solution of Simultaneous Equations where One is Quad- ratic. — To solve such, substitute in the (quadratic the values obtained from the other equations, so as to get a single quadi-atic equation containing one unknown quantity. Solve this and substitute the roots in the other equations. Two sets of answers will generally result. Examples will make this plain. 1. Given(l) a:^ -f 3a:y = 54, and (2) .r -f 4 y = 23 (20 a. = 23-4y. (3) (23 - 4.yy 4- 3 (23 - 4y) y = r,4 .228, 334 TEXT-BOOK OF ALGEBRA. . (3,) 529 - 184 ^ + 16 >f + 69 ?/ - 12 f = 54 (82) 4y2_ii5^_ _475 (334^ 9 115 . 75 -^ ^ ~ ~4~ ~ ^ X y = 23f , or j _?/ = 5 .\ a:' = — 72, ( ic = 3. (229) f (1; a:^ + / + ^^ = 50 2. Given -<[ (2) 2 .r -f 3 // + .- = 23 1 (3) ^4- 2./ + 3^ = 23 Eliminating x and y (so as to have z only in the quad- ratic e(]uation) from equations (2) and (3), and substituting their values, — (2,) 2ic + 3v/ = 23-.~ (3i) 2 a- + 4 7/ = 46 - 6 z (4) 7/=-23-5.^ (xVx. 2) (2o) 4 .X- + 6 // = 46 - 2 ,^ (32) 3:^- + 6//:= 69 - Si'z ~~jc =1 z- 23 (Ax. 2) (1) (7 z - 23)^ + (23 _ 5 ^)^ + ^^ = 50 (228) 49 z- - 322 ;v + 529 + 529 - 230 z^^z^ \z' = 50 75-2- 552.^; = _ loos 25 -2 — 184 ;^ = - 33() 25^2-( ) + (18.4)v= 2.56 ^,,^' _ 1S.4 = il.6 [ ^ == 4. ,,1. ^ 3= ;>_;>(; ^ -; (4) // = 3 // = ().20 ; Ans. (229) [ (5) ./■ = 5 ./• = .52 j (229) 342. Special Forms for which there are more elegant solu- tions tlrni by sidistitution. — If the two given (Mjuations en- yiADKATIC KgUATlONS. 335 able us to calculate quickly the values of the expressions x- rlz- a*// -|- y^f we liave innnediately, by extracting their square roots, the values of x -j- // and x — y^ from which the values of X and y are readily found by adding and subtracting these equations. (255, 4.) 1. Given (1) x"" + y^ = 74 and (2) x -\- y = 12 (Ax. o) [(20 - (30, Ax. 2] (Ax. 6) (Ax. 1) (Ax. 2; = r> and // = 7. 2. {\) X ~ y =. 12 (2) ./•// = S5 (1,) X- - 2 ./•// + //- = 144 (Ax. .')) (2,) 4xy ^ 340 (3) x^ -f 2 xy 4- //« = 4. — 17 (/. A verificHtion will show in this as in all the ♦'xanii)l«'s of this and the prt'ct'ding article that the answers are ohtained in sets. Tims + 17 and + .'> go together, and — .j and — 17. Hut -I- 17 and — 17 as values of j and y would not satixfy the equations. So with all <]nadratic solutions. (-^) .c2-|-2.r// + /=144 (1) x'' -fy^= 74 (^) 2xy = 70 (30 4 xy = 140 (4) a:"-^-2ry + //■- = 4 i^^) ^- - // = i -> {V ^+.'/= 1-^ 2x= 14 or 10 .r= 7 or 5 !>// = looi 14 // = T) or 7. A US. ./ = 7. // = ."> ; or. ./• 336 TEXT-BOOK OF ALGEBRA. 343. Exercise in the Solution of Simultaneous Equations. — Those which come under 342 should be solved by the method there explained. 1. 2. 3. x^y = l x' -{- 2 y' = 34.. 6 x-\-4.y = 23 x^ + 3 xy = 54. 7. 5x — y = 17 xy = 12. 8 \3x - y = 11 I 3 x' - / = 47 i x-\- !/ = !'"> I xy = 3G. (3^-2y = 2 }9x' + iif = 394. \ xtj = 923 (.• + 2/ = 84. xy = - 2193 x + !j=-^. < J' ' y 10. \2x-\-y = l )4^^ + / = 25. 11. '^^ + f + ^'.y = 20S ( ^ + y = 16. SuGGESTiox. — First find the value of xy and then solve as in the last article. x — y = 3 x^ -3xy-^tj^= - 19. Suggestion. — Divide the first equation by the second in Ex. 13. 12. 13. 14. 15. ic2 _ ^2 _ 16 x — y = 2. 4. (xhj'') = 13 xy X — y = 6. y ' X ^ x-\-y = 6. 16. 17. 18. [ ,1 2x-^y x-{-y 3 4 X — y [ * + ^^^±1^0 x + 3 [lx-^3y=l I s + ^ = ^ QUADRATIC E(JUATIONS. 337 19. jc , 21. ' '', .. ... y ^^- ^ {X - 2) - (// - 3) = b. 22- ( ry^ + r/y^ = f (333,3) (1.) ^ =py i V«=' ^pY (Ax. (>) (20 y^ + 2 q {py Jz V ^^ + y>V) // = f' (228) i 2yy V«^ + pV = A - / (1 + 2pq) (328) 4 qh/ («« -f 2>y) = ^.^ - 2 hi/ (1 4- -' />v) + // (1 4- 2 pqY (Ax. 5) which is an equation of the fourth degree and cannot in general be solved. a. The general equation of the second degree in two unknowns, i.e., one which contains every possible term, is ax2 + hxy + nj'^ + (Ax. 6) (4) ic'^ - 2 a-y + / = a- 2 /> (Ax. 2) (40 0^ - y/ = ,i, V^^ - 2 ^> (Ax. 6) i V«T^J7^ i Va - 2 /> .-. X = -F= y = ^ Remark. — The sign ± (read "minus or plus") means minus fit\st. 3. {I) xy + Qx + 1 ij = 50. (2) 3 if?/ 4- 2 ^ + 5 v/ = 72 (li) 3 .ry + 18 ic + 21 y = 150 (2) 3^y+ 2.^+ 5y= 72 (3) 16 a; + 10 2/= 78 (3.) .... = ^;-i-^ (l)^«^^.. + ??iii^ + 7, = 50 (228) 39 y - 8 y-^ + 234 - 48 // -^my = 400 8 y2 __ 47 y = _ 166 47 ■ 166 ir^=--8- ^ ~ 8^ + (^16J ~ 256 ~ 256 47 , 1 / ^ = 16^16^-^1^^- (3) ^ = +f^=p^V^=-3103. (229) 4. (1) 2ic2 3^^_|_y2^63_^ (2) ;r2_^2ify + 6//= 174. Put y =z rx, r being unknown. Tlien substituting the value of y in each equation. y^-^y = Ql'ADliATIC i:gl A ri<>N.>. (1) 2x^-^3rx'-\- rV^= 63 (2) x^ -\-2 rx^ + (•) r-y- = 174 (1.) (2-h.Sr + r^)u-^ = G3 .. + ;-2 330 (97) (Ax. 4) ,1.\, ,^1 -f-L',--^ (•,/-) .•^=174 (97) > 174 ,, ,, 03 1^4 ^ 2T37T^ - i+2r + 6r2 ('*'J1, Ax. < ) 63 + 126 r 4- 378 r*^ = ;U8 + 522 r + 174 r* (Ax. 3) 204 r* - 396 r = 285 68 r^ - 132 r = 95 (Ax. 4) GM = 4 X 17 ; .-. 4 X 17 X 17 is a perfect square = ^Ul 17 X m r' - 17 X 132 r = 1615. (34 ry - ( ) 4- (33)-^ = 1615 + 1089. 34 /• - ,\3 = i 52 .. . 2 "'^l* _ 1 ^5 — ^ (229, .-. a: = 4: 2, or -t V^^^iif* ^^w. y=/-^ = §X(i2)=4,5,and^i|V^^ J//.s-. 5. Given (1) 3.r// — 4.r - 4// = (2) x*^ + y^ + ar + y - 26 = (1,; 2xii-%{x^y)=^0 (2i) a:'^+y^4-(^-hy) = 26 x^ ^2xy + 1/ - ^{x-\-y) = 2^ (Ax. 1) (^ + //)''-j|(^ + y) = -'6 (^ + yy - .«^ + y) + {^y ='26 + il^ = ».V (335j 340 TEXT-BOOK OF ALGEBRA. (^ + y) - ;} = i V X -{- t/ = (j, or — ';•' a; = 6 - //, or - \-' — 1/ (1) 3(G-^)./ + 4y/-?4-4y = (228) yy2 _ y/ = - 8 y-2 _ 6^ + <) = 1 //-o=il w y = 4 or 2 j ir = 2 or 4 \ A?is. (229) (1) ^(-¥-y)y+¥+42/-4y/ = (228) - 13 ij- 3 f -\- %^ =0 36 y' + 12 X 13 n = 208 (334, Ax. 3) 36 7/^ + ( ) + (13)'- = 169 4- 208 = 377 6 y + 13 = -[- V377 - 13 -t V377 2/ = — 6 (4) - 13 =f V377 >Ans. 1 6 J (229) 345. Classes of Simultaneous Equations, two or more being Quadratic, and the others, if any, simple (344). Methods of solution. 1. Special forms akin to those explained in 342 can be solved by a similar process. See Ex. 2, 344. 2. When each equation contains only one term of the second degree, and that term has the same product or square of the unknowns in all of the equations. See Ex. 3 of the preceding article. 3. When neither of two given equations contains the first power of either x or ?/. This is usually spoken of as the homogeneous case, and to solve it y is put equal to rx, or x equal to ry. See Ex. 4. QUADKATK IJjl A I'loN.s. 341 4. When an equation can be framed which can be solved tor \x function of the unknowns. See Ex. 5 wliere the func- tion regarded as unknown was x -\- //. 5. Miscellaneous methods. — Various substitutions, and expedients of other kinds, are often of use in effecting solu- tions in i)ai'ticular problems. Suggestions will be made in the proper places to meet such cases. a. The methods just explained are not mutually exclusive. Some- times an exercise can be solved in a number of different ways. In some instances particular methods have decided advantages over others. 346. Exercise in the Solution of Simultaneous Quadratic Equations, two or more of which are of the second degree. Jx'^-hy^=17() ( a:^-f 3a-y + r = 5 "■' \ xij = i;i \2x^-\-xy-\-2 y" = 5. ( 3a;«-a-y-|-8y2 = 13. ( 8 x// = 336. •■l'(?+')-y = 4. o J.2 _ 3 ^y -}_ 2 ;/2 _ 25 = 0. 342 * TEXT-BOOK OF ALGEBRA. 14. 15. 16. 4:Xf/ — 3 v/^ = 3 a. 2i/ — 4:Xi/-\-3x'' = 17 y^ — x^ = 16. ( (1) :r2 + 4 f + 80 = 15 :r H- 30 7/ I (2) xi/ = 6. Suggestion. — Multiply (2) by 4 and add it to (1). Then ar- range in the form (x + 2 yY — 15 (x + 2 2/) = — 56. Solve this quad- ratic for X -\- 2y^ getting two values for it. j x^ -\- 3 xij = 54 ( xij + 4 f = 115. Suggestion. — Add the two equations and solve for x + 2y. 9 x'' + //2 — 63 X- -\-21y -\- 86 = xi/ = 4. r 1 1 485 x^~^y~W76 11 _23 17. 18. 19. <^ 1/ 24 20. Suggestion. — Solve such exercises in the reciprocal form, i.e., without clearing. ' x'^ ~\- y = Ax y-2 -f .T = 4 //. Suggestion. — Subtract one equation from the other and divide through by x — y. r(i) .^2 _^ ,/ -f .^--^ = 30 {{?y)x-y-r. = 2. Suggestion. — Add 2 times (2) to (1) and extract the square root of the result. 22. 23. £C2 4- y2 _j_ 4 V^--^ + If' = XI xy = 12. j £c2 -f 2 J-// + 7/ + 3 x = 73 QUADKATIC EQUATIONS. 343 28. 9A \ '■ + '^ -f . Suggestion. — Divide (1) through l>y Vx — Vy. U a-* -i- s^ = i [ y^ + ^2 = c. Suggestion. — Add the three equations and divide through by 2. From this equation subtract each of the equations in turn. ^ [ X = fl Var + y Suggestion. — Square each equation and tluMi sul)traet one fr(Mn the otlier. Tlu* resulting etiuation is divisible by .r + //. Also divide one of tb«* «M|U!iti()ns by tli«* oiIht. SECTION III. — Problems. 347. Problems Involving the Solution of Simultaneous Quad- ratic Equations. 1. Find two numbers wliicli nuiltii>li«'(l ^ivc nTd and divided the one by the other give 2\. 2. Two nnml)ei-s are to eacli other as 11 to 1,'^, and the sum of tlicir sfpiarps i< 14210. Wlint are tlif niinil»Ts? 344 TEXT-BOOK OF ALGEBRA. 3. Find two numbers whose sum, product, and difference of their squares are all equal to each other. 4. What number being divided by the product of its digits gives the quotient 2, and if 27 be added to the number the digits will be inverted ? 5. What numbers are there whose sum is 100 and the sum of whose square roots is 14 ? 6. A and B have each a small field in the shape of a square, and it requires 200 rods of fence to enclose both. The sum of the contents of these fields is 1300 sq. rods. What is the value of each at $2.25 a square rod ? 7. AVhat two numbers are those whose sum multiplied by the greater is 120, and whose difference multiplied by the less is 16? • 8. Two farmers together drove to market 100 sheep, and returned with equal sums. If each of them had sold his sheep at the price the other actually did, the one would have returned with $180 and the other Avith $80. At what price per sheep did they sell respectively and how many sheej) had each ? 9. The small wheel of an ordinary bicycle makes 135 revo- lutions more than the large wheel in a distance of 260 yds. ; if the circumference of each were one foot more the small wheel would make 27 revolutions more than the large wheel in a distance of 70 yds. Find the circumference of each wheel. 10. A tailor has noticed that broadcloth on being wet shrinks up ^ in its length and ^\ in its breadth. If the surface of a piece of broadcloth is 5^ square yds. less, and the distance round it 4| yds. less than before it was wet. what was the length and width of the broadcloth originally ? 11. A cistern which is half full can be filled by one of two pi])es in a certain time and eni])tied by another in a dif- QUADKATK i:(,)r ATIONS. 345 ferent time. If both pipes be left open, the cistern is emp- tied in 12 hours. But now if the opening of both pipes be macle smaller so that the one needs an hour longer in tilling and the other also an additional hour in emi)tying, then if both pipes are left oi)en, 15^ hours are needed to empty the cistern. In what time can the empty cistern be filled by the first pipe alone, and in what time can the full cistern be emptied by the other acting alone ? 12. A certain number of laborers remove a heap of stones from one \)\aee to another in 8 hours. Were the number of laborers 8 more, and did each carry 2]^ kilograms less at a time, then the heap would be transported in 7 hours. But if tlie number of laborers were 8 less, each carrying 5 J kilo- grams more, then the heap would be removed in 9 hours. How many laborers were there, and how many kilos did each carry at a time ? 13. Find two numl)ers whose sum added to the sum of their squares is 42, and whose product is 15. 14. Find two numl^ers such that their product added to their sum shall l)e 47, and their sum taken from the sum of their squares shall leave 62. 15. Find three numbers such that when the sum of the first and second is multiplied by the third, the product is 63; when the sum of the second and third is multiplied by the first, the product is 28 ; and when the sum of the third and fii-st is multiplied by the second, the product is 5o. Sl'goestiox. — Solve in a manner similar to Ex. 28, 346. 16. The diagonal of a lx)x is 125 inches, the area of the lid is 45(M) sq. in., and the sum of the conterminous edges is 215 inches. Find the lengths of these edges. 346 TEXT-BOOK OF ALGEBKA. CHAPTER XXTII. QUADRATIC EQUATIONS AND EQUATIONS IN GENERAL. 348. Properties and Solutions of Quadratic Equations and Equations of Higher Degrees. Some of the properties of quadratic equations obviously belong to equations of other degrees. It will be found also that many equations of other degrees can be solved like quadratics. It will be con- venient besides to introduce into this chapter the general discussion of problems and the validity of processes of solution. SECTION I. Properties. 349. The Sum of the Roots of a Quadratic Equation. — When the coefficient of x^ is unity, the coefficient of x taken negatively is equal to the sum of the roots. If in the type form of quadratic equation, ax'^ -}- 2 bx == c, the right member be transposed, and the equation divided b c through by a, there results x^ — 2 - x — ~ = 0. To simplify b c the equation we replace — 2 ~hy p, and — - by q, and let us call the new ecpiation the normal for7n. x'^ -\- px -\- q =0 4a;2 + 4^^^ = _ 4 ^ (334) {2xy + ( ) + ;>2 _ ^2 _ ^ (335) 2 £c -f ;; = _j_ Vi>^ — 4 ^ ^=-hp-\-\ V > ' — 4 y OT X = — ^j) — \ yj p'^ — 4 5' QUADRATIC ?:QUATIONS. 347 If now the two values of a*, which are the roots of the e() nation, be added, the radical parts cancel and the sum is — yj, i.e., the sum of the roots is equal to the coefficient of x in the normal equation taken with opposite sif/n. Q. E. D. 350 The Product of tbe Roots of a Quadratic Equation. — The product of the roots is equal to the known term in the normal equation. Multi])lying together the two roots of the last article, {;-^v/. 4 - \ 0'' - 4 7) = '/ Q K. D. 4 4 351. Solving a Quadratic Equation by Factoring. — By com- paring the theorems of the last two articles with 116, 8, we learn tAxat fartorifuj the normal form of any quadratic equa- tion gives its roots. 1. Given the ecjuation x- -f- 7 ar- 4- 1- = <> to Hud its roots. Ftictoring the left member. (X 4- 4) (.r + :\) = 0. (135, 2) By 349 and 350 the roots are — 4 and —?>. For, (— 4) -(- (— .S) = — 7 which is the coetficient of x taken neya- tirt'h/ : while —4 X — ''^ = + 12, which is the known term. Hence, after factoring, tiike the second terms of the factors with their siijns rhaufp'd for the roots of the ecjuation. 2. x'—Q, -\-h)x -!-///> = () (:r — a)(x - h) =0. (135,2) li«Mice the roots are a and //. XoTK. — Compare this solution with that <;iv<'ii l>v the rrmilar process, Ex. 32, 338 3. Solve x' — 7 r -f- rj = by iiicloriug, and al.s«» Uy completing the square. 348 TEXT-BOOK OF ALGEBRA. 4. x^-9x-\-20 = 0. 9. :c2_20a: -300=0. 5. ^2 _^ 11 ^ 4- 30 = 0. 10. x' -Sx = — 15. 6. a;2_^13a; + 12 =0. 11. x' - llx - 50 = 160. 7. ^2 + 21 ^»^ 4- 110 ^2 = 0. 12. x^- |a.T + f4 =0. 8. a-2 — G a; — 27 = 0. 13. 6 a^^ _ 17 ^ ^ jL2 = 0. Remark. — Factoring as in 135, 3 we get (3 a; — 4) (2 x — 3) = 0. Or (x — |) (x — |) = by dividing the equation through by 3 X 2 (Ax. 4.) Hence the roots are 4 and |. 14. a • + 4 + ^- ^^^ = 13 17. 6 ic2 _ ^ _ 12 = 0. X 15. ^ili + ^^H--« = 7 18. ^2_o3^^o.7. X x^ 16. ^+3i=y^ + 8. 352. Solving Equations of Any Degree by Factoring. 1. Let us verify some of the examples of the last article, substituting the values not in the original but in the factored form. Putting X = — 4 in Ex. 1, we have (-4 + 4) (-4 + 3) =0, or, (- 1) = 0. (Ill, a) Thus when one factor of a product is zero the product is zero, and so the equation is verified. Substituting x = — 3 will make the second factor zero and so satisfy the equation. (189, a.) Likewise in the second example, either x — a, ov x = h, will make one of the factors zero, and so satisfy the equa- tion, and they are therefore roots of the equation. 2. But this reasoning will hold for any number of factors as well as for two. Thus the equation (x — 5) (x -\- 2) (x — 1) = (oi^, x^ — ^ x^ — 1 X -\- 10 = 0, by multiplication) is satisfied by either .r — 5 = 0, a* -f- 2 = 0, or a^ — 1 = 0. .-. ./' = 5, a* = — 2, X = 1. And these are the three roots of the equation. (,»l AI)1;ATI<' K(,>rATInNS. 349 Again, taking the tujiuition x*- -j- 5 u: — 4 = 0, we may write it in the form CJ ./• - 1) (.S ^ 4- 4) = 0. (135,3). Placing each factor in tmn ripial \n /cid. 'J J- — \ ={) .-. ./• = 1. : an. I ."I .<• -}- 4 = .-. ./• = — * 3. When the unknown, a*, divides out of an equation, one root of that equation is zero. Thus in the equation 2x^-nx' -\- V2x^x{x-4) C2x-'A) =(), placing each factor ecjual to zero, ^ = : r — 4 = (I. .-. ./• = 4 ; 2 J- — .'J = i). .'. x = [\. Manifestly r = satisfies the ec^uation 2x^ — li x'^ -\- 1- r = (). foi- it makes every t«'rm (Mjual to zero. 353. Rule for Solving Equations of Any Degree when they can be Factored. 1. 15y ti-ans position if necessary make the right member zero. 2. Factor the left meml>er into binomials the first term of each of which is a multiple of the unknown. 3. Set each fiu-tor equal to zero, thus obtaining as many values of the unknown as there are binominal factors. The' following corollaries of this method are highly im- |K)rtant in the genei-al theory of equations. Their truth is merely suggested here. a. An equation has as many roots as then' are units in its yl»ose right member is zero, then a in a root. Hence if we can find one or more factors wlicthir we can find the others or not, tlicsc factors give roots. c. If one or mun- untoi-s an* known, the equation divided by these leaves an equation which contains the other factors, or the other roots. 350 TEXT-BOOK OF ALGEBRA. 354. Exercise in the Solution of Equations by Factoring, 1. x~ -f 23 ax -f- 90 a' = 0. 2. G x'-" + 17 X- -f 12 X = 0. • • 3. V2x^ — 9 X- — 8 it- -[- 6 = 0. 4. cf3 4- a;2 + a; + 1 = 0. 5. ic (:z; - 6y + 10 a: — 60 = 0. Suggestion. — This equation is evidently divisible by x — 6 giving a quotient x- — 6 x + 10 = 0. Hence G is one root and the others must be found from the equation x- — G r + 10 = 0. It is not pos- sible, however, to resolve this equation by factoring. Turning then to the old process of completing the square the roots are readily found to be X = 3 ± V — 1. So in all cases. If an equation can be reduced down to the second degree, the solution can always be perfected by the old process. 6. x^ + Ax' — ix — 16 = 0. 7. x^ — 2x^ — 16 x^ + 32 X = 0. S. x^ — 31 x' + S^x -1 = 0. 9. x^ — 5 x^ - 6x2 ^ 15 ^. _^ 9 == 0. Suggestion. — Arrange into x^ — G x- + 9 — 5 x (x^ — 3) = 0. 10. a;* -f 7 ax^ - 2 a^ x^ — 2S a^ x — 8 a\ Suggestion. — Arrange as follows : — X* - 2a2 x2 - 8 a* + 7 ax (x^ - 4 «2) = o. 11. One root of the equation x^ — 3x'^ — 14: x -\- 42 = is 3 ; find the other two. 12. Two roots of the equation x^ — 2 x^ -{- 1 x'^ -\- 2 'x ~ 5 = are 1 and — 1 : find the other two. 13. Solve the equations 27 x^ — 1 = ; x^ -\- 1 =0. 355. Conversely. — To Construct an Equation when its Roots are given : subtract each root from x and iuulti])ly all the resulting binomial factors together. ((. Thus far we have always found irrational and imaginary roots occurring in pairs, one with the — sign before its radical term and (,haih:a ric i.(,»i a rioNs. 851 the other with the + sij^ii before its mtlical t«'nn. They are some- times deseril>e(l as eonjiijjate surds or imaginaries. Uixm multiplying two sinli liinoniial factors a trinomial with real and rational coetticients results, i'hus taking the two imaginai7 roots of example 5 of the preceding article, a; - (3 + V -1) g - (8 - V - 1) x2 - 3 X - ar V^^ „ Z 3 X -f- a V^^ 4- 9 - ( -- 1) x*-6i + 10. Furthermore it is easy to see that if the radical terras did not dis- appear from this proiluct, they would certainly not disappear when multiplied by dissimilar radicals in the other roots, and so would not disappear in the final product, which is the equation. Ilejice we conclude that if an equation liaviiii: real coetticients does have imag- inary or irrational roots they (><( m in pairs, and should be taken toyether in reconstructing the e"" — iq. There are two classes, real roots and imaginary roots. 1. Equations having real roots. In order that the roots may be real the radical quantity ■y/p'^ — 4:q must be zero or positive. (1. li p^ — 4y = 0, the radical in the roots disappears, and X = — i^>, or X = — Ip, i. e., the two roots are equal. 7>^ I P In this case, x^ -^ px -\- q ^=i x^ -\- px -\- jr-^^ ( •^' + 2 X + '^ ) =0, thus giving two roots each equal to — \-- (351) (2. If p'^ — \q'> 0, the roots are real. Let us suppose a and h to be the two roots. The signs of the roots depend upon the signs of p and q. Compare articles 116, 8, 135, 2, and 351. (1) x'-\-yx J^q^ix-^ a) (x + ^)= 0, or, when 2^ and q are both positive, the roots are both nega- tive. (2) x^ -px-\-q ^ {x - a) (x -b) =0, i.e., when p is negative and q positive, the roots are both positive. yL'Ai>i; \ Tie kl .\I>1:A IK I <,»l .\ 1 1(»NS. 355 2. Given y as tlie implicit lunctioii of x as contained in the ev substituting we Icaiii that then* is l)otli a niaxiimiiii and a niiniunini. 11 .r == — '2 or any greater negative number, y is imagi- nary. ^lfa- = — 1 j- = x-\ x-'l then >/ = \ +11/= lO. 2/= i— Ah y = \— .4.> , ( /or, + 2' (4.' ( + 4.4.-)' j + 4.4;"> a- n :j J. = 4 \r i + 25 and ic = 5 or any greater posit i\f immbtM-. y is imaginary. To find the exact value ,/- :{- a:^ = .'{ •'• + 4 y X' - 3.r + 5 =4y~yH - 5 (335) ^ - i = i V^ + 4 // - //^ Then ? -|- 4 // — //- = (See previous exam})le) //•- — 4 // = ; //•^ - 4 // + 4 = V y ^ 4.5 or — .5. Returning to the series of values given above we see that 4.5 is the maximum vahu^ between 4.4.") and 4.45 (the func- tion was increasing at the tii-t ainl d' • i. -ising at the second, so that it must have reached a maximum at scnne point l)etween), Avhile — .5 is the minimum between the values — .45 and — .45. Substitutini.c (229), the ('(HTcspoiKliiig value of ./• is lomid to 1k' 1.5. These values may be inserted in the above series, a- = l r= 1.5 a; = 2 Thus, ( - .4.-) ( -.5 ( - .45 ^= ( I.4.V'/= \ \.ryy= \ 4.45 356 TEXT-BOOK OF ALGEBRA. 3. Divide a given number a into two parts such that their product shall have the greatest possible value. Let X = one part a — X = the other part then X (a — ic) = ?/, is a maximum. x^ — ax = — y cr (t^ -y X- - «-^- + 4 = T X 8o that for the maximum ?/ = whence y = —. Corresponding to this, x = |, i.e., the maximum product is obtained when the two halves of the number are multiplied. 4. To inscribe in a triangle a rectangle of maximum area. There is such a rectangle ; for if we conceive of a rec- tangle at the bottom of the figure having a base nearly as long as the base of the triangle, BC, (and consequently a very small altitude) have its base diminish in size and its altitude increase, it will finally come to have an altitude nearly equal to the altitude of the triangle, but with a very small base. Evidently if the rectangle first in- crease in size and then diminish, there must be some posi- tion for which it has a maximum area. By a theorem in geometry, since the triangles ABC and ADE are similar, AP : AQ : : BC : DE : Call BC, h ; AP, h ; and y, the area of the rectangle. Required to find the maximum value of the last. (.•lAhUA I'M l.f^tr A rioNs. Substituting in the precedm-- proixtitioii, h:h~.r:: h : \)K. // unci since the area of a reetuni^U* = imse x Jiltitnde. V - h hij = hhv - hx- (Ax. 3) \lSij,i _ 4 h^hx = - 4 hlnj (334) 4 yijc'^ _ 4 U'hx + Irh^ = h'h'- - 4 bhf/ 1> f,x — bh = i VV>'//- - 4 hinj. Now i*/('^ — 4 bhif = t^'ivcs tlu' iiiaxinuun. Hence 2 />.r — hh = ; or ./• = !; : //-A- — 4 b/t i/ ^ : or // Tluis the niaxiniuni rectangle lias for its altitude ha/f the altitude of the triangle, and for its area ^ of the btise X the altitude, which is half the area of the whole triangle. 5. Determine the rainimuni and maximum values of the function y = ^—^ — ^ ti" *^^ ^®^^ values of x. •^ (>:r — 14 a 358. Discussion of the Equation x = t- for Limiting Values of '/ and //. Infinites and Infinitesimals. a. As stated in 62, the symbol is often used to denote any ex- ceedingly small number less than any that can lie named. Sueh quantities are calle')>< TKXT-r.OOIv OF Al.CKHP.A. 1. To find the value of y when the denominjitor h is finite^ but the numerator is infinitesimah The result may be arrived at perhaps better by an exam- ple than in any other Avay. Suppose b is any number, say 500, and a assumes the values 100000, 10000, 1000. 100, 10, 1, .1, .01, and so on without end,- 500 ^'^y 500 -^^h 3^0 — -^ ? 5 00 — 5? JJL = J„ - 1„_ = -1^ _l_ = _J _^(LL — _^1 5 5 0? 5 5 05500 500055 500 0? and so on without end. Evidently the value of the fraction gets smaller and smaller, or what is the same thing ap- proaches the limit zero. Stating this ])rinciple generally, if in the fraction j, the numerator approach the limit 0, the fraction approaches the limit 0. Stated in the form of an equation : a (1) AVhen a -j= 0, x = 7=0. 2. Similarly the value of j. wlien (/ x^i, finite and i> iiifi iiit<\ a])])r();icli(^s the limit zero. 3. To find the value of j when a is finite and b infinites- imal. This is the reverse of the first case. AVhatever finite value the numerator may have, as the denominator dimin- ishes the value of the fraction increases, and ultimately when the denominator has become infinitesimally small, the value of the fraction has grown infinitely large. (2) WhenZ»:^0, ^^x. 4. AVhen b is finite and a is infinite the fraction ap- proaches the same limit, cc . ( M'AI>l: \ ! 1< 1 • 'I A 11. "NS. 359 5. lo find tli»' valiu' (»t ./• == when lx)th a and h are inhnitesiinals. In the first exuniph* given, how t'\ ci' small the (U'noiniiialor might become, the value d iIm- traction wtis always 5^^^ of it. Thus of these two infinitesimals one was 500 times greater than the other. In general when h is finite and a infinitesimal the value of the fraction is an infinitesimal which is -J times the first. Now, if two quantities grow small together and we have no means of knowing their rel- ative size, ihe mJin' of the frurfltni Is huli'frrttnntifp, (3) AVhcn (I ' o. and A " »». ./ is ui(h't(>rw'niate. h However, in such a fiat lii mi i ^ , \i x be infinitesimal, the value of the fraction does not become indeterminate, since the some infinitesimal is in l)oth terms and so divides fl2 _j_ i/t out. giving T). Likewise the fraction j- does not be- (M»m<' 1 ndeterminate when . / - xf m n X -^a-x ma — ')nx — -1- nx (m :^ n) X = ma ma (Ax. 6) m m ^ — , I.e. j — a, or a. m -J- w 7n-\- 11 r/i — 7i From the v^alues of cc, we learn 1. That there are two points equally illuminated. 2. One position of the point of illumination may always be found ivithin the line PQ. For, since by the nature of the problem m and n can only be positive numbers — , — ■^ -^ 7n-\-n will always be a proper fraction, and therefore ; , - <^ is less than a, and positive. Now we started by assuming that a position of I would be found to the right of P, and the positive result verifies our assumption. 3. The other position of the point of equal illumination is either to the right of Q or to the left of P according {,n ADKATic i:<,)rA'ri(>Ns. 361 as w is greattM- or less tlian n. For, is positive ami m — n (jreater than a when in — n is positive; i.e., when nt > n. Ill tliis case the second point of equal illumination lies ma beyond Q. And is negative when n > 7w, and the second point of equal illumination lies to the left of P. Moreover, all this is in accord with the conditions of the problem ; for the two lights will illuminate equally a point between them situated close enough to the weaker light to make up for the stronger illumination of the other. Besides this point there will be another, situated on the side next the weaker light, so that here again the closer proximity of the one light is offset by the greater intensity of the other. 4. Certain special cases are worthy of attention for the analytical results they give. (1. Suppose m = 0; then x = + or — 0. (358, 1) Here while the light at P is infiiiitesimally weak, the two points of equal illumination are infinitely close to it on either side, making up over the other light in closeness what is lost in intensity. This can be made clearer by an example. Suppose at the distance of one foot Q is very bright and P is tery weak. Tlien at J ft. P is 4 times brighter; at \ ft., 10 times brighter; at j^ ft., 10000 times brighter; and so on. Thus by going closer the illuminating power of the weak light may be made to increase indefinitely. Of course this reasoning assumes the source of the light to be a mathematical point. (2. Suppose m = n, then ar = J a, aiul x . (358, 3) Here one position of I is midway; the outside position is at infin- ity, for it is only at such a distance that the advantage one light has over its equal is reduced to an infinitesimal. (3. Suppose a = 0, then ])<)tli values of ac = 0. This is the same problem as the general case, only in miniature. The value of a and the two values of x are by no means equal. What is conveyed by representing all three by the limit 0, is that all 362 TEXT-BOOK or ATXJKP.IIA. are infinitesimals, and that the two points of equal illumination, one between and the other outside of the infinitesimal distance between P and Q, are infinitely close to them. (4. Suppose a = 0, and ni ^^ », then x = 0, and ^ a. (385, 5) Here the lights are infinitely close to each other, and the one point of equal illumination is midway, and of course, infinitely close to P and Q, while the other position is indeterminate, since we do not know the laws by which the quantities approach the limits. If the lights are practically equal, and at practically the same point, they will illuminate any and every point on either side alike. 360. Validity of Processes of Solution. There are certain ]»i'0('es.ses in the solution of equations which deserve special consideration lest the student be led into serious error in using them. 1. I?i ihc erf I'iicf hill (if fill' xdiiii' root of fliP tiro memhers of 'III equation. (1. Thus, while (5 — 2)'^ = (± :])-, 5 — 2 = 4-3 only and not — 3; that is, the roots of both members must be taken with the same sign. (2. Fiom ./■■- — 2 us +(/- = ./• - r/ = ± and there are iiKo values of ./• each equal to a. (3. The Equation 12 .r — .5 + 2 V 30 x- — 27 .r + 5 + 3 a* — 1 = - 2. (See Ex. 3, 327) gives V 12 .<■ - .-) + v^ 3 .r - 1 = V 27 .r - 2. (Ax. (5) As shown in 327, while x = 3^ satisfies the first equation, it does not satisfy the second, except when the signs of the radicals are taken in a certain way. This anomaly is caused by the presence of imaginaries. 2. An equation ohfaiiwd hy squavinr/ or cnhing, etc.. tlw tiro Diemheri^ of o n ("/luifioii irlll, in f/e)i.eraf hare root.^ ir/iic/i (to im^ s(iti>N>. 8«)8 X =X' (Hyp.) x-^ = X'^ (Ax. 5) V2_X'^ = (Ax. 2) V -■_>;■-• _(X+X')(X - -\')=0. r*\ the ilieoiy ot (•quations, to tiiul tlie roots of this equa- tion the two fiictors are in turn set e(iual to zero. (353) .-. X — X' = 0, and X -f X' = o. Now the hrst of these is the orhjindl equation, while tlie second is a new equation with new roots. Thus, Ex. 0. 329. \ 4 or -H 5 — ViC = Vcc •+- 8 Squaring and transposing, and afterwards factoring 4 jt + 5 - 2 VlWVhx ^ X - (x + 3) -- ( V4 J- -I- 5 - \/.r + \ s +~3) (Via* + 5 - V? — y/x + A) = 0. the second factor bein.!? the oriijinal equation, uliilc tlu' first fur- nishes new roots. (2. To show tlu' sain(^ for cultni- ;iii t'(pi;ii ion, X = X' (Myp.) X-' = X''^ (Ax. T)) X3 _ X'« = (X - X') (X- + X>^' + >^"',^ = <>• ^134, L>) Here, besides the old equation, tluM-e is the factor X- -|- XX' + X'- = i) whi(;h lias new roots of its own. And so for other jKJwers. To ilhistrate the principle further, let us solve the (M| nation, v^TsT+lo = i (x + 8 ) ( V^ - 1). x« + 5 X + 10 = x^ + 9x2 + 27 X + 27 (Ax. :,) 9 x* + 22 X = - 8 9.x2-f-22x+ (V)'^ = V 3x+ V = ±i X = - 2, or - ^. Upon substituting these values in the original equation, wo readily find that neither of them satisfies it. They do satisfy, however, the S64 TEXT-BOOIv OF ALGEBRA. equ ation framed as X^ + XX' + X'^ = 0, viz., V{x^ + 5 a; + 19)2 + i ly or divide by any known finite quantity is therefore always adniissil)le. i)X — a X _4x — 2a4x — n TTV ■•" xTJ) ~ ~~x + i( ■*■ "j-TlT "^ ^• Clearing and simplifying, we have (x + a)x = ix + a){4x- a). (Ax. ;J) (x+ «)x-(x + «) (4x-a) = 0. (Ax. 2) {x-\- a)(x — (4x-a^) = (x + a) (—Sx-\-a) = 0. (136,3) Hence x + a = 0, or x = — « is a root of the equation. But the first, third, and fifth teniis of the given equation taken together vanish. Consequently, it was not necessary to nuiltiply through by X + «, and when it was done, the extraneous root x = — a was wrongfully inserted. y +- 1 +i 2/ V (.'). Clearing the equation r H j = 2 of fractions we get •^ y 1 1 11 11 clearing again, collecting and arranging 4 y2 + 4 y = 8 .'. y = 1 or — 2. But while y = — 2 satisfies the original equation, y = I does not. To find the reason for this, we simplify the left member (See Arts. 183, 162), obtaining y--\ y-\ cancelling, we have left, 2 2 1/2 \ The second factor — j_- j +2 = gives y = — 2 as above. The first factor -^Z\ equals « when y = I. (368, 3). Thus in clearing of 366 TEXT-BOOK OF ALGEBRA. fractions the equation was multiplied through by a factor (y — 1)2, which for a particular value of y equals x . Hence we have no right to expect that this value of y will satisfy the given equation. Tlie principle stated at the beginning of 3 of the present article is of general application and explains the apparent difficulty in 2, as well. It makes clear also the fallacy in the pretended proof of the arithmetical absurdity that 1=2. The " proof " is as follows : — If a = X, then ax — xr = cfi — x^ X (a — x) ^ (a + x) {a — x) x = a + X m (Dividing out by {a — x)) x = x + x = 2x (Since a = x) 1=2 (Dividing out by x). Here the second equation is divided through, according to the hypothesis by zero, and it does not follow that the next equation is true for the same value of x. Thus, X (any number) = X (any other number). It may be said in closing this article that, if a student is at any time unable to tell a 2>^"^ori whether a certain value of the unknown belongs to the initial form of the equation, he has this recourse, that he can substitute in that equation and find out. SECTION III, Equations of Higher Degrees Solved like Quadratics. 361. Equations containing but one unknown solved like incomplete quadratic equations. — Binomial Equations. — Incomplete quadratic equations and similar equations of other degrees are often called Binomial Equations, because when reduced they consist of but two terms. 1. Solve 2 .t4 _ 9 = 7 -|_ a;4^ x'^ = 16 (Ax. a) :r2 = JU 4 (Ax. 6) X = \/T"4 = i ^ ; or V^=^ = -t 2 V-1 . (Ax. (;. 353. :x^a = ia V-3 These two roots along with <( form the three roots of the equation (353, a). These values. — a -{• a V— .') — (I — (t V — :; «, 2 ' 2 ' are sometimes called the three cube roots of the quantity a^ or, if " — 1. "f iniity. 3. ./• " = " ; iHMpiired to tind x. x»* = a" (Ax. 5) n x s=sa m. (Ax. 6) NoTK. — This is the general ease. If the exponent of the un- ktiov.ii is negative, make it positive by transference to the opi)osit(' tt nil of its fraction. 4. Given px~^ = q- SOLUTKIN '\ = U \ p^nx^'. x^ = ^ ', X = ^ . A/fS. 6. X^= 1. SrG(iEsTiox. (.r« — 1) = (./••■* -\- 1) (.!•* — 1) = 0. 6. X* = — a*. Suggestion. —Transjwse — n\ and then factor .r* + «* (134, 5). The factors set equal to a^ro will give the four roots. 368 TEXT-BOOK OF ALGEBKA. 7. Find only the real roots of this and the following • 9 x'' 8. 4.*-' = 500. 9. ci-^==81. 10. ^x ^ = '2 V2. .T- — na nx^ — b 11 '^ -r '^- -r o I -^t^ 1- ♦^ — o ^^ 2 12. 13. a;'- — a x'^ — b a b x^^ — b x^ — a' 14. a;S — 32=0. 15. x^ = 390625. 16. (ic2 + 3 ^ + 2)2 = (cc2 „ 5 ^ .|.. 10)2. 17. (ic2 _ 3 ^ _ 10)2 _ 15 ^^ _ 5^2^ 362. Equations containing but One Unknown Solved like Complete Quadratic Equations. There are two type forms of these equations : ax^^ -\- bx" = G, the simple form, [f(^)Y"-\-b[f(^)T = ^5 tl»e complex form, f(x) in the latter meaning any f miction of x. 1. Solve 3 ic« + 42 :r« = 3321. x'' ^\^x} ■= 1107. (Ax. 4) a;6 _f_ 14 a;3 _|_ 49 _ 1155 ^335^ £C3 _|_ 7 =3 _|_ 34 x^ = 27 or — 41 X = a/27 or v'il. If in the three answers of Ex. 2 of the last article we put a = V27 = 3, or, V— 41, we get the six roots of the given equation, it being of the sixth degree. 2. QUADUATIC i.C^U Allocs 6. 87/ '^4-1 _ 14 y2 = _ 1 369 (Ax3. 3 and 2) — 14 y2 + 49 = 48 -7= ijN/48 = 7 i V48 _ // = i (-* i V3). (323) 3. Given (3x'-\'Sx^- 20)^ _ 9 (3 a;* -f 8 a;^) - 2880 = 0. (Ct. 345, 4 and Ex. 16 in 346.) (3 a;* + 8 y' - 1^0)- - 1) (3 r* -f 8 .r2 - 20) = 3060. To save writing, put z for 3 x* + 8 ic'^ — 20 : then, «2 _ 9 ;g = 3060 4«2-.,%;^ = 12240 4 «2 _ ( ) 4- 81 = 12321 2;s: = 9illl « = 60 or — 51 .-. 3a; * 4- 8 x'^ - 20 = 60 9 a-* + 24 a;-'' + 16 = 256 3a!=^-h4 = ±16 aj^* = 4 or — Y a; = i 2 or f V^^IF. (334) (334, 335) The other four values of x may be found by using the second vabie of z, i. e., from 3 x* + 8 a;2 — 20 == - 51. 4. 5a-4-|-8a-J = 13 25 a:4 + 40 a-l + 16 = 81 6x14-4 = 4-9 13 5 28561 (334, 335) xl = 1 or ~ a; = 1 oi 62r) 370 TEXT-BOOK OF ALGEBHA. 5. ^G _ 7 ^3 ^ 8^ 7^ x^ -74:x''= - 1225. 6. z"" -5z^-\-4:=0. 8. 17?/* -1=16/. In this and some of the following only the real roots are given in the answer. 9. 3 7/4 - 7 ?/2 = 25. 10. x^ - 24 V^= 81, ' 11. a;2-3 = 6 + V:c--3. 12. 5 ic = 39 H- 2 Vx. 13. aic2» -f- te" = c. 14. V^ + 4 a/^ -= 21. 15. -y/x + l -\/^=3oi). 16. 12aj-i-a;-i -2-* = 0. 17. ic + Vx^ = 6 Vic. 18. ^ V8 = 2 ic Vl^^'. 19. Vic2 _ 8 i» + 31 + (x - 4)2 = 5. Suggestion. — Solve like Ex. 3. 20. 2 Vx^ - 3ic + 11 = ic2 _ 3^ ^ 8 21. L + 8y^^^42-| 22. 2 ic2 _^ 3 ic - 5 V2 x'^ + 3 cc + 9 + 3 = 0. 112 23- (2^-4)2-8 + (2:z;-4)*' Suggestion. — Put z ^ {2x — 4)2. 1 17 24. Given 0)2 _|_ __ _. Suggestion. — Add 2 to both members and apply Ax. 6 Note. — This equation is one of a class called recijjrocal equa- tions, whose roots are reciprocals of each other. Thus in this example the roots are 2 and ^, and — 2 and — |. QUADRATIC EQUATIONS. 371 25. l*rove by substitution that -, — r.aud — are roots of T )' the equation w' -}- .. = a if r is a root, i.e., if /•- -| ^ = «. 26. k; //■= 4- \, = i^«. 27. (u; - a) (./• _ ^) (a; — c) -|- f?/>r; = 0. 28. (a-j-b-\-jr) (a-\-b-x) (a-O-^x) {a-'b-x)=^ i). After solving this juoblem by nmltii)lying it out, solve it by 352. 29. x* + 5a-2 + 4 Vic*-f5a:^=60. 30. a:»-6x-9 = 0. SLOGE.STION. — Put y + z — j[, then (y + zf — 6 {y + z) = 9. Developing this we have y* + 2^ + (3 yz — 6) (y + 2) = 9. Now, since the original equation contains but one unknown, and we have just introduced •//- + xy = 24 ( ,r_//- + .-/• = 40. Suggestion. — Factor the left members, and eliminate .r by com- parison. 18 < '^■'' (^ - //) = ^^- ' 19 ^ (1) ^// + '-r = 12 ( .,.2 (2 X + 3 //) = 28. ■ ( (2).r + xy'' =18. St^ggestion. — Divide one equation by the otlier. 20 ^ ^ ^// = ^^^ - •^'^//' 21 ^ "^^^ + y = 21 ( ,r + y = 0. ■ i xY + 2/' = '"33. SrfJfiKSTiON. — Solve first for xy~ = .r', and ?/. QU ADliATlC EQUATIONS. 375 22. ( J-' -f ,/3 = ISi). < ., > _ ,/i ^ (J Va:* — / = 16 ^ - + ''7/ + y^ = 2. (x- + 1) i^ = o^y - 744. Suggestion. — Eliminate the left members, and solve for xy. 23. 24. 2^- t(2)^!/(a^ + 2/) = 70. SuoGESTiox. — Add three times (2) to (1), and extract the cube root of the members of the new equation. / u-« - 7/« = 56 26. 3 16 oci/ 27. + /r = 13 Xt/ = 28. (a- + // = 3 <.,.4_^ // = 17. J«»i:ft(;KSTiox. — These equations are symmetrical in x and y. Let X == M + r, autl ?/ = ff — r. Then x + y = 2 u — :), and ?i = 3. Thei-efore x — I + r, and ?/ = 3 — r, and u-» + y* = (3 + r)^ + (3 — r)» = 17. Upon developing the terms of the last equation it is readily solved as a quadratic. 29 f^Z/=«(^H-2/) Divide these equations through by xy and x*y* respectively, and solve for - and ^ • X y 30. {\) X - y =2 {a •- z) (2)x^ - if = ii (a^ - x'^ (3) a-' _y8=7(««-';i«). Suggestion. — Divide (2) by (1), giving (4); solve for x and y from (1) and (4), and substitute in (.i) -f(l). 376 TEXT-BOOK OF ALGEBRA. 31. (1) X -h y = 15 (2) xij-\-xz-\- yz=2H (3) xyz = 756. 756 Suggestion. — P'iiid z = , .. _ s , and ?/ = 15 — .t, and substitute these values in (2). 364. Problems Depending on the Solution of Equations of Degrees other than the First or Second. 1. A bill to contain 100 bu. of grain is made 3i^ times as long as wide, and 2f times as deep as wide. If one cu. ft. contains |.bu., what ninst be the dimensions. 2. Two cubic vessels together hold 407 cu. inches. When one vessel is placed on the other the total height is 11 in. ; find the contents of each. 3. The product of the sum and difference of two numbers is 8, and the product of the sum of their squares and the difference of their squares is 80. What are the numbers ? 4. One number is 8 times another, and the sum of their cube roots is 12. What are the numbers ? 5. Pind two numbers whose product is 15, and the dif- ference of whose cubes is */ times the cube of their differ- ence. 6. The sum of two numbers is 6, and the sum of their 5th powers is 1056. Find them. (See Ex. 28, last article.) 7. The product of two numbers is 10, and the sum of their cubes is 133. Required the numbers. 8. A man draws a certain quantity of vinegar out of a full vessel that holds 256 gals. ; and then, filling the vessel with water, draws off the same number of gallons as before, and so on for four draughts, when there were only 81 gals, of pure vinegar left. How much vinegar did he draw at each draught ? 9. B has 33 more than the square of the number of dol- lars A has, and the product of the numbers expressing the fortunes of each is 1330. How many dollars has each ? (,)l .\i>i;.\'l'ic KC^lATIO.NS. 377 10. A box measure has a scjuare bottom, and a bin lias for its respective dimensions the S(]uares of the dimensions of the measure. If now the bin is full, and 7 measures full be taken from it, 1710 cu. in. will remain. Moreover, the num- ber of square inches in one of the sides of the measure diminished by the number of linear inches in one side of the bottom is 12. Find the dimensions of the measure. 11. What two numbers are those whose difference, multi- plied by the difference of their squares, is .32, and whose sum, multiplied by the sum of their squares, is 272? SuooESTiON. — Call H llu'ir sum and v their diflference. FIFTH GENERAL SUBJECT. -TOPICS RELATED TO EQUATIONS. (INEQUALITIES, PROPORTION, EXPONENTIAL EQUATIONS, LOG. ARITHMS, PliOGllESSlONS, AND INTEREST.) CHAPTER XXIV. INEQUALITIES. 365. An Inequation, as the word itself suggests, consists of two ineiiibers, one of which is greater or less than the other but not equal to it. We now propose to study briefly the inequation in the same way the equation itself has been investigated. a. The signs > and < indicate that the quantities between which they are placed are unequal, the opening of the character being toward the greater. By (jreater is meant higher up in the scale in the positive direction. . Thus 6>4, 3>-2, - 11 > — 15, etc. Hence if a>h, then a — h is always positive. In the examples just given, 6 — 4 = 2, 8 - (- 2) = 5, - 11 - (- 15) = 4, etc. 366. Properties of Inequations. The inequation has sev- eral of the properties of the equation, hut not all. 1. If the same number be added to or subtracted from both members of an inetpiation, the new inequality will be like the old ; i.e., will retain the same sign. Thus, 7 > 5; and by adding 4, 11 > 9 ; 2 > _ IK), and by adding - 12, - 10 > - 42 : — ^2 > — 13, and by adding 17, 5 > 4; - (•) < 0, and by adding — 18, — 24 < — 18. and so on. It is evident from these examples that both members will be clianged aiiL-f, or merely shoved along in the scale whether by adding or subtracting, and the new inequality will be of the same kind as the old. A general demonstration of the theorem is as follows : a > h (Hyp.) a — h is ])()sitive. (a of last Article.) w i c — (^ -I- c) is positive. (the c's cancel.) .:a ^c> b^c. Q. E. 1). Scholium. — We learn from this that a term can l)e transposed from one side of an inequation to the other by (^hanging its sign (210). 2. An inequation multi})lied or divided through by the same positive multipliei* or divisor will stand as it did be- fore; whereas if the multiplier or divisor be negative the new inequation will subsist in the routmnj sense ; i.e., if the left meml)er were greater before, it will be less now, and conversely. A little consideration will show that this theorem is true for the positive multiplier or divisor, and then also for the negative, since the greater a negative number is numeri- cally, the less it is algebraically. As a special case of this j)rinciple, if the signs of the terms of an inequation be changed throughout, the sign of the inequation nnist be changed. 3. If the two members of an inequation are both j)ositive numl)ers, when they are raised to any integral power the inequation will still hohl as before. If the two members of an inequation are both negative numbers (or at least the 380 TEXT-BOOK OF ALGEBRA. less number is negative), when they are raised to an odd power the inequation will still hold as before. But if the two members of an inequation are both negative numbers, Avhen they are raised to an even power the new inequation will subsist in the contrary sense. Let the student test these cases by various examples. Eaising the members of an inequation the lesser of which is negative to an even power gives rise to an amhiguity. Thus, 5 > — 2, upon squaring gives 2^ > 4, Avhile, 5 > — 8, upon squaring gives 25 ^ 64. Evidently in this last case the greater member must also be the greater numerically. 4. If any root of the two members of an inequality both of which are positive is extracted, or any odd root where both are negative, the resulting inequality will be like the old, providing in the former case the root of the greater be taken as positive. Test this witli examples. 5. Two or more like inequalities may be added, just as equations are, and the resulting inequality will subsist in tlie same sense. 6. But if one inequality l)c taken from another like in- equality, the result is am'oiguous. Thus, 12 > 5 12 > 5 2 > 1 11 > 1 10 > 4 1 ;2^ 4 7. Like inequalities whose members are all ])ositive may be multiplied togetlier, and tlie new inequality Avill subsist in tlie same sense. To the theorems given, still others niiglit be added. Tlie foregoing are the most important, and sufficiently illustrate the elementary treatment. TOPICH iiKl.ArKD TO EQUATIONS. 381 367. Special Theorems in Inequations. 1. To prove that the sum of tlie squares of any two real and unequal nuniljers is greater than twice their i)roduct. Let a and b be the numbers. Now a square number is always positive. 'V\wn (a -by>o (except when a = b, when (a — b)'^ = 0) a^ — 2ab-[-b''> U (114. 1>) a^ -\-b^> 2 (lb (J. J V. />. (366, 1) Remark. — This theorem is much used in the demonstratioii of other special ones. 2. Prove that any positive fraction whose terms are unequal plus its reciprocal is greater than 2. Thus, to prove - + -- > 2. 3. Show that a* -j- ^* > a% + ab^, when a -{-b is positive. SuooESTiox. — Divide the inequation through by a + h. 4. l*rove that a*b + «// > 2 a%% when a > b > 0. 5. Prove that x^ -{- 1 ^ jc' -{-x, when O! -f ^ i^ positivi; and X :^ \. 6. Show that x'^ -\- y- -|- ^^^ > xy -\- xz -\- yx, if .'•. y. and z are unequal. 7. Prove that — < —^-J—\~ < -jf ^ < L < tlie n n -\- q -{■ s s n y .s- denominators ti, q and s being all positive. 8. Which is the greater, a^* + 3 ^* or 2 ^» (a + /;), the let- ters being unequal and positive ? 9. Which is the greater, a* — Z** or 4 a'* (a — b), wlien a> b > 0. SuooESTioN, -r-fftctor the left member and tlivide through by 382 TKXT-BUOK OF ALGEBRA. 10. Griven that a, h, and c, are positive and unequal, to show that (« -{- b) (b -\- c) (c + a) > 8 ahc. Suggestion. —Divide the three factors of tlie left member by «, A, and c respectively, thus cancelling the same factors in the right member. Then multiply out and apply the theorem of Ex. 2. 368. Examples and Problems in the use of Inequations. 1. Given '2 ,r -f i .x — 4 > 6 to fnid the limit of x. Sr(;(;i-:sTi()\. —Solve as an equation by clearing and transposing, v.iience it will appear that x > 4; ^x 5 11 7x 3. Find the limit for both x and y in ( (1) ^x + 4 y > 30. ^ |(2)3.r4-2y = 31. X X \'t -\-h > - -\- X to find the limit of .r, 4. Given •<' X X a — h 7- — - > — ;; — when a'y b y- (). \b a. b 5. Show that ^^ — 8 £c -f 22 is never less than 6 whatever may be the value of xx (See 357.) 6. The double of a certain number increased by 7 is not greater than 19, and its triple diminished by 5 is not less than 13; required the number. 7. A certain positive whole number plus 23 is less than 6 times the number minus 12 ; and nine times the number niiruis 54 is less than twice the number i)lus 9. What is the number ? 8. Prove that the arithmetic mean of two luimbers is always greater than their geometric mean. Or, as a formula it is to jjrovethat ~.> > ^ ab. TOPICS UKLA'PKI) To KQUATlUNS. 383 CHAPTKK XXV. liATIo AND rUOPORTION. 369. The Ratio oi two quantities is the (jiiotient of th(^ first divided by tlie second. The first quantity is called the antecedent, and the second the consequent. Thus, the ratio 3 to 6 is g or \ ; the ratio of a + ?« to m + >i2 is ^-^^, = (a + w) ^ (ni + n') ^ (a -\- }n) : ()n -f- 7i^. a. Ratios are usually denoted by a colon. Thus, a :b is read the ratio of a to h. The colon is supposed to represent a division sign with the horizontal line omitted. h. Some writers regard the second quantity as the dividend and the first as the divisor. But this usage is out of harmony with the idea that the colon represents a division sign, and should be aban- doned. 370. Reduction of Ratios. Since; a ratio is a fraction it (111 1m I'duced to other t* i ms w ithout altering the value of the f taction, i.e., the ratio. (^159.) _, , a ma ^ ^ • i - Thus, a:b = -i=: —J = j^ In i)articular, 12 : 4 = <) : L\ lit QUEUV. — "Will it change a ratio if the two terms be increased or diminished by any number? How, if they are increased or dim- inished by like parts of each ? SuaGESTlox. — Multiply both li-rms by f l -f "- J or by [ J — "'-\, 384 TEXT-BOOK OF ALGEBRA. 371. Ratios can be compared in Value by reducing them to a common denominator. Tlius, to compare the values of 10 : 11 and 23 : 2o. 10 23 250 253 .^. ^^. 10 23 11 > ' = ' < 25' 275 < 275 (^^^' ^^ •"• H < 25' 372. To Compound Ratios is to multiply their correspond- ing terms together. Thus, compounding a:b, c:d, and e :/ we have ace : hdf. a. When the terms of a ratio are squared it is said to be the du- plicate ratio of the first ; when cubed, the triplicate. h. Examples of the compounding of ratios may be seen in com- pound i)roportion in arithmetic. 373. A Proportion is the equality of two ratios. Thus f = 4 ; OY, a:h = c\d-, ov a:b : : c:d using a double colon instead of the equality sign. a. The colon form is read " a is to 6 as c is to d!," while the frac- tional form f = ^ inay be read in the same loay, or as an equation, " a divided by h equals c divided by cZ." Indeed, both forms may be read in either way. ?). The first and fourth terms of a proportion (a and d) are called the extremes, and the second and third terms (b and c), the means. 374. Method of Treatment of Proportion. — To accomplish any desired reduction the proportion is first written as an equation, and then transformed by axiomatic processes. As regards the colon form of writing proportions it need never be used. In truth, it simply duplicates the other notation. If any advantage at all attaches to it, it is due merely to the fact that the proportion is Avritten in a single line. Nothing of importance would be lost and simplicity would be gained if the colon form were entirely discarded. In the study of proportion we seek to familiarize ourselves with the_ different forms in which the same proportion can appear, and the new proportions which can be constructed out of it. * Toi'irs i:klati:i) to kc^katk )NS. 385 375. Transformations of a Proportion. — Thkiujkms. 1. In a pi'oj/ortion f he product of the ejcfn'ines is equal to the prodiif't of the means. This is the usual test for a proportion. Given a '.h::r:(l to prove ad = be. ad = hr Q. E. 1). (Ax. 3) 2. CoxvER.SK Thkohkm. — If the product of two numbers is equal to the product of two others^ one pair may be made the extremes and the other pair the tneans of a proportion. For if the equation be divided througli by any pair of opl)Osite terms, a proportion results. Taking the equation ad = be, or be = ad (1 ) J = -j i-iiui'tii>n,fhe difference of thr ji rsf mid second is fo flir jirsf or seroiid as the difference of the third mid fourth is fo the third or fourth. It o : f) : : r : d, then , , , . ( 1/ — If : h : : ■ — ii : il For '' - 1 =' - 1 (Ax. 2) If d ■'''='ZL± Q.E.D. (167) '/ ll (7) 'LszJL = '-—± o.k.d. ,375./-) 388 TEXT-BOOK OF ALGEBRA. 3. The Compositiox axd Division Theorem. — If four quantities are In proportion^ the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. If a : b : : c : d, then a -\- b : a — b : : c -{- d : c — d For, taking equations (2) and (6) of this article, a -\- b a — b / \ rr^ — J_ = . (Ax. 7) c-\-d c — d ^ ^ 1±1='-±1 Q.E.D. (Z75,b) a — b c — d 4. The Like Powers Theorem. — If four qua^itities are in proportion, like powers of the terms are also in pro- portion. ■ li a\b::c'.d, then a»* : b"' : : c"» : c^'" For 1 = 1 (Hyp.) %- = %- Q.E.D. (Ax. 5) 5. The Like Koots Theorem. — If four quantities are in proportion, like roots of the terms are also in propoi'tion. This is proved in the same way as 4 ; instead of Ax. 5, use Ax. 6. 377. Combinations of Two or More Proportions. 1. In a continued iwoportion (i.e., one in ivhich three or more ratios are all equal to each other) the sum of any ante- cedents is to the .sum of their consequents as any one ante- cedent is to its consequent. If a'.b'.\c\d\\e\f\\(j\h\\ etc., then a -\- c -^ e -\- (j -\- etc, '.b -\- d -{- f -\-h -^ etc. :: a-.b:: etc. To prove it let us suppose r to be the common ratio ; then a c e a , .*. a — br, G = dr, e = fr, (j = hr, etc. (Ax. 3) TOPICS RELATED TO EQUATIONS. 389 Adding these equations (or any set of them) member by member ,, 4.r + 6 + <7 + etc. = {h + d +/+ h + etc.) r (Ax. 1) g-f c-f g + /7-Fetc. _a _c eg h + ,/ +/-f h -h etc. ~ ' ~ b~ d-'f- h~ ^^^' ^'^'^• (Ax. 4) a. Continued proportions are often written more compactly, thus, a :c :e : lb :d : f instead of a : 6 : : c : (Z : : e :/. The former is therefore not one, but two different proportions. 6. The above theorem finds its most important application in geometry, where by means of it the perimeters of similar polygons are found to be proportional to any two homologous sides. 2. Tke correspondltig ten/is of two or more proportions may be multiplied together, giving rise to a new proportion which holds If the given proportions Jwld, If a:b::c:d a'.b'.'.c'.d' a"'.b":'.c"'.d" then aa'a" : bb'b" : : cc'c" : dd'd!' For, writing the given proportions in the equation form, and multiplying them member by member, a c a! c' a" c" aa'a" cc'cf' bb'b" dd'd' Q. E. D. (Ax. 3, 178) and so for any number of proportions. Moreover, it is evident that the terms of two proportions may be divided. Thus, if r-^-'^y df a b c d a'''b'''''^'d'' 390 TEXT-BOOK OF ALGEBRA. • 378. Special Forms in Proportion. — 1. A 7)iean py^oportlonal between two quantities is one whose square is equal to their product. Thus^ if a:b::b\c, h is a mean proportional between a and c. For by the first theorem, h'^ = ac (or b = ^ac). The third quantity e is called the third ■proportional to a and b. a. A mean proportional is not at all the same as an arithmetical mean. Thus, the arithmetical mean of 6 and 4 is 5, or half their sum; while the geometrical mean, the mean proportional, equals \/6 X 4 = V2A, which is manifestly less than 5. 2. Two mean proportionals. — If a \b::b:G::G:d, b and c are two mean proportionals between a, and d. Prove (1) a'.G::a^:b^ (2) a:d:: a^ : P 379. Exercises in Ratio and Proportion. 1. Which is the greater ratio, 3 : 4 or ,3^ : 4^ ? 2. Write the ratio compounded of 8 : 7 and 5 : 6. Which is less and which is greater than the compounded ratio ? 3. The last three terms of a proportion being 4, 6, and S, what is the first term ? 4. Find a third proportional to 25 and 400. 5. Find a mean proportional between 2f and /\j. 6. Arrange in order of magnitude 4:3; 9 : i< > i-.< >i A ri( »ns. -W)! 11. If a :b:: rid prove that ra^ :sb^:: rc^ : sd\ 12. Find two luiiubers in the ratio of .S : 4, (suggestion I] X and 4r) of which tlicir sum is to tlie sum of their sfjuares as 7 to ;">n. 13. Solve X' — Aix- — \)::x- — ."> j- + (> : x' -\- -i x -\- 'X 14. Solve for x in the proi^rtion (a — x) : (x — ^,) :: ,, -, /,. 16. Solve (1) x: f/::3:5 (2) ar : 4 : : 15 : t/. 16. \\ hat (juantity must be added to each of the terms of the ratio in : n so that it may become equal to that of j^ : q ? 17. Four given numbers are represented by ?w, ii, j), and q ; what (piantity added to each will make them proportional ? 18. If four quantities are already proportional, show that there is no number which \mng added to each will leave the resulting four numbers proportional. 19. If a: ft : : f) : c prove that a -{- h : h -{- r ■ : a : h. 20. \i a\h\\c:d prove that 3 a -\.h\h\ : 3 f -\- d : d. 21. \i a:h:.c\d jirove that 2a-\-^h.'^a - 4 /> : : 2 ^ + .3 ,Z : 3 c - 4 :: ^' : ^'Z is {mn -j- itl> i : ( //" ^^ qb) : : {mr -[- nd) : {pc -t- qd) ? 24. If a:h\'.\\ o, d :/:: r, : _'. - . . : C, ; 7, d :b::7 : .'5, and/: r : : 4 : 3 to what iuunl»ti - i> t lie continued ratio a : h : *':d:e:: proportional ? 25. The product of two numbers is 112, and the differ- ence of their cubes is to the cul)e of their difference as 31 : 3. What are the numbers ? 26. Given Va; -j- a : y/x — a::m: n to find x. 27. Find two numbers whose sum, difference, and product are proportional to m. w, and />. 392 TEXT-BOOK OF ALGEBRA. 380. Variation. — One quantity is said to vary with another when there is some law connecting changes in their values. If 3/ is a function of x, then as x changes in valne y also changes. The simplest Cases of variation (the only ones the word as used technically in algebra refers to) are closely connected with ratio and proportion. Indeed, direct variation is nothing other than a new aspect of proportion. 1. Direct Variation. — One quantity varies directly with another when as one changes the other changes in the same proportion. Thus, the distance a man travels in a day varies as the number of miles he travels. If he travel 5 miles per hour he will go over | times as much ground as when he travels only 3 miles per hour ; and if he travel 8 hours per day, he will journey only | as far as when he travels 10 hours per day. When one quantity varies directly as another, or simply varies as another, they are to each other in a constant ratio. This is involved in the definition. For, if a varies as h, and a' and h' , a" and h" , etc., are other values of a and h, then _== — == —^ = etc, ; i.e., the b b b ratio is the same in each case, or is constant. When one quantity varies jointly as each of two (or more) others, as in the example given above, it varies as their product. As an illustration : if one train runs 30 miles per hour and 22 hours per day, while another train runs 25 miles per hour, and 23 hours per day, the distances they will pass over in any given time are proportional to the 'products of these numbers, or the rate and time. 2. Inverse Variation. — One quantity varies inversely as another when as one increases the other decreases, so that their product is constant. ToiMrs IIKI.ATEI) TO KQUATIONS. 393 Tims, if 5 iiu'ii can do a job of work in 16 days, 10 men can do the same work in 8 days, 20 men in 4 days, and so on. If a vary inversely as b, and a' and b' are corresixjnding values, then a: a! \\ b' : b, since this giv'es ab = a'b' which is true by definition. If one quantity varies directly as a second and inversely as a tliird, it varies as the quotient of the second dividinl by the third. Thus the number of bushels of wheat elevated in a mill varies directly as the amount of work i)erformed, and in- versely as the height to which* it is raised. 3. Other Kinds of Variation. — One quantity may vary as any function of another. It is shown in mechanics that if 5 is the distance in feet which a body falls in t sec- onds, then s = 16 f^. In other words, the distance varies directly as the square of the time. Xewton's law of gravi- tation states that the attraction of the heavenly bodies for each other varies inversely as the square of the distance. Works on elementary algebra usually confine their state, ments and exercises to the first two kinds of variation. 4. ExERriSK. (1. If X x. II and // = 5 when a* = 14 liiid ./• when y = 20. * ' • (2. \i X X y and a; = 7 when ?/ = .'U find y when x = 20. (3. If ^ oc i?Cand ^ = 6 wlien // = 4 and C = 15, find B when A = 100 and C = 10. (4. If 7n QC i, and when in = >/. n = />. find ;/ when m = c. (5. If the cube of x varies as the scjuare of y, and if x = 3 when y equals 5, find the equation between x and y. (6. If a -\- b cc a — b prove that «^ -)- b'^ x «/»; and if a cc b prove that a^ — b^ x ab. SrooESTioN. —Let m = the common ratio, then in the first part a + 6 = m (a — 6). 394 TEXT-BOOK OF ALGEBRA. CHAPTER XX Vr. EXPONENTIAL E(,)UATrOX8. — LOGARITHMS. 381. Exponential Equations are those in which the un- known appears as an exponent. Tims, 5^' = 25, in wliieli plainly, x = 2. 8-^ = 500, in which plainly, ic > 3 and < 4. a^ = Z* is the general form of the equation. a. Except in the case of exact powers, such as 2-* = 16 (in which •^ — 4); (1)^ = \ (in which a: = |); etc., there is no process like those for the solution of simple and quadratic equations by which this class of problems can be solved. However, tables of logarithms enable us to find approximate solutions very readily. We turn, therefore, to the subject of logarithms. 382. Tlie Logarithm of a number is the exponent of tlie power to which a fixed number must be raised to produce it. The fixed number is called tlie base. a. Evidently three different numbers are concerned. First, the fixed number; second, any number; third, its logarithm. Thus, if the base is 10, and the number 105, its logarithm is 2,0212, i.e., a little more than 2. For 10'^ = 100, and 10^""^' = 105. And so for any other number. As another example, the logarithm of 865 is 2.9o70 nearly. b. Logarithms, except in a relatively small number of instances, are found approxiniatchj and not exurtly. Thus, the logarithm of 105 is not exactly 2.0212, in which four decimal places of its value are given, nor is it 2.021180, in which six places of its value are given, but these are approximations cor- rect as far as they go. TOPICS lM:i,\ri.I» TO KQUATIONS. nil") 383. Systems of Logarithms. Tables. Any positive nuiii- Ijer t'xcept unity may lu* taken as tho baso of a system of logaritlims. 2i*88o^ means the 21189th power of thr l(XX)000th root of 10; and if these operations were actually perfoniuMl the result wouhl be 1.05. In other words, the logaritlfin of 1.05 in the system whose base is 10 is .021189 correct to six numl>ers he used as the base of a system of real logarithms ; for we saw in this article that the root to be extracted is even. Hence, this root would be imaginary. Then when 396 TEXT-BOOK OF ALGEBRA. raised to the various powers giving the corresponding num- bers, some would be imaginary, some plus, some minus, and all in confusion. Therefore, a minus number cannot be the base of a simple system of logarithms. c. The base of a system being taken positive and greater than unity, the logarithms of all fractions between and 1 are negative (see table in this article) ; the logarithms between 1 and the base of the system are between and 1; and the logarithms of numbers greater than the base are greater than 1. Negative numbers, as such, are not supposed to have logarithms, though numerical calculations in which negative numbers occur are often made irrespective of signs, or just as if all were positive. 384. Terms and Notation. The integral part of a log- arithm is called the characteristic. The fractional part is called the mantissa. " Logarithm " is usually abbreviated into "log.," and the base of the system used may be written as a subscript to it. Thus, log. looO == 1.6990 means that the logarithm of 50 to base 10 is equal to 1.6990. In this logarithm, 1 is the characteristic, and .6990 the mantissa. 385. The Briggslan or Common System of Logarithms.^ The system in practical use, like the Arabic notation, has 10 for its base. 1. Logarithms of fractions in this system (383, c) are negative, and grow larger as the number grows smaller. Thus the logarithm of .001 is - 3 ; that of .0001 is — 4, and so on ad infinitum. At the limit when the number is infinitely small (^^ 0), the logarithm is an infinitely large negative number. Furthermore, it may be seen that the logarithm of a fraction say between .001 and .0001 must lie between — 3 and — 4, one between .0001 and .00001 between — 4 and — 5, dependent on the powers of -^^ ; and so on. 1 Logarithms were invented by Lord Napier about 1614 A.D. Henry Briggs, a great admirer of Napier's logarithms, saw the practical advantage of using 10 as the base, and constructed his tables accordingly. KM'K > i;i:i..\ ii;i) ro i:«.»r.\Tin\s. 397 2. Logarithms between 1 and 10. — The h)garithm of 1 is 0, and of 10 is 1 ; hence the logarithms of numbers between 1 and 10 lie between and 1, i.e., are proper fractions. 3. Logarithms of numbers greater than 10. We have 10^ = 10, 10'^ = KM), 10 « = 1000, 10* = 10000, 10*^ = 100000, 10« = KXMHKM), etc. Here we may see that the logarithm of a number between 10 and 100 is between 1 and 2 ; the logarithm of a number between 100 and 1000 is between 2 and 3, i.e., is 2 -|- a deci- mal, and so on, ad injinitum. Hence, 10* = oo . Here as elsewhere the two x 's are not equal. To further illustrate the preceding, log. 07 = 1.8261 log. 345.25 = 2.5381 log. 1 ( )r,0.45<) = 3.0292 log. 5.678 = 0.7542 386. Briggsian Mantissas. — In the Briggsian or common system tlir rnluc of tin- M.VNTissA wUl uot chduge if the num- ber be multiplied or divided by some power of 10. In other words, the value of the mantissa is indej)endent of the position of the decimal point. Thus, log. 1.27 = 0.1038 log. 12.7 = 1.1038 log. 12700 = 4.1038 etc. etc. The reason for this 'follows from the law of exponents. Let us supi)ose that the logarithm of 1.27 is known t(> be 0.1038, or that lO'*-^^'^ = 1.27. We have, (287. 2) (1) 12.7= 1.27 X 10 = 10^1088 X 1(>1 = 1()0.108H+1 _ 1()1.1088 (2) .(H27 = 1.27 -^ \m = lo^^**** -^ W'= l()o »o»8-^ = |(,iio88 (3) 1270= 1.27 X 1000= 10«»0-^8 X 10' = l()0.108».+8 _ |()8.I0«« (4) .(MM)127 = 1.27 H- 1000() = 1()0 108« ^ 1(>4_ J^(,0.1088-4 __ j()4.1088 398 TKXr-nooK OF ALGKI'.ILV. 6 ju 7 M 8 D 9 0212 4110253 41 0204 +» o;J34i40 15 1701 bi 16 2041 ■>: 17 2304 2.; 18 {2553!24 19 12788 d 1790 H 1818 2008 27:2095 2330 2577 2810 25: 2355 24! 2601 23{2833 29j 1847 27; 2122 25 2380 24 2(525 23 2856 :^^8,1875 2J 2148,2; 2405 2. 2648 '2. 2878 22 0607 -■»{ 0645 ;37 0682 09()9'a5{1004|34 1038 1303 32! 1335:32 1367 1614 :3o| 1644 29 1673 1903 », 1931 i28 1950 217512612201 20 2227 25! 2480 2430 25 2672 23 2900 123 2455 2695 2923 23 2718 2945 0719j36 1072134 1399 b 1703 29 0374140 0755 137 il06 1430 1732 2811987 27 2014 26|2253 26 2279125 24 2504 25 M-2742 2967 2529124 2765 !23 2989 '21 20 3010 22 21 ;3222!2i 22 ^•U24'2(. 23 3()17 i;i 24 , 3802 i 18 3032 3243 3444 3636 3820 26 27 28 29 30 31 32 33 34 3979 4150 4314 4472 4624 3054 3263 3464 19 3655 18 3838 21! 3075 21 3284,20 348319 3674] 3856^18 3096 !2i 3304 3502 3692 3874 3118 3324 3522 3711 3892 3139 21 3345 120 3541 3729 3909 18 3160 3365 3560 3747 3927 18 3181 3385 3579 3766 3945 20 3201 19 3404 19 3598 18 3784 17 3962 18 3997 17 4014 |i7 4031 16 4166 17 4183 17 4200 10 4330 10; 4346 1.; 4:562 15 4487 15 4639 15 4502 jioi 4518 15 4654 id 4669 i7 4048ji7 164216 16 4378 4533 4683 477 1115 4786 :i4| 4800 4914!i4 4928 14 4942 5051 14 5065 14 5185 13 519813 531513 5328,1 5079 5211 5340 1414814 '4955 35 5441 12 36 55(53 1 12 37 5(582 1 38 39 5708 5911 5453 12 5465 13 5478 12 5575 12 5587 112 5599 12 5694 11 5705 1 12 5717 12 5809 12 5821 115832 11 5922 11 5933 111 5 944 11 40 (5021 10 6031 iij (3042 |ii (5053 11 41 (5128 10(5138 11 (5149 |n; 6160 10 42 6232 u 6243 10 6253 10 6263 1 5092 13 5224 5353 15 4829 4969 5105 5237 13 5366 6335 6435 10,634510 916444 10 6355 6454 10 6365 1(» 10 6464 10 5490 5611 5729 5843 5955 6064 6170 6274 6375 6474 4065 4232 4393 4548 4698 4843 4983 5119 5250 5378 5502 5623 5740 5855 5966 4082 ii7 4249 116 4409 16 4564 15 4713 15 4099 |i 4265 16 4425 15 4579 15 4728 !i4 4116 i7j 4133 4281 17 4298' 4440 16 4456 4594 Ii5 474215 4857 4997 5132 5263 5391 5514 5635 5752 5866 5977 6075 6180 6284 6385 (M84 6085 6191 6294 6395 6493 4871 5011 5145 5276 5403 5527 5647 5763 5877 5988 4886 5024 5159 5289 5416 12,5539 115658 12 5775 11; 5888 11 599!) 4609 4757 4900 5038 5172 5302 5428 5551 5670 5786 5890 (ioin 116096' 10 6201 10 6304 106405 10 6503 ,10(5117 10 (3222 11 610 11 6212 10 6314 10 6415 iio 6425 10 6513 9 6522 6325 45:653210 46 1(3628 6721 (3812 6902 50 (3990 51 7076 52 7160 53 7243 6542 6637 6730 (3821 6911 6551 6646 6739 6830 6920 10 6561 10(3656 10 6749 9 6839 816928 6571|9 6665 10 (3758 jo 684819 6937! 9 6580 10 6590 (567519 '(5684 67(57 !> 6776 ()857i9 (58(36 6946 1 9i 6955 9:6599 9 1 6693 9 6785 9! (5875 9 1 6964 54 7;;24 8 6998' 9 708419 7168 9 7251 Is 7332:8 7007 9 7093 8 717718 7259 .'^ 7340 s 7016 7101 7185 72()7 7;ns 7024 9 71108 7193 7275 9 735(5 s 703319 7042 7II818 7126 7202 8^7210 7284 8:7292 73(54 8 7372 6609 6702 6794 6884 6972 6712 6803 6893 6981 {8.17050 9 1 7135 817218 8 1 7300 |8! 7380! s' 7388 1817396 917059 8 j 7067 87143 917152 817226 9 '7235 S7308 8 1 7316 ATIO^S. 3119 N O 1. 1 >. 2 I. 3 i» 4 ■> 5 1) 6 I) 7 1' 8 i> 9 1) 55 7404 s 7412;7 7419 « 7427,H;74;jr) 8 7443 8 7451 « 7459 r 74(50 8 7474 « 56 i7482's!74{M) 7 7497 8|7')()5 751:5 7 7520.8 7528 ;« 75:50,7 7543 18 j 7551 8 57 1 7551) 7 750<5 s 7574 8 { 7582 7580 8 7597 7 700418 7012 j 7 i7010|8 7(527 7 58 i7G;U ^'7042 r 7<{4()«, 7(557 7004 8 7072 7 7070 7 7080 j 8 7(594 1 7 7701 8 59 77o«» : 771 ; ; 772:1 18 77;il 77:J8!7 7745 7 7752 is V700!- I7707J7 7774 18 00.77.S2 r 77s'.i : 770(5 7 7803 7810!8 7818 7 7825 7! 78:32 '7 j 78.30 7 7840[7 61 i 785.1,7, 7800 |H 7808 7 7875 7882 j 7 7880 7 7800 7 7003 7 i7010 7 7017 1 7 62 7024 7!70:ir7 71KJ8:7 7045 70521? 7050 7 70(50 7! 7973 17 '7080 7 7087 « 63 IW.] 7 8000 7 S007 7 8014 80211 7 8028 7 80:l5i«8041 7 8048 17 18055 '7 64 8i)t52 r 80()i> (i 8075 i 7 8082 80801 7 8000 Gi8102;7 8100 7i8110i6j8122:7 «5i8120,7 8i:J(5«|8142 718149 8150 « 8102 7 8100 7 8170JC 8182|7J8180 <•. 66 i 8105 7 8202 7 8209 6I8215 8222 6 8228 7 8235 fi 8241 1 7 8248 « 8254 7 67 H2(51 c;82(57 7 8274 8280 8287 6 h-j9;5 t; ^299 7 8300 c 8312!7!8310 68 h:]25 c: 83.31 '718; J;{S « 8:544 8351 6 KW,1 ,; .s:;(;3 7 8:J70 ( 8:170 c' 8.382 « 69 .8:188 7 8;}95;fl'8401 j«;8407!7 8414 6 8420 1; 8420:6 84:52 i 7 84.30 fii 8445 :« 70!845l|fl'8ir,7l«!84(5.3 7|8470|6|8470 6 8482!«,8488ic 8404 1 « j 8500 C: 8.506 1 7 71 851 • ^525 (i 8531 6! 85:57 6 8543j«|8549j6 8555 |r.i 8501 « 8567!fi 72 m:,7.; --)85 6 8591 .; 8597 6 8(503|6 8000:g 8015 c 18(521 « 802710 73 •■ ..;» .; ?S(>45 6 8(551 16! 8(J57 6 8(5(53 6 8000 in 8(575 6 j8081 5 8(58(5 i 6 74 ■.'.i8i6J8704 6 87106|8710 6 8722 5!8727|6 87.33 fi ! 87:10 c 8745ic 75 >:01 • >7.')»5 <^os •; ,ssi4 <> SS20 5 8825 8831 6 88:57 A 8842 i6 88481 « 88.54 i 51 8850^6 77 HS«i.') .: ,sH7I .', XH~() c;8H82 r, 8887 6 889:l 6 8800i«;8904 (i 8010 518015 « 78 W2\ »; H<>27 .'. H«»;52 fj 8!>:I8 5 894:5 6 8040 5,8954 6;89(;o .', 8905 6 8071-5 79 s.iT,: .: .'ts-i ., 8987 |6i 89931 5! 8998 6 0004 fil (MKM) 6|0()15 15 9020 i«| 0025 6 HU ' ;t5 <: 9042 « 0047 6 {)053|« 0058 « 0003 6 0060 5 0074 6 1 0070! 6 81 .-'- ' !>09f5 0101 5 0106 16 0112 a 0117 5 0122 6 0128 .'i!oi:33:.'. 82 913- 1149 0154 5 0150 6 0105 5! 0170 5 0175 5 9180 <; 918(5 :, 83 0191 ... . 1)201 0206 6 0212 5 0217 5 0222 5 0227 5 92.32 i«| 92:18 .'. 84 9243,5 9248 15! V>25:} 0258 5 02(53 e 0200 5I0274 5 0279 1 5 0284 |5i 9280 i.i 85 9294 .1 i 9290 5 i 0304 ' '. ! 0:J00 6 0315 5 0320 «| 0.325 -J 1 0.3.30 1 5 ;9.3.35!5 0340 .'. 86 9.545 V0350 .-. 9355 .■; 9:5(50 5 9:5(55 5 0370 5 9375 5 9:580 5 9:585:.'; 9:590 .-. 87 9:59.'. . '.t»n;> -. .141).-, .-. 04 10 6 9415 5 0420 .'. 9425 5! 94.30 5 94:55 .5 9440 .-. 88 944'. 155 .5 94(50 5 94r,5 4 94 '!9 -. !I474'.'. 9479 5 948415 948915 89 9491 504 15 950914 9513 5 95 IS .',9523 .',,9528 5 95:53 15 95.38 4 J><) i>54- '552 5 9557!* 05(52 4 9.500 5| 0.571 1«| 0570 1 5 0.581,.'.; 9.580 4 91 959n »'A)0 •> mo'y i 9«}09 5 0(514 A (;.-,2 ■-. 9*557 4 iM5(;i •hI 4 <»(;75 .-. 9(580 .V 93 iM58., Mi04 .'. iM5<»1> 4 970:} .5 07(^^ • 717 :, 9722 .'. 0727 4 94 ; 07:51 '741 14, 07451 5, 0750 4 0754 .-4 97 59 ; 4 1 9703 i 6 1 0708 r, \ »7 73 4 95 ' 077 7 t780 ! '. 0701 < 0705 a 08 ' «»s(i.-.'4'u800 5 9814 '4 9818 '. 96 98_>; 18:52 4 08:K5 .'J0841 4 9S4.-, '■>:,4 r. 985<> 4 98(53 .'. 97 98«;s i.s77 4 0881 5! 0886 4 WShi 1 .•>.9!> 4 990;5 .V9908 * 98 9912 .', 9917 4 1>021 .-. SK>2(5 4 IKKIO 4 oo:i4 5 5^939; 4, 5)043 5 9948 4 9952 4 99 JH)5<5 5 00(51 4 •HX55 4 1KW59 -. 0074 4 9978 .5 1)08:1 4 0087 4 9iM)i .'. oihm; 4 400 TEXT-BOOK OF ALGEBKA, a. Mantissas are always taken positlvel)/. Otherwise they would change when the characteristic became negative. Thus the loga- rithm of .127 is 0.1038 — 1 = —0.8962. But by considering the characteristic alone as negative, we avoid the introduction of a new decimal. In such cases the minus sign is written over the character- istic^ and not before the logarithm. Thus, 4.1038 means the binomial — 4 + .1038 = — (3.8962). 387. Explanation of the Accompanying Four-Place Table. ^ — The table gives tJis mantissas only of all numbers from 100 to 999. The first two figures of the numbers are found in. the column marked "X," the third is one of the ten figures at the top of the page. Thus 487 is found by taking the 48 in the '' N " column, and the 7 from the upper row of figures. The mantissa of the logarithm of 487 is found in the 7 column, opposite 48, and is .6875, At the intersection of lines and columns are found the 900 mantissas corresponding to the 900 numbers, tlie tw^o first figures determining the line, and the last the column. A decimal point before each mantissa is understood. The columns of numbers marked D are merely the differ- ences between the mantissas. Thus the difference between .31)32 and .3054 is 22 ; and between .4133 (corresponding to 259) and 4150 (corresponding to 260) is 17. It may be added (see 386) that this table gives also the mantissas of all numbers consisting of three or less than three figures, preceded or folio ired, hi/ any number of ciphers, (I lid Irrespective of the posit ion of the decimal point. Thus t'.e mantissa of 31 is the same as that of 310, and is .4914; t'le mantissa of 8 is the same as that of 800, and is .9031. 388. To Find the Logarithms of Numbers. 1. To find the characteristic of the logarithm of a number. (1. Numbers greater than 1. (1) The logarithms of all numbers consisting of one 1 riu' six-place tables prepared by Professor (i. W. Jones, of Cornell University, Iliiaca, \.V., are among the be^t of tliosc imblisbcd in tliis coiuifrv. T()i'i( s i;i:latei) tu KgUAiioNs. 401 integml fi«,'ure (i.e., between 1 and 10) liave for etween and — 1, and the mantissa is always added. Thus .2, .35, .6978, .934258, all have - 1 as characteristic. (2) The logarithms of all numbers between .1 (= !()-') and .01 (=10-^) have — 2 for their characteristic, since they must lie between -. 1 and — 2, and the mantissa is added. 63) The logarithms of all nuinbcrs between .01 (= 10-^) ;iii not exactly cont'it, but sutticiently so tor a Four-place table. In the exanii)le before us the rate will change from * ' ^ to * '* which is a very small difference. Consecjuently intermediate values of logarithms may be found by proportion with only very small errors. We have. Log. 2r)(). = 2.4(KS2 (Froui table.) Log. 25H.8 = 2.4()1)() (See explanation.) Log. 257. = 2.4099 (From table.) To interpolate the logarithm of 2r>().«S proi>ortionally we say, the difference of the numbers : its fractional i)art = the difference of the logarithms : its fractional part. Here, 1 : .8 = 17 : a; .-. a: = 17 X .8 = 13.f) = 14—. (Imusmuch iis the table being used contains 4 decimal places only, 14 is taken instead of 18.(), since 13.() is nearer to 14 than to 13.) Adding 14 to 2.4082 gives the result 2.4096. (2) For a greater number of figures. Thus, to find the logarithm of 52.7298. Log. 52.7 = 1.7218 (From table.) Log. 52.7298 = 1.7220 (Ky interpolation.) Log. 52.8 = 1.7226 (From table.) Tabular difference = 8 ; 8 X .298 = 2.384 = 2 + : 18 + 2 = 20. The reasoning is the same as before. From these examples it is clear that to find the mantissa » In order that this change may be «inall, logurithinic tables begin with W) or 1000. Four-phice tables include the numbers from 100 to V99. Five and Six place tables have four figures in their numbers, and extend from 1000 to 9«9l». 404 TEXT -BOOK OF ALGEBKA. of a number consisting of more than three figures, proceed as follows : — Find the mantissa corresponding to the first three signifi- cant figures of the given number. Multiply the tabular difference following (column D) l)y the remaining figures of the number regarded as a decinuil ; and add the product, taken to the nearest unit, to the mantissa of the first three figures. Remaimv. — It is convenient to regard the differences between the mantissas as whole muuberH instead of retaining their decimal places. No mistakes need arise in so doing. (o. Write for practice the logarithms of (1) 321.6 (4) 2.5675 (7) .08742 (10) .0004523 (2) 14165 (5) 2379.6 (8) 2!)()r)(;24 (11) 7.0633 (3) 1416.5 (6) .200375 (9) .00001328 (12) 1202800 (13) 11^, carry the decimal to 3 places (14) H (15) 5^. 389. Conversely : To Find the Significant Figures of a Number from its Mantissa. 1. When the given mantissa is the same as one in the table. The first two figures of the number are found in the same row at the left in the column marked '' N." The third is at the top of the column in which the given mantissa is found. Find the number corresponding to 1.3118, its logarithm. Turning to the table, the mantissa is found opposite 20 and 5. Hence 20.5 is the number. (388, 1.) 2. When the given mantissa lies between two in the table. Here the number is interpolated in the same way that the logarithm was before. TOPICS KKLATED TO Ei^t'ATlONS. 405 Thus, given ,3.847«S to find tlic huiiiIkm- coirespondinii:. R«'ferring to the table, we find :i847() (next less) = log. 7()4(> (See table.) .S.H478 (given log.) = log. 7043.33 (By interpolation.) 3.S482 (next greater) = log. 7050 (See table.) Tabular difference = G ; 78 — 70 = 2 ; (> : 2 : : 1 : a; .-. .r = 2 -f- = .a-^i The charaxiteristic being 3, there are 4 integral figures in the number (388), which places the decimal after the first 3. And so in all ca.ses. Hence to find a number when its logarithm is given. (1. From the given mantissa take the next less found in the table, and divide the remainder by the tabular differ- ence following the latter. (2. The resulting decimal figures are to be annexed to the three figures corresponding to the lesser mantissa. (3. The decimal point is then inserted between the figures in accordance with the rules of the j)receding article. 390. Exercise in Finding Numbers from their Logarithms. 1. Find for pracvtice the number corresponding to 3.201(). In the table we find the next less mantissa to be .2014, cor- responding to 159, and the tabular difference 27. Subtract- ing and dividing, 2 -j- 27 = .074 -f- • Annexing these figures to 151), we have 159074 -|- . Now. tlie given characteristic is .'J. Hence 4 figures prectuU' tlic decimal ]H)int. ;nid fho numl)er is 1590.74 + . 2. 4.8016 6. 2.0095 8. 4.7320 11. 9.8423-10 3. 2.1144 6. 4.248S 9. 1.0410 12. 7.0453-10 4. 0.44HS 7. 1.94SS 10. 3.0210 13. ().52()9 — 10 391. Uses of Systems of Logarithms. — Logarithms wcic invented to abridge the lalKu- of multiidication and division. They are used to multii»ly. divide, raise to ]K>wtMs. and 406 TEXT-BOOK OF ALGEBRA. extract roots, as well as in the solution of exponential equations. 392. To Multiply Numbers by the Use of Logarithms. 1. Let us lind tlie product, for example, of 155, 207 and 939. 155 = W'-'^'' (See table, log. 155) 207 = 102-3i«> (See table, log. 207) 939 = 10-^-«'-^^ (See table, log. 939) .-. 155 X 297 X 939 = 10' -^^^^ (Ax. 3) (Adding the exponents of 10 by the rule for the multipli- cation of monomials.) By Definition, 7.4790 is the logarithm of the product sought. Referring to the table to find the number corre- sponding, it is found to be 30128571 +. Hence 155 X 207 X 939 = 30128571, approximately. (= 30127815 exactly.) Thus multiplication is made to depend on addition, a process much more easily performed. 2. Hence, to multiply together two. or more numbers, first find their logarithms and add them, and then find the number corresponding to this logarithm. 3. Multiply together the following numbers by loga- rithms, obtaining the corresponding approximate results. (1. Multiply 29 by 39. log. 29 = 1.4624 (Table, log. from number.) log. 39 = 1.5911 (Table, log. from number.) log. 1131 4- 3.0535 (Table, number from logarithm.) .-. 29 X 39 = 1131 + approximately (= 1131 exactly). (2. 19 X 37. (4. 85 X 10 X 21. (3. 43 X 68. (5. 122 X 133 X 144. (6.12 X 13 X 14 X 15 X 16 X 17. TOPICS ItKLATF.n TO Kgi'ATlONS. 407 393. To Divide Numbers by the Use of Logarithms. 1. o hiid the quotient, e. g., of UU7 divided by l.\)o. 207 = lO-2-8i«> (Table, log. from number.) 1.93 = loo-^^'^c (Table, log. from number.) ... 207 -r- 1.93 = 10=^'«*^* (Ax. 4) (Subtracting the exponents of 10 by the rule for the division of monomials.) Now, 2.0304 is the logarithm of the quotient. Turning to the table, we find for the number corresponding, 107.25. Thus, division is made to depend upon subtraction. 2. Hence, to divide one number by another, first find their logaritlims and subtract the logarithm of the divisor from that of the dividend, and then find the number corre- sponding to the difference of their logarithms. 3. Perform the following divisions by logarithms : (1. Divide 35 by 7. log. 35 = 1.5441 (Table, log. from number.) log. 7 = 0.8451 (Table, log. from number.) log. 5 = .6990 (Table, number from log.) (2. 91 -^ 13. (5. 2345 -4- 163.5. (3. 85-^17. (6. 796.325 -r- 196275. (4. 198 -^ 2.67. (7. .00367^2.61. (8. .01917 -4- .00021. 4. Rule for evaluating Com|X)und Expressions, such as 9 X 13 X 103 , . , .^, — — — — by means of logarithms. to X 87 (1. Find in turn the logarithms of the numerator and denominator by adding the logarithms of their respective factors. (2. Subtract the latter sum from the former, and the number corresponding is the quotient sought. 408 TEXT-BOOK OF ALGEBUA. Find tlie value of (1. '^^J^^J^-^ . Model Solution. ^ 73 X 87 log. 9 = 0.9542 log. 7:3 = 1.8633. log. 13 = 1.1139 lo g. 87 = 1.9395 log. 103 = 2.0128 denominator = 3.8028 numerator 4.0809 denominator 3.8028 . 9 X 13 X 103 _ ^ ^^^ 1.897 0.2781 * * 73 X 87 ' ' * ^2 336.8 X 37 .^ 212 X 6.13 X 2009 ^ ' 7984 X 22 • ^ ' 365 X 5.31 X 2.576 ' .o .07654 .^ .0062 X .0007 X 2 (6. 83.947 X 0.8395 ' 3.6 X .00005 X 9.764 (2^- X 61) -^ U (271-206)81 (7. Find x in 28.035 : 3.278 = 3114.27 : x. 394. To Raise Numbers to Powers by the Use of Logarithms. 1. liaise 13 to the liftli power. 13 = 10i-"2» 135==105-^««^ (Ax. 5) (Multiplying the exponent of 10 by 5 in accordance with the rule for raising to powers.) Finding the number corresponding to the logarithm 5.5695, we have 13^ = 371090.9, approximately. 2. To raise a number to a i)ower multiply its logarithm by the index of the power and find the number corresponding. 3. Perform the operations indicated in the following exercise. (1. Kaise 6 to the tliird power. log. 6 = 0.7782 (Table, log. from number.) log. 216 + = 2.3346 (Table, number from log.) .'. 6^ = 216 +, approxim.ately. TOPICS i:klati:i) to hquations. 409 (2. :*=? (7. (TA)*. (3. 5a*. (8. (.9975)". (4. (W- (9. (33.9 X 43.4 -r- 3814)". ((>. (l..%72)". (10. (W)'. .. /. 000106 y ■ ■ \'My X .07/ * 395. To Extract the Roots of Numbers by Means of Loga- rithms. 1. Since the laws of exponents hold for fractional as well as integral i)owers, the rule of the preceding article holds good here : hence, multiply the logarithm of the given number by the fractional exponent of the power to which the numl>er is to be rais + -'.«'^^'^) _ log. .3697 Ans. = 1 -(- .."')(;78 = 1.5678. 410 TEXT-BOOK OF ALCJEBRA. (3. 8^ (7. 11^ (4. (8. (11 ^ V(f) /^49 ,,^_ ^73567 V 1991 ^ (12. ^ /m X V117 (13. (6. 906.80* (10. 2.5673" 16^ X 15^ X 14* 13^ X 12i X 11 29.62^ 396. Accuracy of Results obtained by using Logarithms. — A little consideration will show that in getting a number from its logarithm (4-place table) one cannot be sure of more than 3 figures. As a general thing the fourth and often the fifth will be correct, but the latter is very un- certain. Besides, the logarithm Itself may be somewhat in error from the circumstances under which it was obtained. Thus, dividing 771 by 119 FOUIt-PLACE. log. 771 = 2.8871 log. 119 = 2.0755 SIX-PLACE. 2.887054 2.075547 log. 6.480 = 0.8116 log. 6.479 = 0.811.507 Here four-place logs, are in error by about a half unit of the fourth place, as six-place values show. These errors are superadded in the quotient logarithm, and, as a result, the four-place table gives 6.480, while the more accurate six-place table gives 6.479. In raising to powers an error may be considerably increased. Thus, (3g2)t worked out by a four-place table gives rise to an error of about 4 units in the fourth place, the reason being that the log of 32 is too small, and that of 6 too large in the table, and the error doubled by subtraction is then multiplied by 4. We subjoin a little table of errors PUOBLEM ANSWEK BY LOGAHITHMS TliUE VALUE. 155 X 207 X 939 207 ^ 1.93 9 X 13 X 103 71 X 75 290^ 301285714- 107.25 2.2621-h 2.5729 30127815 107.254 2.2631 2.5728 nearly. TOPICS i;i:i..\ ri;i) !'( > i:(.)rA'ri()Ns. 411 Tlie student is strongly urged to verify by actual nuiltiplications and divisions all problems which can be so solved without too great an expenditure of time and labor. A fair idea of the accuracy of logarithmic computations, and the advantage derived in their use may thus be gained. If other (six- or seven-place) tables are at hand let them be tested in the same way. Every endeavor should be made to get a practical as well as theoretical undei-standing of the sub- ject at the start. Constant use is made of logarithms in trigonometry and other branches of mathematics. The student might be led to suppose that if logarithms give results correct only to a certain number of places (depending on the number of decimal places in the table used), not much practical use could be made of them. But such is not the case. In actual measurements, as with a rule, it is not possible to get lengths any closer than hun- dredths of an inch, and often not so close. Hence results obtained by logarithms may be made correct within the limits of error by choosing a table having a sufficient number of places. 397. * Logarithms to other Bases can be derived from the Briggsian Logarithms. — To show this the following theorem must be proved : — Theorem : If the logarithm of any number he taken to two different bases, thejirst logarithm equals the second mul- tijdied by the logarithm of the second base in thejirst system. Let n be any number and b and b' two bases. Put log.^ /I = jc, and log.^ b' ^y\ then by definition, b' = /^'^ and n = V = {Jyyy = ^^, i.e., log.ft n = xy = log.ft n X h)g.,, //' n. h\ />. Thus, log.,0 5 = log.2 5 X log.,,, L'. Referring to the values as given in 383. wo have .69*)0 = 2.3223 X.3010 which is easily verified by multiplication. If log.io n = log.^. n X log.,0 b' log.^ n = log.,0 V 4- log.,0 b' (Ax. 4) 412 TEXT-BOOK OF ALGEBRA. Hence, to find the logarithm of any niimhev to a new base, divide the member'' s lofjarithm to base 10 by the logarithm of the new base to base 10. The student can easily verify that this is true in the examples of 383. 398. Solution of Exponential Equations by Means of Loga- rithms. 1. To solve the equation a^ = b. Since the two members of the equation are equal, their logarithms are equal (general axiom), and since log. (a^) = X log. a (395), we have X log.io a = log.io b .^_log^ (Ax. 4) log.io ^ Remark. — This result may also be derived by means of the theorem at the end of the last article, x, b, and a equaling respec- tively, log.yii, n, h\ 2. Exercise. (1. Solve the equation 13^ = 129. log.io 129 2.1106 , „,,, , X = / ^" , ., = zr-pr^ = 1.895 A71S. log.io l':5 1.1139 (2. 2^ = 64. (6. 9.1^=2467.2 (3. 2(K = 100. (7. 3'^'' + i = 27. (4. IP = 3. (8. .052^'- ^1 = 396.2 (5. 3-- =.8 (386, a) (9. lO^-^ + 2 X 10-^ = 80. Suggestion. — Solve first as a quadratic for 10*. (10. Given a'^^b"^ = e Taking logarithms of both sides log. «"'^ H- log. b"^' = log c (392) or, mx log. a + nx log. b = log. c (394) log. c . . ,x .-. X = ^i -^ — 1 y (Ax. 4) m log. a -f n log. b ^ -^ TOPICS IIKLATKIJ TO K(M ATIONS. 413 (11. L'^ ;>i^ = LMMM). Suo«KsTiox. — Put r- — 4x 4- 5 .==r z Jind solve for 2. 414 TEXT-BOOK OF ALGEBRA. CHAPTER XXVII. ARITHMETICAL AND GEOMETRICAL PROGRESSIONS. 399. An Arithmetical Progression is a series of terms which increase or decrease by a common difference. E.g., in 4, 7, 10, 13, 16, etc., the common difference is 3 ; and in 8, 6, 4, 2, 0, — 2, — 4, etc., the common difference is 2. In the study of progressions there are two problems which stand out prominently; viz., to find the nth term, and to find the sum of n terms. 400. Formula for finding the 7ith Term of an Arithmetical Progression. Let a be the first term, d the common difference, and n the number of terms considered. Then, by definition, «, a -j^ d, a ^2d, a ^'dd, a ^4:d, etc., is the arithmetical progression, the signs being taken all positive or all negative according as it is an increasing or decreasing series ; for each term is greater or less by d than the one which precedes it. We are to find the nth term. Let us write several terms with the ordinal of each above it. a, a + d, I* ) l.(M AlIoNS. 415 Calling / the /itli or last term, (1) ^ = a _1_ (n - 1) d. Hence, to find the last term, multiply the common diifer- ence by one less than the number of terms, and add the product to the first term if an increasing, or subtract from the first term if a decreasing progression. 401. Formula for finding the Sum of // Terms of an Arith- metical Progression. As lu'fore, call a the first term, d the common difference, ?i the number of terms, and / the last term. Denote by &• the sum of 7i terms. If rf be added each time, we may write the last term /, the next to the last I — c?, the second from the last I — 2d, etc. But if d is subtracted each time, the next to the last term would be d (jreater tlian /, or I -f- d, the second term from the last / + 1' '/, etc. (See 45, i, and remark on =p Art. 344.) Consequently we may write a,il^^d,a^^2d,a^^^d, . . l^'Sd,l^2d,l^d,L Trying to sum this series, a little study suggests the plan of adding the first and last terms, the second and next to the last terms, the third and third from the last terms, and so on, since the sum in each case would be the same, viz., a + /. It will l>e a little plainer, however, to add the series to itself, writing the seccmd in reverse order, as this also will give the sum (f -\- I for each addition, and it is clear that tliere will l>e just n sudi additions. N = (/-}-« i uin l»y lialf the number of terms. 416 TEXT-BOOK OF ALGEBRA. 402. Other Problems in Arithmetical Progression. — The progressions furnish excellent exercise in the solution of such equations as were given in 44-50 of Art. 216. Thus, in equation (1) above, Z = a -j- (71 — 1) d, there are four quantities involved (s not appearing), three of which being given, the fourth may be calculated. This gives four prob lems. Then, similarly, equation (2) gives four more involv- ing a, n, I, and s, (the fifth quantity, d, not appearing). Next, by eliminating n between (1) and (2), we have an equation involving a, d, I, and s. Eliminating a between (1) and (2), the resulting equation contains d, n, I, and s. Last of all, eliminating /, the resulting equation contains a, d, 71, and s. Consequently there are five equations, each one of which gives four problems. As an example, suppose /, d, and s given to find a. Since ?«- is the omitted quantity, and (1) and (2) both contain it, it must be eliminated. From (2) ^1 = "- (^^-^^ Substituting this value in (1) J , (2 s — a — I) d l = a + ^ , , ^ a -\- I in which a is supposed to be the unknown. al + P =a'^-^al-{-2 ds - ad - Id (Ax. 3) a^ - da = l^^ld-2 ds (Ax. a) 4 a'' -^da-{-d' = 4:P + 4.ld-i- d' - S ds (335) 2a — d=J^ V(2 l-^dy- i 9 V(2 l^dy-Sds Atis. 1. To find I, being given (1. a, d, 71 (2. a, 71, s (3. a, d, s (4. d, n, s. TOI'KS KKLAIKl) In I .(,H A T l< iN S. 417 2. To find .v. Ijeiiig given (1. '/. /, // i'J. c I. d (.'1 a, d, n (I. /. . o, /, n (4. /. f/. .V. 5. To find eillg given (1. d. n, I (1*. d, //, .s (.'> /, //. n. XoxE. — Of these problems only one is ooninionly given special coiisideration. It is that in which a, /, and n are given to find d. It may be stated in MOrtls as follows: required to insert it — 2 arith- metical means (« and / being the other two terms) between any two quantities n and /. After having found the common difference, it is an easy matter to write down tin vc mtans. Tin* formula for f/, / ,/ as given among the answers, is , 7t = 20. To find /, / = a + (7? - 1) rf = 5 -f 19 X 2 = 43. Am. To find N, .^ = I {a 4- /) ^ 10 (5 -h 43) = 480. Ans. 2. Find the 17th term in the series 10. 11.^, 13, etc. 3. Find the last term and the sum m ."», 9, 13, etc.. to 19 4. Find the L'tHh terni of the series T), 1, — 3, etc. 5. Find the wth term of 2, 2^, 2§, etc. 6. Sum 7 -h V -|- V' + ^*<'v to 16 terms. 7. Find the last term and the sum in — ]. — ;]. etc. to iti) terms. 8. Sum 2 a — 5 h, 7 tf — 'J b, etc., to 9 terms. 9. (liven ^/ = 5, c/ =4, and I = 201, required /i, and /f. 10. Insert 12 means between 12 ;tn1M( S Ll.l.A ri:i) i«> iAji ATIONS. 410 24. Two huiidied stones beiug placed on tlie ground in a straight line at the distance of *2 feet from each other, how far will a person travel who shall bring them separately to a basket whicli is placed 20 yards from the first stone, if he start from the spot where the basket stands ? 26. A body falls l(),».j ft. the iii-st second, and in eacli succeeding second 32^ ft. more than in the next i)receding one. How far will it fall during the Kith second, and what will be the whole distance fallen through in 16 seconds ? 26. If a person saves $1(>(> and puts it at simple interest at r}f/c at the end of each year, how much will his property jimount to at the end of 20 years ? 27. A man wius paid for drilling an artificial well 3.24 marks for the first meter, 3.29 marks for the second, 3.34 marks for the third, and so on. The well had to be sunk 500 meters. How much was paid lor the last meter, and how nmch for the whole ? 28. A travels uniformly 20 miles a day ; B starts 3 days later and travels 8 miles the first day, 12 the second, and so on in arithmetical jjrogression. In how many days will H overtake A ? 29. Show that if the same quantity be added to every term of an arithmetical i)r()gressi()n the sums will be in a. p} 30. Show that if every term of an a. p. be multiplied by the same (quantity, the products will be m a. p. 31. The sum of three numl)er8 in a. p. is 12, and the sum of their scjuares is 06 ; find them. 32. Find four integers in a. p, such that their sum sliall be 24, and their product 1)45. 33. The sum of five numbers in a. p. is 45, and the pro- duct of the first and fifth is five-eighths of the product of the second and fourth. Find the numl)ers. > For brevity n.p. will be used for aritliineticul progression and g. p. for geometrical progression. 420 TEXT-BOOK OF ALGEBRA. 404. A Geometrical Progression is a series in which the terms increase or decrease by a common ratio. E.g., 3, 6, 12, 24, 48, etc., the common ratio being 2; and, 5, 1§, |, /y, ^\, etc., the common ratio being ^. 405. Formula for finding the 7ith term of a Geometrical Pro- gression. Let a be the first term, ?' the ratio, w the number of terms, and I the last term. Then, by definition, - 1st 2^ 3^ 4*^ 5^^ 6^h, etc. a, ar, ar^, ar^, ar*, ar^, etc. Here it is evident that the exponent of r in any term is always one less than the number of that term. Conse- quently the exponent of r m the nth term would be 7i — 1. Hence, we have, (1) i = ar''-\ Or, to find the nth term of a geometrical progression, raise the ratio to a power whose exponent is one less than the number of the term, and multiply the result by the first term. 406. Formula for finding the sum of 7i terms of a Geometrical Progression. We use the same letters, and as in a. p. call s the sum of 71 terms. To sum the geometrical series an artifice is em- ployed analogous to that used for summing an a. j)- The value of s is first written down, and then, underneath, the same equation multiplied through by the ratio. By sub- tracting the first equation from the second, all the inter- mediate terms on the right go out, leaving only the first and last terms. s = a -\- ar -\- ar^ -f a7^ -f- ar^ -f- -f" o^^"*"^ rs = ar -\- ar^ -\- ar' -\- ar^ -f- ar^ -\- . -j- «r"~^ -|- a.r^ rs — s = ar'' — a (Ax. 2) [(Ax. ?>) Tol'M l;i:i,ATl.l) TO KgLATlUNS. 421 (r - 1) s = a (/•" - 1) Or, to find the sum <>t // itims of a ^. 7>., multiply the first term by the quotient ol the ratio raised to the ?ith jx)\ver less one, divided by the ratio less one. A formula in which n is replaced by I is readily derived from (2) and is convenient for reference. (3) 5=^^:^ By this formula, to iind the sum. multiply the last term by the ratio, from the product -iilitiact the first term, and divide the remainder by the ratio less one. Remark. — When the ratio is less than 1, the use of negatives can be avoided by changing the signs of both numerators and denom- inators in (2) and (8). ™. « ( 1 — r") , a — rl Then, s = , - , and .h = 1 — /• 1 — r 407. Other Problems in Geometrical Progression. — The three eipiations found in the last two articles are really equivalent to only two indej)endent ones, since one Avas Ul.l.A lKi» 1<» i.(.>r A IK »NS. 4'28 3. l*r()l)leins which lend to the solution of equations higher than the second. (1) For example, given o. ,i. s. to tiiul /. Here r must be t'liniinatcd. As shown above, r = ^ _ . . Substituting this value in (1), / = a ^^-^^:y"\ or / (s _ /)»-! = KgiATioNS. 425 15. Suiu to luHnity the series .IV.V.V.^ 16. Sum to inrtnity .37878; also .11^1351:^5 17. Show that the reciprocals of the terms of a //. p. are also III fj. p. 18. Find the Mim ot V«, Vrt^ V«*, etc., to er of years, and adding the product to the sinij)le in- terest, we have for the annual interest sought f,r i n Jr.n + ^^^(/' -h lY-l ^ = pr ( n +/+ '^ (n -f !>/'_ 1) V KxAMTLK. — Calculate thr amount of $1.0(10, for 1() years inid 1? months at annual interest. su))posing the rate to be Ans. ^^lnL^ 415. In Compound Interest all unpaid interest is added to the principal a,s fast as it becomes due. Call p the principal, a the amount. ;• the rate expressed decimally, and n the number of years. We may have the following ]U"oblpms : — 428 TEXT-BOOK OF ALGEBRA. 1. To find the amount ivhen the principal, rate, and time are given. The amount of the principal at the end of the first year is p (1 -\- r). Putting this amount at interest during the second year, the amount is ^ ( 1 + r) (1 -j- r) = j) (1 -|- r)^. Again, putting this sum at interest during the next year, the amount becomes p (1 -^ ry, and so on through the n periods. Hence, (1) a=p(l + ry If the interest be compounded q times a year, there will be q7i periods and the period rate will be - and the formula becomes (2) a^p{l + '^ If i is the interest, then i =p (1 -\-ry—p =p [(1 + ry — 1]. Example. — Calculate the amount of a note for $400, which has been standing for 8 years, allowing 7 per cent interest to be paid semi-annually. log. a = log. 400 -f log. (1 + • Y-)'' (392, 2) = log. 400 + 16 log. 1.035 (394, 2) = 2.6021 + .2384 = 2.8405 ... a = $692.67. Remark, — It must be borne in mind that a four-place loga- rithmic table gives only approximate results. Here the exact amount is $693.60. In this chapter, in order to secure the proper degree of accuracy, a six-place table should be used. 2. To find the present worth of a sum payable n years hence, allowing compound interest. Solving from (1) above a ^ . 1 P = (1 ^ rY Letting j--py = v, p = av\ TOPICS KKi.AlLl* i<> Kl^HATIONS. 429 XoxE. —The values of the expression (1 + r)" are given in com- pound interest tables for ditferent values of the arguments r and )i. Using these lessens the labor of calculation. The logarithms of the c'.v may be found by subtracting those of (1 + r)" from 0. It should be observ'ed that the above formula? compound the in- terest in the last fractional period. The rules commonly given direct to calculate simple interest for the odd fraction of a year. Compound interest in this case favors the borrower. 3. To find the time ivhen the amount^ priiicipal, and rate are given. Taking logarithms of both sides of a — j){\ -\- /•)" log. a = log. 7; 4- log. (1 + ry ( Ak. 8) log. a = log. p-\-n log. (1 4- r) log.(l + r) • ExAMi'LE. — Fin4 in how many years $1(M) will aniount to $1050 at T) jxT cent eomiKmnd interest. Ans., 4S. 17 years. 4. To find the rate when f/tr nmointf, in'inrifml^ (hhI finic are f/iren. This is left as an exercise for the student. SECTION 11. ANxrrni-> < 1.1:1 \in. 416. An Annuity Certain is a sum of money jxiyable at the end of each year for a fixed i)eriod of years, or forever. In the one cit.se the annuity is said to l>e fprminnhh'. in the other perpetual. a. Contingent aunuilirs (a inni m uic iu.sui.um r i an- «»uiii> i»a>a- ble as long as siH'citied |M*rsons remain alive or for other uncertain IH^riods. and their values are governed by the laws of probability. They cannot therefore bi' taken up at tliis stage of the student's progress. 430 TEXT-BOOK OF ALGEBRA. The problems of annuities are analogous to those of com- pound interest. We shall investigate two cases; viz., to find the amount of an unpaid annuity, and to find the cost of an annuity. 1. To find the amount of an unpaid annuity. Let V represent the amount of an annuity of $A which has run n years. Either simple or compound interest may be taken. We shall use the latter, and we will let R = 1 + r, where r is the rate expressed decimally. The first pa3anent of f A should have been made at the end of the first year, and therefore runs for n — 1 years, its amount being AR"~^ ; the next runs for n — 2 years, its amount being AK"~"^ ; the third runs for 7i — 3 years, its amount being AR"""^; and so on. The last payment of $A being made at the end of the period bears no interest. Thus we have for the amount of the annuity, AR"-i + AR"-2 + AR"-« 4- . . . + A. But these amounts form a geometrical progression whose ratio is R. Summing it (406, (3)), we have for the value of V ^ AR-^ - A _ A (R " - 1 ) ^~ R-1 ~~~^ (1. If the annuity be not paid until m years after the last payment is due, it will amount to A (R'* - 1) r Example. — Let it be required to find the amount of an annuity of ^10 to run 12 years at the end of 15 years, interest compounded semi-annually at 5 per cent. Substi- tuting in the formula, ^, _ $10 (1.025^^ -1) .,^^^^ (1.025^) - 1 ^ ' '' TOPICS i:i:latkd to ki^ations. 431 Calculating this by logarithms, starting with tlu* log. of l.OLC) = .0107. w.* have log. 1.02.r* = 0.1^508 .-. 1.025-* = L.SUC). l(.g. H) = 1. log. .SOG = i.o()(;;^ log. (i.o:>r))« = ao(U2 .97();"i h.g. A)rAH'> = 1^.7041' 2.2G(>3 giving V = f 184.r>2. (2. Another way of looking ^t this problem is to- con- sider the amount of the annuity as an obligation to be met at a certain date in the future and the payments as sums to be set aside annually as a Sinh'iii'j Fund with which we }>ay off this de])t when due. Solving for A. wc have A ^''• R"-! As an example, suppose a town borrows $10,(HX), agreeing to pay it back at the end of 15 years. What sum will have to be raised by taxation every half year to meet this obli- gation, supposing interest compounded semi-annually at 4 per cent ? A = ^"<^(-"^> = $24r,..iO. 1.02«> — 1 2. To find the rosf of (in onnifif//. Let I^P be the cost of an annuity of $A to run n years, 1 and V l+r Then the present value of the first payment of $A due one year hence is Av ; the present value of the second pay- ment due two years hence is Av% and so on. The present 482 TEXT-BOOK OF ALGEBRA. value of the last, due n years hence, is Ai?". Adding, to find the present value of all, we have P = Ai; + kv" + Ai;3 + . . . . -^ Ai;« = A(^ + i/^ + t;«+ _^,,«), Summing the parenthesis as a g. p.. whose ratio is v, we have 1 1 - V — 1 V — 1 r (1. If the annuity \^ perpetual, n — rj^ , and = 0, (358, 2) so that here P = A. ! + >■ Example. — Required the cost of a perpetual annuity of $50, supposing the rate of interest to be 5. It is evident that this is calculated in the same way as a problem in simple interest in which the annual interest and rate are given to find the principal. (2. To find the cost of an annuity to begin at the end of in years and i-un for n years. Here all that is necessary is to find the present worth of the preceding result due m years hence. This gives ~ r (1 + tY' {?^. If A instead of P is regarded as unknown, the al)ove formula gives the value of n annual payments which will reimburse a lender for a loan of f P. Example. — A city borrows $200,000, which it agrees to repay in 20 annual installments, interest and princi})al to- gether. How much will have to be paid each year, assum- ing a 5 per cent rate for interest ? T(>ri< i:i:i.A ri:i) lo i:(,>uati()Ns. 433 Vr 2U ()()0() X .05 A = J = ^£ = {|»1()()44. (1 + /•!" 1.05* NoTK. — For more extended infonnalion on the subject of inter- est and annuities certain the student is referred to Part Firat of The Institute of Actuaries Text-Itonk, publislied by the Laytons, London. .SK( rioN III. 417. Bonds, like ])romissory notes, dniw interest at stated intervals, and the i)rinc'ii)al itself becomes due at maturity. Let p be the face of a bond, n the number of years it has to i-un, r the rate (expressed decimally) named in it, and / the rate the investor desires to realize on his investment (also regarded as the current rate of interest). We proceed to investigate the different forms in which the problem may appear. 1. 7V> calculate the price x ttt In- imhl for a bond at the time of its iasue in order to realize a certain per cent on the investment. We solve this problem by balancing income and outgo. The price ])aid for the bond at comj)()und interest at r' per cent will amount at the time of maturity to (1 -|- /)"x dollars. The following amounts come back to the investor : first, the first interest payment %pr, which w411 therefore run n — 1 years at r \^\ cent comjKmnd interest, amount- ing at the maturity of tlve bond to %pr (1 + r)""^ ; next, the second interest j)ayment %yr, which will run for n — 2 years amounting to %pr {\ -}- '•)"~"'^; Ji"d so on for all the payments, the last being paid at the time of settlement. ' The author is indebted to Kmory MoClintock, LL.l)., F.I.A., actuary of the Mutual Life Insurance Company of New York, for important practical sug- gestions. 434 TEXT-BOOK OF ALGEBRA. Then the face of the bond is also paid at maturity. Hence we have . (1 + >•')•' . X = pr [(1 + r'y- ' + (1 + r'y- ^' + (1 + »■')'-" + . . . +l]+y^. Summing the bracket (quantity, which is a geometrical progression whose ratio is (1 + r'), we get, = P ('•(1 + 0"- '• + '■'). Dividing through by the coefficient of x. r\ (1 + /)'* Example. — What will be the cost of a bond for $1000 payable in 12 years and bearing 5 per cent interest, if the purchaser desires to realize 4 per cent on his investment ? . = 1000 / ^. _ .05 - .04 \ _ 25000 ( M - ^^ .04 \ 1.0412 ) \ 1.04^ Calculating the value of -^^4^ by logarithmic table, it equals .00625. Whence x = f 1093.75. The formula gives the price to be paid for a bond on the assumption that all interest payments are made annually. But they are usually paid oftener than once a year. 2. If the ifiterest payments are'7nade q times instead of once a year. The period rate will now be -, and the terms in the amount series will be r ,x"-i pr ,.„_! pr ,.«_•! r TOPICS RELATED TO EQUATIONS. 435 Hciv the ratio is (1 -h r)i . Hence we have (406, 167), L ,^ (1 + /yl _ y J ... a- = t \' ' ^^ -^'''y^ + y (^ -t - ''')\ - w + ^-^l This and the preceding result were calculated on a basis of annual compound interest, l^ut regular investors expect to compound their interest semi-annually. Tables of bond values are usually constructed in accordance with this practice. 3. If the hiteresf payments are made q times a t/ear ami the interest is compounded semi-annually. In order to compound the interest semi-annually, instead ol (1 H- /•), we substitute (1 -|- ^Y^ 1 -f- r,. This latter (/•,), is often called the effective annual rate corresponding to the nominal rate /. ^ _ /T r (1 4- r,y + y (1 + r,)T ^ (y + r) ! Example. — Required the present value of a mnnici})al bond for $2CK)f) to run for 20 years having six per cent coupons, payable (juarterly, to net purchaser 4^^ per cent. Here / = .045 ; whence r, = (1.0225)'^ - 1 = .0455 -f ; log. 1.0455 = .0193; .0193 X 20 = ..3800, giving (1 -f /•,)- - 2.4.322 ; \ of .0193 = .0048, giving (1 + r,)i . . . = 1 (H 12. X = ?!?^ r .06 X 2.4322 + 4 X 1.0112 - 4.0() 1 4 [_ 2.4322 X. 0112 J 500 X .1307 " 2.4322 X .0112 436 TEXT-BOOK OF ALGEBRA. log. 500 = 2.6990 log. 2.4322 = 0.3860 log. .1309 = 1.1163 log. .011 2 = 2.0492 1.8153 2.4352 2.4352 log. X = 3.3801 .-. x = 2400. 4. If interest pay merits have been made on a bond like a promissory note, it virtually becomes a neiv bond, to run for the remainiiig time, and a jyrospective purchaser may calcu- late from this standpoint. If bought at some time bettveen interest payments, the discount is from the time of sale to maturity. Thus, instead of n as the exponent of (1 -|- /), or (1 + ^f)? ^'^ the denominator of the value of x, the exact discount interval must be substituted. Example. — Let it be required to find the present value of a bond for $1000, drawing 7 per cent in|:erest, payable semi-annually, dated Jan. 1st, 1895, and due Jan. 1st, 1910, on June 18th, 1898, supposing the purchaser is willing to take 5 per cent for his money (to be compounded semi- annually). Here we may calculate the bond as if it had been issued on Jan. 1st, 1898. It will therefore have but 12 years to run. Also the exponent of (1 + r^ in the denominator will be 11^8^. The effective rate, r, = .050625. Log. 1.0506 = .0214; .0214 X 12 = .2568 ; (1 + r^^ = 1.8062 ; .0214 X 11-1% = 0.2466 ; \ of .0214 = .0107; (1 + r,)7 = 1.025. _ 1000 (.07 X 1.8062 -f 2 X 1.025 - 2.07) _ ^~ 2 X 1.0506"A X .025 ~ *^^ ^^• 5. To calculate the rate of interest tuhich can be realized when the cost of a bond is given. In order to make this problem soluble by a formula, we distinguish between the current rate of interest and the Tones i;i;i.A iKD so equations. 487 rate to be realized, iiy tlie current rate of interest we mean the rate at which the interest payments can be promptly re-invested. Let us denote this rate by r^. Then the for- mula in 2, above, becomes P \ r(l + r,) - 4 - g (1 + Te)^ - (y 4- r)\ Solving for /, which is now regarded as unknown, Example. — A $500 bond bears 8^ interest payable semi-annually, and is due 18 years hence. If the current rate of interest is 3J, what will a purchaser realize who Ijays .f 800 for it ? Here r, = .0353. Log. 1.0353 = .0150; .0150 X 18 = .1>700; (1 + r,)- = 1.862; \ <,f .{)\m = .(K)75; (1 + r,)l = 1.0175 Remark ox the Last Case. — Properly speaking only two rates of interest are recognized in bond calculations ; that named in the bond itself, and the rate to be realized by the purchaser. However, we may substitute a probable value of the latter on the right side of the al>ove equation, and thus find an approximate value of the rate to lie realized. This answer could be used in a second similar calcula- tion to find a second approximation. However, a simpler and preferable method of solution is to assume two rates for r« (one too large and the other too small) in the formula in 3, and calculate the corresponding values of x. Then the desired value of r, may be found from the given value of x l>y interpolation. 418. Exercise in Interest, Annuities, and Bond Calculations. 1. What will $1 amount to in 20 years at 5 per cent, interest beinj^ compounded semi-annually ? 438 TEXT-BOOK OF ALGEBRA. 2. What sum of money will amount to ^1000 in 8 years and 4 months at 4 per cent compound interest, interest being compounded quarterly ? 3. At what rate per cent per annum will $500 amount to ^760.17 in 8 years and 6 months, the interest being com- pounded semi-annually ? 4. How many years will it take a note for $100, bearing 4 per cent interest payable quarterly, to amount to $150 ? 5. In how man}^ years will a note, whose face is $300, amount to $400, allowing 8 per cent, payable quarterly ? 6. In how many years will $967.80 amount to $1269.00 at 5 per cent compound interest, the interest being com- pounded semi-annually ? 7. The population of the United States in 1880 was, in round numbers, 50,000,000; in 1890, 62,500,000. What was the annual gain per cent in this decade, supposing it to have been the same throughout ? 8. At the same rate of growth how many years would it require for the population to reach 100,000,000 ? 9. Allowing 6 per cent interest, what will be the cost of a perpetual annuity of $500 ? 10. Calculate the annual interest and amount of a note for $500 which had remained unpaid 5 years and 3 months. Rate named 6 per cent. 11. A gentleman wishing to establish a free scholarship of $300 to be in force for fifty years desires to know how much it will cost at a 5 per cent rate of interest. What will he have to pay ? 12. An annuity of $20 ran for thirty years. How much would it have amounted to in all had the payments been withheld until the last became due '/ Assume the rate to be 6 per cent. TOPICS KKLATKl) TO IX^rATlONS. 4od 13. What will an annuity of $100 cost to begin in 10 years and run for 25 ? Rate 4 per cent. Note. — If any annuity begin, say Jan. Ist, 11K)1, tlie lirst pay- ment on it will not be made till Jan. 1st, 1902. 14. A town borrowed $18,000, agreeing to repay it in 25 annual installments. What sum had to be raised annually, if 5 per cent was the rate named in the contract ? 15. An annuity of $50 was to be paid semi-annually, $25 every six months, and to run for eight years. Not being paid until eight years and six months had elapsed, and the current rate of interest being G per cent, what was the amount to l)e paid ? 16. A gentleman wishes to purchase a $1000 bond bear- ing 7 per cent interest, payable quarterly, due in ten years, so as to realize 5 per cent on his investment. What can he afford to pay for the bond if he compound his interest semi- annually ? 17. A $1,000 bond bears 6 per cent semi-annual interest and is to run 12 years. It is offered at $1025. What an- nual rate will be realized from it. by a purchaser, allowing s 'nii-annual interest? SuooEHTiox. — Assume .05 J as the value of r«. 18. On a bond for $1(MM) payable by 6 per cent semi- annual coupons, and which was to run 24 years, the sixth payment of interest has just been made. What sum can be paid for it to net the purchaser 4^ per cent, computing all the interest semi-annually ? 19. A town issued a series of 12 bonds of $500 each dated May 1st, 1802, and bearing 6 per cent interest payable semi- annually. The first was to become due May 1st, 1900, the second May 1st, 1901, and so on, the last to become due May 1st, 1911. What can a purchaser afford to pay for, say, the 440 TEXT-BOOK OF AI.GEBKA. first and last of the series on Sept. 15, 1892 desiring to realize o ^, interest compounded annually ? 20. A bond for f 2000 bears 5 per cent semi-annual in- terest and is to run 30 years. What rate (allowing semi- annual compound interest) will be realized if it be purchased 3 months after its issue for f 1911 ? SuGGESTiox, — Assume .05|^ as the vahie of Vg. ANSWERS. ART. 18. 2. +;"); 3. "7; 4. -'JO; 6. Mr>; 6. "l.'J: 7. +4: 8. " :i : 9. +1'; 10. -r>; 11. "41 ; 12. -4(;. 28. 1. (.'}. -f 111^7: 5. - :\()\;i; 6. r>i).228 ; 7. + 34 ets. 31. 9.-7; 10. 'J'A: 11. - ."iC* : 12. — r>S'J : 13. — 1^001 ; 14. — IT);""). 33. 1. -f 221^2; 2. -1-19: 3. —'.Ml; 4. — .VUf; ; 5. -{- VM\ 6. + 17°. 37. 1. - KJ: 2. -h LMC; 3. - r.7 : 4. — 42; 5. -f- I'SS : 6. - S«. 41. 1. - \i\r>n : 2. — .SS: 3. r)7 ; 4. - .".',: 5. .(MrjKi; 6. - r.-.. 42. 2. (1. 2r)li; (L\ jr>{\', {^:\. Aim>: (4. - lai; (5. - 2744 ; ((>. 12.S.LM ; (7. — LMl)7. 44. 1. _ r,; 2. - T): 3. — 10: 4. — 1 : 5. — .77."); 6. -.ll.';; 7. .().S2. 46. 1. i (), i 13, i 7, i i:>, J- 1 1 : 2. 2. - :;. - 4. r,, -7; 3. ^2,^3,^4; 4. i 2, i :;. 80. 1. '/ -h ^' + '• : 2. 2 A ; 3. . .,»-h.S/'^-f- W: 6. ^'^ , 4A_^/ A ^ ;5 .. 24 82. 1. 1 2 : 2. 1 1 1 : 3. - 178 : 4. 4 ; 6.^ 12 ; 6. i 22. or ^tz 1<>. <>r ik <*»• <•»* 0. 441 442 TFXT-BOOK OF ALGEBRA. 83. 1. 144 ; 2. 840 ; 3. 198 ; 4. 200. 85. 2. a; = 7; 3. 12; 4. 6; 5. 10 ; 6. 5; 7. 4 ; 8. 5 ; 12. 24, 16 ft.; 13. $7, 70, 28; 14. 50; 15. 9, 14, 27; 16. 45 17. 4, 6, 9 cts. 89. 1. (1. 9a; (2. 7am; (3. 0; (4. -7a^; (5. -2m^n; (6. 2 axt/ ; 2. 20 «i^ — 2 a^^, ^ 3. a- - (w/ + 3 // - 2 ; 4. 5a-\-Ab — Sc — e; 5. 6 ; 6. ; 7. 6 ab -\- 7 ax^ — 4 a^x -\- ax^ — 5 a%', 8. 5 cc^ + 4 ic^ + 3 o;-^ — 4 a; — 9. 93. 1. (1. 2a; (2. 5b', (3. - 17c; (4. -x'-, (5. 31a%^: (6. a^b — ab^ ; (7. — 15 ad^ ; (8. 24 a^b^ ; (9. 6 m ; 2. a?^ _ 8 ^2^2 + 12 ; 3. 2 ax - 7 ^*?/ + 8 c^ ; 4.19ab — 2c — 7d — 2x^-\-3x'^ — x — l', 5. 2 ax -{- 2 x'^ : 6. a-3 — 2 .t2^v — 2 a-V' + ?/' ; 7. 2 .ry — (; rz + 6 w : 8. 2 a^ - 2 a'c + Sac'^. 96. 2. 6 (a-\-b) ; 3. - 2 (;r + z) ; 4. 16 (a + by ■ 5. 5(x-\- y) -\- (x — //) ; 6. - 9 {a + 2 bY - 8 (a - m). 97. 2. (a + 3) ic?/ ; 3. (« + ^ + ^ + //) ^ + (^^ + d -h/4- A) 2/; 4. (a + 2 c + 4 cZ) £r. 98. 2. (a _ Z,) (.r + 7/ + ;.) ; 3. {a + m) (.X + 2/) + (^ - n) (ir - y). 101. 1. 3a;2-2y + 2a;- 1; 2. 2x^'dy; Z.4.xy; 4.2b - Ac; 5. ; 6. 3 m. 102. 2. 3a + 3c;3. 5a — 6a;; 4. a-^ — »•; b. 2a — Ax; 6. _ rt + 8 ^^ + 7 a- — 6 ?/ - 2 ; 7. 12 a; — 8 y. 103. 1. (a + Z,) + (, + ^) ; 2. (a^ + 2 aZ* + Z/'^) - (c^ -f- 2 c^ + d^) ; 3. (a + Z<) + c. 108. 1. —16; 2. -12a2; 3. 45 ^/i^^^. 4 _24a^»c; 5. 6 a^^/^a:^ ; 6. — 2 aa;^?/ ; 7. 14 a%c ; 8. 6 a^^ ; 9. a%'^ ; 10. 4 a;^?/» ; 11. — 20 a%-^x^y ; 12. a;^^^,-^ ; 13. 2 a;"* + ^ ; 14. —?/« + *. ANSWERS. 443 112. 1. — 288 rtV.» ; 2. 8 ac -f \2 be ; 3. _ ^ ahxy -f ^ acx'^ — 21 nx ; 4. 78 a^mVy — 60 a*//*'///V/ + •'<> " '//'*^y ; 6. m* -|- m» + m« -|- /wc ; 6. — 14 ar -f 28 />6" -|- 42 c^ ; 7. 2 «2^,c — 2 a^»-^c — 2 a^.c'^ ; 8. —^(i*b^-\-2\a^b*-\-A0a%^\ 9. o-'^ + 5ar + 6; 10. x^-\-Qx - 10 ; 11. />^2 _j_ ^23.2 _ ^,2^2 _ ,^2^2 . 12. .!•< — y* ; 13. ar» + ar- - 4 a- + 80; 14. a-^ — 14 ar + 45 ; 15. ./- 4- 2 ^/.r -f- a"-^ ; 16. m'^ + (« + b) m + m 4- 4 np 4- 2 « + 4/> ; (m a!» 4- 6 xhj 4. 12 a*//* 4- 8 //». 117. 1. (3. (2). 9 a-y, Z»-^S 9 c^x\ 4 «'^//V, K; r/«^;»V^ «'^"', a-^''; (3). 8«W4», -'27«V>», — ^a%hnhi\ 34.S;A/V. ff""^*"' (4). 768 x'\ - 32 x"//*, a^^x'K pY'. - 2 A', o^\ 121. 1. 27; 2. —4; 3. 24^/-/.: 4. ./-./•-: 5. -4oVr: 6. 5 .v V : 7. _ 1 ,.y»r-; 8. - ^^ ; 10. '/""V; : 12. ;; (^^ - />)•'. 2 .r// 127. 2. 3 m^ (a 4- c)-*; 3. 3 a-^" ; 4. ff'"'; 6. — 5 /w/' — 4 /« // 4- 2 // ; 6. 3 //» — 4 «//•' ; 7. — 3 y>r.s» 4- 6 jor<2 4-7 ra--'; 8. 2a4-4c4-6i; 9. 3 (a - ^)< — 2 (a — i)=' 4- (a — b); 10. 2ar 4- 3//; 11. c — d; 12. 3 — a; 13. 2 a^ -{- 3 ab -{- b^; 14. a: 4- // ; 16. a;*— .^ 4- ^; 16. 3a:« — x 4- 2; 17. a* -\- a^x 4- wa-* 4- a-'. 444 TEXT-BOOK OF ALGEBRA. 128. 3. (1. 1 ; (2. 1 ; (4. 81, 27, 9, 3, 1, ^, i, ^^, ^ ; 129. 1. (3. (1) i4«&2; (2) i3«^.^; (3) 2^/; (4) ^4; (5) - 3 7^i ; (6) - 2 ac' ; (7) ^ 25^/^' 5 (^) i 3 ic^ 131. 1. a + ^^ ; 2. (wi,-^ + wi^i + 71^) ; 3. 52 + 5 X 2 + 2^ = 39; 4. 1 + y + t/^; 5. (2 a + 3 b) (4 a^ _ ^^, _|_ 9 ^,2^) . q ^12 _ ^6^12 _^ ^24. 7. (J^^ -f ?/;s^) (x' - yz^) ; 8. (a* - /y^) («,8 _|_ ^4^3 ^ ^6^ . 11. (4^4-10 h) (16 ^;^ - 40 aZ» + 100 li") ; 12. (2 n - h) (32 rA^ + . . . 4- />5), (2 ^t 4- ^.) (32 ^/^ — . . . — h^) ; 14. (m2 + n^) (m^ — m^n^ -\- n'^) ; 17 — 20. Not divisible. 133. 1. (2. 2 X 7m - m- m- 71-, 3. 7 X 13 (( ■ a • a • a ' b • c • c • c ; 4. 33 (^6 4- b) (a -h /v) (^r -\- b) -, 2. (2. 5 a; (4 ic^ _ 9 ,^2) . 3. ./'-^ (12 r^ — 3 b -\- 1) ; (4. xY (^'^ — a-^ 4- 1) ; r>. 7 br'x (2 6'2 - 3 bf + //^) ; (k 3 //f2 (2 x - T) c — b). 134. 1. (2. (^/.-^ 4- f''') (" 4- ^>:) {^( - b) ; (3. {nb + c^) (r/// - cd) ; 2. (2. (./; - v/) (.r^ 4" •''// 4" //') ; (3. (.r-2)(a;-^4-2.r + 4); 3. (2. (x 4- ;//) (./-^ — r^// 4- ..•>//' — .''-//•■' 4- •'■.'/' - r)^ (.''• — !/) (x'' 4- xhj 4- xhf 4- J-!/' 4- .z-,//^ 4- //'v ; (3. (a 4- 2) {a^ — 2 r/-^ 4- 4 a' - 8 ^//^ _|_ k; ^, _ ;>2). ^,, _ i>j («5 4- 2 a^ 4- 4 (i^ 4- 8 • 4- 4)^ (4. (.. 4- 6 ^)^ 2. (2. (^ 4- 3) 0. 4- 2) ; (3. (^ 4- ^) (^ + <^) ^ (9. {rs -f 5 .^) (I's 4- 18 .^) ; (12. (./• - 5) (.r - 4) ; ( 19. {n 4- 5) {a - 3) ; (20. (x - (>) (x + 5) ; 3. (2. C^x 4- 2) (3a^ 4- 1) 5 (3. (3 a^ 4- 7) (X 4- 2) ; (4. (4.r-l) (.r4-3); (5. (9.7-4-1) (.r 4- 7) ; (0. (3 x-2 y) (x+Aij)', (7. (2 .r 4- 1 ) (.r - 1 ) ; (8. (3 ;/• 4- 2) (./' - 7) ; (9. (2 r - ./) (r - (J ./) ; ANSWERS. 44.") (10. (2 m + ij) (//i - 2 y) ; (11. (12 x + 5) (j- - 3) ; (12. (15 « + 1) (« - 15) ; ( l:5. (.S a- - 4 y) (8 .r + y) ; 5. (2. (5 a-2 - .r - 12) (r, ./• + 1) ; (3. (4 a-'-^ - 2 ^- - 5) (2x-5). 136. 1. (a (2) (a + 1)»; (3).(4^» + 1/; (4) (2;r - 5y)»; (^.■-) (<; X- - y)»; 2. (2. (2 a + :U - c) (2 tt - 3 ^ 4- r) ; (3. (2 a + 3 ^» - 2c) (2 « - 3 ^» + 2 c) ; (4. (2a + c + 3/0(2a + r-3^); (5. (/ + m — 7i) (/ - m -f w) ; (6. (« -}- ^ + ^0 (« -\- 1> - •') ; (7. (a - i + 1) (a - ft - 1) ; (8. (4 a' - 4 ft^) (2 a' + 2 ft^ ; (19. (4 a-2 + 3 .r// - y^) (4 x^ -?ixy - f), or (Ax' + 5a-y + y^) (4x^ - 5«// + y'^) ; (20. (3 x« + 2 xy + 7 y«) (3 x« - 2 .ry + 7 //-) ; (24. {2x^-{-(Sxz-\-^z'^ (2ir*-6a-« + 95;2); 3. (3. (1) {a J^h){e-d)', (2) {a' + 1) (« + 1); (3) (ar -3) (y + 2); (4) (3a; -2y) (2a -7y); (5) {m _ 3 w) (9 a — 4 ft) ; (6) (m + »«) („<2 _ w) ; (7) (x^-,/) (x-y); 4. (3. (m^ + 4 7/^ + 1) (w+ 1); (6. (a-2 — 4 a: + (>) (4 a- - 5). 137. 2. (2. («-|-^,-|-c + J) (aJ^b-\-c -d)', (3. (2 7/^ - 3 7* +/> - 2 *7) (2 w — 3 w - ;> + 2 y) ; 3. (2. (« - 3 ft + 5 r)2. 142. 1. 3 a- ; 2. 2 « ; 3. b'c ; 4. 4 x^y'u^ ; 5. 2 (« + ft) ; 6. 7 xy ; 7. 2 a'-^afy (3 x — y)\ 8. a- — y ; 9. Triine to eacli other; 10.3a(a; + 4); 11.3(a;-l); 12. ./'3_r>; 13. ./"" -f-fl; 14. r (a — ft): 148. 1. 31; 2. 12G; 3. 2; 4. a; + 7 ; 5. x - 2y; 6. 2 a -I- 3 a; ; 7. c (a^ — ?») ; 8. 3 a* + 4 ^/ ; 9. a- (ar — 1) ; 10. a- (a; — 2) ; 11. 3 y - 7 ; 12. ax - fty ; 13. 2 a: -|- 1 ; 14. x' (3 a- + 2) ; 15. x^ - 5 a: -f ^>) ; 16. (« + ^f- 153. 1. (>48 ; 2. 720 ; 3. 21(>0 ; 4. 252 axh/\ 5. x^ (a + a ) ; 6. X (x^ - 1) ; 7. 3(> aVA'^d' ; 8. a-* - y^ ; 9. «*ftVa-* ; 10. SOx'Yz'; 11. 54a»ftV; 12. 72 aW; 13. 210^2/,^; 14. 12 xy^ (x^ — y«) ; 15. 8 (1 - x^) ; 16. ab (a + ft) ; 446 TEXT-]JOOK OF ALGEJiltA. 17. 12 x^ (x^ + 2) ; 18. xi/ (4 x^ — 1) ; 19. xhf {x^ — /)-^; 20. 24 ah {ii^ — l^) I 21. abc (x — ft) {x — h) (x - c). 160. 2. i, ^L, . ; 3. A, ,^\, ; 4. :^ ; 5. 7 ^// ; 12. t. 13. ^' . c- ' «'^ + Z*-^ 162. 2. 31, 71^, 5H, 12ftV; 3. Z> + ^; 4. x + y; 5. x^A 6. 2 — 2x^ 9. 2 r//-' bdf bdf ' * 30 abc 30 «^»c' 30 a^c ' 7. 15 (fiTia? 4 nx 10 a^ lOa'^/i' 10 aV 10 ahi ' g 567 a 98 Z> 198 (/a; 882 {a + &) ■ 504 X 504 ic' 504 x ' 504 ir ' 175. 2. 2A ; 3. t'^^ -1 ; 4. |o .^ ; 5. ^^- + ^^+ ^^^ ; 10 />"+^ 8. .+l>. + ^!-^^ • 10. ■' ; 11. -= ; 12. 17. tF^^; 21.0. ft^ — J"* 179. 2. ./• : 3. ; 4. — .) oh ; o. — '- ; 6. : nl \br 1 — x 7. — -^ ; 8. ///- - /r ; 9. — — ; 10. -^— ^ ; « ' 3 (a; - y) ' X ^ ' 10 a - 5 «■' 181. 2. - 2^^ 3. - J!il= 4. »*; 5. i^N 6. ^V^ : ;!r/ 4.t* « 4'/ 1(1. r-' 7.i^';8. -L-;9.fL±*; 10. ^ 7> (r — 1 X -{- // (f -|- b 183. 2. ^_(^^-±i); 3. H<^^ + f^) , 4. ?li:^^ 5.' rtft*-' ^ (y — ^') « (^^-^P — «) 1<> ^'^'' 4 7. *^ ;8. -^V;9-^^^%^; io.-L_; a -f- bi/ m-\-bn a (m^ -j- 1) /// -f- 1 •l0(3x-l)* 209. 2. I : 3. 1(); 4. 2,7^ ; 5. — ; 6. — : 7. 7 : 8. — a .H (I n 9. 2H ; 10. ,'; ; 17. ^?-±4±/; 19. a - /. : 21. 4. " -\-b-\- c 211. 3. 1 : 4. ;ij; 5. -^^ ; 6. 2; 7. m — a a + 6 + c 44S TEXT-BOOK OF ALGEBRA. 8. ^ "" " ; 10. 1 ; 11. 1 ; 12. 4; 13. 1. m -\- ~{-p 213. 2. 11^; 3. 7; 4. 7; 5. 275§§ ; 6. 420; 7. 7 ; 8. ^ ; 9.1; 10..; 12.^. 214. 2. 6 ; 3. 25 ; 4. 6. 216. 1.4^; 2. 4; 3.45; 4. ^p^ ; 5.4; 6. 7; 7. 12; 8. 7; 9. i ; 10. 3 ; 12. ''''' ~ ''^' - ; 17. ,\ ; 28. 4 ; 33. - 9; a-\-h — 111 — n 49. a = (iLZlM, h = d{a-\-c)^ d — h a-\-d 224. 4. 7 ; 5. 1()(>, 75 ; 6. 60 ; 7. 58^2^ ; 8. 6 ; 9. 10 ; 11. 25 ; 12. 22, 10 ; 13. 6, 8 ; 14. 14, 12 ; 15. 15, 5 ; 16. 3, 9, 15, 21 yrs. ; 17. 21, 7, 14 cts. ; 18. 25, 75 ; 19. 20 of $100 . . . 1280 of $1 ; 20. 34, 17 gals. ; 21. 35, 90 ; 22. $42, 28, and 18 ; 23. 175 mi. ; 24. 30 ft. ; 25. ^ oz. ; 26. 5; 27. 54; 29. 54; 30. 143^1 mi.; 32. $12000; 33. $25200, $3000; 34. lOfc ; 35. 30; 36. $45; 37. 3; 38. 37^ ft. ; 40. 10 da. ; 41. 4 min. ; 42. 40 miu. ; 43. 21 ; 44. 12i§ min. ; 45. 60 ; 46. 120. 230. 3. X = 2, ?/ = 1, or 2, 1 ; 4. 3, 5 ; 5. 8, 2 ; 6. 25, 15 ; 7. _ 1 0, — 60 ; 8. 1, 3 ; 9. 8, 6 ; 10. 12, 8 ; 11. 6, 15 ; 232. 2. 8, 7; 3. 8. 1 ; 4. 6, 3; 5. 21, 35; 6. 5, 5: 7. 3, 5. 234. 4. 5, 2 ; 5. 20, 15 ; 6. 4, 6 ; 7. — 6, 80 ; 8. 20, 60 ; 9. 3, 5 ; 10. 7\, 5 ; 11. 3, 4 ; 12. 9.\, 7 ; 13. .4, .1 ; 14. ^Jl^zJ^\ P — y ; 15. — ^"^ _!!!!!_, a (h — f/)' b — d^ (m -f- an)^ w -\- an 236. 2. 30, 9 ; 3. 33i^, 21] ; 4. 6.3. 39.2 ; a - li' ah — 1 5. a^ — i; a^ — h 237. 2. 15. 3 ; 3. 11, - 1 ; 4. 2, — 1 ; 5. 7, 11. A.NSWKijs. 449 239. 1. (>, 3; 2. 10, 7; 3. .4. .1 ; 4. 20, 4; 5. 57. \n:\: 6. (5, 10; 7. 12, 30 ; 8. 2, (^ ; 9. 1,4; 10. 19, 3 ; ^^ ;< (r - d) m{c-d) , j2. r>. 4 ; 13. -^'^ + ^\ '^^JJ^^-Z^, mi -f- hnt on -f- />/// 7 — 2 b 7 — 2 A 240. 2. 90, (JO fts. ; 3. 7142Sr), 142Sr>7; 4. .1i;420, $200; 6. 180, 145; 6. 92800000, 07100000 lui. ; 7. $12, 10; 8. $2r>0, 320; 9. 2.322, 11.03. 244. 1. 1, 2, 3; 2. 7, 8, 9; 3. VKS.7, 05.4, 32.1 ; 4. 24, 9, 5; 6. (•), 2, 1 : 6. 3, 9, 15; 7. 4.5, 10.8, 11.7; 8. 7, 4, 3; 9. /, a, '"' "' *'' " ' ■ (a' 4- A'*) r ' (a^ + //-*) c ' a' + b' 245. 1. 1, 3, 5 ; 2. $200, 300, 840 ; 3. 1, 2, 3, 4 cts. ; 4. $40, 30, 24, 2(>; 5. $122|.i. 07 j.;. liorsrs: 32,',. V2^ saddles ; 6. 65, 9, 49 years ; 7. 3 rt — 2 r, 2 A + 2 c — 3 ^r, 3 a - 2 A. 249. 1. 11 . 1 ; 8, 3 ; 5, 5 : 2. 7 : 2. 0. 1 ; 2. ; 3. 5. 1.1; 2. 2, 1 ; 3, 1, 2; 1, 1, 3; 7. 7, 8, 0; 9. Nu values. 250. 1. 55, 10; 15, 50; 2. 8. 2: l."i. 1 : 22. 0. etr. : 7. 123, 224, 325, etc. 729. 255. 2. . "^ , ; 3. ''i^l^: r». .\ - • ; 4. 45, 10; f -\- m -{- n n — 1 lO4i,0O.U 300,-0; 5, '"'-"'' -^ ^' : 3; 6.—^ : T, ; // -|- 1 ni -\- H b'r — be' or' — u'l- -,...,.. -.., ., „ nbc '■ laTT^b' W^^.' ii-'id.-'lr,: 8. -^-^; 9. P^'' : .; ; 10. ^ (^^^ — ^) ' 2 m 2 m 11.^/-^; 6, 479. 257. 1. r/V^ 121 /;*.-«. 10 w'A^V^j' ./^r«. -i^, '*— — , 9.rV-' 4 .,,„,..,„ fM)a*\ . 1 10 , ',.. \^b'^ ) 2i« 225 ^ (^_ „)'' ( — //)" (— r/, !>"'. ^,». ^b'^c^% 9, 729, Y^oS (—1)'" t^^'"^^'V"\ 260. 1. i?2^2 — 2 ^^(/r + r2 ; 2. 8 w^^ — 12 /M'^yy + 6 z/^/- — 7/^ ; 3. ic« + 12 xY + 48 ic^y^ + 64 / ; 4. 125 a^ — 75 6i2^,c + 15 ab'e'' — Pc^ ; 5. 16 a' + 32 a.^;r + 24 a^^^r^ + 8 a^x^ + a^x* ; Q, I - S b -\- 24.1'' — 32 b^ + 16 b' ; 7. 6t« + 9 a« + 36 (t' + 84 a« + 126 a' + 126 a^ + 84 a^ + 36 (^2 + 9 « + 1 ; 8. a* - f + I a^b"" — If ab^ + if ^*; 9. 625 — 2000 X + 2400 ^r^ — 1280 x^ + 256 x' ; 10. 8 ?-^ — 72 wr^ + 216 7n^' - 216 m^ 261. 3. xY + 2/'-' + ^'^'' + 2 ic V- + 2 xi/z + 2 icy;^'^ ; 4. 1 + 3 a;2 + 3 £P^ + x« + 2 £C + 4 ic^ + 2 ic^ ; 5. ^3 _^ 3 ^^2^ _j_ 3 ^^,2 _^ ^3 _ 3^2^. _ 6 ^^^ — 3b''c + 3 ac-^ _|_ 3 jjc'^ _ c3; 6. 1 — 3 ri + 5 a^ — 3 a^ - a^; 7. 1 _ 6 :r + 15 x^ — 20 £t;3 + 15a^* _ 6 ^^ + ic« ; 8. 8 (t^"' — 12 a^^'b"" + 6 aH'^'' — &«" + 12 a^m^i^ _ 12 «'"?/''c^ + 3 ^''"c^' + 6 a"* c^P — 3 b^'c^-P + c^^ ; 10. a* + 16 ^^^ + c^ + 4 (2 a«^» + 8 ab^ - a^c — ae" - % b\ — 2 bc^) + 6 (4 a%^ + a^c^ + 4 Pc'') + 12 (2 a^^c^ - 2 a^^^c - 4 ab-'c). 266. 1. ^t «^^^ i ^^ ^*5 110 I'^al root, — 4, -t- 3 x** ; 2. §, i -^, i ^ , - '^^ imaginary, i — ; 3 Af^ _ 6; 4. — 7 a'^x^, — — ^,110 real root, no real root: 6xY ^y K '^^^ .,8 2 1 9 A>2^4^ ;r^ y (vb^ 1 269. 2. /« + 3 o^^ 3. ^ (6^ + % n' ; 4. a -- ; ANS\VKi:t3. 451 5. ^'^ + b'y\ g ^, _ ^ ^. 1 . 9. ^^ 1 _2; 14. „ _^ .:':i - — 4- -^, +, etc. 272. 1. T)!, 217, 42.1, 20.82; 2. 6.42, .Sl.()8, 4.1<)4, .(>;i21 ; 275. l.'Ja — Sb; 2. 2 a — 7 ./• : 3. .r- + u; + 1 ; 4. ,,-' _ afj -i-O'^i 6. - — // ; 9. 1 -'' — — — . etc // ' ^ '-^ 278. 1. 2.S, 2;U, 11.4, 5.51 ; 2. .r)().S, .2()(). i ; 3. 8.O20 - ,* 2.755 + . 1.710 — , .585 - , :5..S.S2 + . 281. 1. 'Jo — S ./• ; 2. 1 — ./ -f ./ - : 3. 5 ./• — 8 // ; 4. 2 — X ; 7. 2 ./• - 1 . 288. 1. 11 -i- 7 = 18, or 4: 2. — 'rV..--^: 3. 15 „th^', 4. <; (fr — //)*; 7. 7 «x*'^ _ 7 /,<• — 2 ///* + 6 r " 4- 5 x>/ ; 8. 8 c«(/« + 10 c^rfJ + 3 r*c/i + 11 r8c/i ; 9. - 327(y4. 10. 216, 823543, ^i^, ^i,, ^ ; 11. — 18 y^^, 18 ««//" + '•; 12. 2401 ; 13. a«A a^/j»h. 14. x^m^; 11^^; 15. a^ -\- a^ — «-* -f- a-V — a^^'^ — a^^; 16. «-'^»Wif/-?*; 19. /// — y/ ; 21. .r* ; 26. <^* — 3aV; -- 26. a" -h «T^-"'.-|-//-"; 31. u"'b-^ aibh>n (-^ + y) '^ ; (x- - y) , 33. j-iy- ', * ; 452 TEXT-BOOK OF ALGEBRA. 301. 1. i (:^)% i (10)^ 1 (6)^ ^j (21)^ 1 (22)^ ^ (2) 2 :^ r\>\i- 1 u: ^•'^^ ,,.. 1 2. -(6 a^^c) ^ J (6 ao-) ^ -^ (b)^ , ~ (3 a^»)S I (10)^ 304. 1. (_125a«^^)'; 2. (36)% {J^a'b'')^ , [ ±] , (49 //^.V-^)^ ; 9 3- (i^j . (0^ - -')^)^ (216 .^«^-«)^ (^^^^ j ; 4. (r/'^)«, (a;6«)^, (a^f, (a% ^07. 1. (350)^ (1)^ (j^y, (804)3, (40)^ (i)^; 2. (5^»)^ (12a^ic2)^, (ct^^/)^, (o'« + 2^;'"+2y«5 310. 1. (64)^, (81)^ (6)«; 2. (««)-, (^')^'5 3. (25)^ (64)^^; 4. (a')^, {b)i', 5. (6561) ^ (512)^ (15625)"; 6. (a'''y\ {a^y^, (a^)^'. 313. 1. 48 (5)^ 107 (3)^ ; 2. 55 «// (3 af — 33 a// (2 a^ ; 3. (20 a^ + 15 ^^2 + 4 c'^) (7 ic)^ ; 4. 16 (11)^ 20 (3)^ - 13 (2)^ ; 5. 12i (3)^; 6. a (3 a)^; 7. 3 (2)'; 8. 9 (2 «)^. 316. 1. 14 (6)-^, 12 (3)^, 10 a (3)^ 14 (9)^ ; 2. a^»" (a&)^ ^ {2f, 30 o^t/^ ; 3. ^ (6/i»)^ 2 «^ 140 ; 4. 2 (5)^ 6 (3)^ V- (4)^; 5. 4 (80000) ^S (648000)"; 7. (2 Z.'^c)*, f (12)^; 8. (^^j ,10 + 4 (35)^ + 30 (2)^ 9. 5 (9 a«)^, 20 b {aH>f, (432 a'c')"; 10. 6 a/^V ; 11. (ic"/'-^?)"^, y\h • 12. of-y (a%^c^x'^i/*z'^y^ ; 15. 3 (20)^ - 12 (3)^ - (180)^ + r2, ab + "^ , f a [ bd\ , , ANS\VKl:S. 4:>:^, 319. 1. -^-^ , ^^^, • j^^ .- ^,^^^ - , G^l:>5)i .^ (25000)* (a^^)^ ,,. r> ' ' 10 ' ^ 4.3[(0)»-f(fi>4-f(4)i]; 5. ^ -f 3 M - (1 -- a;V\^,._^,,.)i__,. 9. _ (1 7 + L' • *j5;ii + S . 'Al -\- ('» • L>« + 4 • ;^5 4. i> . iji.sS). 320. 1. "..■' (.^r. - 10 (^- I0)i ; 2. ,^, ^ (3)* ; 3. - ^l^- (2 :ry«)i, - ^ (^^)* ' ^- " '' (^)*' »>'' «>'' ' « ' 6. 125 (5) J, (4)*, (2 a)- ; 6. 3 (1 -h 2 x + «'), 26 - 15 (3)^ ; 7. x - 2 a^V"' + y"*. 49 - 20 (6)*; 8. .r* — 3 a-// -|- 3 a-*// — //*, ^ - ^ ^— ^- ; a' 9. (^ + 27 !/)x^ + (9 ^ + 27 //)//*, - a* + (/..•)* ; 10. {a')K a^i 11. 2 ^r- (2 r)'**, C,rV/)* ; 12. 2 «x' (12)i , (7 «)* . 324. 1. (l.')i - 1, 5 - 2 (6)*, 3 (7)* - 2 (6)*, G^ - 2, 2 (7)* + 14* ; 2. 3 (7)* + 2 (3)*, 2 - ^ (3)*, 1 + ^ (2)* ; 3. (a; _ 1)* -l,x}- (a-//)* , (1 + a)^ - (1 - ^)* , ai [(x - a)*- «»), (a- + //)* + -*; 4. 1 + 3», 1 - 5*, 2* - 7^ 4 - 7*, 1'^ + 3. 5 - 3* ; 5. 1 + 'JK 1 + 5*, i (1 + 5i), 5()i + 1«*. 326.^ 1. 14 (- 3)* - 18 (- 2)i ; 2. 5 - 7 /; 3. 3 (_ 1)1, (2 t)* 4- (^ 0* ; 4. (- 2)*, (/>» - .4) /; 6.-8 (6)*, — 210 ; 6. - 60 (42)* i, — 12 ; 7. - // '/. ^/^ ; 8. 5 (3)*, 2 - i (2)*i ; 9. 4 ; -f- 5*, ^l CA)h . 10. ;, — 2-2(^-3)*; 11. 7() _ ;; (3()^* — [10 (5)* + 21 (6)*] /, - 46 - 43 (-3)*. ' The lettiT i is ofti-n written for y/— I. 454 TEXT-BOOK OF ALCKIUIA. 329. 1. 14 ; 2. 3 ; 3. 4 ; 4. 10 ; 5. 7 ; 6. 144 ; 7. .\ ; 8. T) ; 9. 1; 10. ^il+i!; 11. 1; 12. 1; 13. ^f^H^; 14. V- a 2a — 332. 1. i 2 : 2. i 8 ; 3. i 3 ; 4. (62)^ i 5. ^ 4 ; 6. ^ 7 ; 7. -7^-T-; 8. i 1; 9. (ahy--, 10. ^ M- 338. 1. G, and — 10 ; 2. 3, and - 25 ; 3. 9, - 8 ; 4. 8, 2 ; 5 1, _ D ; 6. 6, ; 7. 31, — 11 ; 8. 15, 8 ; 9. 14, - 2 ; 10. — X — i; 11. 6, — 5i; 12. 70, 50; 13. 2, .48; 14. 1 (3 i (5)i) ; 15. A- (- 19 i 58li) ; 16. i (5 i 67^) ; 17. ,V (8.^) i 451.96^) ; 19. 2^, - 7| ; 20. |, f^ ; 21. ,%, - f. 339. 1. =k 2, i 10 ; 2. =t 7 ; 4. 7, 3 ; 5. 225, 121 ; 6. 27 ; 7. 112 and 76 ; 8. 4 mi. per hour ; 9. 3, 4, 5, and — 1, 0, 1 ; 10. 259, 481 yds. ; 11. 8; 12. i 6; 13. $7950; 14. 15, 17, 8 ft. ; 15. 12 ; 16. 140, 120 ; 17. 160, 90 ; 18. 9 pence. 343. 1. 4, 3, and 5\, If ; 2. 3, 5, and ^ 72, 23| ; 3.4,3; -1,-20; 4.4,1; 7,10; 5.^12,3; 3,12; 6. 5,6^',— Jjf , — 71 ; 7.t 71, 13 ; 8.t 43, — 51 ; 9.t 1, 1 ; 10. 2, 3; I, 4; ll.t 12, 4; 12. 7, 4; - 4, - 7 ; 13. 5, 3. 346. l.t il3, il; 2.t il, il; S-t i2, ^1; 4. ilOl, i4; i6, i7. 5. i-a(i5^il), ^./(i5^=pl); 6. i (i(6 ahY^ U^fby^ y it. (i (6 aiy =p (a/>)i) ; 7.11,5; 6,10; 8. 9^, 3i; - 2, - 4; 9. f (1 ± 5^), o^„(l i 5^) ; 10. i2,i:l; i4(7)^=FH7)^; 11. i 2, i 1 ; i ^ , i c>c , (See 358, 2) ; 12.t i |, Azh', 13. ±7,i2; ±3^^3(3)^ 347. 1. Az 36, i 16 ; 2. i 77, :t ^1 ; 3. 1 (1 i 5^) ; i (3 i 5^) ; 4. 36 ; 5. 64, 36 ; 6. $2025, $900 ; 1 This problem is symmetrical in x and y; i.e., tliey can cliange places without altering the equations (2-30, «)• This the answer indicates, since the vahies for X and y will satisfy the equation when interchanged. Hereafter, instead of writ- ing these double results, the answer will be marked with a dagger to show that the values can be taken the other way. ANSWERS. 455 7. i 10, i 2 ; i 6 (2)i, i 4 (2)* ; 8. $2, $3 ; 60, 40 ; 9. 4 ft., 13 ft. ; 10. 10 yds., 2 yds. ; 32 yds., 1 yd. ; 11. S Ins., lirs. ; 12. 30, 38^ ; 28, 22^. 351. 3. 3. 4 ; 4. 4, T) ; 6. — T), — ; 6. — 1, - 12 ; 7. _l()i,, -11^; 8. +y, -3; 9. -10, +30; 10.3,5; 11. 21,-10; 12. ^, i. 354. 1. -r>./, -18«; 2.0, -3, -^; 3. ^iH«)*; 4. - 1, i i; 6. i 2, - 4 ; 7. 0, i 4, 2; 8. 1, 2, i- 355. 2. x« - 7 a! + = ; 3. a;* + 4 x» - 7 j-2_ 10 u- = ; 4. .1- - 15 a-2 -h 60 X — 46 = 0. 357. ,/ = 2, a minimum, y = — V, »'*' maximum. 361. 5.1. -^^j-3)\ -l.L^i^; 6. ./ 2-* (i 1 i ; 7. ; 8. 025 ; 9. 18.72 ; 10. 32 ; 11. + 2 362. 5. 2, - 1 ; 6. i 2, i 1 ; 7. ± 7, i 5; 8. i 1, i i; 9. [i (7 i 349 *)]» ; 10. 9, 9* ; 11. 12*, 7* ; 12. 9, 6.76 ; 13. / -^:^(^' + 4qc)*) y. ^^ 81,2401; 15. 343, -J^^*^^; 16. 256, (- 24)' ; 17. 0, 4, 9 ; 18. (§)* , (i)* ; 19. 5, 3, 4 i 10* ; 20. 2, 1, H3i(-31)*); 21.4,2, H- 7 ±17*). 363. l.t 5, 2; 2.t ±2(2)*, ±5*; S.f ± i, ± i ; 4.t±3, ±1; 6. 1944*,72»; 6.t7,4; 7.t5,3; ll.t +1,-2; 12.t 7, 2 ; 13.t 27, S ; 14. 81, H ; 16. 8,32; 2«, 4*; 16.t ±4, ±3; 17. 32, i; 4, -3; 18. 2, 1 ; 19.2,2; 16, i; 20.4,2; 3 ± 21*, 3 ^ 21* ; 21.2,3; Ti«,18; 22.t r,. 4 ; 8 ± T^a (- 2505)*, 8 =f yij (- 2505)*. 364. 1. 8J, 2i, 5^ ft. ;' 2. 343, 64 cu. ft. ; 3. ± 3, ± 1 ; 4. 64, 512; 6. ± 6, ± 3; 6. 4, 2 ; 3 + (- 19)*, 3- (- 19)*. 367. 8. The first; 9. The second. 456 TEXT-BOOK OF ALGEBllA. 368. 2. .T < 14 ; 3. X < 8, y > 3i ; 4. ^ > a, and {ah — a + h) x < ah'^ ; 6. ; 7. S. 379. 1. I ; 2. If ; 2"*^, l^t ; 3. 3 ; 4. 6400 ; 5. § (3)^ ; 6. h I, e, II ; 7. 2 : 1 ; 9. - i\ ; 12. 6, 8 ; 13. | ; 14. 1^- 15. J, 6, i 10; 16. !!i2Lzi^; rt^ + ^^ p — q m — n — i> + ^ 380. 4. (1. 56; (2. 88f ; (,3. 100; (4. ^; c (5. 25 x^ = 27 ?/=^. 388. 2. (1. (1). 2,6599; (2). 2.5378; (3). 2.5647; (4). 2.6532; (5). 1.6532; (6). 1X021; (7). 0.6021 ; (8). 0.4133; (9). 2.5682 ; (10). 3.6990; (11). 5.8319; (12). 3.5623; (3. (1). 2.5073; (2). 4.1512; (3). 3.1512; (4). 0.4095; (5). 3.3765; (6). 1.3018 ; (7). 2.5731 ; (8). 6.4632 ; (9). 5.1232. 390. 2. .63329 ; 3. .013015 ; 4. 2.8107 ; 5. 102.2 ; 6. 17733 ; 7. 88.88 ; 8. .0005395 ; 9. .1099 ; 10. .001051 • 11. .6955 ; 12. .00111 ; 13. .0003318. 392. 3. (2. 703 ; (3. 2924 ; (4. 28556 ; (5. 2337200 (2336544, exactly). 393. 3. (2. 7; (3. 5; (4. 74.167; (5. 14.342; (6. .004057: 5. (2. .07094; (3. .001086; (4. 523; (5. .004939. 394. 3. (2. 2401 +; (3. 418333333; (4. 2.051; f5. ^''>34.1 (6. 22.8; (7. 429.6; (8. .941. 395. 2. (3. r:>.iS6(j ; (4. .8806 ; (5. 146.76 ; (6. 5.4S75 : (1. 1.6155; (8. .70717; (9. 4.957; (10. 1.8217. 398. 2. (2. 6 ; (3. 1.537 ; (4. .4581 ; (5. — .2031 ; (6. 3.537; (7. 1 ; (8..- .4319; (9. .9031; (11. 3.011 : (12. -t: 2.26; (13. 4.336. - .336. 402. 1. (2. — - a ■ C3. - 1. ^/ i S2 (Is + (a — i- d)H^', n ' ' ANSWERS. 457 (;j. i ,. J- a 4- (,^ - I) ./;; (4. ^ ;. yil - {n - l)d\; /i I — a. /o 2(j; — aw) . ,« l^ — a'^ . ,. 2(w/--«) . ^- (1- ;73T' (^- 7(;r3T)' (^- 25-/-^' ^^' 7(73T)' 4.(1.^^+1 (2. ^fr ^i ((2« - rfr + 8rf*)4 - 2« + ^i; (3. -j^; (4. ^ }2 / + ,/ ± [(2 i + rf)' -8rfs]» i; 5. (1. l-{n- 1) rf; /. 403. 2. ;U ; 3. 77, 779 ; 4. - 71 : 5. ^ (ii + 5) ; 6. 142 ; 7. - V» - 1<>5 ; 8- 198 a + 6.S /y ; 9. 50, 5150; 10. d = 5 ; 11. 2, 2.3 . . . 7.7, 8; 12. 43,-1; 13. 7, 6, a'*; 14. 25'^ 16. 106'\ 407. 1. (2. (1) ?L+i!lZllii; (2)(^-^|"^^""N ^•^- '^ ^ ;:^' (2)^^;^^^ C'^) ^-z- (— i)-s-; a. (i) Q^; ^-^ :^7 ' ^'' ^$^; 2. (2. (1) ^-^-;-/;^'- + 1 ; / H - 1 _ ^^ n - 1 (1>) lo g- (ff + ( r - 1) s ) - log, ff . log. r (.S) lo g- / "- log. (Ir - ( r - 1) ^) _^_ -^ . log. r 3. ( 'J. o (s — ft)"-^ = /(s —/)'•- ^ : = 0. 458 TEXT-BOOK OF ALGEBRA. 410. 1. 6561, 9841 ; 2. ,|^, 11H| ; 3. 1, ^1^, 4. 243 (6)^ 364 (6^ + 2'^) ; 5. ^\, 12^ ; 6- Uh fill ; 7. a = 10,l== 1000000000 ; 8. 1953.1 ; 9. _ 28, 14, - 7, I, - t, I ; 10. (a/j)h • 12. 27 ; 13. f ; 14. f|. 418. 1. 12.68; 2. $719; 3. 5% ; 4. lOi, nearly; 5. 3 yrs. 7 mos. 17 da. ; 6. 5^ ; 7. 2i, nearly ; 8. A little more than 21 years from 1890 ; 9. $8333 ; 10. Int. $177.75 ; 11. $5477.40 ; 12. $1580.40 ; 13. $1055 ; 14. $1276.76; 15. $516.22; 16. $1160; 17. .0576; 18. $1183. INDEX. Addition — Of Algebraic Xumbers, 15-17, 28-30 Of Algebraic Quantities, 50, 76, a ; 86-89 Of Fractions, 174 Of Quantities with Fractional Exponents, 287, 1 . Of liiidicals, 311 Of luiaginaries. 325 Of Equations, 233 Of Inequations, 366, 5 ADFKCTED QlADKATK , 330 Alokbka — As related to Arithmetic and (;eonietr>', 2, 3, 4 Definition, 7. See 189, r Annuity Certain, 416 Antecedent, 369 Approximation, 396, and Re- mark, 417 Arithmetical Progression, 399 A HR AX Can be um.i i.- » the i>tuROCAL. 128, h Reduction, 158 Of Radicals, 292 Rationalization, 317 Of Repeating Decimals, 409 Root — Of Quantity, 60. Sec Poircr. Root — Of an Equation, 193 Satisfy an Equation, 189, a Scholium, 204 Signs, 20, 49, 45, b ; 64, 358 Similar — Terms, 76, 287, 1 Radicals, 311 SlMPLPJ — Term, 68, a Equation, 196 Simultaneous Equations. See Elimination. Sinking Fund, 416, 1 (2 Solution of an Equation, 193 Square, 57, h ; -Root, 60, a Substitution — Elimination by, 228 Principle of, 113. This prin- ciple is much used in al- gebra. It constitutes one of the advantages of the literal notation. Refer- ences, 95, 112, 113; 208, 254, a; 258, 259, 3; 344, 4; 362, Exs. 1, 3, 11, 19, 23, 30, 39; 363, Exs. 13, 16, 29; 397; 398, 13 Subtraction, 90. See Befer- ences for Addition. Surd, 290 Symbol, 61. Classification, 62-67 Symmetry, Def., 230, a Examples, 363, 1-4, (>, 7, 10, 11, 12, 13, 1(), 20, 22, 2.5, 28 System — Of Equations, 225 Of Logarithms, 383 Terms, 68, 69, Sec. lY., Chapter ill. Of a Fraction, 154, a Theorem, 201 See 114, 116, 128, 130, 258, 322, 366, 367, 375-377, 397 INDEX. 463 In E(i nations, 210 In Inequalities, 366, 1 TniNOMIAL, 71 Square of, 116, 1 Cube of, 261, 1 Square, 114 Factoring of, 135 Tripi.u ATK Hath), 372, a Type-Fokms. See Equation. UXDETKKMINKI) MVI-THM-IKKS, 236 Unity — As Coefficient, 75, h As Exponent. 56. o L'mtv — As Denominator of Integer, 169 Unlike Signs. -See Like Siijns. Value, Num. 81 ; Max. 357 Variation, 380 Verification, 189, a Vinculum. -ts f(1 3-7 5- ^' 3 hr ^At " ^i'^ 911281 aAl53 THE UNIVERSITY OF CALIFORNIA LIBRARY