1 ^'^. (/fj[/l^C^^^ <2^^^ lA Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofalgebrOOtaylrich ELEMENTS OF ALGEBRA BY JAMES M. TAYLOR, A.M., LL.D. PROFESSOR OF MATHEMATICS, COLGATE UNIVERSITY 3>»ic Boston ALLYN AND BACON 1900 COPTKIGHT, 19 00, BY JAMES M. TAYLOR, NorfajootJ 5|«ss J. S. Gushing & Co. - Berwick & Smith Norwood Mass. U.S.A. T3 ' PREFACE The author has aimed to make this treatment of algebra so simple that the pupil can begin the book to advantage immediately upon completing an ordinary course in arith- metic j and, at the same time, so scientific that he will have nothing to unlearn as he advances in the study of mathematics. Great care has been taken to develop the subject logically, yet the immaturity of the pupil has been constantly kept in mind, and every legitimate aid has been given him. Simplicity has been attained not by using inexact statements and mechanical methods, but by avoid- ing many of the outgrown phrases of traditional algebra, by giving demonstrations and explanations in full, and by making fundamental concepts clear and tangible. An introductory chapter explains the meaning and advantages of the literal notation, and illustrates the use of the equa- tion in solving arithmetic problems. In Chapter II real numbers are first considered, and are defined as multiples of the quality-units, + 1 and — 1, and the pupil is drilled in the use of particular real numbers before he is required to represent general real numbers by letters. General principles are first illustrated by particular ex- amples, the study of which prepares the pupil to grasp the meaning of the formal statement of the principles, and iv PREFACE makes it less likely that lie will memorize without com- prehending the demonstrations which follow. With this arrangement the reproduction of the demonstrations may be left for the review; but the pupil should become familiar with each principle and definition before a new one is considered. When the demonstrations are not re- produced, it is recommended that the proofs be carefully read and discussed in class, so that the pupil may be fully convinced that the principles are true. He should then be required to state the authorities for each step in the proof when the steps are given. The identity and the equation are sharply distinguished. Two groups of principles are stated, the first for proving the identity, the second for solving the equation. The need of the principles of the equivalency of equa- tions and systems is clearly shown. These principles are fully illustrated and proved, and upon them are based the methods of solving equations and systems of equations. In the chapter on factoring, the formation of equations with given roots serves as an introduction to the converse problem of finding the roots of a given quadratic or higher equation, and to the method of making factoring funda- mental in the study and solution of quadratic and higher equations and systems. The graph is used to illustrate the meaning of equations in two unknowns, of systems of equations and of equivalent systems; it also serves to make clear some of the general properties of equations in one unknown. The theory of limits is given as briefly as is thought to be consistent with clearness. It is used in proving the PREFACE V laws of incommensurable numbers and in evaluating ex- pressions which assume the indeterminate form 0/0. The treatment of imaginary numbers affords a good illustration of the advantages derived from regarding alge- braic numbers as arithmetic multiples of quality-units. When a pupil understands that the quality-units V— 1 and — V— 1 include the idea of the arithmetic one and that of oppositeness to each other, that (V— 1)^ = —1, and (V— 1)*= +1, he has mastered all that is new in imagi- naries, and can then state the general laws for products and quotients of imaginary and real numbers (§§ 274, 276). This concept makes for simplicity, for it enables us to express general laws which are true for real, imaginary, and complex numbers, and it clearly separates the problem of finding the arithmetical value of a result from that of finding its quality. Graphic representations are used to illustrate the meaning and reality of imaginary and com- plex numbers. Special attention is invited to the brevity and complete- ness of the demonstrations of the, principles of proportion, the early introduction of the remainder theorem, the use of type-forms in factoring, and the treatment of fractional and irrational equations. The methods of working examples have been chosen for their simplicity and the scope of their application. The problems are varied, interesting, well graded, and not so difficult as to discourage the beginner. Many exercises contain easy examples which, especially in the review, should be used for oral work. Suggestions as to the method of attack are freely given ; rules are stated only Vi PREFACE for the most difficult operations, but not until after these have been illustrated by particular examples. The author has sought to treat each subject with suffi- cient fulness to meet the college entrance requirements, and more subjects are given than are ordinarily considered as a part of elementary algebra. The author is indebted to many teachers for valuable suggestions, but especially to his assistant, Mr. C. D. Kings- ley, who has carefully read all the manuscript and most of the proof sheets. JAMES M. TAYLOR. Colgate University, June, 1900. CONTENTS CHAPTER I Introduction SECTIONS PAOB 1-8. Symbols of Number and of Operations .... 1 9-18. Equalities ; Identities and Equations .... 8 19. Problems solved by Equations 15 CHAPTER II Positive and Negative Numbers 20-31. Real Numbers and their Symbols 20 32, 33. Principles of Identities 28 CHAPTER III Addition, Subtraction, and Multiplication op Bbal Numbers 34-38. Fundamental Laws of Addition .... 39-44. Definition and Fundamental Laws of Subtraction . 45-50. Definition and Fundamental Laws of Multiplication 51-53. Bases and Integral Positive Exponents . 54, 55. Change of Quality of Expressions .... 56, 57. Two Uses of + and — . Abbreviated Notation 58-60. Coefficients. The Distributive Law 30 33 36 40 42 43 44 CHAPTER IV Addition and Subtraction of Integral Literal Expressions 61-64. Addition of Similar Terms 49 65, 66. Addition of Polynomials 52 67. Subtraction of One Expression from Another ... 54 68-70. Removal and Insertion of Signs of Grouping ... 56 vii viii CONTENTS CHAPTER V Multiplication of Integral Literal Expressions SECTIONS PAGE 71-73. Degree of an Expression. Homogeneous Expressions 61 74-76. a- 0=0. 0» = 0. a'«-a" = a"»+" 62 77-81. Product of One Integral Expression by Another . . 63 82. Multiplication by Detached Coefficients ... 69 CHAPTER VI Division of Integral Literal Expressions 83- -89. 90, 91. 92- -94. 95- -97. 98, 99. Definition and Laws of Division 71 Definition and Product of Fractions .... 74 «»"/«"= rt"*-". Distributive Law 75 Division of One Expression by Another ... 77 0/a=0. Division by Detached Coefficients . . 84 CHAPTER VII Linear Equations in One Unknown 100-103. Definitions, Degree, and Root 87 104-110. Principles of Equivalence of Equations ... 88 111-114. Solution of Linear Equations 94 CHAPTER VIII Problems solved by Linear Equations in One Unknown 115, 116. Problems solved by Equations. Interest Formulas . 99 CHAPTER IX Powers, Products, Quotients 117-119. Powers of Powers and of Products .... 109 120,121. Square of Any Expression Ill 122, 123. The Product (a -f &) (a - b) and (x -\-a) {x + b) . 113 124-128. Powers of a -f 6. Like Powers 115 129, 130. The Quotients (a" - 6«) ^ (a - 6) and (a'* ± &") -^ (a + b) 118 131-133. The Remainder and the Factor Theorem . . .122 CONTENTS IX CHAPTER X Factors of Integral Literal Expressions SECTIONS 134-136. Monomials and Common Factors . 137, 138. Perfect Squares 139, 140. Type-forms x"^ -\-px+ q and ax"^ + 6x + c 141. Type-form a" — ^", where n is Even 142, 143. Type-forms a*' T 6", where n is Odd 144, 145. Type-forms a* + ha'^b^ + &* and a" + 6», n Even 146, 147. Perfect Cube. Summary. Special Devices 148, 149. Formation and Solution of Equations . PAGE 125 127 130 134 136 138 139 144 CHAPTER XI Highest Common Factor and Lowest Common Multiple 150-153. H. C. F. by Factoring 154-157. H. C. F. by Division 158, 159. L. C. M. by Factoring 160-163. L. C. M. by H. C. F. 150 152 158 160 CHAPTER XII Fractions 164-169. Definition and Quality of Fractious . 170-178. Reduction of Fractions .... 179. Addition and Subtraction of Fractions 180-184. Multiplication and Division of Fractions 185, 186. Complex Fractions. Powers of Fractions 163 165 170 177 181 CHAPTER XIII Fractional Equations 187-190. Equivalence of Fractional Equations. Problems 187 CONTENTS CHAPTER XIV Systems of Linear Equations SECTIONS PAGE 191-195. Equivalent Equations in Two Unknowns . . 197 196-199. Systems and Equivalent Systems . . 199 200-203. Elimination by Substitution .... . 201 204-207. Elimination by Addition or Subtraction . 205 208, 209. Systems in Three or More Unknowns . . 213 210,211. Systems of Fractional Equations . CHAPTER XV Problems solved by Systems . 216 212. A Determinate Problem .... . 220 CHAPTER XVI Evolution. Irrational Numbers 213-219. An nth Root. A Rational Number. Number of Roots 230 220-222. Principal Root. Evolution 232 223-225. Root of a Power, of a Product, of a Quotient . . 232 226,227. < y dollars for the horses. Here y denotes any whole number, x any whole or fractional num- ber, and XX y their product. Ex. 3. If 60 dollars is divided equally among 5 boys, each boy re- ceives 60-4-6 dollars. If x dollars is divided equally among n boys, each boy receives x -^ n dollars. Ex. 4. li m men earn n dollars in one day, each man earns « -5- m dollars in one day, and therefore x men will earn w -f- w x as dollars in one day. INTRODUCTION 3 In these examples the reasoning is the same whether the numbers are denoted by figures or by letters. When letters are used in its statement, each problem is a general problem, and includes an unlimited number of par- ticular problems. 4. The following signs, or symbols, of operation, with which, as has been assumed, the pupil is already familiar, are common to all branches of mathematics. The sign of addition, +, read 'jylus/ indicates that the number after the sign is to be added to the number before it. E.g., S -^ i means that 4 is to be added to 3. a + 6, read 'a plus 6,' means that the number denoted by b is to be added to the num- ber denoted by a ; or, more briefly, it means that b is to be added to a. The sign of subtraction, — , read ^minus,^ indicates that the number after the sign is to be subtracted from the num- ber before it. E.g.., a-\-b — c, read 'a plus b minus c,' means that 6 is to be added to a, and then c subtracted from this sum. The sign of multiplication, x , or a point above the line, read ^multiplied by, or Unto/ indicates that the number before it is to be multiplied by the number after it. The sign of multiplication is usually omitted between two letters or a figure and a letter. E.g.^ 2ab, read '2a6,' means 2xaxb; 7a6c, read ' 7 a&c,' means 7 • a • 6 • c. The sign of multiplication cannot be omitted between two factors when both are denoted by figures ; for by the notation of Arithmetic, 54 means 60 + 4, not 6x4. The sign of division, -i-, read ^divided by,' or ' by,' indicates that the number before it is to be divided by the number after it. E.g.,, a -i- b X c -i- d, read ' a by & into c by d,' denotes that a is to be divided by 6, the result multiplied by c, and then this result divided by cl. 4 ELEMENTS OF ALGEBRA Observe that in a series of additions and subtractions, or in a series of multiplications and divisions^ the operations are to he performed from left to right. Exercise 1. 1. If a boy has 5 marbles and wins 4 more, how many- marbles has he ? If he has a marbles and wins h more, how many marbles has he ? 2. One part of 25 is 7. What is the other part ? One part of 25 is n. What is the other part ? One part of the number m is n. What is the other part ? 3. The difference of two numbers is 6, and the smaller is 12. What is the greater ? The difference of two num- bers is n, and the smaller is x. What is the greater ? 4. How old will a man be in 6 years, if his present age is 36 years ? How old will a man be in c years, if his present age is x years ? 5. In 10 years a man will be 50 years old. What is his present age ? In & years a man will be m years old. What is his present age ? 6. The length of a room is x feet, and its width is h feet less than its length. What is its width ? 7. One number is x, and a second number is y times as great. What is the second number ? 8. One number, x, is y times as great as a second num- ber. What is the second number ? 9. The number which contains 4 units and 5 tens is 10 X 5 + 4. Write the number which contains x units and y tens. 10. Write a number containing a? units, y tens, and v hundreds. INTRODUCTION 6 11. Of three consecutive whole numbers 6 is the second; what are the first and the third ? If the second is m, what are the first and the third ? 12. Of three consecutive whole numbers 7 is the first; what are the second and the third ? If the first whole number is x, what are the second and the third ? 13. Of three consecutive even integers, 8 is the third; what are the first and the second ? If the third integer is m, what are the first and the second ? 14. If a goat costs x dollars, and a cow costs 4 times as much as a goat, and a horse costs 3 times as much as a cow, how much does a horse cost ? 15. In example 14, how much do a goat, a cow, and a horse together cost ? 16. A is ic years old, B is 17 years older than A, and C's age equals the sum of B's age and A's age. How old is C ? 17. If m sheep cost x dollars, and n cows cost y dollars, what would c sheep and b cows cost ? 18. A travelled a hours at the rate of m miles an hour, and B travelled b hours at the rate of y miles an hour. How many miles did A and B together travel ? 19. A rides his bicycle n yards ; the circumference of each wheel is m feet. How many revolutions does each wheel make in going this distance ? 5. A mathematical expression is any symbol or combina- tion of symbols which denotes a number. If all the symbols of number in an expression are numerals, the expression is called a numeral expression. An expression which involves one or more letters is called a literal expression. The number denoted by a numeral expression is a par- ticular, or a fixed, number. For sake of distinction, the 6 ELEMENTS OF ALGEBRA number which is denoted by a literal expression is called a general, or an arbitrary^ number. By the value of an expression we mean the number de- noted by it. E.g., 4, 5 — 3, and 7x5 + 4x2 are numeral expressions, and each denotes a particular, or fixed, number ; while a, a +4, and ax-\-b — c-i-y are literal expressions, and each denotes a general number. 6. An axiom is a truth so obvious that it may be taken for granted. Two numbers are said to be equal when they bear the same relation to the same unit. E.g., 4x3 and 6 x 2 are equal numbers, since each is 12 times 1. I and I are equal numbers, since each is 6 times \. The axioms concerning equal numbers, which are most frequently used in Algebra, as in Arithmetic, are the fol- lowing : 1. Any number is equal to itself. 2. Any number is equal to the sum of all its j)cirts. 3. If each of two numbers is equal to the same number, they are equal to each other. 4. If equal numbers are added to equal numbers, the sums are equal. 5. If equal numbers are subtracted from equal numbers, the remainders are equal. 6. If equal numbers are multiplied by equal 7iumbers, the products are equal. 7. If equal numbers are divided by equal numbers, except zero, the quotients are equal. E.g., 12 = 8 + 4, and 12 -- 4 = (8 + 4) -- 4. Again, 2x0 = 5x0; but we cannot divide by and say that 2 = 5. 8. The value of a mathematical expression is not changed when, for any number in it, an equal number is substituted. INTRODUCTION 1 7. The following signs of relation are common to all branches of mathematics: The sign of equality, =, read 'is equal to,^ is placed be- tween two expressions to indicate that they denote equal numbers. The sign of inequality, >, read Us greater than/ is placed between two expressions to indicate that the first denotes a greater number than the second. The sign < is read 'is less thanJ E.g., 4 4- 8 > 10 is read ' 4 plus 8 is greater than 10 ' ; and 7 — 2 < 12 is read ' 7 minus 2 is less than 12.' Observe that in each case the small end of the symbol is toward the less number. The sign ^, read 'is not equal to/ is used in stating that two numbers are unequal, without indicating which is the greater. Thus, a^h\?, read ' a is not equal to 6.' 8. The signs of grouping are the parentheses ( ), the brackets [ ], the braces \ \, and the vinculum . Each of these symbols indicates that the expression in- cluded by it is to be treated as a whole. E.g., the expression 12 — (3 + 5) denotes that the sum 3 + 5 is to be subtracted from 12 ; that is, 12 _ (3 + 5) = 12 - 8 = 4. The expression [32 - (4 + 6) -r- 6] -=- 3 denotes that one-fifth of the sum 4 + G is to be subtracted from 32, and the remainder divided by 3 ; that is, [32 _(4 + 6)- 6]- 3 =[32 - 2]- 3 = 10. Wlien one sign of grouping is used within another, to avoid ambi- guity different forms must be used as above. 9. Classification of expressions. A term is any expression in which the symbols of number are not connected by the sign -I- or — ; as 4 x 5 ^ 2 or 3 a6 -h c. 8 ELEMENTS OF ALGEBRA Hence the signs x and -;- indicate operations within a term, and the parts of an expression which are connected by the sign + or — are its terms. E.g.,, each of the expressions 5, a, and 5 aj -r- a is a term. The expression 2 ax + 3 6 -^ c consists of two terms, 2 ax and Sh-^c. In this definition of a term an expression ivithin a sign of grouping must be considered as a single symbol of number. Hence a factor or a divisor in a term can itself consist of two or more terms. E.g., the expression (a + &)(c + (?) is a term in which each of the factors, a + b and c -\- d, consists of two terms. A monomial is an expression of one term ; as 4, 6 xy, or A polynomial is an expression of two or more terms ; as 4 + T or a + 3 a^?/ + 7 6. A polynomial of two terms is called a binomial. A polynomial of three terms is called a trinomial. Observe that all operations within each of two terms must be performed before performing the operation between them. E.g., the binomial 10 - (4 + 2) (7 - 3) -^ (6 + 2) denotes that. 4 + 2 is to be multiplied by 7 — 3, this product divided by 6 + 2, and the resulting quotient subtracted from 10. Exercise 2. Express in its simplest form the number denoted by each of the following numeral expressions : 1. 14 + (7 -4). 5. 18 -(6 -2)3. 2. 18 -(12 -7). 6. (6 + 9)^5. 3. (6 + 2) -(7 -3). 7. 16 -(7 -1)^3. 4. (3 + 8)3. 8. 22-(18-6)-^4. INTRODUCTION 9 9. 12 +[4 -(5 -3)]. 11. 19 -[(2 + 4) -(5 -3)]. 10. 18 -[8 -(4 + 2)]. 12. 22 - [23 - (7 - 4)] - 5. 13. How many terms in each of the expressions found in examples 1 to 12 inclusive ? 14. Find the sum of 5 a; and 7 x. Just as 5 + 7 = 12, so bx+1 x = V2 x. Reduce each of the following expressions to its simplest form : 15. 2x-{-4:X. 17. ^x-\-^x — (dx. 19. \x-ir\x-\-^x. 16. bx-\-lx. 18. ^x — ^x-\-2x. 20. la + ^a — \a. 10. An equality is the statement that two expressions denote the same number. The expression to the left of the sign of equality is called the first member of the equality, and the expression to the right of this sign is called the second member. E.g., (5 + 3)0 = 72 is an equality ; of which (5 + 3)9 is the first meuiber and 72 is the second member. 11. Zero is the number obtained by subtracting any num- ber from itself ; that is, zero is dehned by the equality a - a = 0. (1) 12. To find the value of a given literal expression when each of its letters has some particular value, we substitute for each letter its particular value, and simplify the result- ing expression. Ex. Find the value of the expression (x + y)z -h(a — b), when a; = 6, ?/ = 3, = 4, a = 9, 6 = 2. Substituting, 6 for x, 3 for y, 4 for z, 9 for a, and 2 for b, in the given expression, we obtain (X + ?/)0 -4-(a - 6) = (6 + 3) X 4 -(9 - 2) (1) = 36-7. (2) 10 ELEMENTS OF ALGEBRA In the work above we have three equalities ; by axiom 8, the first expression is equal to the second and the second is equal to the third ; hence, by axiom 3, the first is equal to the third. 13. In working examples the student should give heed to the following suggestions : 1. Too much importance cannot be attached to neatness of style and arrangement. Neatness is in itself conducive to accuracy. 2. It should be clearly brought out how each result fol- lows from the one before it ; for this purpose it will some- times be advisable to add short verbal explanations. 3. Unless the members are very short the signs of equal- ity in the steps of the work should be placed one under the other. Exercise 3. Find the .value of each of the following expressions when a = 5, 6 = 3, c = 4, x = ^'. 1. a-\-h. 6. (a -f h)x. 11. x-^ {a — c). 2. a—b. 7. {a—b)c. 12. {a-\-b)(G -\- x). 3. a-\-b — c. 8. {a-\-b)^x. 13. (a — b) (x — c). 4. abc. 9. (a — b) -^ x. 14. [x — (b -{- 1)] a. 5. ab ~- c. 10. x^(a-\~c). 15. [a; + (a — c)] -j- a. 16. [3 6-(aj-a)]--c. 18. (9-a)(2b-c)(2x-3b). 17. (3a-2b)-h{x-b). 19. (3x-4.c)(3b-2 c)^(x-c). 20. [2 a - (3 6 - 2 c)] -- [(3 c - 3 6) (2 a - 3 6)]. 21. [3ir-2(a-6)]-f-[(2ic-3&)(a-c)]. 22. (2x-3b)(Aa-3x)-^{3x-3c-b). 14. A proof is a course of reasoning by which the truth of a statement is made clear, or is established. INTRODUCTION 11 15. Identical expressions. Two numeral expressions which denote the same number or any two expressions which de- note equal numbers for all values of their letters are called identical expressions. E.g., the numeral expressions 3G -^ 4 and 13 — 4 are identical, for each denotes the number 9. Again, the literal expressions 3 a; + 7 x and 6 x + 4 x are identical, for each denotes the general number 10 x. To prove that two expressions are identical, we reduce one to the form of the other, or we reduce both to the same form. Ex. Prove that the expressions 7x + 3x + 2x and 14 x — 2 x are identical. 7x + 3x + 2x denotes 12 x, and 14 x — 2 x denotes 12 x ; hence, by definition, the two expressions are identical. An equality whose members are identical expressions is called an identity. The sign of identity, =, read Us identical with,'' is often used instead of the sign = in writing a literal identity, i.e., one whose members involve one or more letters. E.g., 9 + 6 = 6x3, (1) or 3 X + 7 X = 8 X + 2 X, (2) is an identity, (1) being numeral and (2) being literal. Any equality which involves only numerals is an identity. The sign = points out the fact that equality (2) is an identity. The pupil should now prove the identities in Exercise 4. 16. Letters denoting unknowns. Any problem involves one or more numbers whose values are given, and one or more numbers whose values are to he found. Numbers given are called knowns, numbers to he found are called unknoivns. An unknown is usually denoted by one of the last letters of the alphabet ; as x, y, z. The following simple problems illustrate the advantage of denoting an unknown by a letter. Prob. 1. The sum of two numbers is 80, and the greater is 3 times the less. Find the numbers. 12 ELEMENTS OF ALGEBBA Let X = the less number ; then, since the greater is three times the less, ^x = the greater number. Hence their sum = a; + 3x = 4a;. Therefore, by the conditions of the problem, we have 4 a; = 80. (1) Divide by 4, x = 20, the less number. Multiply by 3, Zx = 60, the greater number. Observe that the numbers 20 and 60 satisfy the conditions of the problem ; that is, 20 + 60 = 80, and 60 = 20 x 3. Prob. 2. A farmer bought a horse, a cow, and a goat ; the horse cost 3 times as much as the cow, and the cow 4 times as much as the goat, and all three together cost 255 dollars. What was the cost of each ? Let X = the number of dollars the goat cost ; then 4 ic = the number of dollars the cow cost, and 12 x = the number of dollars the horse cost. Hence the number of dollars all three cost = x -}- 4 X + 12 a; = 17 ic. Therefore, by the conditions of the problem, we have 17 X = 255. (2) Divide by 17, x — 15. Multiply by 4, 4 a; = 60. Multiply by 3, 12 a; = 180. Hence the goat cost $ 15, the cow f 60, and the horse •$ 180. 17. Equations. Any equality which is not an identity is called an equation, as (1) or (2) in § 16. A value of x in an equation in x is any number which when substituted for x makes the equation an identity. An equation in one unknown as x restricts x to one value or to a definite number of values. INTRODUCTION 13 E.g., if in the equation 2 a; + 2 = a; + 8, (1) we put 6 for x, we obtain the identity 2x6 + 2 = 6 + 8. Hence 6 is a value of x in equation (1); and, as will be seen later, 6 is the only value of a; in (1). An equation, as (1) or (2) in § 16, expresses in symbols the conditions of a problem ; and it restricts its unknown to such values as will satisfy these conditions. Thus th« equation 4 a; = 80 restricts x to the one value 20 ; and the equation 3 « = 15 restricts x to the one value 5. The two kinds of equalities, equations and identities, must be clearly distinguished the one from the other. An equation states a condition, and the values of the unknown which satisfy this condition are to be found ; while an identity states that one of two expressions can be reduced to the other, and is to be proved. Exercise 4. Prove each of the following identities : 1. 7x3x2 = (10-3)x6. 4. 2X+ 7 a; = (27 --3). a;. 2.88-4 =(7+4)x2. 5. 9a + loa =( 6x4) • a. 3. 204-6 =(12+5)x2. 6. 106 + 8 6 =(36^2) • 6. By inspection find a value of x in each of the following equations, and verify it by substitution : 7. a;-4 = 0. 10. 4a; = 20. 13. 3x + l = 10. 8. 2a; = 14. 11. 2a; + l = 7. 14. a; - 1 = 6. 9. 3a;-15 = 0. 12. 2a;+4 = 8. 15. 2a;-4 = 4. 16. 2 a; + 1 = a; + 3. 17. 2 a; - 1 = a; + 2. 14 ELEMENTS OF ALGEBRA 18. The following principles, which are proved in Chap- ter VII, are used in finding the values of the unknown in an equation : (i) If the same number is added to or subtracted from both members of an equation, the unknown has the same values in the derived equation as in the given one. (ii) If both members of an equation are multiplied or divided by the same known number (except zero), the unkjioivn has the same values in the derived equation as in the given one. Ex. 1. Find the value of x in the equation 2a: + 5 = ll. (1) Subtracting 5 from each member, we remove all the known terms from the first member, and obtain 2 X = 6. (2) Dividing each member by 2, we obtain x=3. (3) By principle (i), x has the same value in (2) as in (1); and by (ii), X has the same value in (3) as in (2). Hence 3 is the one and only value of x in equation (1). Ex. 2. Find the value of x in the equation 4x-2 = a; + 4. (1) Add 2, 4x = x + 6. (2) Subtract ic, 3 a; = 6. (3) Divide by 3, x = 2. (4) By principle (i), x has the same value in (2) as in (1), and the same in (3) as in (2) ; by (ii), x has the same value in (4) as in (3) . Hence 2 is the one and only value of a; in (1). Check : Putting 2 for x in (1), we obtain the identity 4x2-2 = 2 + 4. Hence 2 is a value of x in (1). IN TB OD UCTION 1 5 Ex. 3. Find the value of x in the equation fx-fx = |. (1) To clear (1) of fractions, we multiply both its members by 8, I.e., by the least common multiple of its denominators. 12 X - 10 a; = 7, or 2 x = 7. (2) Divide by 2, x = |. (3) By (ii), X has the same values in (2) as in (1), and the same in (8) as in (2); hence | is the one and only value of x in (1). The foregoing examples illustrate the method of finding the value of the unknown in a simple equation. Exercise 5. Find the value of x in each of the following equations : 1. 3a;-7=2aj + 3. 11. 5.x- - 2 = 3a; + 4. 2. 3a; + 4 = a;-hl0. 12. Tx- 9 = 17 + 2a;. 3. 4a; + 4 = a; + 7. 13. |ic-4 = 5-Ja;. 4. 7a; + 5 = a; + 23. 5. 8a; = 5a;+42. 6. 6.T — 5 = 4a: + l. 7. 18a;- 7 = 43 -7a;. 14. 4a;-3 = 7-^a;. 15. l^-\ = \-\^- 16. i^-\ = i-\x. 17. 7a; + 21 = 45-5a;. 18. ^ + f = H-i.'^'- 19. \x + l=^\x + \. 20. A^ + it = A^ + 4. 9. 19a; -11 = 15 + 6a;. 10. 3 a; + 15 = a; + 25. 19. Problems solved by equations. Read the problem care- fully to find out exactly what it means; then state in algebraic symbols just what it says. To do this, let x denote the unknown number; or, if there are two or more unknown numbers, let x or some multiple of x denote one of them, and then express each of the others in terms of x. 16 ELEMENTS OF ALGEBRA By an equation express the condition which the problem imposes on x. Then find the value of x in this equation. Exercise 6. 1. A line 30 inches long is divided into two parts, one of which is double the other. How long are the parts ? Let X = the number of inches in the second part ; then 2 aj = the number of inches in the first part. Hence the number of inches in the two parts = 2 a; + a; = oic. Therefore, by the conditions of the problem, we have 3x = 30. Divide by 3, x = 10, number in second part. Multiply by 2, 2 a; = 20, number in first part. 2. A, B, and C together have f 90. B has twice as much as A, and C has as much as A and B together. How much has each ? Let X = the number of dollars A has ; then 2x = the number of dollars B has ; hence 3 a; = the number of dollars C has. .-. ic + 2 .r + 3 X = 90. 3. The sum of the ages of A and B is 67 years, and A is 17 years older than B. What is the age of each ? Ans. 42 and 25 years. 4. Three men, A, B, and C, trade in company and gain ^ 600, of which A is to have 3 times as much as B, and C as much as A and B together. What is the share of each ? Let X = the number of dollars B is to have, etc. 5. A farmer bought 3 cows for $ 180, and the prices paid were as the numbers 1, 2, and 3. What was the cost of each ? INTRODUCTION 17 Let X — the number of dollars paid for the first ; then 2x = the number of dollars paid for the second, and 3 X = the number of dollars paid for the third. 6. Divide 500 into two parts which are as the numbers 1 and 4. 7. What number is that whose double exceeds its half by 27? 8. Divide $ 575 between A and B so that A may receive $ 75 more than B. Let X = the number of dollars B receives ; then X 4- 75 = the number of dollars A receives ; hence 2 x + 76 = 675. (1) 9. Divide 105 into two parts whose difference is 45. 10. What number is that to which if 40 is added the sum will be 3 times the original number ? 11. Divide $84 among A, B, and C, so that B shall have $ 13 more than A, and C $ 16 more than B. 12. Three men, A, B, and C, contribute to an enterprise $ 2400. B put in twice as much as A, and C put in as much as A and B together. How much did each contribute ? 13. Find two numbers whose difference is 10, and one of which is 3 times the other. 14. If two men, 150 miles apart, travel toward each other, one at the rate of 2 miles an hour, and the other at the rate of 3 miles an hour, in how many hours will they meet? 15. A horse, carriage, and harness together are worth % 625. The horse is worth 8 times as much as the harness, and the carriage is worth $ 125 more than the harness. Find the value of each. Ans. $ 400, $ 175, and $ 50. 18 ELEMENTS OF ALGEBBA 16. A man bought a cow, a sheep, and a hog for $ 80 ; the cow cost $ 32 more than the sheep, and the sheep $ 6 more than the hog. Find the price of each. Ans. f 50, $ 18, $ 12. 17. The sum of $6000 was divided among A, B, C, and D; B received twice as much as A, C as much as A and B together, and D as much as A, B, and C together. How much did each receive ? Ans. $ 500, $ 1000, $ 1500, $ 3000. 18. A man has two sons and one daughter. He wishes to divide $ 12,000 among them so that the younger son shall have twice as much as the daughter, and the older son as much as both the other children. How much must he give to each ? 19. Divide 90 into five parts so that the second shall be 5 times the first, the third shall be J of the first and second, the fourth shall be ^ of the first, second, and third, and the fifth shall be 2 times the sum of the other four. 20. A, B, and C enter into partnership to do business. A furnishes 5 times as much capital as B, and C furnishes \ as much as A and B together. They all together furnish $ 18,900. How much does each furnish ? 21. A gentleman, dying, bequeathed his property of $21,840 as follows: to his son 2 times as much as to his daughter, and to his widow 1\ times as much as to both his son and daughter. What was the share of each ? 22. A farmer purchased 100 bushels of grain. He bought 2 times as many bushels of corn as of oats, and 2\ times as many bushels of wheat as of oats and corn. How many bushels of each kind did he buy ? 23. Three candidates for an office polled the following votes respectively : B received 3 times as many votes as A, and C 1^ times as many as A and B together. The whole INTRODUCTION 19 number of votes was 11,000. How many votes did each receive ? 24. A banker loaned to each, of 4 men equal sums of money. One man had the money 2 years, another 2^ years, another 3^ years, and another 4^ years. The entire interest money received was $ 275. How much did each man pay ? Let X = the number of dollars in the yearly interest on the sum loaned to each man. 25. A library contains 9 times as many historical works, and 5 times as many scientific books, as works of fiction. The historical works exceed the works of fiction and science by 10,500 volumes. How many volumes are there of each ? 26. A drover, being asked how many sheep he had, replied that if he had 3 times as many as he then had and 6 more, he would have 150. How many had he ? 27. The expenses of a manufacturer for 5 years were $ 17,500. If they increased $ 500 annually, what were his expenses each of the five years ? 28. A farmer had 590 sheep distributed in three fields. In the first field there were 25 more than in the second, and in the third there were 15 more than in the first. How many sheep were in each field ? 29. Of a herd of cows, 280 are Jerseys, and these are 35% of the entire herd. How many cows in the herd ? Let X = the number of cows in the entire herd ; then ^^^ x — 280. 30. A town lost 7% of its inhabitants, and then had 6045 inhabitants. What was its population before the loss ? 31. What number increased by -J- of 25% of itself equals 315? 32. The annual rent of a house is $240, and this is 8% of its value. What is its value ? CHAPTER II POSITIVE AND NEGATIVE NUMBERS 20. Algebra treats of tlie equation, its nature, the methods of solving it, and some of its applications. 21. In each of the equations thus far considered, the un- known is an arithmetic number. But in many equations the unknown cannot be an arithmetic, or absolute, number. E.g., take the equation ^x = 2x-n. (1) Subtracting 2 x from each member of (1), we obtain a; = - 5, or - 5. (2) Hence the value of x in equation (1) is denoted by the expression — 5, which has no meaning in Arithmetic. If, therefore, such an equation as 3 x = 2 ic — 5 is to be of any use, we must so enlarge our concept of number as to give a meaning to such an expression as — 5. To gain this larger idea of number let us first consider opposite concrete quantities. 22. Positive and negative, or opposite, quantities. Two quantities are said to be opposites, if, when combined (or united as parts into one whole), any amount of the one destroys, or annuls, an equal amount of the other. / Of two opposite quantities, we call one positive and the other negative. JS.g.f debts and credits are opposites ; for when they are com- bined, any amount of debt annuls an equal amount of credit. If we call credits positive, debts will be negative. 20 POSITIVE AND NEGATIVE NUMBERS 21 Two forces acting in opposite directions are opposites ; for when they are combined, any amount of the one annuls an equal amount of the other. If one of these forces is called positive, the other is called negative. Distances measured or travelled in opposite directions are oppo- sites ; for when they are combined, any distance travelled in the one direction annuls an equal distance travelled in the opposite direction. If one distance is called positive, the other is negative. The sign -f- or the sign — is often written before the measure of a concrete quantity to denote its quality, as positive or negative. When thus used, the signs + and — are read ^positive' and ^negative/ respectively, and are called signs of quality. E.g., if we call credits positive, + $5 will denote $5 of credit, and — $4 will denote .^4 of debt. If + 8 inches denotes 8 inches to the right, — 9 inches will denote 9 inches to the left. If + S° denotes 3° above the zero point, — 7° will denote 7° below that point. If 4- 400 years denotes 400 years after Christ, — 300 years will denote 300 years before Christ. In this chapter and the next we shall use as signs of quality the small signs ■•" and ~, which, by their size and position, are clearly distinguished from the signs of opera- tion, + and — . Exercise 7. I 1. If credits are regarded as positive, what is denoted by +^ 8 ? By "$ 11 ? By +$ 125 ? By -^ 175 ? If debts are regarded as positive, what does each of the above expressions denote ? 2. If degrees above the zero point are regarded as posi- tive, what is denoted by +1°? By +22°? By "5°? By -20°? 3. If distances measured from the point to the right are regarded as positive, what is denoted by '1 inches? By +14 inches ? By "13 inches ? 22 ELEMENTS OF ALGEBRA 4. If distances north of the equator are regarded as posi- tive, what is denoted by +300 miles ? By ~700 miles ? State in symbols each of the following in two ways : 5. $45 gain and $ 2^ loss is equal to $ 20 gain. +$45 +-|25 =+.$20, gain being positive ; b. (1) +a + -b = -{b - a), when a < b. (2) Proof. AVhen a>b and 'b is added to +«, the b nega- tive units in "6 annul b of the a positive units in +a. 26 ELEMENTS OF ALGEBRA There remain a — h positive units; hence, +(a — 6) is the sum. When a < &, a of the h negative units in ~h annul the a positive units in +a. There remain h — a negative units j hence, ~(h ~ a) is the sum. Exercise 9, 1. To make the sum zero, what number must be added to +3 ? To -7 ? To -31 ? To +14 ? To +a ? To 'b ? 2. Find the sum of +8 and "6. Of +5 and "7. Of "8 and +4. Of +11 and "15. Of "5 and +17. 3. Find the sum of "(i) and +(f). Of "(f) and +(^). Of +(f) and -(A). Of +a) and -(,4). What is the vahie of +a + -b, 4. When a = 43, & = 23 ? 6. When a = 23, 6 = 43 ? 5. When a = 63, & = 43 ? 7. When a = 43, 5 = 63? 8. Write six different sums each of which denotes zero. What is the vahie of x-\- y, 9. When x= -J, y= +9? 10. When x = +14, y = -19? 11. When x = -(-J^), y = -^{i)? 12. When a. = +(i-i), 2/ = -(!)? 30. The sign of continuation is ... or ---, either of which is read, ' and so on,' or ' and so on to.' Thus, 1, 2, 3, 4, ..-, is read, '1, 2, 3, 4, and so on' indefi- nitely ; 2, 4, 6, 8, ... 32, is read, ' 2, 4, 6, 8, and so on to 32.' The sign .-. stands for hence or therefore. The sign •.• stands for since or because. POSITIVE AND NEGATIVE NUMBERS 27 31. The integers of arithmetic number make up the series (1). 12 3 4 5 6 7 8 9-.. /j^a 1 1 1 i 1 1 1 1 1 \ ^ ^ Writing the positive and the negative integers in opposite directions from zero, we obtain series (2). ... -4 -3 -2 -1 +1 +2 +3 +4 ... (2\ 1 i 1 1 1 1 1 \ 1 ^ ^ If the divisions of the lines in (1) and (2) be taken as units of length, then each number in (1) expresses simply its distance from the zero point ; while each number in (2) expresses not only its distance, but also its direction, from the zero point, distances to the right being regarded as positive. Note. Arithmetic numbers are not positive numbers. An arith- metic number has no quality. If to any number in series (2) we add +1, we obtain the next right-hand number. E.g., -4 + +l = -3, -2+n = -l, and so on for the entire series. Hence, if we say that a number is increased by adding to it +1, the numbers in series (2) increase from left to right ; that is, ..., -3<-2, -2<-l, -1<0, 0<+l, +K+2.... We have, therefore, the following properties of positive and negative numbers : (i) Any positive number is greater than zero; while any negative number is less than zero. (ii) Of two positive numbers the greater has the greater arithmetic value; while of two negative numbers the greater has the less arithmetic value. E.g., +4> by +4, -4 < by +4, -7< by +7, +4 > +2 by +2, -4 < "2 by +2, -7 < -3 by +4. 28 ELEMENTS OF ALGEBRA Note. If we agreed to say that a number was increased by adding to it -1, tlien the numbers in series (2) would increase from right to left ; positive numbers would be less than zero, and negative numbers greater than zero. By common consent, however, it is agreed to say as above that a number is increased by adding to it +1, the primary unit. Exercise 10. Which is the greater, and how much the greater, 1. +3 or +7 ? 4. or +1 ? 7. "7 or +3 ? 2. +2 or -8 ? 5.0 or "1 ? 8. +2 or "3 ? 3. -11 or +2? 6. -5 or "9? 9. '5 or "11? 10. When is the product of two arithmetic fractional numbers greater than each number ? Less than each num- ber ? Greater than one and less than the other ? Can the product of two arithmetic integral numbers ever be less than either number ? 11. When is the sum of tw^o algebraic numbers greater than each number? Less than each number? Greater than one and less than the other ? Is the sum of two arith- metic numbers always greater than each number ? 12. Multiplying by an arithmetic fractional number in- volves what two operations with arithmetic whole numbers ? Addition of algebraic whole numbers involves the one or the other of what two operations with arithmetic numbers ? 32. In proving and using identities, the following princi- ples concerning identical expressions will be useful. These principles clearly follow from the definition of identical expressions in § 15 and the axioms in § 6. (i) Any expression is identical with itself. (ii) If eoAih of two expressions is identical with a third, they are identical with each other. POSITIVE AND NEGATIVE NUMBERS 29 (iii) If two identical expressions are added to or subtracted from tivo other identical expressions, the residting expressions are identical. (iv) If two identical expressions are multiplied by two other identical expressions, the products are identical. (v) If two identical expressions are divided by tico other identical expressions, not denoting zero, the quotients are identical. (vi) If, for any expression in an identity, an identical ex- pression is substituted, the resulting equality is an identity. 33. The converse of an identity is obtained by interchang- ing its members ; that is, the converse oi A = B is B= A. li A = B, then, from definition, B = A. Hence, the proof of an identity proves its converse. ^.<7., in proving +a ++& = + (a + &), we prove +(a + &)=+a ++b. CHAPTER III ADDITION, SUBTRACTION, AND MULTIPLICATION OF REAL NUMBERS 34. The positive and negative numbers defined in Chai^- ter II are together often called real numbers. In performing any operation with real numbers, we must keep in mind that any such number is simply an arithmetic multiple of the quality unit ^1 or ~1, and that arithmetic numbers are added, subtracted, multiplied, or divided in Algebra just the same as in Arithmetic. 35. Addition. Observe that, by § § 27 and 29, the addition of one real number to another is reduced to the additiortr of one arithmetic number to another, or to the subtraction of one arithmetic number from another. To find the sum of three or more numbers we add the second to the first, to this sum we add the third, and so on. Ex. 1. +8 + - 5 + +6 + -7 = +3 + +6 + -7 = +9 + -7 = +2. Ex. 2. -7 + +5 + -3 + +9 = -2 + -3 + +9 = -5 + +9 = +4. 36. The two following laws of addition are constantly used in Arithmetic and Algebra : The commutative law. The sum of tivo or more numbers is the same in whatever order they are added. That is, a -{-b + c = b -\-c -\-a =c -}- b -\-a= "'. (A) 30 ADDITION OF REAL NUMBERS 31 Thus, we can commute siimmands {change their order) to suit our convenience or purpose. E.g.^ in Arithmetic we write ^ + 3 + ^ + 2 + ^ = | + i + i + 3 + 2. (3) Here by a change of order we can add the fractions first. Prove each of the two following particular cases of {A) : Ex.1. +4+-5++0=+4++6+-5. §14 Ex. 2. +2 +-3 ++4 =+2 ++4 +-3 =+4 +-3 ++2. Proof of law (A). This law holds true for arithmetic numbers, as is learned in Arithmetic; hence the total number of positive units in a, h, c, etc., will be the same in whatever order these summands are written. For the same reason the total number of negative units in a, 6, c, etc., will be the same in whatever order their summands are written. Hence the sum will be the same, however we change the order of the summands; for equal numbers of opposite units always annul each other. The associative law. The sum of three or more numbers is the same in whatever way the successive numbers are grouped. That is, a + 6 + c = a+(6+c). {B) Thus we can associate successive summands (group them) to suit our convenience or purpose. Prove each of the two following particular cases of ( JB) : Ex. 1. +4 +-6 ++6 =+4 + (-5 ++6). Ex.2. -5 ++4 +-7 =-5 + (+4 +-7). Proof of law (B). a -{- b +c = b -\- c + a by(^) = (6 + c) + a by notation = a-h(b-{-c) by (A) A similar proof would apply to any other case. 32 ELEMENTS OF ALGEBRA The rules for addition in Arithmetic are based on the commutative and associative laws just given. E.g., to add 45 and 23, we have 45 + 23 = 40 + 5 + 20 + 3 by notation = 40 + 20 + 5 + 3 ^J (A) = (40 + 20) + (5 + 3) by (^) = 60 + 8 = 68. Writing one number under the other and then grouping the vertical columns, as we do in Arithmetic, is but a convenient way of applying laws (A) and (5). 37. Since, by the laws of addition in § 36, we can change the order of summands and group them to suit our purpose, we have the following rule for adding three or more num- bers, some of which are positive and some negative : Add all the numbers of one quality, then add all the numbers of the opposite quality, then add the two resulting sums. • Ex. -5 ++9 +-11 ++6 =-5 +-11 ++9 ++6 by (^) = -16++15=-l by (J?) ' In practice, the rearrangement and regrouping of the summands should be done mentally and simultaneously. Exercise 11. By § 37 find the value of each of the following sums: 1. +19 +-7 ++5. 5. +4 +-5 ++6 +-8 ++7. 2. -12 ++9 +-4. 6. -9 ++6 +-11 ++12 +-4. 3. -22 ++5 ++7. 7. +15 +-9 ++7 +-8 ++11. 4. +42 +-9 +-3. 8. -(|)++(2)+-(|)++(|). Find the value of x -\- y -\- z -{- v. 9. When a; ="25, 2/ ="^32, 2 =-45, v =+28. 10. When a; =+94, 2/ ="75, ^=+82, v =-Qb. SUBTRACTION OF REAL NUMBERS 33 38. From the definition of zero it follows that a + = a. That is, any number plas zero equals the number itself. E.g., 7+0 = 7, 8 + = 8. Also, 9+(2-2)=9, 6+(5-5) = 6. 39. Subtraction is the inverse of addition. Given a sum and one of its two parts, subtraction is the o^^eration of finding the other part. As in Arithmetic, the given sum is called the minuend, the given part the subtrahend, and the required part the remainder. Hence, to subtract any subtrahend from any minuend is to find a third number, the remainder, which added to the subtrahend gives the minuend. Ex. +9 = +9 + (+5 + -5) §§ 28, 38 = (+9+ +5) + -6. §36 Hence, +9 ++5 is the number which must be added to -5 to obtain +9 ; that is, +9 --5 = +9 ++5. Here the remainder +9 ++5 is obtamed by adding to the minuend +9, the subtrahend -5 with its quality changed. This example illustrates the following rule : 40. To subtract one real number from another, add to the minuend the subtrahend with its quality changed from + to ~, or from ~ to '^. That is, Jtf - +a = 1/ -f "a, (1) M--a = M+^a, (2) when M is any real number. 34 ELEMENTS OF ALGEBEA Proof. If to the second member of (1) we add the sub- trahend, '^a, we obtain the minuend M; that is, (M-^-a)-^+a = M+(-a-^+a) by (5) = if §§ 28, 38 also, (M-^+a) +-a = M+ (+a +"a) = M. Hence, by § 39, the second member of (1) or (2) is a remainder. Ex.1. -4-+7 =-4+-7 =-11. Ex.2. -5 --8 =-5 ++8 =+3. Thus, subtracting any real number gives the same result as adding its arithmetically equal opposite number. E.g., subtracting §200 credit from an estate is equivalent to adding $200 debt; and subtracting $300 debt is equivalent to adding .^300 credit. Subtracting $100 income is equivalent to adding $100 expenditure. Exercise 12. Perform each of the following indicated subtractions : 1. +19 -+7. 4. +6-+7. 7. -20 --25. 2. -23 -+12. 5. +12 -+20. 8. "68 --98. 3. -16 --30. 6. -214 -+25. 9. -118--120. What is the value of a — b, 10. When a =+5, 6 =+4? 12. When a ="4, 5 ="7? 11. When a =+7, 6 =+9? 13. When a ="14, 5 =-11? 14. From +4 +-8 ++9 +-3 subtract +7 +"2 ++9 +-8. 15. From -10 +-7 ++15 +"3 subtract +7 +"11 +"17. 41. When a monomial or the first term of a polynomial is preceded by the sign of operation + or — , zero is to be understood before this sign of operation. SUBTRACTION OF REAL NUMBERS 35 Thus, — +a = — +a = ~a, —~a = — ~a= '^a. Again, _ +5 + +7 = - +5 + ^7 = "5 + +7. 42. Successive subtractions or successive additions and sub- tractions can be performed from left to right, one at a time in succession. Ex.1. +8--3-+2--6=+ll-+2--6 =+9 --6 =+15. We can, however, express each term to be subtracted, as a term to be added, and then apply the principle in § 37, for finding the sum of three or more numbers. Ex.2. +8 --3 -+2 --6 =+8 ++3 +-2 ++6 (1) =+17 +-2 =+15. 43. Commutative law of subtraction. Since each term to be subtracted can be expressed as a term to be added, the commutative law holds for subtraction as well as for addi- tion, provided the sign of operation -[- or — before each term is transferred with the term itself. E.g., +7_-8+-9-+4=--8++7-+4+-9 = _+4_-8+-9++7. Exercise 13. Find the value of each of the following expressions : 1. +6-f-2-+3. 2. -14-+9-f"4. 3. +32 +"5 -"16. 4. +6--24-+3. 5. +4--2-+3+-2-+5++3--6. 6. +25 -+14 +-10 ++14 --5 -+18 ++16 +"18. 7. -35 +-5 --32 ++24 --14 +-28 --8. 44. From the definition of zero it follows that a-0 = a. That is, any number minus zero equals the number itself 36 ELEMENTS OF ALGEBRA 45. Multiplication. As in Arithmetic, the number multi- plied is called the multiplicand, the number which multiplies is called the multiplier, and the result the product. Ill Arithmetic the product 9 x 3 is obtained by taking the multipli- cand 9 three times, and the multiplier 3 is obtained by taking the pri- mary unit 1 three times. The product 9 x |- is obtained by dividing the multiplicand 9 by 3, and multiplying the result by 2, and the multiplier | is obtained by dividing the primary unit 1 by 3, and multiplying the result by 2. In each case we obtain the product by doing to the multiplicand just what is done to the primary unit to obtain the multiplier. Hence, we define multiplication as follows : To multiply one number by another is to do to the multi- plicand just what is done to the primary unit to obtain the multiplier. The multiplicand and the multiplier together are called the factors of the product. 46. Multiplier any arithmetic number. Let a and b be any two arithmetic numbers ; then b times a units of any Mnd is equal to ab units of that Mnd; that is, ^axb = +{ab), (1) and -a xb = -(ab). (2) E.g., +4x5 = +20, -7x4 = -28, -(f) x 8 = -12. 47. Multiplier any positive or any negative number. To obtain ^b from the primary unit +1 we take that unit b times ; hence, by definition, to multiply any number by +6 we take that number b times; that is, +a X +6 = +a X 6 = +(ab), (X) and -a X +6 = -a X 5 = ~(ab). (2) Hence, to multiply any number by +1 is to take that number once ; that is, "*"a x "^1 = +a ; ~a x ■*"! = "a. MULTIPLICATION OF REAL NUMBERS 37 To obtain ~b from the primary unit +1, we change the quality of that unit and multiply the result by 6 ; hence, to multiply any number by ~b, we change the quality of that number and multiply the result by b ; that is, +a X ~6 = ~a X 6 = ~{ab)j (3) and ~a X "6 = +a X 6 = ^(a6). (4) Hence, to multiply any number by ~1 is to change the quality of that number; that is, "^a x ~1 = ~a; ~a x ~1 = ^a. From identities (1) and (4) we have the law, Tlie product of two real numbers like in quality is positive. From identities (2) and (3) we have the law, The product of two numbers opposite in quality is negative. These two laws together are called the law of quality of products. From identities (1), (2), (3), (4), it follows that The arithmetic value of the product of two real numbers is the product of their arithmetic values. E.g., +5 X +7 = +35, -6 x "8 = +48. +4 X -9 = -36, -7 X +8 = -56. Exercise 14. Find the value of each of the following numeral ex- pressions : 1. +2 X +4. 3. +9 X -8. 5. -21 X +3. 7. +22 x "6. 2. -2 X -7. 4. -11 X -1. 6. +31 X -1. 8. -32 x "4. 9. +10x-3 + -8x+2. 11. -6x-5+-^8x-4-+12x-5. 10. +14x-2 + -6x-5. 12. -9x+2 + +16x-4--14x+3. 88 ELEMENTS OF ALGEBRA When a = +2, 6 = "4, m = "3, n = +9, x = +6, find the value of each of the following literal expressions : 13. ab -f- mx. 16. ax — nb. 19. (a + b) (n -\- m). 14. ax + bm. 17. (a — b) x. 20. (a — b)(n — m). 15. am— bx, 18. (m — n)b. 21. (6 —a?) (771— n). 48. Continued products. By § 47, we obtain +a X "^6 X +c = +(a?>) x +c = -^(abc). +a X'^b x~c = +{ab) x~c = -(abc). -^a x~b x~c = ~(ab) x~c = +(abc). ~a x~b x~c = +(ab) x^c = ~(abc). From these and similar identities we have the following laws which are more general than those in § 47 : Aj^roduct which coyitains an odd number of negative factors is negative; any other product is positive. The arithmetic value of a product is the product of the arith- metic values of its factors. Ex. Find the value of +3 x "2 x +4 x -6 x -5. The product is negative, since there is an odd number, 3, of nega- tive factors ; its arithmetic value is 3 x 2 x 4 x 6 x 5, or 720. Hence, +3 x -2 x +4 x "6 x - 5 = "720. Exercise 15. When a = -2, & = +4, c=-6, x = -S, y = ~5, find the value of each of the following literal expressions: 1. abc. 4. (6 + a) ex. 7. axy — be. 2. abxy. 5. (x — y) abc. 8. xy — abc. 3. abcxy. 6. (b 4- c) axy. 9. x-\- obey. MULTIPLICATION OF MEAL NUMBERS 39 10. Prove +lx-lx-lx-l = -l; -lx-lx-lx-l = +l. 11. Prove +a X "6 X ~c = (+1 x "1 X ~1) (ahc). 12. Prove "a x "^6 X ~c x ~x = {-l x +1 X ~1 X ~1) (obex). Examples 11 and 12 illustrate that the product of two or more num- bers is equal to the product of their quality-units multiplied by the product of their arithmetic values. 49. The two following laws of multiplication are con- stantly used in Arithmetic and Algebra : The commutative law. Tlie product of two or more num- bers is the same in vjhatever order the factors are multiplied. That is, abc = acb = cba="'. (A') Prove each of the two following particular cases of (A') : Ex. 1. +2 X -3 X +4 X -5 = -5 X -3 X +2 X +4. Ex. 2. -3 X +7 X -2 X -1 = -2 X +7 X -1 X -3. Proof In Arithmetic we have learned that this law holds true for arithmetic numbers. Hence, by § 48, the arithmetic value of a product of real numbers is the same in whatever order the factors are multiplied. From the law of quality, in § 48, it follows that the quality of a product of real numbers will be the same in whatever order the factors are multiplied. Hence, a change of order of factors affects neither the arithmetic value nor the quality of their product. The associative law. Tlie product of three or more num- bers is the same in whatever way the successive factors are grouped. That is, abc = a(bc). (B) Prove each of the following particular cases of {B') : Ex. 1. -3 X +4 X -2 = -3 X (+4 x "2). Ex. 2. +5 X -0 x -1 X +2 = +5 X ("6 x -1 X +2). 40 ELEMENTS OF ALGEBUA Proof, abc = hca by (A') =\bc)a by notation = a(bc) by (A') Exercise 16. By using the commutative and associative laws, find in the simplest way the value of each of the following ex- pressions : 1. +33 X -21 X -4. 4. +144 x "3 x -16|. 2. -123 X -33^ X +3. 5. "371 x "7 x "4. 3. +142 X -121 X -8. 6. -333i X -5 X -7 X +3. 50. A x>ii'oduct of two or more factors is multiplied by a number if any one of the factors is multiplied by that number. Proof (ab) x c = (ac)b = a(bc). § 49 51. Powers. A product of two or more equal factors is called a power. Any number also is often called the first power of itself. E.g.j the product aa is called the second power of a. The product bbb is called the third power of b. aa is written a^ ; aaa is written a^ ; ooa ••• to n factors, written a% is read 'the nth. power of a.' a^ is often read ^the square of a/ and a^ 'the cube of a.' In a*", a is called a base. Thus in 3^*, 3 is a base ; in a^, a; is a base ; in y'", ?/ is a base. 52. A positive integral exponent is a whole number which (written to the right and a little above a base) indicates how many times the base is used as a factor, as 3 in a^, or n in a". MULTIPLICATION OF REAL NUMBERS 41 To avoid ambiguity, a base which is not denoted by a single symbol must be enclosed within parentheses : E.g., (-3)2 = -3 X -3 = +9, while -3^ = -1 (3 x 3) = -9. Again, (4 x 5)2 = (4 x 5) (4 x 6) = 20 x 20 = 400, while 4 X 52 = 4 X (5 X 5) = 4 X 25 = 100. The meaning of fractional and negative exponents will be deter- mined in a later chapter. A power is said to be odd or even according as its expo- nent is odd or even. 53. Quality of a power. An odd power of a negative base is the only power which involves an odd number of negative factors ; hence, by the law of quality in § 48 it follows that An odd power of a negative base is negative, and an even power positive ; any power of a positive base is positive. E.g., any power of +1 is +1 ; any even power of -1 is +1 ; any odd power of -1 is '1. Exercise 17. 1. What number is the base and what the exponent in -3^? In {Sy? In 3xy"? In (Sxyy? In (a + 6)"? In a + b^? In (x")* ? Find the value of each of the following expressions : 2. -3\ 3. (-3)*. 4. 17-32. 5. (17-3)2. Express each of the following products by a base and exponent : 6. (xy) (xy) (xy) ••• to 8 factors. 7. (a + 6) (a + &)(« + &)••• to 12 factors. Express in symbols : 8. The sum of the cubes of x and y. 9. The cube of the sum of x and y. 42 ELEMENTS OF ALGEBRA 10. The sum of the squares of a, b, and c. 11. The square of the sum of a, b, and c. 54. If, in any one of the identities in § 48, the quality of one factor is changed, the quality of the product is changed, but its arithmetic value remains the same. This illustrates the following principle : TJie quality of any product is changed by changing the qual- ity of one, or of any odd 7iumber, of its factors. Proof By changing the quality of an odd number of factors, the number of negative factors in the product is changed from odd to even, or from even to odd ; hence, by § 48, the quality of the product is changed. Note. When for brevity we speak of the quality of an expression, we mean, of course, the quality of the number which the expression denotes. 55. The quality of an expression is changed by changing the quality of each of its terms. Proof Changing the quality of a term does not affect its arithmetic value. Hence, changing the quality of each term of an expression will simply change a positive sum into an arithmetically equal negative sum, or vice versa. This principle is illustrated by the fact that if in a business account we change debts into credits, and credits into debts, the balance will not be changed in amount, but it will be changed from credits to debts, or from debts to credits. Ex. 1. Change in four ways the quality of -4 x +3 x -2. Of -3x-6x-7. Of +ax+bx-c. Of +ax-bx-c. Of +xx+yx-2. Ex. 2. Change in two ways the quality of -4 x +3 - -2 x +7. Of +a X -b + -c X +x. Of -a X -X — +6 X +c. MULTIPLICATION OF BEAL NUMBERS 43 56. Two uses of the signs + and -. Hereafter the larger signs + and — will be used, not only as signs of operation, but also with numerals as signs of quality. To avoid ambiguity, parentheses will be used when needed. Thus, in the expression, (+4)-(+7) + (-3)-(-4), each sign within parentheses denotes quality, and each without denotes an operation. Again, (- 3)ax -(+ 4)6y + (- 5)c2; = -3 ax - +4 6y +-5c5r. A letter with the small sign + or " will continue to be used to denote a general positive or a general negative number. 57. Abbreviated notation. The sign — is never omitted. But, for the sake of brevity, the sign -f has been omitted, and is to be understood in the two following cases : (i) When no sign is written before a monomial or before the first term of a polynomial, the sign -f is to be under- stood. (ii) When only one sign is written between two successive terms of a polynomial, the sign + is to be understood either as a sign of operation or as a sign of quality. E.g.^ 2 denotes + 2, 3 a denotes + 3 a, and a denotes + 1 a. Again, 6 — 5 denotes the difference (+6) — (+5) or the sum (+ 6) + (— 5) ; in each case the sign + is understood between 6 and 5 ; in the first case as a sign of quality^ and in the second case as a sign of operation. Since (+6) - (+5) = (+6) + (-5), 6 — 5 denotes the same number whether it is regarded as ex- pressing the difference (+6) — (+5) or the sum (+6) + (—5). Again, 7-5-f8 = (4-7)-(+5)4-(-l-8), or (+7) + (-5) + (+8), 44 ELEMENTS OF ALGEBRA according as we regard the written signs in the first expres- sion as signs of operation or as signs of quality. Hence, in the abridged notation, the written signs in any- polynomial can be regarded either as signs of operation or as signs of quality. When all the written signs are regarded as signs of qual- ity any polynomial becomes a sum. E.g., -_5 + 3-2=(-5) + (+3) + (-2). or the sum of the terms — 5, +3, and — 2. Again, 7ac-4x + 3y = -i-7ac+(-4)x + (+ 3)?/, or the sum of the terms + 7 ac, — 4 x, and + 3 y. In general formulas, such as {A), (B), etc., it is usuall}- better to regard the written signs as signs of operation ; but in most other cases it is preferable to regard the written signs as signs of quality and, therefore, to regard every polynomial as a sum. 58. Coefficients. If a term is resolved into two factors, either factor is called the coefficient, or the co-factor, of the other. E.g.^ in 4 a&c, + 4 is the coefficient of a&c, + 4 a of &c, + 4 a& of c, dbc of + 4, and 6a of + 4 c. A numeral coefficient is a coefficient expressed entirely by numerals, and a sign of quality written or understood. A literal coefficient is a coefficient which involves one or more letters. E.g.^ in —4xy, —4 is the numeral coefficient of xy ; x is the literal coefficient of — 4 ?/, y of — 4tx, and —4x of y. When in a term no numeral factor is written, 1 is understood, e.g., a denotes -h 1 • a and — a denotes — 1 • a ; abc denotes + 1 • abc and — abc denotes — 1 • abc. MULTIPLICATION OF HEAL NUMBERS 45 Exercise 18. Find the value of each of the following expressions : 1. 15-9. 5. (-Il)x7. 2. -9+7. . 6. (-7) -(-4). 3.-8-6. 7. 9-74-4-3 + 5. 4. (-3) (-4) 8. 18-(-3)x(-4)-8. 9. 35_j.(_7)x6-r-15x(-2). Find the value of a-[-h — c-\-d and a—{—h + c — d). 10. When a = 2, 6 = — 4, cz=^—Q>, d= —7. 11. When a= —7, b = — S, c = 5, d = — 6. Find the value of x(y — v-\- z). 12. When x = 6, y = — 7, v = — 9, 2 = 8. 13. When a; = — 5, y = l^j v = — 4, 2 = — 7. Find the value of x-T-{y — v — z). 14. When x = -10, y = -Q, v = -9, z = S. 15. When a; = -16, 2/ = - 10, ^ = -12, z = 6. 16. What is the coefficient of a in a? In —a? In -7ay? 17. In the expression —Sab(x — y\ what is the coeffi- cient of a; - 2/ ? Oib(x-y)? Of 8a? Oi -S(x-y)? 18. If the sum (x — y) -\- (x — y) -{- (x — y) -\ to a sum- raands is expressed as a product, what is the coefficient of x-y? 59. Having given a product and one factor, division is the operation of finding the other factor. That is, if n is one factor of m, m -^ n denotes the other factor ; whence (m-i-n) X n= m. (1) 46 ELEMENTS OF ALGEBRA 60. The distributive law. The product of a polynomial by a monomial is equal to the sum of the products obtained by multiplying ea,ch term of the polynomial by the monomial; and conversely. That is, {a-{-b-\-c-{-"')x = ax-{-bx + cx+"' (0) The distributive law lies at the basis of multiplication in Arithmetic, e.g.^ if we wish to multiply any number as 248 by 7, we separate 248 into the parts 200, 40, and 8, multiply each of these parts by 7 and add the results. Thus, 248 X 7 = (200 + 40 + 8) x 7 (1) = 200 X 7 + 40 X 7 + 8 X 7 . (2) = 1400 + 280 + 56 = 1736. We pass from (1) to (2) by the distributive law (C). Prove each of the following particular cases of (O): Ex. 1. (4-3 + 5). (-2) = 4(-2) + (-3).(-2)4-5(-2). Ex. 2. (- 4 + 2 - 6)(- 3) = (- 4).(- 3)+ 2 (- 3) + (- 6).(- 3). Ex. 3. (« + 6 + c)-3 = 3a + 36 + 3c. Proof Let the multiplicand be any binomial a-\-b. The proof involves three cases : when the multiplier is (i) a positive integer, (ii) a positive fractional number, (iii) a negative number. (i) Let m be any positive whole number ; then (a -h 6) m = (a + 6) -h (a + 6) H — to m summands § 47 = (a + a H — to m summands) + (6 + 6 H — to m summands) § 36 = am-\-bm. (1) (ii) Let m and n be any positive whole numbers other than zero; then — will denote any positive fractional number. MULTIPLICATION OF REAL NUMBERS 47 (a + b) (m ^ n)7i = (a + 6)m §§ 49, 59 = am -\- hm. by (1) = a{m--rn)n-\-h(m^n)n §§ 49, 59 = [a {m -T-n)-\-b (711 h- n)'] n. by (1) Dividing the first and last expressions by n, by (v) of § 32 we obtain (a + b) (m -J- n) = a(m-i- n) -\- b (m h- n). (2) Let r be any positive number, whole or fractional ; then, from (1) and (2) we have (a -\-b)r = ar + br. (3) (iii) If the quality of equal numbers is changed from -f to — , or from — to +, the resulting numbers will be equal. Henee, changing the quality of both members of (3) we have (a + 6) (- r) = a (- r) + 6 (- r), §§ 54, 55 where — r is any negative number, whole or fractional. A similar proof would apply to any polynomial as well as to a + 6 ; hence the law as stated in (C). Ex. 1. Multiply 3 a2 - 5 a + 3 6 by 2 X. (3 a2 - 5 a + 3 6)(2 a;) = (3 a2)(2 a;) + (- 5 a)(2 x) + (S 6)(2 x) = 6a2x- 10ax + 6bx. Observe that in applying ( C) we regard a polynomial as a sum. Ex. 2. Multiply 2 x^ - 3 x2 - 2 x by -3 a. (2x3-3x2-2x)(-3a)=(2x8)C-3a) + (-3x2)(-3a) + (-2x)(-3a) = - 6 ax3 4- 9 ax2 + 6 ax. 48 ELEMENTS OF ALGEBRA Exercise 19. Multiply : 1. a; + 2 by 3. 6. 2 cy -Ax hj -a. 2. 6a-75by-2. 7. 2 a- 3 & -c by - 2 a;. 3. 2 a; - 6 by - 5. 8. - 3 a; + 2 ?/ - 5 2; by 3 a. 4. 2x-5by-3a. 9. a;^- 3 a; + 4 by -2 a. 5. ax-3b hj -2c. 10. ar'-2 ?/-3 2; by - 5 a. 11. -2x'-hSxy-4:i/-x-{-2y-7 hj -3a, CHAPTER IV ADDITION AND SUBTRACTION OF INTEGRAL LITERAL EXPRESSIONS 61. An integral literal expression is an expression which involves only additions, subtractions, multiplications, and positive integral powers of its letters. Any expression which contains a literal divisor is called a fractional literal expression. E.g., a'^ + f and 4 x^ — | 6< are integral literal expressions ; while - and are fractional literal expressions. y 4-6 A letter can, in general, denote any integral or fractional number ; hence, any literal expression can have any integral or fractional value. E.g., wlien x = I and y = |, the integral literal expression x + y = ^+^ = f, a fractional number. Also, when a* = 2 and y = 3, the integral expression | xy = ^. Again, when x = 10 and y = 2, the fractional expression ? = 5. y Tlie pupil must clearly distinguish between integral and fractional expressions and integral and fractional numbers. 62. Like or similar terms are terms which do not differ, or which differ only in their coefficients. E.g., 4 ab and 4 ab are like terms ; so also are 4 ab and — 10 ab. Again, 6 axy and — 4 bxy are similar terms, if we regard 6 a and — 4 &, respectively, as the coefficients of xy in the two terms ; but if 6 and — 4 be taken as the coefficients, these terms are dissimilar. 49 60 ELEMENTS OF ALGEBRA 63. Sum of similar terms. The converse of the distribu- tive law in § 60 is ax-\-hx-\-cx-\- '•• = (a-\-'b -{- C + '^•)x. (O) That is, the sum of two or more similar terms is equal to the sum of their coefficients into their co7nmon factor. 1. Find the sum of 7 a, — 5 a, 4 a. (+ 7)a + (- 5)rt + (+4)a = (7 -5 + 4)a^6a. 2. Find the sum of 3 ah"^, - 5 ah"-, - 8 al)^. (+ 3) a6--2 + (- 5) a62 + (- 8) a62= (3 - 5 - 8) ah'^= - 10 db^. 3. Find the sum of 7 (a — 6) , — 5 (a - 6), 4 (a - &). ( + 7)(a-6) + (-5)(a-5) + (+4)(a-&)=(7-5+4)(a-6) = 6('?'-6). 64. By § 57 the sum of two or more terms is indicated by writing them in succession, each term being preceded by the sign of quality of its numeral coefficient. The sum of unlike terms can only be indicated. E.g., the sum of — 5 c, 7 a, and — 9 6 is -5c + 7a-9&, or 7rt-5c-9?). Again, the sum of — 3 ax, — 5 by, and 6 cz is — ^ax — ^hy-^Qcz, or 6 C0 — 3 ax — 5 by. Exercise 20. Find the sum of : 1. 2 a, -3 a, 5 a. 6. 4 aft^, -1 ab\ 3ab\ 2. —4:X,2x,—x. 7. —Sx% 5x% —4 a;". 3. ab, -2 ah, 3 ah. 8. 2 oc^, - 5 ac^, - 8 acl 4. 2 a*, -3a^ 1 a\ 9. -5aV, -3aV, 9 aV. 5. a;", —2 a;'*, 4 a;". 10. 4&"2/'"^ — Td^y™, 9 5"2/'"' ADDITION AND SUBTRACTION 51 11. 7aa^, — 5aic^, 4aa^, —9aa^, — 14aaj^, 25 aa?^. 12 9aa^, — aar', 4aic^, —Ihx^, —14 car'. 13. -3^2/2!, f a^2;, -^xyz, 6xyz, -^xyz, -^^-xyz. 14. (x-af, -2(x-a)\ 4.{x-af, -5{x-a)\ 12{x-aY. 15. (a^+2/2), -5(a^+/), 9(a:^+yO, -3(a^4-2/^, -l{^+fr 16. (aT^-2/'), -4(ar^-.v«), _3(a:«-r), -7(a^-2/«), 8(a^-2/3), Simplify each of the following expressions by combining like terms : 17. a^_7a^+4a^-5a^. 19. o?y--^a?f+4.a?f-l^y\ 18. a;"— Sx^+Saj"— 7 a;". 20. aar'— 7a.-2+6a:2_5^ 21. _9ic2_^i7^_^3a.o_^^_^^^2_5^2 22. Sab' -7 aV + 8 a/>- - 4 ah'' -\- 7 ca^ - 11 ca^. 23. -12a^-^4:a^-9x^-\-7a^ + Sa^-9a'' + 7a\ 24. 7 aftcc? — 11 abed + 41 a6cd + 7xy — 20xy. 25. _5aj2_2a:2 + |a^ + 8/--|/-f/. 26. 7ar' + 2a2-5.'c2-3a2. 7 a:2 + 2 a2 - 5x2 - 3 a2 = 7 a;2 _ 5x2 + 2 a2 - 3 a2 by (^) = 2x2-a2. 27. 7a6 — 5a;2/ + 3a6 + 2a^ — 6a5 — a^. 28. -9ax'-\-5bf-\-7ax^-3bf+llaa^ + 4:bf. 29. — 7 c/ — 4 a6 + 9 a.-2; 4- 11 C2/2 + 10 a6 — 5 a;2; — a6. 30. 2(a^-l) + 3(a2 + l)-4(x2-l)-5(a2 + l). 31. 3(a' + b')-4.(x-\-y)-7(a'-\-b')-^5(x-\-y), 32. Review this exercise, solving each example mentally. 52 ELEMENTS OF ALGEBRA 65. Addition of polynomials. Ex. 1. Add - 3 x^ + 7 X to 5 a;2 _ 4 a;. (5 a:2 - 4 x) + (- 3 x2 + 7 x)= 5 x2 - 4 x - 3 x- + 7 x by converse of {B) = 2x2 + 3x. by (^), (S) Ex. 2. Find the sum of 4 x2 - 3 xy + y2, _ 2 x2 - 5 xy - 6 2/2, and 2 x?/ - x2 - 3 62. In adding polynomials, it is convenient to write them under each other, placing like terms in the same column. Thus, (4 x2 - 3 xy + 2/2) + ( _ 2 x2 - 5 x?/ - 6 2/2) + (2 ccy - x2 - 3 h'^) can be written 4 x2 — 3 xy + 2/^ - 2 x2 - 5 xi/ - 6 2/2 - x2 + 2x2/ -3 62 x2 - 6 X2/ - 5 2/2 - 3 62. Here the rows of terras are the groups of terms as given, while the columns of terms are the groups of similar terms obtained by rearrang- ing and regrouping by laws {A) and {B). Since there is no carrying as in Arithmetic, the addition can be performed from left to right, or from right to left. 66. When in a polynomial the exponents of some one letter increase or decrease, from term to term, the polyno- mial is said to be arranged in ascending^ or in descending^ powers of that letter. This letter is called the letter of arrangement. E.g., the polynomial x^ + 2 x^y + 3 xy'^ + 4 2/^ is arranged in descending powers of x, x being the letter of arrangement ; or, in ascending powers of 2/, y being the letter of arrangement. In arranging a polynomial in ascending or descending powers of any letter, we must first combine all the terms which contain the same power of that letter. In adding polynomials, it is usually convenient to arrange them in ascending, or descending, powers of some letter, as below: ADDITION AND SUBTRACTION 63 Ex. 1. Find the sum of 2 x^ - 3 x^ + y, _ 4, 7 X - 4 x2 4- 5 x3 + 5, and 7 x^ - 4 x^ + 2 x - 1. Arranging each polynomial in descending powers of x, we have -3x3 + 2x2+ x-4 6x3-4x2+ 7x + 5 -4x3 + 7x2+ 2x-l - 2 x3 + 5 x2 + 10 X Exercise 21. Find the sum of : 1. a + 26-3c, -3a + 6 + 2c, 2a-3& + c. 2. -Zx-\-2y-\-z, x-^y-\-2z, 2x + y-Sz. 3. -15a-196-18c, 14a4-1564-8c, o + 56 + 9c. 4. 5 aa; — 7 6?/ + C2;, aa; + 2 61/ — 02;, — 3ax + 2&2/ + 3c2;. 5. 20 p + q-r, p-20q + r, 2^ -\-q- 20 r. 6. — 5 a& + 6 6c — 7 ac, 8 a6 — 4 6c + 3 ac, — 2 06 — 2 6c 4- 4 ac. 7. pq + qr — pTf — pq -\- qr -{- pr^ pq — qr -{- pr. 8. 2 a6 4- 3 ac 4- 6 a6c, — 5 a6 + 2 6c — 5 a6c, 3 a6 — 2 6c — 3ac. 9. x^ + xy — y-, —z^-\-yz-\- /, xz -\- z^ — 0^. 10. 5a»-3c» + d3, fo3_2a3 4-3^3, 4c3 -2a'»- 3d». 11. a,^ + 2/2_2ic?/, 2z2_3/-42/«, 2 ar^ - 2 2:2 - 3 ii'2;. 12. a.'3 + 3ar^2/ + 3a^', - 3 ar^3/ - 6 a^/ - a^, 3 a;^^, + 4 a^. 13. x'-^x'^y-hx'f, 3a;^+2a.V-6an/*, Zs^f+Qxy'-f. 14. a«-4a26 + 6a6c, a-6 - 10 a6c + c^, 63 + 3a26 + a6c. 15. 3a2-1062 + 5c2-76c, - a2 + 462- IOC24- 3a6, c2+116c+8ac-2a6, 4c2-46c + ac, -2a2+662-9ac-6c. 54 ELEMENTS OF ALGEBRA 16. 4ic* + 12ar^-a;-10, llx" -2x^ - x' -\-'d, ^x" -3x' + 4.X, 4ar^-x^-5, Q,x^ - si? -\-2x- -1. 17. i:^-\x + \, -^x' + lx-^, Ix^ + lx^^. 18. la' + ^ah-^h^ lo?-ah-\b'', - a^- f a6 + 2 6=^. 19. _2^_a;2/ + 2/^3x2-|a?y-i2/', - f a^ + 2x'?/- |/. 20. -f a^-|a;/ + 22/^, |aj22^-|_aj?/2_|_i.2^a |ar''-2a;2^-f ^Z'. 21. a^-3ax2_|_5^'j^_^3^2a^ + 4ax2-6a2a;, Gaa^-Sa^a? + a^, — 2ar^ + 4a^a; — 5al 22. 3a;2 4-/-3?/2-2^ 2 a;^/ - 3 2/^ + 3 2/^, -4.x^-2xy-{- 23. Given a; = 6 + 2c — 3a, y = c + 2a — Sh, and 2; = (x-f26 — 3c; show that x-\-y -\-z = (). 24. Given a = 5a; — 32/ — 2 2;, b = 5y — 3z — 2x, and c = 5z — Sx--2yj show that a + 6 + c = 0. 67. To subtract one expression from another, change the sign before each term of the subtrahend from + to ^ or from — to +, and add the result to the minuend. Proof. Changing the sign before each term of the sub- trahend changes the quality of the subtrahend (§ 55) ; and by § 40 the minuend plus the subtrahend with its quality changed is equal to the remainder. Ex. 1. From — 5 x'^y take 4 x'^y. - 5 ic2?/ - ( + 4 a;2i/) = - 5 ic2y + ( _ 4 a^Zj/) = - 9 x'^y. Ex. 2. From 5 x- + a;?/ - m take 2 x^ + 8 xy - 7 y2. Changing the sign before each term of the subtrahend from + to - or from — to +, and adding the result to the minuend, we have 5 a;2 + xy — m - 2 x2 _ 8 xy + 7 1/2 3 x2 — 7 xy — m + 7 ?/2^ Remainder. ADDITION AND SUBTRACTION 55 Note. The signs of the subtrahend need not be actually changed ; the operation of changing the signs ought usually to be performed mentally, as in the following example. Ex. 3. From 2x* -Sx^ + 7 x-S take x* -2x^-9x + ^, 2x*^ -Sx^-\- 7x- 8 x^-2x^ -9x+4 x* + 2 a:8 - 3 a;2 + 16 a; - 12 Exercise 22. 1. From 4 a — 3 6 + c subtract 2a — Sb — c. 2. From 15x -\-10y — 18z subtract 2x — Sy + z, 3. From — 10 be -\- ab — A cd take — 11 ab-\- 6 cd. 4. From ab -\- cd — ac — bd take ab -\- cd-^ ac + bd. 5. From m^ + Sn^ subtract —4:m^—6n^-{-71x. 6. 7xy-(-Sxy) = ? 7. -9x'y-(-\-6x'y)-(-20x'y) = ? 8. 32a^-(-122/^-(-hl4a^)-(+92r) + (-2/) = ? 9. 28 a'b' - (+ 17 a'6-) - (- 19 ar^y) - (+ 15 a^y) -(-5a'b')=:? From 10. -8a^f-\-15x^y + lSxf take 4:a^i/^-{-Txh/-Sxf. 11. a^bc + b^ca + c^a6 take 3 a^bc — 5 &-ca — 4 c^a6. 12. -7a264-8aZ>2 + cd take 5arb -7 ab' + 6cd. 13. 10a262-|.i5a?>2_^3^25 take - 10 a-^^ + 15 oft^ - 8 a^fe 14. hs^c^-2 abc take a^-{-b^-S abc. 15. 7abc-Sa^-\-5b^-(^ take a^ + 6^ + . In solving example 20, under the values of J., B, and (7 write that of D with its quality changed, and then add the results. 25. From 5a^ + 3a; — 1 take the sum of 2x — 5-{-7 a^ and 3aj2-f4-2a^ + a;. 26. From the sum of 2 a^ - 3 a^ + a - 2 and 2 + 8 a^ - a^ subtract 3 a — 7 a^ + 5 al 27. From the sum of 4:a^ + 3x-7, 2af ~Sx + 2x^~l, and — 5a^H-2a7 — ar^ + 9 take the sum of 2 aj^ — 11 ic and 9a^ + 5a^ + 3-2a;. 68. Removal of signs of grouping. The converse of the associative law for addition in § 36 is a-^(b-\-c) = a + b-\-c. (1) That is, a sign of grouping preceded by the sign + can be removed if each enclosed term is left unchanged. Observe that the sign + is understood before b within the parentheses. Ex. 1. a + (4 a - 7 1/ + 5 «)= a + 4x-7y4-5«. Ex.2. z+(^-Sx-{:2y -ia)=z-Sx-\-2y-ia. ADDITION AND SUBTRACTION 57 By the rule for subtraction in § 67, we have A sign of grouping preceded by the sign — can be removed, if the sign before each enclosed term is changed from + }y^x^ — hx — ax^ — cx^ — h(?x — l x^. Simplify the following expressions, and in each result add the terms involving like powers of x\ 13. a:(?—2cx—{bQi?— \GX—dx—il)x^-\-Z c^)\ — {c^—hx)\. 14. 5 aa^ - (7 6aj - 7 ca;-) - J6 6a^ - (3 aa;^ + 2 aa;) - 4 cx'X. Express in descending powers of x the sum of, 15. a^ci?—5x, 2ax^—5aa^, 2x^ — bx'^ — ax. 16. aa?^bx — c, qx — r—py?, ar4-2a; + 3. 17. pa? — qx, qa? —px, q — x^, jyy? + qm?. 18. 2aa^-3ca?2+^a;, 3y?a;- ma^-2ca^, a;-2a^-3a?'. 19. bx — ay? — bo?, Sx^ — 4:nx — 2ma?, 2a?— po?. 20. coi?^2ax-\'mo?, 4:X^'-bx^, 4:7ix-\-2px\ 3a?—2rx^~x. CHAPTER V MULTIPLICATION OF INTEGRAL LITERAL EXPRESSIONS 71. The degree of an integral term is the number of its literal factors. But we usually speak of the degree of a term in regard to one or more of its letters. E.g.^ 5 ax is of the second degree, and 7 a^^a is of the fifth degree. Again 4 abx-y^, which is of the seventh degree, is of the first degree in a, of the second degree in x, of the third degree in y, and of the fifth degree in x and y. 72. The degree of a polynomial is the degree of its term of highest degree. E.g., the trinomial ax"^ + bx -\- c is of the first degree in a, 6, or c, and of the second degree in x. The binomial ax^y + by-, which is of the fourth degree, is of the second degree in x or y, and of the third degree in x and y. The trinomial ax^ + 2 bxy -f cp^ is of the second degree in x, in y, and i?i x and y. 73. An expression is said to be homogeneous in one or more letters when all its terms are of the same degree in these letters. E.g., 2a^ -\-Sab + 4b^ is homogeneous in a and b ; 5 x^ + 3 x:^y -{-Sxy^ + y^ is homogeneous in x and y ; and 0x2 ^ 2 bxy + cy^ is homogeneous in x and y. Exercise 25. What is the degree of the term 3 arba^y*, 1. In a? 3. In a;? 5. In a and 6? 7. In a, a?, and ?/ ? 2. In 6? 4. In y? 6. In x and y? 8. In b, x, and y? 61 62 ELEMENTS OF ALGEBRA What is the degree of the trinomial a'x'-\-7a'b'i^y'-5abxy% 9. In a;? 10. In a? 11. In a; and?/? 12. In 6 and y? Write two trinomials of the third degree and homogeneous, 13. In a and b. 14. In x and y. 15. In a and x. 74. A product is zero when one of its factors is zero. That is, a . = and . a = 0. Proof a'0 = a(b-h), §§ 11, 32 = ab-ab = 0. §§60,11 Similarly, • a = (6 — 6) a = 0. Conversely, when a product is 0, one or more of its factors isO. That is, if a • 6 = 0, then a = 0, or 6 = 0, or a = and 6 = 0. 75. Any positive integral power of is ; that is 0" = 0. Proof 0" = • . .. . to ?i factors = 0. § 74 76. Product of powers of same base. Ex. 1. 23 X 22 = (2 X 2 X 2)(2 x2) = 2 x 2 x 2 x 2 x 2 = 2^. Ex. 2. a^a^ = (aaa) (aa) = aaaaa = a^. These examples illustrate the following law of exponents. The product of the mth poiver and the nth power of the same base is equal to the (m + n)th power of that base, and conversely. That is, a'" ' a" = a'"+". Proof a'^a'' = (aaa • • • to m factors) (axxa • • • to w factors) § 52 = aaa ••• torn + 91 factors §49 = a"'+\ § 52 MUL TIP Lie A TION 63 Ex. Multiply 3 a-x^ by — 4 a^x^y. (3 aV) ( - 4 a%2y) = s a'^^^ {- i) a^x^y § 49 = 3(-4).a2«4.a;3a;2.y §49 = - 12 a^xSy. §§ 47, 76 This example illustrates the method of finding the 77. Product of two or more monomials. Using the commu- tative and associative laws, we have the following rule : Multiply together their numeral factors, observing the law of quality ; after this write the product of their literal factors, observing the law of exponents, Ex. Multiply together — 5 ay^, — 2 a^x^, and — 9 az^y. (_ 5 ay2)(_ 2 a2x8)(_ 9 axh,) == (_ 5)(- 2) (- 9) aa^Q;x?xhpy = - 90 a*x5y3. Exercise 26. Find the product of : 1. a^ and a*. 7. —Za% and 12a5«. 2. a^ and a*. 8. — abed and — 3 aC-b'^c. 3. rf, jf, and y^ 9. 7 x^y'^7^ and — 5 3?yh. 4. aa; and 3 ax. 10. — 3 a^6V and 8 a^b^c*d. 5. — 2a6a; and —Tab. 11. 2 a^, — 4 a^ft, and 5ab^ 6. 6 ic^y and — 10 axy. 12. — 5 ax, — 7 a% and 2 aic^. 13. 8 xy^, — 3 ar^i/, and — 3 xy. 14. - 7 aft^, _ 3 a'b% and - a^ftl 15. a^b% 2 aft'^c, and — 5 a6c. 16. — 7 x^y^, ay^y*, and aa^. 17. — a^bx, aWx, and — aa^. 18. — a^ar, — b'^x, and — a5?/. 64 ELEMENTS OF ALGEBRA 78. Multiplication of a polynomial by a monomial. The dis- tributive law of multiplication is (a -f- 6 -f c -|- •")x = ax -\- bx -\- ex -\- •••. That is, to multiply a polynomial by a monomial, multi- ply each term of the polynomial by the monomial, and add the several products. Ex. Multiply 2 ^2 _ 4 2,3 ijy _ 3 y^, = _ 6 ?/^3 _!_ 12 yiz. Writing the multiplier under the multiplicand, ^ the work can be arranged as at the right of the ~ ^^ page. —Qyz^-\- 12 y^z Exercise 27. Multiply : 1. 4a2-5a + 3& by 2 a\ 2. 2a2 + 3a6 + 262 by -Sd'b\ 3. bc-{-ca — ab by abc. 4. 2a^-3x^-\-5x-4:hj -Bx". 5. -4:X^-\-3a^-Sx^ + 4L hj -6i^. 6. 9gh-12ga-3gb by Sgh. 7. — a^6c + b^ca — c^ab by — ab. 8. — 5 £C2/^2; + 3 xyz^ — 8 x?y%— 7 a;^/^ by — 2 a;^^;. 9. ci^ftV -~ abc — ax —by — cz by — 5 abcxy. 10. f aV — f aa^ + ^aa; by — |a^a;. 11. -|i»?/2^iaaJ2/-2a/ + |a2^ by -faa^. 12. i^a^y-fa^/H-iaa^-^a^/by -i^a;^2/. 13. (x + yy-2a(x-\-y)-\-5a^ hj 2(x + y). 14. (aj + l)«-4a(a; + l)'-2a6 by -5a6(aj + l)'. 15. (a'-hiy-Sx{a' + iy-4:xy by _ 3 aj^^/ (a^ -f 1)4. 16. (a^ + yy- a(a^ -^yf + 3 a'b^ by - 4 a=^6* (a^ + 2//. MULTIPLICATION 66. Remove the signs of grouping, and simplify each of the following expressions : 17. (a + h)c-(a-h)c. 19. ^(6 - 2 c) + |(c- 2 6). 18. 2(a-6)+4(a + 6). 20. 1 aQ) - c) -2h{a- c). 21. a^h^c^-dF) + (^d?{a?-W) + hh\d'-aF). 22. 2\Zab-4.a{c-2b)\. 23. 7ac-252c(a-36)-3(5c-26)a[. 79. To multiply one polynomial by another, Multiply each term of the multiplicand by each term of the multiplier J and add the resulting products. Proof Let x-\-y-^z be the multiplicand, and a + & the multiplier ; then by successive applications of the distribu- tive law, we have {x-\-y + z)(a + b) = x(a-\-b) + y{a-^b)+z{a-^b) = xa -\- ya -{- za -{- xb + yb -\- zb. § 36 Similarly when each factor has any number of terms. Ex. 1. Multiply -2x + 3y by 4x-7y. =(-2a;).4a;+3y.4a;+(-2a;)(-7y)4-3y(-7y) (1) = - 8 x2 + 12 xy + 14 xy - 21 2/2 (2) = - 8 x2 + 26 xy - 21 y2. (3) Performing tlie steps in (1) and (2) mentally, we can arrange the work as below : -2a; + 3y 4x — 7y - 8 a;2 + 26 ary - 21 2/2 Observe that the first and last terms in the product are the products of terms in the vertical lines, while the second term is the sum of the products of the terms in the diagonal lines. In this way solve the first 15 examples in Exercise. 28. 66 ELEMENTS OF ALGEBRA Ex. 2. Multiply a;^ - 2 jc + x^ + 1 by 2 - x -\- x^. Arranging both multiplicand and multiplier in descending powers of ic, we have x^ + x^-2x +1 (1) x^-x +2 afi + x^-2x^+ a;'2 (2) _ a4 _ a;3 + 2 a;2 _ x (3) 2 x3 + 2 a;-2 - 4 X + 2 (4) x5 -x^ + 5x'^-bx + 2 Expression (2) is the product of (1) multiplied by x"^ ; Expression (3) is the product of (1) multiplied by — x ; and Expression (4) is the product of (1) multiplied by 2. The sum of these partial products is the required product, by § 79. A vertical line, or bar, is often a convenient sign of group ing. Its use is illustrated in the next example. Ex. 3. Multiply x^-2x^y + ^ xy'^ - y^ by x2 - 3 xy + ?/2. x3 - 2 x2y + 3 x^-Sxy + XJ/2 - y3 y^ x5-2x*.v+ 3 -3 +6 + 1 x^y^- 1 - 9 - 2 x2|/3 + 3 + 3 xy* -y' x5 - 5 x^y + 10 x^y^ - 12 X22/3+6 xy^-y^ The sum of the numbers before each bar is the coefficient of the literal factor after it. In this example the multiplicand and the multiplier are both homo- geneous. Observe that the product is homogeneous also. This illustrates the following principle. 80. The product of two or more homogeneous expressions is homogeneous. Proof If the homogeneous multiplicand is of the nth degree, and the homogeneous multiplier is of the mth degree, then each term in the product will be of the (m + n)th degree ; that is, the product will be a homo- M UL TIP Lie A riON 67 geneous expression of the (?>i + n)th degree ; and so on for any number of factors. When the multiplicand and the multiplier are homogeneous, that fact should be noted in every case by the pupil ; and if the product obtained is not homogeneous, it is at once known that there is an error. Exercise 28. Multiply : 1. x-{-2y by x — 2y. 10. —a; + 7 by a; — 7. 2. 2x-\-Syhy3x — 2y. 11. —x—16 by —x-\-16. 3. a — 3 & by a + 36. 12. — a; + 21 by a; — 21. 4. x + 7bya;-G. 13. 2a + |& by 3 a + J &. 5. 3a;-7 by 2a;-l. 14. ^a- ^b hj ^a-^b. 6. 2.0; — 4 by 2a; + 6. 15. ax — by by ax-\-by. 7. 22/ + 5^> by 3?/ — 46. 16. a.'2 + a; + 1 by a; — 1. 8. 2 7/i2 + 5n2 by 2 m- — ir. 17. cr + a6 + 6- by a — b. 9. 3m2-l by 3?n- + l. 18. a'--a6 + 6- by a + 6. 19. x^-xY-\-y^ by a^ + 2/^. 20. a^-ab-^b'' hy o^-{-db-\-b\ 21. a^ _ 2 ax + 4 aj2 by a^ + 2 aa; + 4 x^. 22. 10 a- + 12 a6 + 9 6^ by 4 a - 3 b. 23. a-x — a^ -\- si? — a? by x-\-a. 24. x^^x-2 by x--\-x-Q,. 25. 2a:3_3aj2^2a; by 2a^ + 3a; + 2. 26. a^ + 2a?b+2ab^ by a^ - 2 ab -\- 2 b\ 27. Qi? — 3xy — 'ifhy—^-\-xy-\-y'^. 28. ar - 2 a;?/ + 2/- by ar' + 2 a.-?/ +/. 29. 27a.-3-36aa:2^48a2a;-64a3 by 3a; + 4a. 30. ab + cd + ac + bd by ab + cd — dc — bd. 68 ELEMENTS OF ALGEBRA 31. x^^ — x^y^ -{- x^'y* — xY' -\- y^ by x^-\-y~. 32. -23?y + y^-^^x'y-^-x'^-2xi/ hy x'^-\-2xy + f. 33. a^ -\- h^ -\- c^ — he — ca — ab by a-\-h + c. 34. a;"+2 _|_ 2 a;"+i — 3 aj'* — 1 by a; + l. 35. — ax^ + 3 aajy^ — 9 ay^ by — aa; — 3 ay^. 36. —a^y + y^ -\- x^y^ + x* — xy^ by ic + 2/. 37. ia^ + ia + i by ia-|. 38. |a^-2aj + f by \x^\. 39. f a^ + ic?/ 4- f 2/^ by i a; - 1 2/- 40. Ja^_|a^-f by ^a;2+|a^-|. 41. I a^ — aaj — I a^ by I a.-^ — 1^ aa? + I a^ 42. laaj + fa.'^ + ^a^ by fa^ + f a;2_|^^ 43. 3 a?"* - 2 x"^-^ + 4 a^'^-^ by 2 aj'^ + 3 x"*"^ - 4 a;'"-^ 44. 3 a;'*-^ + a;'*-2 - 2 x^^-i - 4 a;'' by 2 a^"-^ + 3 a;"-'*. 45. 4 a V* — a^a^" 4- 5 a;" by a^a^^^-^ + 6 a;"-^ 46. 3 a^'-'^y? — a"- V + a'* by aV'^ - 2 a;""^ - 3 ax''^"^. 47. 4 a^"*+i — 3 a^*" — 2 aj'"+^ + J a^'""^ by 1 a^-+i - 2 ar'^'+i - a^'"-^ 48. 3(a4-?>)'-2(a4-&)'(aJ-2/)-4(a+6)(aj-2/)'+7(a;-2/)^ by 2{a^lSf{x-y)-Q>{a^h){x-y)\ 81. Removal of signs of grouping. Ex. Remove the signs of grouping, and simplify, 42_5[-12a;-3{-15x + 3(8-7-3 (k)}]. The expression = 42 - 6 [- 12 x - 3 {- 15 x + 3 (3 ic + 1)}] = 42 _ 5 [- 12 X - 3{- 6 a: + 3}] = 42 - 6 [6 X - 9] = 87-30 X. MULTIPLICATION 69 Exercise 29. Eemove the signs of grouping, and simplify : 1. 36-S5a-[6a + 2(10-6)]S. 2. a — {h — c) — la-h — c — 2\h-\-c\~\. 3. 8(6 + c)-[-Sa-6-3(c-6 + a)J]. 4. 2(36-5a)-7[a-6J2-(5a-6)j]. 5. Q>\a-2[b-^{c + d)']\-^a-^[b-4.iG-\-d)^]\. 6. 5[a-2[a-2(a + a.')]|-4Ja-2[a-2(a + a;)"|5. 7. _l0Ja-6[a-(6-c)]5 + 60J6-(c + a)i. 8. _3J-2[-4(-a)]K5J-2[-2(-a)];. 9. _2S-l[-(x-2/)]S + S-2[-(a;-2/)]J. Multiply together the following expressions, and arrange each product in descending powers of x : 10. ax^-\-bx-\-l and ex + 2, 11. ax^ — 2bx-\-3c and x — 1. 12. ic* -h aaj^ — bx — c and ar' — a.^•^ — 6u; + c. 13. aa^ — i»2 -f- 3 a; — 6 and oar* + a^ -f- 3 a; + i^. 14. x'^ — €131? — bx^ -\- ex -{■ d and x^ 4- aar^ — 6a^ — ca; -f d. 82. Multiplication by detached coefficients. The labor of multiplication is lessened by using the method of detached eoefficients in the two following cases : (i) When two polynomial factors contain but one letter. Ex. 1. Multiply 4x^-Sx^ + 2X-5 by 5 a;^ + 3 x - 1. Writing coefficients only, we proceed as below : 4_ 3+ 2- 5 6+ 3- 4 20 - 15 + 10 - 25 + 12- 9+ 6-15 - 16 + 12 - 8 4- 20 20 - 3 - 15 - 7-23 + 20 70 ELEMENTS OF ALGEBRA Inserting the literal factors, whose law of formation is seen by inspection, we have for the complete product, 20x^-Sx^-l6x^-7x'^- 23 x + 20. (ii) When each of two polynomial factors is homogeneous and contains only two letters. Ex. 2. Multiply 5 a* + 4 a^^ _ 3 ^fts + 2 6* by a^-2 h\ 5_l_4_}_0— 3 + 2 I^ th^ first expression, the term con- 1 4. — 2 taining a^h^ is lacking ; that is, its co- efficient is zero, which is written in "^ "^ ~ "^ the line of coefficients. In the second ~ ~ ~ expression, the term containing ah is 5 + 4 — 10 — 11 + 2 + 6 — 4 missing; hence its coefficient is zero. In the method of detached coeffi- cients, the zero coefficients must evidently be written with the other coefficients. Inserting the literal factors, whose law of formation is seen by inspection, we have for the complete product, 5 a6 + 4 a55 _ 10 a*62 _ n a%^ + 2 a'^h^ + 6 a&s _ 4 56. Observe that the entire number of coefficients (zero coefficients being included) in the product is one less than the number of coefficients in both the multiplicand and multiplier together. Exercise 30. 1. Multiply ^J^2x^-x'^^x-l by x^-2x-3. 2. Multiply 3a^ + 2a2_5a-|.4 by 2a^-^a-2. 3. Multiply a^ + ^x'y-A.xif + ^Tf by 2x^-3xhj + f, 4. Multiply 3 a^ - 2 a^6 - 4 a^h^ - ab' by a^-2 h\ 5. Multiply ^x'' -Z:^y -^1 xy^ ^-2f by y?-\-'6f. 6. Rework by detached coefficients those examples in ex- ercise 28, from 19 to 42, to which the method is applicable. CHAPTER VI DIVISION OF INTEGRAL LITERAL EXPRESSIONS 83. Division is the inverse of multiplication. Having given a product and one factor, division is the operation of finding the other factor. That is, to divide one number by another is to find a third number which multiplied by the second number gives the first. Thus, - 12 -- 3 = - 4 ; for - 4 X 3 = - 12, and _12h-(-3)= 4 ; f or 4x(-3) = -12. As in Arithmetic, the given product is called the dividend, the given factor the divisor, and the required factor the quotient. 84. Law of Quality. In each of the following identities the third number multiplied by the second gives the first; hence by definition the third number in each case is the quotient of the first divided by the second. +(a6)-^-a=-6; "(aft) --+« =-6. J ^^ From identities (1) it follows that, The quotient is positive 2vhen the dividend and the divisor are like in quality ; and negative when they are opposite in quality. Tlie arithmetic value of the quotient is equal to the quotient of the arithmetic value of the dividend by that of the divisor. 71 72 ELEMENTS OF ALGEBRA Any number divided by ^1 is equal to the number itself. Any number divided by ~1 is equal to its arithmetically equal opposite number. Exercise 31. Perform each of the following indicated operations : 1. -25 -5. 5. 75 -(-25). 9. 21 - (- 1). 2. 36 ^(-6). 6. -72 -(-6). 10. - 36 -f- 4. 3. _5i^(_3). 7. _ 105 -(-21). 11. -1^|. 4. _33^(_l). 8. -144-24. 12. l-^(-|). Find the value of (a? + ?/) — z, 13. When x = — 15, y = — S, z = 6. 14. When x = — AS, y = 6, z = — Find the value of (x — y)-r- (a 4- b), 15. When 05 = 22, y = -2, a = 5, & = 3. 16. When a? = - 21, y = G, a = -7, b = 6. 85. From the definition of division we have quotient x divisor = dividend. That is, since the quotient of N divided by a is A^ — a, we have, (N^a)xa = N. (1) 86. The reciprocal of a number is 1 divided by that number. Since their product is + 1, any number and its reciprocal have the same quality. E.g., the reciprocal of 4 is ^ ; the reciprocal of — 4 is 1 -4-(— 4) or — ^ ; and the reciprocal of — | is 1 -4- (— |), or — f. DIVISION 73 87. Dividing by any number except zero gives the sams result as multiplying by the reciprocal of thai number. That is, N-^a = Nx(l^a). (1) Proof. The second member of (1) multiplied by a is, by § ^5, equal to N\ hence it is the quotient of N divided by a. Ex. 1. 16 H- 4 = IG X ^ = 4. Ex.2. 16-(-4)=10 x(-i) = -4. 88. The commutative law for division. Ex.1. _40^(-2)-(-5) = -40x(-^)x(-^) = -4. (1) Ex.2. ^-^(_|)^(-|)=^x(-3)x(-^)=f (2) Since we can change the order of the factors in the second member of either (1) or (2), we can also change the order of the divisors in the first member of either identity ; this illustrates that, The commutative law holds for division as well as for multiplication, provided the sign of operation, -r- or x, before each number is transferred with the number itself. That is, N xb^c = N^cxb. (1) Proof Nxb-^c = Nxbx(l-hc) § 87 = Nx{l^c)xb §49 = N-r- cxb. § 87 Ex. (-60)x(-22)-f-(- l5) = (~60)-(- 15)x(-22) = 4 x(-22) = -88. 89. A product of two or more factors is divided by a num- ber if any one of the factors is divided by that number. Proof {ab)-^c = a^cxb={a^c)b, §88 or {ab) -i- c = b -T- c X a = (b ^ G)a. § 88 74 ELEMENTS OF ALGEBRA 90. Any indicated quotient is called a fraction. A quotient is often indicated by placing the dividend over the divisor with a line between them. E.g.^ a -^ b, -, and a/b are but different ways of indicating that b a is to be divided by b. Each of these expressions is a fraction, a being the dividend, and b the divisor. The dividend and divisor of a fraction are often called its numerator and denominator respectively. When the dividend or divisor consists of more than one term, the horizontal dividing line in a fraction serves as a sign both of division and of grouping. E.g., in the fraction ^ ~ the horizontal dividing line takes the c + d place of both the sign of division and the two parentheses in the form (a-6)^(c + c?), or (a- 6)/(c + d). In § 1 any fractional number as 5/6 was regarded as (1/6) X 5 ; but it can also be regarded as 5 -r- 6 ; for ]Sr-r-a = Nx(l-^a) = (1/a) x JV. §§ 87, 49 91. The product of two or more fractions is equal to the product of their dividends divided by the product of their divisors; and conversely. mi- J. • a b c abc ,^. That IS, _._._ = (1) X y z xyz ^ ^ ^ . a b c a b c ..„ Proof. -'~'-'X-y-z = ~'X'-'y'-'Z §49 '^ X y z ^ X y ^ z = abc. § 85 Dividing each member by xyz, we obtain (1). Ex 4 ^ 3 ^-2_ -(4x3x2) ^ 8 _ 5 _ 7 _ 3 -(5x7x3) 35 DIVISION 75 92. Quotient of powers of the same base. Ex. 1. a^ ^ a- = a^-'^ = a^ ; for a^ x a'^ = w'. Ex. 2. a? -^ a^ = o?-^ = a* ; for a'^ x a^ = a?. These examples illustrate the following law : If /n > /I, the quotient of the mth power of any base divided by the nth power of the same base is equal to the (m — n)th power of that base; and convex That is, a'" -T- a" = a'"'". Proof a"*-" X a" = a*"-''+'' = a"*. § 76, 83 Ex. 1. Divide 20a*6S by -bah^. 91 20q^ft5 _ 20 a^ 6^ — 6 ab^~ — 5 a 6* = - 4 a362. §§ 84, 92 91 Ex. 2. Divide - ba^H^ by lla262a;2. -ba%^x^ _-b a^ b^ 7? 11 a262a;-^ ~ 11 ' d^' b'i' ic2 =-/x • 1 . 62 .x = -^bH. These examples illustrate the following section. 93. The quotient of one monomial by another. By the con- verse of § 91 we have the following rule : Divide the numeral factor of the dividend by that of the divisor, observing the law of quality; after this write the quo- tient of their literal factors, obset^ving the laiv of exponents. Ex. 1. - 84 a^xs -f- 12 a*x=- 7 ax"^. Ex. 2. 77 a^x^y^ ^ ( - 7 ax^y) =-11 axyK Check. Multiplying the obtained quotient by the given divisor, we obtain the dividend ; hence, the division is correct. T6 ELEMENTS OF ALGEBRA Exercise 32. Divide : 1. - 72 a^ by - 9 a. 6. 84 affz' by - 7 icyV. 2. 84a3by-7al 7. 28 a^d^ ^y _ 4 ^3^^ 3. - 35 a^ by 7 ic^. . 8. - 35 a%^ by 5 ab. 4. 4a^6V by —ab^c\ 9. — 16a^/ by — 4ic/. 5. - 12 a^ft'^c^ by -Sa^ftc^. 10. 36 m%^2 ^^y 9 ^6^9^ 11. 96a*afz* by 12 aV;^^ 12. - 256 xyz^' by - 8 a;y;2«. 13. SAaWc' by 14a&V. 16. - 144 aV by -24aV. 14. - 16 %a;2 by - 2 a;?/. 17. - 3 x'^+^ hj 5 x'^+\ 15. 50 yV by —Ba^y. 18. _ 4 a7'«+«2/'»+" by 7 x*"?/"*. 19. 5 a;"+V'+^ by - 8 aj^^/"*- 20. - 7 a;^+'^"'+2 by - 2 a;"- V"^- 21. — 42 x«+3a"*-i by — 7 a^'-^a'^-l 22. — 50 a7"+«?/'"+* by 25 a;"-*^"*-". 94. Distributive law for division. The quotient of one ex- pression divided by another is equal to the sum of the results obtained by dividing the parts of the first expression by the second; and conversely. That is, «±At£_±^^«_l_i + £+.., (o X XXX Principle (C) lies at the basis of division in Arithmetic; e.g., to divide 894 by 6 we separate 894 into the parts 600, 240, and 54, divide each of these parts by 6, and add the results. Thus 8M^0Og^24O^54^j00^,„^9^j^g_ 6 6 6 6 DIVISION 77 Proof. ±±±±SL±^=(a-^b + c+:.)^ §87 X X = a- + /)- + ci+... §60 :- + -+- + -. §87 95. To divide a polynomial by a monomial. By the distributive laAv for division, in § 94, we have the following rule : Divide each term of the polynomial by the monomial, and add the resulting quotients. Ex. 1. Divide l2x^-6ax^-2 a^x by 3 x. 3a; ~ 3x 3a; 3x ^ = 4a;2- 5rta;-|a2. Ex. 2. Divide 12 a" + a^ _ 6 gS by - 3 a2. 12q« + 9a*-6a5 _ 12ff« 9 a* -6a^ „q. = - 4 a - 3 a2 + 2 a8. Exercise 33. Divide : 1. 5x^-7 ax-i-4:X by X. 6. -24 a^-32 a;^ by -Sir^. 2. a;'''-7ar^ + 4aj*by a.-^. 7. a^-a-ft-a-ft^ by al 3. lOx^—Safi-^Sx* by a;^. 8. a^ — a6-ac by —a. 4. 27 3^-363.*^ by 9ar\ 9. .r^-a^-oa; by -a;. 5. 15 ar' - 25 a;* by -5. T^. 10. 3a;«-9.Tyby -3aj. 11. 4a^6^~8a%^ + Ga6^ by -2a6. 12. — 3a^ + f a^ — (j^2; by — f a;. 13. _5a^ + |.^y + j^0a, by -fa;. 14. ia.V-3.r^y-5.Ty by -^sc^y^ 78 ELEMENTS OF ALGEBRA 15. J a^x — Jg- abx — | acx by | ax. 16. - 2 a^or^ + 1 aV by | a^x. 17. 25(aj + 2/)'-3a(a; + ?/)2 + 106(a^ + ?/) by 5(aj + 2/). 18. -S(a-by -12 x(a -by- 16 y (a -by by 4(a-6)2. 19. Ga^'" — 4 a-'" by 2 a". 20. 10 2/"+V - 15 r^'^' by - 5 y^z. Divide 12 a;2«+y - 16 x^'^+Y - 20 aj^^+y by : 21. 4 a;". 22. — «'»+y. 23. — Sa^^n^^. 24. ia;2n-3^2^ 25. Divide 4 a;-'^+y — 16 a;-"+V +i by 4aj2«2/«. 26. Divide - 15 a^+Y+^ + 21 x''+Y+'^ by 3 a;*+y+2, 96. To divide one polynomial by another. Let it be required to divide 2 x^y"^ — x^y + x^ — xy^ + y^ by y"^ — xy + x^. First arrange dividend and divisor in descending powers of «, for convenience placing the divisor to the right of the dividend as below : a* — x^y + 2 x'^y^ — xy'^ + y* x^ — x^y + x'^y'^ x^ — xy + 2/2 Divisor x^ + y'^ Quotient x^y'^ — xy^ + 2/* x^y^ — xy^ + y* From the law of exponents we know that a;*, the term of the highest degree in x in the dividend, is the product of the terms of highest degree in the divisor and the quotient ; hence, the first term of the quotient is x^ -r- x^, or x"^. Multiply the divisor by x^ and subtract the result from the dividend. The remainder, xV - xy^ + y'^, is the product of the divisor by the other terms of the quotient ; hence, x^ij'^, the first term of the remain- der, is the product of the first term of the divisor and the second term of the quotient. Therefore the second term of the quotient is a;2y2 ^ a;2^ or y"^. Multiplying the divisor by y^ and subtracting the result from x^y^ — xy^ + ?/*, we have no remainder. Hence the required quotient is x^ + y\ DIVISION 79 Observe that by the above process the dividend was separated into the two parts ar* — x^y + x^y'^ and x'^y^ — xy^ + y^ ; hence, by the distrib- utive law for division, we have x^ — x^y + 2 x-y'^ — xy^ + y* _x* — x^y + x-y'^ x^y"^ — xy^ + y^ x/^ — xy + «/2 x^ — xy + y^ x^ — xy -{■ y^ = x2 + y2. If the dividend and divisor were arranged in ascending powers of x, the quotient would be obtained in the form y^ + x^. Hence, to divide one polynomial by another, we have the following rule : Arrange the dividend and divisor in descending powers of some common letter. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the divisor by this first term of the quotient, and subtract the resulting pj-oduct from the dividend. Divide the first term of the remainder by the first teiin of the divisor, and write the result as the second term of the quotient. Multiply the divisor by this second term of the quotient, and subtract the resulting product from the remainder previously obtained. Treat the second remainder, if any, as a new dividend and go on repeating the process iintil the remainder is zero, or is of a loiver degree in the letter of arrangement than the divisor. Ex. 1. Divide 2 a - 4 a^ + 3 a^ - 1 by 1 - a. Arranging dividend and divisor in descending powers of a, we have 3a8-4a2H-2a- -1 -a + 1 3a'-3a2 - a2 + 2 a - a2-|- a - 3 a-2 + a - a - a - - 1 -1 80 ELEMENTS OF ALGEBRA Ex. 2. Divide x^y'^ + x^ + y^ by y"- — xy + x'^. Arranging dividend and divisor in descending powers of x, we have ic* + x-y'^ xl^ — x^y + xhf' x^y x^y — x-y'^ + xy^ + 2/* X?/ + y2 yp- + xy ^ xhf- — xy^ + y^ Ex. 3. Divide 16 a* - 1 by 2 a - 1. 16 a4 -1 16 a4. 8^3 8a3_ -4a2 4a^ 4a2_ -2a 2a-l 2rt-l 2a 8 rt3 4 4 ^2 + 2 a + 1 Exercise 34. Divide : 1^ a;2 + 3a^ + 2 by a; + 1. 4. 3a;2 + 10a; + 3 by a; + 3. 2. a2_ii^_^3o by a-h. 5. oa^ + lla; + 2 by x^2. 3. aj2_7^_^i2 by a;-3. 6. 5a^ + 16a;H-3 by ic + 3. 7. 2x2-t-llx+5 by 2a.'-}- 1. 8. 2a^H-17a;-f 21 by 2ic-h3. 9. 4a)2-f-23ic + 15 by 4aj + 3. 10. 6a^-7aj-3 by 2a;-3. 11. 12a2-7aa;-12a.'2 by 3a-4a;. 12. 15 a^ -h 17 aic — 4 a^ by 3 a -f- 4 a;. 13. 12 a2 _ 11 ac - 36 c^ by 4 a - 9 c. 14. 60a;2-4ic?/-45?/2 by lOic-Oi/. DIVISION 81 15. -A.xy-loy-^-^ijx' hy 12x-6y. 16. 10()a:3-3.T-13a;2 ^^^ 3-|_25x. 17. 16-96ic4-216ic2-216a:3 + 8i«^ by 2-3a;. 18. x'-x'-'^x-nhy x'-\-^x-\-^. 19. 22/3_32^2_gy_l by 22/--52/-1. 20. Q>m^-m'^-l^m-\-S by 3m2H-4m-l. 21. 6a'-13a^ + 4a3 + 3a2 by ^a^-2o?-a. 22. a.'* + .T3 + 7iv2-6a; + 8 by x' + 2x-[-^. 23. a^-a3-6a- + 15a-9 by a-4-2a-3. 24. a^ + Oa^ H-13a2 + 12a4-4 by a2 4-3a + 2. 25. 2a;^-ar»4-4a^ + 4a;-3 by ar^-a; + 3. 26. a;5-5a;^ + 9ii-3-6a.-2-a; + 2 by a^-3a; + 2. 27. ar'-4a;*4-3it'3-|.3a.-2_3a' + 2 by a^-a;-2. 28. 30a-*4-lla.'3-82x2_i2a; + 48 by 2.T-4 + 3a^. 29. 69?/-18-71 2/3 + 28^-352/2 ^^ 4^y2_x^yj^Q^ 30. 0^--15^•^ + 4^^ + 7fc2-7A: + 2 by 3A:3-A: + 1. 31. 2x^-^x + x^-\-12-lx' hy x' + 2-3x. 32. a^-2a;^-7x'3 + 19x2_i0^. 1,^ ^_r'^_j_5 33. Ux^ + ^^a?y + l^^f-{-A^xi/-{-Uy' by 20.-2 + 50^2/ + 72/'. 34. x^ -{- x^y - a?f -\- Qi? -2 xy'^ -\- f by a^ + a^-/. 35. x^ -2y^* -7 afy* -7 xy^'- -\-Uxy by x-2y\ 36. a^ + 63 _^ c^ _ 3 ((6c by a + 6 + c. 82 ELEMENTS OF ALGEBRA a3 _ 3 a6c + &^ + c3 a + h + c aS + a'^b + a^c a'^ - ab ~ ac + b'^ - be -\- c^ - a% - a2c - 3 abc + b^ -\- c^ -a^b-ab^- abc - a^c + a62 _ 2 abc + b^ + c^ — a^c — abc — ac^ ab^ - abc + ac^ + b» -\- c^ ab-^ + b^ + b^c - abc + ac2 - 6% + c- - abc - b'^c - bc^ ac2+ 6c2 + c3 ac2 + ?)C2 + c^ Here we arranged the dividend and divisor in descending powers of a, and gave b precedence to c throughout. 37. a^ -\- if — z^ -\- 3 xyz hj x-\-y — z. 38. S x^ — y^ -{- z^ -{- 6 xyz by y — z~2x. 39. 27a^-Sb^-^(^-\-lSahchj 3a-2b-\-c 40. a^ + 3a26 + 3ai!>2 + 63 + c3 by a + & + c. 41. a' -\-b' + c' -2b'<^ -2 a'c" -2a'b' by a + ^ + c. 42. ^:^-^-^^^xf^ + j\fhjix-^ly. 43. |a3_ 9^2^_^_2_7^^_27ar^ by ia-3a;. 44. ^aS-^i^a^ + ^-Va-eV by ^a-i- 45. I^V H-^a^ by ia^ + iac. 46. ^9^a4_|a^-7a2 + .|a4.i_§ by f a^-f-a. 47. 36a52+i2/' + i-4a^-6aj + i2/ by Gic-iy-i 48. ^8_o^5_2 4|ci,a;4by |a-fa;. 49. 43 a^'"-^ + 6 «2-+i _ 29 aj^- - 20 aj^--^ by 2 a;'" - 5 x'^-\ 6 aj^'w+i - 29 x-^ + 43 a;2'«-i - 20 x^""-^ | 2 a;'" - 5 a;"*-^ 6 a:2"*+i - 15 x^"* 3 x'^+^ - 7 a;"* + 4 x""-^ - 14 x2« + 43 ic2'»-i - 14 a;2»» + 35 a;2'»-i 8 a;2»»-i — 20 a;2»*-2 8 a.2m-i _ 20 x^""-^ DIVISION 83 50. ic'*" — a^"^/"' + x'^if"^ — y^"" by x^ — 2/"*. 51. 6a3'»-25a-« + 27a''-o by 2 a'* -5. 52. 6a'"'-lla^ + 13a2« + 23a»"H-2-3a" by 3a"4-2. 53. 12 a;"+i + 8 a?" - 45 a;"-^ + 25 x''-'- by 6 a - 5. 54. i^ia' - Li3.ab -{-9 ac + 2b' - be by Ja-36+fc. 55. (5 + c) a^ — bcx -\- x^ — be (b -\- c) by a^^ — 6c. 56. a.'^ + (a + 6 + c) ar 4- (a6 + ac + 6c)a; + a6c by x-\-b. 57. a?^ 4- (a H- 6 — c) a^ + (a6 — ac — 6c) a; — a6c by x— e. 97. When, as in each example given above, the division is exact, the quotient is the same whether the dividend and divisor are arranged in deseending or in aseending powers of any common letter. But when the division is not exact, the partial quotient obtained with one arrange- ment is not the same as that obtained with the other. E.g., ^l±l = x-l + -^; (1) while 1+^^=1 -a; + 2x2 -2x8+^^. (2) 1 + x 1+x ^ ^ Here the partial quotients x — 1 and 1 — x + 2 x^ — 2 x^ are evi- dently unequal. The entire quotients, or the second membei-s of (1) and (2), are, of course, identical. In (1) the remainder is of a lower degree than the divisor. In (2) the division can be carried to any number of terms. When arranged in ascending powers of some common letter, an expression of a lower degree can be divided by one of a higher degree in the letter of arrangement. E.g., -l_ = l + a; + aj2 + aj3 + ...+aj"-i+--^^. 1 — aj 1 — X 84 ELEMENTS OF ALaEBRA Exercise 35, Divide : 1. a^ -^ y^ hj X -\- y. 5. x^ — a^ hj x -\- a. 2. x^ -{- if hj X — y. 6. 1 by 1 + ic to 4 terms. 3. x^ — y^ bj X -\- y. 7. 1 + a; by 1 + ic^ to 5 terms. 4. x^ + y*hj x-}-y. 8. 1 -f- 2 a? by 1 — 3 a; to 4 terms. 9. Divide 2 by 1 -}- a; and thus reduce the second member of identity (1) in § 97 to the form of the second member of (2). 98. Zero divided by any number, except zero, is equal to zero. That is, when a^O, 0^a = Q. §§ 85, 74 Conversely, if a quotient is zero, the dividend is zero. Zero as divisor will be considered in Chapter XXVII. Prior to that chapter it will be assumed that any expression used as a divisor does not denote zero. 99. Division by detached coefficients. In § 82 we considered two cases in which the work of multiplication could be shortened by using the method of detached coefficients. In the same two cases the labor of division can be lessened by using detached coefficients and an arrangement of terms known as Horner's method of synthetic division. This method is illustrated by the following examples : Ex. 1. Divide 2x^-1 x^-\-2x^- x^-Qx-^ 20 by 2x3-3x2+4aj-5. cS i 3 S " 5 Quotient 1-2-4J Remainder Inserting in the quotient the literal factors, whose law of formation Is seen by inspection, we have for the complete quotient x'^ — 2x — A. Explanation. The modified divisor, or the column of figures to the left of the vertical line, consists of the coefficients of the divisor, the 2 -7 + 2 - 1 - 6 + 20 3 -4 5 -6 + 8 -10 -12 + 16 -20 DIVISION 85 quality of each coefficient after the first being changed; this change of quality enables us to replace the operation of subtraction by that of addition at each successive stage of the work. Observe that the number of coefficients in the quotient will be one more than the number of coefficients in the dividend minus the num- ber of coefficients in the divisor, in this case 1 + 6 — 4, or 3 (§ 82). Thus, the numbers to the left of the vertical bar are the coefficients of the quotient^ and those to the right of this bar are the coefficients of the remainder. Dividing the first coefficient of the dividend by the first coefficient of the divisor, we obtain the first coefficient, 1, of the quotient. Multi- plying the modified coefficients of the divisor (3, — 4, 5) by this first coefficient of the quotient, we obtain line (1). Adding the coefficients in the second column to the right of the divisor, and dividing the sum by the first coefficient, 2, of the divisor, we obtain — 2, which is the second coefficient of the quotient. Multi- plying the modified coefficients of the divisor by this second coefficient of the quotient, we obtain line (2). Adding the coefficients in the third column and dividing the sum by the first coefficient of the divisor, we obtain the thirds or last, coefficient of the quotient. Multiplying the modified coefficients of the divisor by this third coefficient of the quotient, we obtain line (3). Lines (1), (2), and (3) are evidently the coefficients of the three partial products obtained by multiplying the divisor by each term of the quotient, the first term of each product being omitted and the quality of the others being changed. Hence by adding each of the vertical columns after the third, we obtain the coefficients of the remainder. Here the coefficients are all zero, and the division is exact. Ex. 2. Divide 2x^-1 7^y+\2 v^y"^ - 8 x^y^ ■\- a; V by 2 x^ - 3 x^y - y^. 2 2 -7 +12 -8 + 1 + + (1) 3 + 3 + 1 (2) - 6 -2 (3) 1 + 9 4-3 (4) + 3 + 1 (5) 1 _2 + 3 +1 I +2 +3 +1 Inserting the literal factors, we have for the quotient x^ — 2 x'^y + 3 xy"^ + ?/'^, and for the remainder 2 x'^^/* + 3 a;?/^ + y^. Explanation. The terms in xy^ and y^ are missing in the dividend, and the term in xy"^ in the divisor; hence their zero coefficients are 86 ELEMENTS OF ALGEBRA written with the other coefficients. The sums of the vertical columns after the fourth give the coefficients of the remainder. To find the remainder after one term of the quotient, add lines (1) and (2) after the first vertical column ; to find the remainder after two terms of the quotient, add lines (1), (2), and (3) after the first two vertical columns ; to find the remainder after three terms of the quotient, add lines (1), (2), (3), and (4) after the first three vertical columns. Exercise 86. Divide : 1. a-^ - 4 or' + 2 a;2 + 4 a; 4- 1 by ^ - 2 a; - 1. 3. x^^-{-x^-\-l by 0^ + .1^ + 1. by 2x^-3xy-\-Ay^ 5. Sf-22xy'-^20a^rf^xy-7x'y + 6x^ . by 4:y^ — 3xy -^2x^0 6. a' -3 a'b^ + 8 ab^ - 5b^ hj a^ -4.ab -{-b' to four terms in the quotient. CHAPTER VII INTEGRAL LINEAR EQUATIONS IN ONE UNKNOWN 100. An integral equation is an equation all of whose terms are integral in the unknown. (Review §§ 10, 16, 17.) E.g.^ 2 a;2 + 3 = 2 X and ^ -}- — = x + 2 are integral equations. 2 101. The degree of an integral equation in one unknown is the degree of its term of highest degree in the unknown. A linear equation is an equation of the^rs^ degree. A quadratic equation is an equation of the second degree. A higher equation is an equation of a higher degree than the second. E.g., 3 ic + 1 = 4 and ax + 6 = are linear equations in x. 6 x2 — 7 X = 1 and ax^ + 6x + c = are quadratic equations in x. 6 x^ — 4 x^ + 3 X 4- 4 = is a higher equation in x. 102. A root, or solution, of an equation in one unknown is any value of the wiknown ; that is, it is any number which when substituted for the unknown renders the equation an identity. E.g., 12 is a root of the equation 2 X - 5 = X + 7. For, putting 12 for x in the equation, we obtain the identity 24 - 5 = 12 + 7. Any root of an equation, since it satisfies the condition expressed by the equation, is said to satisfy the equation. 87 88 ELEMENTS OF ALGEBBA 103. To solve an equation in one unknown is to find all its roots. In solving equations we use the principles of EQUIVALENT EQUATIONS. 104. Two equations in one unknown are said to be equiv- alent, when every root of the first is a root of the second, and every root of the second is a root of the first. E.g.^ the equations 4a;-8 = 2-x (1) and bx= 10 (2) have the same root, i.e., are equivalent ; for 2 is a root of each equa- tion, and, as will be seen later, 2 is the only root of either. In solving equations we need to know what operations on the members of an equation will make the derived equation have the same root, or roots, as the given one. Of such operations the most elementary and important are found in §§ 105, 106, 108, 109. 105. Identical expressions. If in the equation 4(x-l)-(3x-2)=3, (1) we substitute for the first member the identical expression a; — 2, we obtain the equivalent equation X - 2 = 3. (2) For, as is easily shown, 5 is a root of either equation ; and, as will be seen later, 5 is the only root of either. This example illustrates the following principle : If , for any expression in an equation, an identical expres- sion is substituted, the derived equation will he equivalent to the given one. INTEGRAL LINEAR EQUATIONS 89 That is, ii A = B denotes any equation in one unknown, as X, and A = A'\ then the equations A = B (1) and A' = B (2) have the same root, or roots. Proof. To prove that equations (1) and (2) have the same root, or roots, we must prove that every root of (1) is a root of (2) ; and conversely that every root of (2) is a root of (1). Since A and A^ are identical expressions, any value of x which when substituted for x will make either one identical with By will make the othei? identical with B (§ 32). Hence, any root of (1) is a root of (2), (§ 102) ; and con- versely any root of (2) is a root of (1) ; that is, equations (1) and (2) have the same roots, i.e.y are equivalent. E.g., since, 3(a; - 1) - {3 x - (2 + x)} = x - 1 ; the equations 3(x - 1) - {3 x - (2 + x)} = 5 (1) and x-l = 5 (2) have the same root ; that is, we neither lose nor introduce a root by- substituting for 3(x — 1) — {3 X — (2 + x)} in equation (1) its identical expression x — 1. 106. Addition or Subtraction. If to both members of the equation 2x-8 = 7-8 (1) we add 8 + x, we obtain the equivalent equation 3x = 15. For, as is easily shown, 5 is a root of each equation ; and, as will be seen later, 5 is the only root of either equation. This example illustrates the following principle : If identical expressiojis are added to, or subtracted from, both members of an equation, the derived equation will be equivalent to the given one. 90 ELEMENTS OF ALGEBRA That is, if 31= M', the equations A=B (1) and A±M=B±M' (2) have the same root or roots. Proof. Any root of (1) makes A = B. § 102 But, when A = B, A±M=B±M'. § 32, (iii) Hence, any root of (1) is a root of (2). Conversely, any root of (2) makes A ± M= B ± M'. B\xt,when A ±M=B±M', A = B. § 32, (iii) Hence, any root of (2) is a root of (1). Therefore, equations (1) and (2) are equivalent. § 104 If, to each member of the equation ax — b = ex — d, (1) we add — ex and + 6, we obtain the equivalent equation ax — ex = b — d. (2) Adding — ex to both members of equation (1) removes the term + ex from the second member, and transfers it, with its sign changed from + to — , to the first member. Likewise, adding + 6 to both members of (1) removes the term — b froili the first member, and transfers it, with its sign changed from — to +, to the second mem- ber. This example illustrates the following important application of the principle proved above. If any term is transposed from one member of an equation to the other, its sign being changed from -\- to —, or from — to +, the derived equation has the same root or roots as the given one. 107. An expression is said to be unknown, or knoivn, ac- cording as it does, or does not, contain an unknown number. E.g., if X is an unknown number, a: — 2 is an unknown expression ; if a is a known number, 9 + 5 a is a known expression. INTEGRAL LINEAR EQUATIONS 91 108. Multiplication. If both members of the equation 2 4 3 12 ^^ are multiplied by 12, we obtain the equivalent equation 6 X + 9 = 4 X + 13. (2) For, as is easily shown, 2 is a root of each equation, and, as will be seen later, 2 is the only root of either. This example illustrates the following principle : If both members of an equation are multiplied by the same known expression, not denoting zero, the derived equation will be equivalent to the given one. That is, if C represents any known expression, not denot- ing zero, the equations A = B and CA=CB have the same roots. (1) (2) Proof Any root of (1) makes A = B. § 102 But, when A = B, CA= CB. § 32, (iv) Hence, any root of (1) is a root of (2). Conversely, any root of (2) makes CA = CB. § 102 But, when CA = CB, A = B, since C ^ 0. § 32, (v) Hence, any root of (2) is a root of (1). Therefore, equations (1) and (2) are equivalent. Ex. 1. Solve the equation (5 x - 12) -4- 6 = (x - 3) ^ 3. (1) Multiply by 6, 5 x - 12 = 2 x - 6. (2) Transpose terms, 5 x — 2 x = 12 — 6. (3) Unite terms, 3 x = 6. (4) Multiply by 1/3, x = 2. (5) 92 ELEMENTS OF ALGEBRA Proof of equivalency. Equation (2) has the same roots as (1) by §108, 'identical ex- pressions. ' Equation (3) has the same roots as (2) by § 106, ' addition. ' Equation (4) has the same roots as (3) by §105, 'identical ex- pressions.' Equation (5) has the same roots as (4) by § 108, 'multiplication.' Hence the one and only root of each of these equations is 2. Ex. 2. Solve the equation ^^^ ^ ^ = \. (1) Multiply by 12, 3 (x + 1) - 4 (x - 1) = 12. (2) Remove (), 3a; + 3-4x4-4 = 12. (3) Transpose terms, 3ic — 4a; = 12 — 3 — 4. (4) Unite terms, — x = 5. (5) Multiply by - 1, x = — 6. (6) Proof of equivalency. Equation (2) has the same roots as (1) by the principle of ' multiplication ' (§ 108) ; (3) as (2) by ' identical ex- pressions ' (§105); (4) as (3) by 'addition' (§106); (5) as (4) by 'identical expressions' (§105); and (6) as (5) by 'multiplication' (§ 108). Hence the one and only root of each of these equations is —5. The two following applications of the foregoing principle are very important : (i) When, to clear an equation of fractional coefficients, we multiply both members by the L.C.M. of their known denominators, the derived equation has the same roots as the given one. (ii) When the sign before each term of an equation is changed from + to — , or from — to -f- (that is, when each member is multiplied by — 1) the derived equation has the same roots as the given one. 109. Roots introduced or lost. If we multiply both members of the equation 3x-7 = 2x + 2 (1) by the known expression 0, we obtain the identity (3ic-7) x0=(2x + 2) xO. (2) INTEGRAL LINEAR EQUATIONS 93 Observe that (1) restricts x to the one value 9, while (2) does not restrict the value of x at all. Again, if we multiply both members of the equation 6x-l=4x + 3 (3) by the unknown expression a; — 5, we obtain the equation (6x-l)(x-5) = (4x + 3)(x-5). (4) Equation (4) has the two roots 2 and 5, while (3) has only the one root 2. Hence, the root 5 was introduced into (4) by multiplying (3) by the unknoion expression x -^ 5. The two examples above illustrate why the multiplier in § 108 was limited to a known expression, not denoting zero. If we divided identity (2) by 0, we would obtain equation (1). If we divided equation (4) by x — 5, we would obtain equation (3), and one root would be lost by the operation. This illustrates why the divisor in the following article is limited to a known expression not denoting zero. 110. Division. If both members of an equation are divided by the same known expression, not denoting zero, the derived equation will be equivalent to the given one. That is, if represents any known expression, not denot- ing zero, the equations A = B (1) and A^C = B-^G (2) have the same roots. Proof Any root of (1) makes A = B. § 102 But, when A = B, A-^C=B-i-C. § 32, (v) Hence, any root of (1) is a root of (2). Conversely, any root of (2) makes A-t-C=B^C. § 102 But, when A^C=B-^C, A = B. § 32, (iv) Hence, any root of (2) is a root of (1). Therefore, equations (1) and (2) are equivalent. 94 ELEMENTS OF ALGEBBA Ex. 1. Solve the equation (x - 1) (x - 2) + 5 = (x + 1)2. (1) Eemove (), x^-Sx + 2-{-6 = x^ + 2x + l. (2) Tra.nspose terms, — Sx — 2x — 1— 2 — 5. (3) Unite terms, — 6x = -Q. (4) Divide by - 5, x = ^. (5) Proof of equivalency. No root is lost or introduced by any one of the operations performed on the members of the equations from (1) to (5); hence, the one and only root of (1) is |. Ex.2. Solve3(x-l)-{3x-(2-x)} = 5. (1) Remove ( ), Sx -S - Sx + 2 - x = 5. (2) Transpose, 3 x - 3 x - x = 5 + 3 - 2 (3) Unite terms, —x = 6. (4) Divide by - 1, x=-6. (5) Proof of equivalency. No root is lost or introduced by any one of the operations performed on the members of the equations from (1) to (5); hence, the one and only root of (1) is — 6. Observe the difference in the meanings of equal, identical, and equivalent. Equal applies to numbers, identical to ex- pressions, and equivalent to equations. Two numbers are equal or unequal, two expressions are identical or not iden- tical, and two equations are equivalent or not equivalent, i.e., have or have not the same roots. We should not apply the word equivalent to numbers or expressions. 111. Erom the examples given above, it will be seen that the different steps in the process of solving a linear equa- tion are the following : (i) Clear the equation of fractions, if there are any. (ii) Remove parentheses, if there are any. (iii) Transpose the unknown terms to one member of the equation, and the known terms to the other member. INTEGRAL LINEAR EQUATIONS 95 (iv) Unite like terms j and divide both members by the co- efficient of the unknown. Note. In order to form the habit of clear and accurate thinking, the pupil should at first state the operation by which each equation is derived from the preceding one, and note whether by this operation any root is lost or introduced. But as he advances he should perform the simpler steps mentally, and apply two or more principles at the same time. 112. A numeral, or numerical, equation is an equation in which, all the known numbers are denoted by numerals. Exercise 37. Solve each of the following numeral equations : 1. 3a;-5 = 2a;4-l. 8. x- (4:-2 x) = 7 (x-1). 2. 3x-\-^ = x + S. 9. 5(4-3a;) = 7(3-4a;). 3. 4:X-4: = x-7. 10. 4(l-a;) + 3(2 + a;) = 13. 4. 7 a; -5 = a; -23. 11. S(x -2) = 2(x - S). 5. 8 x + 42 = 5 X. 12. 2 x - (5 x -\- 5) = 7. 6. 5 a; -12 = 6 a.- -8. 13. 3(.c + 1) =- 5(a; - 1). 7. 7a; + 19 = 5a; + 7. 14. 7 (a; - 18) = 3 (a; - 14). 15. 2(a;-2) + 3(a;-3) + 4(.r-4)-20 = 0. 16. . 2 (a; - 1) - 3 (a; - 2) + 4 (a; - 3) + 2 = 0. 17. 5a; + 6(.T-f l)-7(a; + 2)-8(a;4-3) = 0. 18. 2a;-[3-|4a; + (a;-l)|-5] = 8. 19. (x - 1) (x - 2) = (x + 3) (x - 4). 20. 3 a^ = (a; + 1)' + (x + 2f + (a; + 3)1 21. (x - 2) (x - 5) 4- (x - 3) (x - 4) = 2 (a; - 4) (a; - 5). 22. 5(x-\-iy-\-7{x + sy = i2(x-^2y. 23. (x - 1) {x - 4) = 2 a; H- (x - 2) (x - 3). 96 ELEMENTS OF ALGEBRA 24. x/5-x/4: = l. 27. 2 x/3 + 5 = 5 x/6 + A. 25. {x-l)/2-\-{x-2)/3 = 3. 28. x/2-\-2x/3 = 5x/6 + 7. 26. 2x/3 + 4. = 5-{-x/3. 29. 3 x/A ~ 5 = 7 x/S - 6. 30. |-(2-a;)- |-(5ir + 21) = a! + 3. 31. J(^ + l)-|-^(a; + 2)+i(a^ + 4) + 8=:0. 32. K^-5)-K^-4)-i(^-3)-(x'-2). 33. ^(a)H-i)_J(2a;-i) + li-0. 34. (3x + 5)/8 - (21 + x)/2 = 5x- 15. 35. (a^ - 2)/3 - (12 - x)/2 =(5x- 36)/4 - 1. 36. (a.- -f 8)/4 - (5 a.- + 2)/3 = (14 - a;)/2 - 2. 37. (x - 15)/4 - (7 - 2 a.')/21 - 3 a;/14 + 1/2. 38. 5[4-(3a?-l)]=6(a^-ll) + 49. 39. (x - 2)/4 + 1/3 -lx-(2x- l)/3]- 0. 40. 3ip8-(a^/8+24)] = 3i(2i + a./4). 41. 5(3x-5)-17 = 8(3x-5)-2 (3 x - 5). Combine the terms involving Sx — 6. 42. 2 (i» + 1) - 3 (ic + 1) + 9 (x + 1) + 18 = 7 (a; + 1). 43. X (x -\- 2) -^ x (x -\- 1) = (2 X — 1) (^x -{- 3). 44. x'-xll-x-2(3-x)'] = x-^l. 45. 3 (x - 1)/16 - 5 (a^ - 4)/12 = 2 (i» - 6)/5 + 5/48. 46. 0.5 a; + 3.75 == 5.25 ^ — 1. To clear of fractions multiply by 4. 47. 2.25 X - 0.125 = 3x-\- 3.75. 48. 0.25 X + 4 - 0.375 x = 0.2 ic - 9. 49. 0.375 X - 1.875 = 0.12 x + 1.185. 50. O.W a; + 1.2 - 0.875 a; + 0.375 = 0.0625 x. INTEGRAL LINEAR EQUATIONS 97 113. A literal equation is an equation in which one or more of the known numbers are denoted by letters. E.g., ax-\-2x + i = and ax -\- b = ex are literal equations. Ex. Solve the equation (2 ~ 5x)/a = (ex + 7)/b. (1) Multiply by ab, (2 — 5x)b = (ex -\- 7) a. (2) Remove ( ), 2b - 6bx = aex + 7 a. (3) Transpose, 6bx — acx = 7 a — 2b. (4) Unite terms, — (5 b + ac) x = 7 a — 2 &. (5) Divkle by - (5 6 + «c), x = ?A=lI_«. (6) 56 + ac Proof of equivaleney. No root is lost or introduced by any one of the operations performed on the members of the equations from (1) to (6); hence, the one and only root of (1) is jj^iven in (6). 114. Any linear equation in one unknown has one, and only one, root. Proof. By transposing and combining terms, any linear equation can be reduced to an equation of the form ax = c, (1) which, by §§ 105, 106, is equivalent to the given equation. Divide by a, x = c/a. (2) Equation (2) is equivalent to (1) by § 110 ; hence, c/a is the one and only root of equation (1), or of the given linear equation. If c = and a^O, the root c/a is zero (§ 98). If c ^ 0, then the smaller a is, the larger arithmetically is the root c/a. Observe that, if b is an arithmetic number, the linear equation x — b = has the arithmetic root b, while the equation x-\-b = has no arithmetic root. 98 ELEMENTS OF ALGEBRA Exercise 38. Solve each of the following literal equations : 1. ax -i- b^ = bx -{- a\ 3. (a -\-b)x + (b — a)x = b^ 2. x/a + x/b = c, 4. 2(x — a)-\-3(x—2a) = 2 a. 5. (a-\- b)x-\- (a — b)x = al 6. (a + bx) (b 4- ax) = ab (x^ — 1). 7. (a — x) (a -\- x) = 2 a^ -{- 2 ax — x^. 8. i(x-\-a-\-b)-{-i{x-{-a-b) = b. 9. i{a-\-x)-\-l(2a-\-x) -\-i(3a + x)=3a. 10. ica -=- 6 + a?6 ^ a = a^ + &^. 11. (a + &aj) (& + ax) = ab(x^ — 1). 12. (a^ + x) (b^ -\-x) = (ab + xf. 13. (a; 4- a + ?>)' + (a^ + a - ^')' = 2 a^. 14. {x -a)(x-b)-\- (a + &)' = {x -\- a) (x -\- b). 15. aic (cc + a) + bx (ic 4- 6) = (a + ^) (a? + a) (cc + b). 16. What kind of a number is the root of a numeral equation? Of a literal equation? See § 5. CHAPTER VIII PROBLEMS SOLVED BY LINEAR EQUATIONS IN ONE UNKNOWN 115. Having learned some of the properties of linear equa- tions in one unknown, we return to the subject of solving problems by equations, which was introduced in the first chapter. Reread § 19. Prob. 1. A has $80, and B has § 15. How much must A give to B in order that he may have just 4 times as much as B ? Let X = the number of dollars that A must give to B ; then 80 — x = the number of dollars that A will have left, and 16 + X = the number of dollars that B will have. But A will then have 4 times as much as B ; that is, 80-a; = 4(15 + ic). (1) From (1) X = 4. Hence A must give ^4 to B. Prob. 2. A man has 10 coins, some of which are half-dollars, and the rest dimes, and the coins altogether are worth $ 6. How many has he of each kind ? Let X — the number of half-dollars ; then 16 — a; = the number of dimes. The X half-dollars are worth ^ x dollars, and the 16 — a; dimes are worth ^^ (16 — x) dollars. Now the coins altogether are worth $ 6 ; hence ix + T^(16-x)=6. From (1) 05 = 11, the number of half-dollars. .'. 16 — X = 6, the number of dimes. 100 ELEMENTS OF ALGEBRA Note. It should be remembered that any letter as x always denotes a number, and not a concrete quantity. Observe, also, that in any problem all concrete quantities of the same kind must be expressed in terms of the same unit ; for example, in each of the above examples all sums of money were expressed in terms of the unit, one dollar. Prob. 3. A father is 7 times as old as his son, and in 5 years he will be 4 times as old as his son. How old is each ? Let X years = the son's age ; then 7 x years = the father's age. Hence (x + 5) years = the son's age after 5 years, and (7 ic + 5) years = the father's age after 5 years. .-. 7aj + 5 = 4(x + 5). (1) From (1) x = 6. .: 7x = 35. Hence the son is 5 years old, and the father 35. Prob, 4. Divide 60 into two parts, so that three times the greater may exceed 100 by as much as 8 times the less falls short of 200. Let X = the greater part ; then 60 — x = the less. Three times the greater part is 3 x, and 3 a; — 100 = the excess of 3 a; over 100. Eight times the less part = 8 (60 — x) and 200 - 8 (60 - a;) = the defect of 8 (60 - x) from 200. But this excess and defect are equal ; that is, 3x- 100 = 200 -8(60- x). (1) From (1) X = 36, the greater number. .*. 60 — X = 24, the less number. Prob. 5. A could do a piece of work in 14 hours which B could do in 6 hours. A began the work, but after a time B took his place, and the whole work was finished in 10 hours from the beginning. How long did A work ? Let X = the number of hours that A worked ; theu 10 — X = the number of hours that B worked. G^ A PROBLEMS SOLVED BY LmEAE EQUATIONS 101 Since A could do the whole work in 14 hours, in 1 hour he would do 1/14 of it; hence ^ x = the part of the work done by A in x hours. Since B could do the whole work in 6 hours, in 1 hour he would do 1 /6 of it ; hence 1(10 — x) = the part of the work done by B in 10 — x hours. But A and B together did the whole work ; hence t1jX + K10-^)=1. (1) From (1) z = 7. .-. 10 - x = 3. Hence A worked 7 hours, and B worked 3. Prob. 6. Find the time between 5 and 6 o'clock at which the hands of a watch are together. Suppose that the hands are together at x minutes after 5 o'clock. At 6 o'clock the hour-hand is 25 minute-spaces ahead of the minute- hand ; hence, while the minute-hand moves through x minute-spaces, the hour-hand will move through a; — 25 such spaces. But the minute- hand moves 12 times as fast as the hour-hand ; that is, in any given time the minute-hand passes over 12 times as many minute-spaces as the hour-hand. Hence a; = 12 (a; - 25). (1) From (1) x = 27^p Hence the hands are together at 27y\ minutes past 5 o'clock. Exercise 39. 1. Find two numbers whose sum is 72, and whose dif- ference is 8. Ans. 40 and 32. 2. Divide 25 into two parts whose difference is 5. 3. Divide 12 into two parts whose difference is 16. Ans. 14 and — 2. 4. The difference between two numbers is 8; if 2 be added to the greater, the result will be 3 times the smaller. Find the numbers. 102 ELEMENTS OF ALGEBRA 5. A man walks 12 miles, then travels a certain dis- tance by train, and then twice as far by coach as by train. If the whole journey is 78 miles, how far does he travel by train ? 6. Find two numbers whose difference is 12, and whose sum is equal to ^ their difference. 7. Find a number such that the sum of its sixth and ninth parts will be equal to 15. 8. Find the number of which the eighth, sixth, and fourth parts together make up — 13. A71S. — 24. 9. Find a number such that ^ of it shall exceed ^ of it by 2. Ans. — 35. 10. Two numbers differ by 28, and one is -| of the other. Find them. 11. A, B, and C have a certain sum of money between them. A has ^ of the whole, B has \ of the whole, and C has $ 50. How much have A and B ? 12. A and B together have $ 75, and A has $ 5 more than B. How much has each? 13. A has $ 5 more than B, B has $ 20 more than C, and they have $ 360 between them. How much has each ? 14. A has $ 15 more than B, B has $ 5 less than C, and they have ^ Q^ between them. Hoav much has each ? 15. A has $100, and B has |20. How much must A give B in order that B may have half as much as A? 16. The sum of two numbers is 38, and one of them exceeds twice the other by 2. What are the numbers? 17. Find a number which when multiplied by 8 exceeds 27 as much as 27 exceeds the original number. 18. Find two numbers of which the sum is 31, and which are such that one of them is less by 2 than half the other. PROBLEMS SOLVED BY LINEAR EQUATIONS 103 19. Divide 100 into two parts such that twice one part is equal to 3 times the other. 20. Four times the difference between the fourth and fifth parts of a certain number exceeds by 4 the difference between the third and seventh parts. Find the number. 21. Fifty times the difference between the seventh and eighth parts of a certain number exceeds half the number by 44. Find the number. 22. A father is 4 times as old as his son; in 24 years he will be only twice as old. Find their ages. 23. A is 25 years older than B, and A's age is as much above 20 as B's is below 85. Find their ages. 24. A's age is 6 times B's, and 15 years hence A will be 3 times as old as B. Find their ages. 25. Find a number such that if 5, 75, and 35 are added to it, the product of the first and third sums will be equal to the second sum multiplied by the number. 26. The difference between the squares of two consecu- tive whole numbers is 121. Find the numbers. 27. Divide $380 between A, B, and C, so that B will have i$30 more than A, and C will have $20 more than B. 28. The sum of the ages of A and B is 30 years, and 5 years hence A will be 3 times as old as B. Find their pres- ent ages. 29. The length of a room exceeds its breadth by 3 feet ; if the length had been increased by 3 feet, and the breadth diminished by 2 feet, the area would not have been altered. Find the length and breadth of the room. 30. 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 dimensions of the room. 104 ELEMENTS OF ALGEBRA 31. The width, of a room is | of its length. If the width had been 3 feet more, and the length 3 feet less, the room would have been square. Find the dimensions of the room. 32. A, B, and C have $ 1285 between them ; A's share is greater than | of B's by ^25, and C's is j\ of B's. Find the share of each. 33. If silk costs 6 times as much as linen, and I spend $66 in buying 40 yards of silk and 24 yards of linen, find the cost of each per yard. 34. If $600 be divided among 10 men, 20 women, and 40 children, so that each man receives $ 15 more than each child, and each woman receives as much as two children, find what each receives. 35. Divide $152 among 5 men, 7 women, and 30 chil- dren, giving to each man $4 more than to each woman, and to each woman 3 times as much as to each child. 36. A sum of money is divided between three persons, A, B, and C, so that A and B have $60 between them, A and C have $ 65, and B and C have $ 75. How much has each? 37. A dealer bought four horses for $1150; the second cost him $ 60 more than the first, the third $ 30 more than the second, and the fourth $ 10 more than the third. How much did each cost ? 38. Two coaches start at the same time from York and London, a distance of 200 miles, travelling one at 9J miles an hour, the other at 9J miles an hour. In how many hours after starting did they meet, and how far from London ? Ans. 10| hours ; 98J miles from London. 39. A man leaves ^ of his property to his wife, l to his son, and the remainder, which is $2500, to his daughter. How much did he leave to his wife and son each ? Let X = the number of dollars which he left in all. PBOBLEMS SOLVED BY LINEAR EQUATIONS 105 40. A man divided his property between his three chil- dren so that the eldest received twice as much as the second, and the second twice as much as the youngest. The eldest received ^3750 more than the youngest. How much did each receive? 41. A third of the length of a post is in the mud, a fourth is in the water, and 5 feet is above the water. Find the length of the post. 42. A flock of sheep and goats together number 84. There are 3 goats to every 4 sheep. How many are there of each ? 43. Find the time between 3 and 4 at which the hands of a clock are together. 44. A can do a piece of work in 30 days which B can do in 20 days. A begins the work, but after a time B takes his place, and the whole work is finished in 25 days from the beginning. How long did A work ? 45. A can do a piece of work in 20 days which B can do in 30 days. A begins work, but after a time B takes his place and finishes it. B worked 10 days longer than A. How long did A work? 46. One number exceeds another by 3, while its square exceeds the square of the second by 99. Find the numbers. 47. Of two consecutive numbers, ^ of the greater exceeds I of the less by 3. Find the numbers. 48. A garrison of 1000 men having provisions for GO days was reinforced after 10 days, and from that time the provisions lasted only 20 days. Find the number in the reinforcement. 49. In a mixture of spirits and water half of the whole plus 25 gallons was spirits; and a third of the whole minus 5 gallons was water. How many gallons were ther% of each ? 106 ELEMENTS OF ALGEBRA 50. At 3 o'clock, A starts upon a journey at the rate of 4 miles an hour, and after 15 minutes B starts from the same place, and follows A at the rate of 4J miles an hour. When does B overtake A ? 51. A fish was caught whose tail weighed 9 pounds; his head weighed as much as his tail and half his body, and his body weighed as much as his head and tail. What did the fish weigh ? 52. Find a number such that if | of it be subtracted from 20, and ^j of the remainder from i of the original number, 12 times the second remainder shall be half the original number. 53. A cistern can be filled in half an hour by a pipe A, and emptied in 20 minutes by another pipe B ; after A has been opened 20 minutes, B is also opened for 12 minutes, then A is closed, and B remains open for 5 minutes more, after which there are 13 gallons in the cistern. What was the capacity of the cistern ? 54. A father was 24 years old when his eldest son was born ; and if both live till the father is twice as old as he now is, the son will then be 8 times as old as now. Find the father's present age. 55. If 19 lbs. of gold weigh 18 lbs. in water, and 10 lbs. of silver weigh 9 lbs. in water, find the quantities of gold and silver in a mass of gold and silver weighing 106 lbs. in air, and 99 lbs. in water. 56. The sum of $ 1650 is laid out in two investments, by one of which 15 per cent is gained, and by the other 8 per cent is lost; and the amount of the returns is $1725. Find the amount of each investment. 57. How many children are there in a family, if each son has as many brothers as sisters, and each daughter has twice as many brothers as sisters ? PROBLEMS SOLVED BY LINEAR EQUATIONS 107 116. Interost formulas. In problems of interest, the quan- tities involved are the principal, interest, rate, time, and amount. Let p = the number of dollars in the principal ; r= the rate, or the ratio of the interest per annum to the principal ; t = the number of years in the time ; i = the number of dollars in the interest for t years at rate r. a = the number of dollars in the amount, or the sum of the principal and the interest ; then /=prt; (1) and a = p-\- prf. (2) Proof. The interest on $p for one year is $pr', hence i, or the interest on $p for t years, is $j^?*i; whence (1). But a=2J-\-i; whence (2). If any three of the four numbers /, p, r, t, or a, p, r, t are given, the fourth can be found by solving equation (1) or (2). Ex. Find the principal that will amount to $ 1584 in 5 years 4 months at G per cent. Here a = 1584, < = 5|, r = 0.06. Substituting in (2), we have 1584 = p 4- P (0.06) (5|) = 1.32 p. .-. p = 1584 - 1.32 = 1200. Hence the principal is $ 1200. Exercise 40. 1. Solve i =prt forp, r, and t. 2. Solve a = p -{-jvt forp, r, and t. 3. Find the interest on $ 4760 for 4 years 6 months at 5 1 per cent. 108 ELEMENTS OF ALGEBRA 4. Find the amount of $ 3500 for 5 years 4 months at 6 per cent. 5. Find the interest on $ 7240 for 3 years 3 months at 8 per cent. 6. The interest on $ 1250 for 8 months is $ 50. Find the rate per cent. 7. The amount of $ 1050 for 2 years 6 months is $ 1260. Find the rate. 8. The interest on $3420 at 6 per cent is $049.80. Find the time. 9. A sum of money doubles in 12 years 6 months. Find the rate. 10. Find the principal that will yield f 262.50 per month at 7 per cent. 11. Find the time in which $ 1350 will amount to $ 1809 at 6 per cent. 12. The interest in 4 years 3 months at 4 per cent is $ 2099.50. Find the principal. 13. Find the time in which a sum of money will double itself at 6 per cent. 14. The interest on $ 1270 for 8 months is $76.20. Find the rate. 15. At 4 per cent how much money is required to yield $ 2500 interest annually ? CHAPTER IX POWERS, PRODUCTS, QUOTIENTS 117. Certain products and quotients are so frequently required in Algebra that the student should prove and memorize the identities by which they can be written out. In this chapter the most important of these identities are proved, and used in obtaining products and quotients. In the next chapter the converses of these identities are used for factoring. 118. Power of a power. The nth power of the mth power of any base is equal to the mnth power of that base; and conversely. That is, (a/")" = a""". Ex. 1. (23)2 = 23 X 23 = 23+8 _ 26. Ex. 2. (a2)* = a^a^a'^a^ = a2+2+2+2 = ^8. Ex. 3. (a2)6 = a-ixo = qjIj . conversely, a^xc = {a^y. Proof (a**)" = a'^a*" ••• to ?i factors by notation — ^m-f-m-f •••ton sunimandg g i^Ct = a"•^ 119. Power of a product. Tlie nth power of the product of two or more factors is equal to the product of the nth powers of those factors ; and conversely. That is, {aby = a"b", (abc)" = a"b"c". Ex.1. {ahcY = {abc){ahc){ahc) by notation = {aaa) (bbb) (ccc) by laws (A') , (B') = a363c3, by notation 109 110 ELEMENTS OF ALGEBRA Proof. {aby'=(ah) (ab) ••• to n factors by notation = (aa ••• to n factors) (bb -■' to n factors) by (A')^ (B') = a^"". by notation Similarly for any number of factors. Ex. 2. (— «)^ = (— 1)^«^= — «^ ; conversely, (— \ya^ =(— a)^. Ex. 3. (2 x'-yy = 2i(x-2) V = 16 x^y* ; conversely, 2*(x'^yy'^= (2 x~yy. Ex. 4. (- 5x'^y^y= (- 5)3(x2)3(?/3)3^ _125iK6^9 Exercise 41. Express as a power or as a product of powers each of the following powers of products : 1. (-a^y. 7. (aa^y. ' 13. {ab-a^yy. 2. {-xy. 8. {-Q^y. 14. (-xh/'zy. 3. (-/)l 9. (-aa;2)^ 15. (-2a5V)^ 4. (-z'f. 10. {-bfy. 16. (-2a26c«)^. 5. [(-a)^^ 11. (-2aary. 17. {-^Ci'x^yy. 6- [(-2)^'. 12. (-a2ic3)^ 18. {-a?Wxyy. 19. (- a)^ (- ay, (- a)^ (- d)\ (- a)«, (- a)^ 20. (- aby, (- a5)3, (- aby, ( - a6)^ (- aby. 21. (-2 a3z>4)2^ (_ 2 a354^^3^ (_ 2 a^by. 22. As a power of tlie base 3^, express 3^ 3«, S^^^ 3^0, S^-. 23. As a power of the base x^, express x'^, x^, x^, x^^, x^. 24. Asa power of the base a?, express a^, a}^, a}^, aP-, oF*. Express as a power of a product : 25. 6^x4^ 27. {-ay{-by 29. {-xy%f{-zy 26. 4^x(— 3)^ 28. (—ayb^ 30. a\x-\-yy. POWERS, PRODUCTS, QUOTIEXTS 111 120. Square of binomials. By multiplication, we obtain (a + 6)- = fl2 + 2a6 + 6-. (1) Tliat is, the square of the sum of tivo numbers is equal to the sum of their squares plus twice their product. Ex. 1. (3 .X 4- 5 vY = (3 xy + (5 y^ + 2 (3 x) (5 y) by (1) = 9 x2 + 25 y^ + 30 xy. § 1 19 Ex.2. (2x-Syy=[{2x-)-{-(-Zy)Y = (2x)2+(-3y)--^ + 2(2x)(-3y) (2) = 4 a;2 + 9 2/2 _ 12 xy. (3) Ex.3. (a_6)2^[«+(_5)-|2 = a2+(-^)2 + 2rt(-6) = a2 - 2 ah + b^. In the examples of this chapter there are two steps : First step. The application of an identity. Second step. The simplification of the result obtained by the first step. E.g., in example 2, the application of identity (1) gives the result in (2), and the simplification of this result gives (3). At first the pupil should write out these steps separately ; later he should apply the identity mentally, and write only the final result. Observe the advantage gained in this chapter by regarding a poly- nomial as a sum. 121. Square of polynomials. If in the identity {a + x) - = a''^x' + 2ax (1) we put b -{-y for x, we obtain (a -h 6 -f /)- = a' ^(b-{-yy + 2a(b+y) = a- + b- +/ + 2 a6 4- 2 a/ 4- 2 6/. (2) And so on for a polynomial of any number of terms. 112 ELEMENTS OF ALGEBBA Hence we infer that The square of any polynomial is equal to the sum of the squares of its several terms, plus the sum of the products of twice each term into each of the terms which follow it. Ex. (x2 + 2?/-3c)2 ^(x2)2 4.(2?/)-^+(-3c)2 + 2a;2(2|/) + 2x2(-3c)+2(22/)(-3c) = X* + 4 2/2 + 9 c2 + 4 ic^?/ - 6 cx2 - 12 cy. Exercise 42. Write the square of each of the following expressions : 1. 2a + x. 18. a«-2?/^ + 3c«. 2. X' + y-. 19. ^x^ — Qx — Q. 3. 3a- 5 6. 'ZO. x + y + z + v. 4. ^d^-Wx. 21. x + y — z-v. 5. -2a2 + 56l 22. x-y-z-v, 6. -aa^^ + V- 23. o.-^ + a^^ - 2 cc - 2. 7. 3a6c-4a;V 24. a + 26-3c + 4:d 8. -2 2;2-a&iKl 25. 1 + a) - a^^ _|_ ^^ 9. -J^a? + ?>cz\ 26. 3a2-a;2 + c--2/. 10. a + h + c. 27. 4x2_3(^_4c_32/2. 11. a + &-c. 28. 3ir^-2a2 + 4 6-2A'. 12. a-6 + c. 29. 8 o.-^/ - 4 a^^s. 13. a-h-c. 30. |-a^2/^ + f^y. 14. a + 26 + 4. 31. 2x-'»-7. 15. £c + 2i/ + 3;2. 32. ijx-y^-4.x^y\ 16. 2 4-2a;-3a^. 33. 4 a'-^^ - 3 a'-'^^H 17. 2a^-3a;-2. 34. 2(a + 1) - 5(6 + c). POWERS, PRODUCTS, QUOTIENTS 113 122. Product of sum and difference. By multiplication, we have (a+b)(a-b) = a'-b\ (1) That is, tJie product of the sum and the difference of the same two numbers is equal to the square of the first number in the difference J minus the square of the second. Ex. 1. (2x-^ + 5 6y3)(2x2-6 6y8) = (2a;2)2_(5 6?^)2 by (1) = 4a:4_25&2y6 §§119,118 By properly grouping terms, the product of two poly- nomials can frequently be written as the product of the sum and the difference of the same two numbers. Ex. 2. (a + 6 + c){a + 6 - c)=[{a + h)+e] [(a + 6)- c] = (a + 6)2-c2 by(l) = a2 + 2a& + Z>2 -c2. Ex. 3. (a + 6-c)(a-6 + c) = [a + (6 - c)J [a -(6 - c)] = a2 - (6 - c)2 = a2 - 62 4. 2 6c - c2. Exercise 43. Write each of the following indicated products : 1. (b + a){a-b). 5. (x' + 4.if){x'-^f). 2. (5 + a;)(a;-5). 6. (3 or^ + 5 r') (3 ar' - 5 /). 3. (l4-3ic)(l-3«). 7. {^by-\-2ax)(2ax-Sby). 4. (62 + a2)(a2_62). 8. (4ca^ + 5 6-V)(4car-5i2^). 9. (a 4- 5 + c) (a - ft - c). 10. (l + 6_c)(l-6 + c). 11. (a-6 4-c)(a-ft-c). 12. (a; + 32/-22;)(.^•-3?/-f-2^). 13. (or^ + a^?/ + /) (x" -xy + f). 14. (2r' + 2/ + 2)(/-?/ + 2). 114 ELEMENTS OF ALGEBRA 15. (3a + 6-3c)(3a-6 + 3c). 16. (a2 + 3a-l)(a2-3a-l). 17. (a^-2a'-\-l)(a'-{-2a'-l). 18. (a^ - 62 _ c2) (a2 -{- b^ -\- c^). 19. (_a^_/ + 7)(a^-2/^ + 7). 20. (ax — hy-\- cz) {ax + hy — cz). 21. (3a; + 9-42/)(3a;-9+42/). 22. (l4-4a; + 32/ + 22)(l+4ic— 32/-20). 23. (a) + 22/ + «-6)(a; + 22/-a + &). 123. By multiplication, we obtain {x + a){x^b) = X' + (a + 6) jr + ab. That is, ^/ie product of two binomials having the same first term is equal to the square of the first term, plus the sum of the second terms into the first term, plus the product of the second terms. Ex. 1. (x + 7) (x + 5) = aj'-^ + (7 + 5) X + 7 X 5 = a;2 + 12 X + 35. Ex. 2. (a: - 7) (a; - 5) = [X + ( - 7)] [a: + ( - 5)] = x2 + (-7-5)x+(-7)(-5) = x2 - 12 X + 35. Ex.3. (a; + 7)(x-5)=x2 + (7-5)a; + 7(-5) = a;2 + 2 a: - 35. Ex.4. (x-7)(a; + 5)=a;2_2x-35. Exercise 44. Write each of the following indicated products : 1. (x4-8)(a; + 5). 4. (a; - 4) (aj + 11). 2. (a; - 3) (a; + 10). 5. (a + 9) (a -5). 3. (a; + 7)(a;-9). 6. (a-8)(a + 4). POWERS, PRODUCTS, QUOTIENTS 115 7. (a -6) (a + 13). 9. (a - 9 6) (a - 8 6). 8. (x-3a)(x + 2d). 10. (3x -2y){Sx -\- y). {Sx-2y)iSx + y) = (Sxy-^(-2ij + y)Sx-^(i-2y)y = 9x^-3xy-2y-2. 11. (a -5 6) (a + 10 6). 17. {xy - 7 ab) (xy - 2 ab). 12. (x'-6)(a^ + 4:). 18. (x-4:ab)(x-h5ab). 13. (a2 + 2a;)(a2-5ic). 19. (xz - 9 ab) {xz -\- 11 ab). 14. (a;?/ - 9) (a.-2/ + 6). 20. (a" + c) (a" - 6). 15. (xy-6ab)(^xy-\-2ab). 21. (a;''+i-3)(ic"+i-8). 16. (ab-5)(ab-\-7). 22. (ic2«-i-6-)(a.-2"-i + c-). 23. (iB2 + 42/ + 42)(a:2_5y_52)^ Regarding 4(y + 2) and - 5(?/ + z) as the second terms, we have [x2 + 4(2/ + 5!)][x-i - 5(?/ + 2)] =x* -(y + ^)x2 - 20(?/ + z)'^. 24. (a; + 2/ + 3)(aj + ?/-5). 26. (x'-y - 9)(a;-y + 8). 25. (a + 6-7)(a + 6-8). 27. (a-4-2 6)(a+6-26). 124. Cubes of binomials. By multiplication we obtain (fl + bf = fl^ + 3 a-6 + 3 a6- + 6^ (1) That is, ^/ie CM6e of the sum of two numbers is the cube of the first, jjIus three times the square of the first into the second^ plus three times the first into the square of the second, plus the cube of the second. Ex. 1. (2x + 3?/)3 = (2x)3+3(2x)2(3y)+3(2x)(3y)2 + (3y)3 by (1) = 8 x3 + 36 x^y + 54 xy'^ + 27 y\ Ex. 2. (2 X - 3 a)3 = [(2 X) + ( - 3 a)]3 = (2 x)8 + 3(2 x)2(- 3 a) + 3(2 x)(- 3 a)2 + (- 3 a)* = 8 x3 - 36 x2a + 64 xa^ - 27 a^. Ex. 3. (a - 6)8= a3 - 3 a26 + 3 ab'^ - b\ 116 ELEMENTS OF ALGEBRA Observe that when the second term has a negative numeral coeffi- cient, each even term in the result contains an odd power of this coefficient, and therefore has a negative numeral coefficient. The same is true in § 120. 125. The operation of raising a number to any required jjower is called involution. Exercise 45. Write out the cube of each of the following expressions : 1. aj-f-l. 6. a — 2 b. 11. 2aa^-\-7n^n. 2. 2x-j-a. 7. ax + hy. 12. i a V - | 6 Y 3. a + 3 6. 8. 2 a — 3 6c. 13. 2 a;" + 5 2/". 4. i» — 1. ^. x^-\-^a. 14. 3.'c'"?/" + al 5. 3ic — a. 10. xy — 4:ab. 15. x"b — 3ay''+\ 126. Powers of sums. By multiplication we obtain (fl-f-6y = a^ + 4a«6+ 6a'b'-{^ ^ab' + b\ {a + 6)^ = a^ + 5 a'b + 10 a'b' + 10 a'b' + 5 ab' -h b'. {a + 6/ = a« + 6 a'b + 15 a'b' + 20 a'b' + 15 a'b' + 6 ab' + b\ The expression obtained by performing the indicated operation in (a-\-by is called the expansion of (a + by. Thus, the second member of each of the above identities is the expansion of its first member. By inspection we discover in each of these expansions the following laws of exponents and coefficients: (i) The exponent of a in the first term is equal to the exponent of the binomial, and it decreases by 1 from term to term. (ii) The exponent of b in the second term is 1, and increa^ses by 1 from term to term. POWERS, PRODUCTS, QUOTIENTS 117 (iii) The coefficient of the first tenn is 1, and that of the second term is the exponent of the binomial. (iv) If in any term the coefficient is multiplied by a^s exponent, and. this jwoduct is divided by 6's exponent plus 1, the result is the coefficient of the next term. E.g., in the expansion of (a + 6)^, from the second term 5 a*6, by (iv), we obtain 5 x 4 -=- 2, or 10, which is the coefficient of the third term. From the third term 10 a^'fe'^ by (iv), we obtain 10 x 3 -4- 3, or 10, which is the coefficient of the fourth term ; and so on. It can be proved that the above laws hold for any power of a binomial. In the expansion of (a + 6)* there are 5 terms, each of the fourth degree in a and 6, and the first two coefficients are repeated in inverse order after the third term. In the expansion of (a + 6)^ there are 6 terms, each of the fifth degree in a and 6, and the first three coef- ficients are repeated in inverse order after the third term. Observe that in each of the above expansions : The sum of the exponents of a and b in any terra is equal to the exponent of tlie binomial. The number of terms is equal to the exponent of the binomial plus 1. The coefficients are repeated in the inverse order after passing the middle terra or half the number of terms, so that the coefficients of the last half of the expansion can be written out from the first half. Each expansion is homogeneous in a and h. Ex. 1. (2x+3 6)4 = (2 x)4+4(2 a;)8(3 6) +6(2 x)2(3 6)2+4(2 a;)(3 6)8+ (3 6)* = 16 X* + 96 x36 + 216 a;262 ^ 216 x62 + 81 6*. Ex. 2. {2x-ay = (2 x)8 + 6(2 xy{- a) + 10(2 x)X- a^ + 10(2 a;)2(- a)3 + 5(2x)(-a)* + (-a)6 = 32 x6 - 80 xowers of any tivo numbers, as a and b, is exactly divisible by the difference of the numbers, taken in the same order ; the laws of exponents and coeffi- cients in the quotients being as follows : (i) The exponent of a in the first term is 1 less than the exx>onent of a in the dividend^ and it decreases by 1 from term to term. (ii) The exponent ofb is 1 in the second term and increases by 1 from term to term. (iii) The numeral coefficient of each term is +1. The number of terms is equal to the exponent of a in the dividend. Or stated in symbols the theorem is °"~f = a" ^+ a"--b + a"-^b'' -f ... + ab"-'-+ b"-\ (1) a — b where n is any positive integer. Proof Multiplying the second member of (1) by the divisor a — b, we obtain the dividend a" — 6". Hence (1) is an identity (§ 83). Ex 1 27 a'^-/>'^ _ (.^> «)«-?>« 3a -6 (3a)- 6 = (3a)2 -f (3a)6 + 62 = 9rt2 + 3a?) + &2. Ex. 2. •^?-^^^^l^^^^= (2 a)4 + (2 a)8(3 a-) + (2 a)2(3 a;)2+ 2 a(3 x)8 + (3 a-)4 (2 a) -(3 a;) = 16 a< + 24 aH + 36 a%2 + 54 ax^ + 81 «*. Observing that each term in the dividend is the fifth power of the corresponding term in the divisor, we write the quotient by taking the proper powere of 2 a and 3 x. The quotient is homogeneous when the dividend is homogeneous. 120 ELEMENTS OF ALGEBRA Exercise 47. Write each of the following indicated quotients : , ^x^-4.f ^ a^-16 ,, 8aV-3437>3 12. 3. -^ 8. ^- 13. ^^ ^^^ ^^^ g^ .^^ . ^^^ 9a;^- -4/ 30^^- -22/ 81a* -16 6« 9a2 -46« xY- -a« xy- ■ a 64aV -8a^ 4a- -2a; l--7292/^ a*- 16 a — ■2 81 a^ '-1 3a -1 a;^- 32 iC — • 2 32 or' '-1 2x -1 243 < :i^-32 h' 1-92/ 3a-26 130. By division, we obtain : ^^^ I (a* - 6*) -- (a + 5) = a^ - a-6 + ^^^ - h^. {a^ 4- 53) ^ (f^ ^ 5) ^ ^2 _ «5 _^ 52. 2aV -Ih 64a«63 -S^f 4a^6 -2x'y ar^"+2 - 4 a;«+i - 2 ^+3_ y,. a,r.+ l_ 2,.. ^4n+4 _ if" ^n+1 _ r {{a + 6^) -- (a -h 6) = a* - a^ft + a^ft^ _ ah^ + &*• From the identities in (i) and (ii) we infer the two fol- lowing theorems: (i) Tlie difference of the like even powers of two numbers is exactly divisible by the sum of the numbers. (ii) The sum of the like odd powers of two numbers is exactly divisible by the sum of the numbers. In each quotientj the laws of exponents and the number of terms are the same as in § 129. The numeral coefficient of any odd term is -f- 1, and that of any even term is — 1. POWERS, PRODUCTS, QUOTIENTS 121 Or stated in symbols, when n is even, the theorem is °"'^f = a"-' - a"-'b + a"-'b' + ab"-' - b"-\ (1) a + 6 and, when n is odd, the theorem is ^"~^f = a"-^ - a"-'b + a"-'b' ab"' -h b"-\ (2) Proof. Multiplying the second member of (1) or (2) by the divisor a -\- b, we obtain the dividend a" — b" or a* + 6^ Hence (1) and (2) are identities (§ 83). j,^ J 16x4-81?/^ _ (2a;)*-(3y)* 2x + Sy 2x + Sy = (2 a:)8 - (2 xy(Z y) -\- (2 x) (3 y)2 - (3 y)» = 8 x3 _ 12 x-2y -1- 18 a-y2 - 27 ys. ^^ 2. i«!^ijt21«je!^=(2a62)2_(2a&2)(6ca;3) + (6cx8)a 2 aft* + C ca:* = 4 a'^b* - 12 aft2cx8 + 36 c'^x\ Exercise 48. Write each of the following indicated quotients : J 1 - a'b ^ ^ x' + l jg a' + 32 ' l-\-ab' ' x + l' ' a-\-2' 2aa; + 32^* * x-\-2' ' a6 + 3 Oa^^-'-lGy^ g 1±A^'. 15 729 + 8 6« 3aj3 + 42/ " * l+2a ' 9 + 2b^ ^ a*b^-a^f' ^Q a^y> + 216y' ^^ a'%'^-\-S2a^ ab^-\-x^f' ' xy-\-6z ' ' a-b^'-^2x a.y-16m« J J (W-hSc^. ^j 16xy-2o6a\ xif + 27n^' ' rt-V + 2c' * 2x^-\-4:d' 6 ?Vzil. 12 tl±l. 18 ^'^'' + ^y" - 19. Make a list of the identities proved in this chapter. 122 ELEMENTS OF ALGEBBA ' 131. The remainder theorem. The result obtained by sub- stituting a for X in any integral expression in x is the same as the remainder arising from dividing the expression by X — a. E.g., dividing 2 x^ — x^ — 6 by x — 2 until the remainder does not contain aj, we obtain the remainder 6, and 6 is what 2 x^ — x^ — 6 equals when x — 2. Again, dividing x^ + a^ by x — a, we obtain the remainder 2 a^, and 2 d^ is what x^ + a^ equals when x = a. Proof Let P denote any integral expression in x. Divide Phj x — a until the remainder does not contain x. Let Q denote the quotient and R the remainder ; then P=q{x~a) + R. (1) Let P]„ (read ^ P for x = a^) denote the value of P when a is substituted for x. Put a for X in (1); then, observing that Q']a{a—a) is zero, and that R does not contain x, we have Pla^R- (2) 132. The factor theorem. If any integral expression in x becomes zero when a is substituted for x, the expression is exactly divisible by x — a. Proof From P]„ =Ri\i% 131, it follows that if P], = 0, the remainder is zero, and the division is exact. Ex. 1. The expression x^ — a^ becomes zero when a is put for x ; hence x^ — a^ is exactly divisible by x — a. Ex. 2. The expression x'^ + y^ becomes zero when — y is put for x ; hence x'^ + y"^ is exactly divisible \)j x — {— y), or x + y. Ex. 3. The expression a" — &« becomes zero when & is put for a ; hence a" — &»* is exactly divisible by a — 5. Ex. 4. When n is odd, a" + 6" becomes zero when — & is put for a ; hence a" + 6" is exactly divisible by a — (— &), or a -\-h. POWERS, PRODUCTS, QUOTIENTS 123 Exercise 49. By § 132, prove that each of the following dividends is exactly divisible by the corresponding divisor : 1. (a^_a._6)-f-(a;-3). 3. (a^ _ 14 a; - 8) -- (a; - 4). 2. (a^-2a;-15)-=-(a;-5). 4. (s^ - Sx" + 4:)^{x-{-l). 5. (2i>:*-3i>^-4:X-\-5)-h(x-l). 6. (x*-2a'a^-\-ia^x-3a*)-i-(x-a). 7. (x^-3a^x'-7a^x-6a'')-ir{x-\-a). When n is odd, by §§ 131 and 132, prove: 8. a;" + a" is exactly divisible by x -{- a, but not by X— a. 9. af — a" is exactly divisible by x — a, but not by x + a. When n is even, prove : 10. x"" — a" is exactly divisible by both x-\-a and x — a. 11. a;** -fa" is not exactly divisible by either a; + a or X — a. 133. The following examples illustrate how the formulas in § 129 or § 130 often aid in writing out the partial quo- tient and the remainder, when a division is not exact. Ex. 1. Divide a'^ -{- b^ by a - b. Adding to the dividend zero in the form — b^ + b^, we have a^ + fe^-a* - &2 + 2 62_^ ^ ^ ^ 2 62 a — b a — b Ex. 2. Divide a^ + 1 by a - 1. Adding to the dividend zero in the form — 1 + 1 , we have a— la— 1 a— 1 Ex.3. ?i±i = 5izii±2 = ^_^. + ^_l+^_. x-^\ x-\-\ x + l Ex.4. «i+3^a»-8 + 11^^3 + 2« + 4+^J- 0-2 a^ + 4 x-2 a;2 + 5 x^l 0.-^ + 3 x-\ y? H- cc" 124 ELEMENTS OF ALGEBRA Exercise 50. Write the partial quotients and the remainders : 1. :._^. 5. t^. 9. t±L x — 2 x-\-l 2. - ■ - - 6. ^^!^1±/. 10. •:?i_±_^. ax — y^ X — a x'^ -{■■ a'^ x^ — a^ X — a x-\- a 4. :^^. 8. ^^ + ^. 12. ^^±^. X — a x-\- a x—2 Simplify each of the following expressions : 13. (l + a)2-(l + a)(l-a). 15. (a + h)\a-h)\ 14. {\-xy+(l+x'){l-x'). 16. (22/-3a)2(2 2/+3ay 17. {x — a){x-^a){x^ + a^){x'^-\-a'^). 18. {x' - a-) (aj« + a«) (a;^ + a^) (a;^ + a^). 19. (a;2 - X + 1) (x" ^x + l){x'-x^ + 1). 20. (x-\-y--\- z) {x + z — y) {y -\- z — x) {x-{-y — z). Write each of the following indicated quotients : 2j a%^'-a^y^ ^^ g^" - 64 a;^»+« ab'^ -x?f' ' a"' - 2 a;"+^ ' 22 64ft^^-729a;^» 512^ _ 729/"+^ 2a2 + 3ar' ' ' b^-^-3y^'+' 23 XV + 128 2^ ^5m^l0n._32a;5n^«+5 it'?/ + 2 * a"*62«, _ 2 a;n^»i+l 2^ a^V"+&^^'" 2g a;^" + (a + &y'"+^ , CHAPTER X FACTORS OF INTEGRAL LITERAL EXPRESSIONS 134. The problem of multiplication is ' given two or more factors, to find their product.' The converse problem, ' given a product, to find its factors,' is the problem of factoring. Reread §§ 33, 117. Certain forms of products which frequently occur are called type-forms, as a- + 2 ab + 6? or a- — b\ 135. Any monomial is readily resolved into its factors. E.g., the factors of 5a; (a + y) are 6, x, and a -\- y. The factors of xy are a; and y or — a; and — y ; but we usually use the factors x and y because of their simpler form, unless there is some special reason for using — x and — y. Again, the factors of x^ are x and x or — x and — x ; that is, x^ is the square of x or — a:. 136. The converse of the distributive law is ax + bx -\- ex -\- " =(a -{- b -h c -h'--)x. (1) Hence, anjj factor ichich is common to all the terms of a polynomial is a factor of the j^olynomial. Ex. 1. Factor 3 ax^ + a^x - 9 a^x^. Here .3 ax is seen tp be a factor of each term ; hence Zax^-\-6a^x-9 a^x^ = a;(3 ax) + 2 a (3 ax) + ( - 3 a*a;*) (3 ax) = (x-\-2a-Sa-^x^)Sax. Hence the required factors are 3, a, x, and x + 2a — S a^7?. In identity (1) the letters x, a, 6, c, ••• can stand for any binomial or polynomial. 125 126 ELEMENTS OF ALGEBBA Ex.2. ractora;(« -3 6)-2?/(a -35). The binomial a — 3 & is a factor of each term ; hence x{a-Zh)-2 y{a -^h) = {x-2y) {a - 3 6). Ex.3. y -x-2a{y - x)^\{y -x)-2a{]} -X) = {\-2a){y-x). Ex. 4. Eactor {y - x){a^ ^h)-2(y- x)(a^ - b). The expression = (a2 _]_ ft) (^ _ a;) _ 2 (a^ _ ft) (^ _ a;) = [a2 + ft_2(«2_ft)](y_a;) = (Sb-a'^){y-x). Ex. 5. a2(n - a;) - 6-(.x - «) = a2(w -x)+ b'\n - x) = («2+ ft'2)(^ _ x). Exercise 51. Factor each of the following expressions : 1. 3 a; + 3. 13. 2 a'*^/" + 6 «"+y +^ 2. a^ + 5 a;. A71S. 2, a% ?/". 1 + 3 a^ 3. ab-\- be. 14. aa;"*+y+^ -f bx'^+^y''+\ 4. 4 a^ - 6 a^d. ^l^^s. 3^"*+^ 2/"^^ « + bxy. 5. 2ax-\-3x\ 15. 6 ?/'«+-- 3 2/^ 6. 7a«-21a2?). I6. 8x2"-4a;». 7. a^-5a^22/ + 20ar^/. 17. 7a;-+i-14a^. 8. 5 aa^ - 10 a^x _ 5 a^a^l 18. x(a + 1) - y(a + 1). 9. 38a«6«-57a^6l 19. y (x - a) - x + a. 10. 3a3&-6a2&2 + 9a263. 20. y (a? - a) - (a - a^). 11. 15 a^^ - 10 a^c + 5 aU 21. 4 (a^ + 1)^ - 6 (a^ + 1). 12. H a'x - 4: a'y- 12 a'b'. 22. x(y -bf - c(b - y). FACTORS OF EXPEESSIONS 127 137. Trinomials of the type-form a- + 2 a6 -f b\ The converse of the identity in § 120 is a^ + 2 a6 + 6' = (a + by. (1) That is, a trinomial, tivo of ivhose terms are the squares of two numbers respectively, and the remaining term is twice the product of these numbers, is equal to the, square of the sum of these numbers. Ex. 1. Factor 9 x2 + 24 x + 16. 9 x'^ is the square of 8 a;, 16 is the square of 4, and 24ic = 2.3x.4. .-. 0a;2 + 24x + 16 = (3ic + 4)2. (1) Or, 9 x2 is the square of — 3 x, 16 is the square of — 4, and 24a; = 2(-3x)(-4). .-. 9x2-f 24x+16 = (-3x-4)2. (2) The factors in either (1) or (2) are correct, but unless there is some reason to the contrary we usually take the simpler factors given in (1). Ex. 2. Factor 36 «» + 6* - 12 a^b^. 36 a* is the square of 6 a- or — 6 «-, and b* is the square of b^ or To obtain the term — 12 a^b^ we must take either 6 a^ and — b^ or - 6 a^ and 6- ; that is, - 12 a'62 = 2 . 6 rt2(- 62)^ or 2(- 6 a2)ft2. .-. 36 a* + 6* - 12 a262 = (6 rt2 _ ^2)2^ or ( - 6 a2 + 62)2. Any polynomial which is to be factored should be first examined for any factors common to all its terms. Ex. 3. - 3 a^ 4- 3 a^b^ - lb a'^b^ = - 3 a3(a2 - 10 ax^ + 25 6«) = - 3 «3(a - 5 63)2. In identity (1), a and 6 can denote any binomial or polynomial. Ex. 4. (x-2y)2 + 2(x-2y)(3?/-2x) + (3y-2x)2 = [(X - 2 2/) + (3 y - 2 x)]2 - (y _ x)2. 128 ELEMENTS OF ALGEBRA Exercise 52. Factor each of the following expressions : 1. a2 + 6a + 9. 13. 4.xY-x^-^y^. 2. ic2_|_i2ic + 36. 14. Sx--4.x'-4:. 3. a'- + 25 + 10a;. 15. xY -\- x\j -^ \ xf. 4. x^ -\- 121 - 22 X. 16. 4.a'x^-\-4:abxy-{-by. 5. a2 + 49-14a. 17. 9 a" + 25 b'^ - 30 ab. 6. a2_|.25-10a. 18. 25 a'^a^ - 30 a^b-x + 9 b\ 7. l-8a; + 16ar^. 19. 25a^ + 25b' - 50a'b. 8. 4a2 + 962_12a6. 20. a^ + 25 ?>« - 10 a5^ 9. 9 a* + 24 a^d^ + 16 6*. 21. ^a' + ^^b^ - ^^a^b. 10. x^-^^y^-^xy. 22. 4 a.y - 4 a;y + ar^. 11. 5a^-10a-6 + 56l 23.* (a -\- bf -\- 2 {a + b) + 1. 12. a^ - 6 a^^ + 9 afel 24. (2x-a)2-8(a-2a;)4-16. (2 X - rt)2 - 8(a - 2 x) + 16 = (2 X - «)2 + 8(2 a; - a) + 16. = (2x-a + 4)2. 25. (aj^4-2aj.'?/ + /)« + (a; + ?/)5l (a;2 + 2 x?/ + ?/2)a + (x + j/) &2 = (x + ?/)2a + (x + ?/)&2. = (ax + ay + &2)(a; + 2/). 26. x\x + 2)-\-2(x-{-2y + 2x{x + 2). 27. 7?i^ 4- 2 m?i + n^ —p {711 + n) . 29. x"" -\-2 x'^y"' + 2/^"*. 28. a{b-c)-Q)'-2bc+c'). 30. 36 a;'^+2 _ 43 ^«+i _l_ Ig ^n 138. A perfect square which contains only two different powers of some one letter can often be reduced to the type^ form o? -{-2ab -\-W by first writing the polynomial in de- scending powers of that letter. FACTORS OF EXPRESSIONS 129 Ex. 1. Factor x^ -h y^ + z- -{- 2xy - 2xz - 2yz. The expression contains only two different powers of x ; hence, we arrange the expression in descending powers of x, as follows • The expression = x^ + 2x(y-z)-\-(y^-\-z'^ — 2 yz) = x^ + 2xiy-z) + (y-zy = (X + y - zy. We could have arranged this expression in descending powers of y or z. Ex. 2. Factor a* + 4 6* + 9 c* + 4 a-b^ - a-c^ - 12 b^d^. Arranging the expression in descending powers of a, we have The expression = a* + 2a^(2 b'^ - 3 c2) + (4 6* -f 9 c* - 12 b^c^) = a*-\-2 a2(2 b^ - 3 c^) + (2 62 - 3 c2)2 = (a2 + 2 62 _ 3 c2)2. Ex. 3. Factor a6 _ 2 a^ + 3 a* + 2 a\b - 1) + a^Cl - 2 6) + 2 a6 + 62. The expression contains only two different powers of 6 ; hence, we arrange it in descending powers of 6, as follows : 62 + 2 6(a3 _ a2 + a) + (a6 _ 2 a5 4. 3 a* - 2 a3 + a2). This expression is a perfect square, if its last term is the square of a' - a- + a. By § 121, we have (a« - a2 + rt)2 = a« - 2 a^ + 3 a* - 2 rt'' -f d^. Hence the given expression is identical with 62 + 2 6(a8 - a2 + o) + (a^ - a'^ + a)2, or (6 + rt3 _ ct2 4. a)2. Exercise 53. Factor each of the following expressions : 1. c^-6c{a-\-b)-\-9(a + by. 2. a* + 6--|-4c2 + 2a6-f-'4«f + 46c. 3. 4a2 4-624-9c^ + 66c-12ac-4a6. 4. 4 a* + 6* + c-* - 2 6-(r - 4 aV + 4 a-ft^. 130 ELEMENTS OF ALGEBBA 5. a^ -\- 4: y^ + 9 z^ + 4. xy + 6 xz-^ 12 yz. 6. 25 a* + 9 &^ + 4 c^ - 12 b'c' + 20 c^a^ - 30 a'b'. 7. 6 aca^ 4- 4 5V + aV« + 9 c- - 12 ^ca^^ - 4 abx\ 8 . - 6 ft^c^ 4- 9 c4 + 6^ _ 12 c^a^ -f 4 a^ + 4 a-6l 9. 6 aft^c _ 4 a^ftc + aW' + 4 a^c^ + 9 b\^ - 12 abc\ Note. The products in exercise 53 can be factored by using the converse of § 121. 139. Trinomials of the type-form jr- 4- yojr + q. The converse of the identity in § 123 is X' + (a -\-b)x-^ab = (x + a) (x + b). (1) Any trinomial in the form x' +^9a; 4- g can be written in the form x^ -\- (a -{- b) x -{- ab and factored by (1), when we know the two factors of q whose sum is p. The two factors of q whose sum is p can often be found by inspection as below : Ex. 1. Factor x^ + 7 x -{- 12. Here p = 7 and q = 12. The two factors of + 12 are both +, or both — ; hence, as their sum is + 7, both are +. The pairs of positive whole numbers whose product is 12, are 12 and 1, 6 and 2, 4 and 3; since 4 + 3 z= 7, 3 and 4 are the two factors of 12 whose sum is 7. .-. a;2 + 7 a; + 12 = x^ + (3 + 4)a; + 3 x 4 = (x+3)(x + 4). by (1) Ex. 2. Factor x^-9x + 20. The two factors of + 20 are both + or both — ; hence, as their sum is — 9, both are — . The pairs of negative whole numbers whose product is 20 are — 20 and — 1, — 10 and —2,-5 and — 4 ; since (— 5)+(— 4) = — 9, — 5 and — 4 are the two factors of 20 whose sum is -9. ,, a;2-9a; + 20= x2 +(- 5 - 4)a: + (- 5) -(-4) = (a:-6)(x-4). by (1) FACTORS OF EXPRFSSIONS 131 Ex. 3. Factor x^ + 6 x - 27. The two factors of — 27 are opposite numbers ; hence, as their sum is + 6, the positive factor is arithmetically the larger. The pairs of whole numbers whose product is — 27, the larger arithmetically being + , are 27 and — 1,9 and — 3 ; since 9 + ( — 3) = 6, 9 and — 3 are the required factors. .-. x2 + 6 a; - 27 = x2 + (9 - 3)x + 9 . ( - 3) = (x + 9)(x-3). by(l) Ex. 4. Factor a^x^ - 6 ax - 84. The two factors of — 84 are opposite numbers ; hence, as their sum is — 5, the negative factor is arithmetically the larger. The pairs of whole numbers whose product is — 84, the larger arithmetically being -, are - 84 and +1, - 42 and +2, - 28 and + 3,-21 and + 4, - 14 and + 6,-12 and 7 ; since - 12 + 7 = 5, - 12 and + 7 are the required factors. .-. (ax)2 - 5(ax) - 84 = (ax - 12) (ax + 7). Ex. 5. 9 x* - 12 X - 77 = (3 x)2 - 4(3 x) - 77 = (3x-ll)(3x + 7). Ex. 6. Factor x^ - 32 xy - 105 y^. The two factors of — 105 y^ whose sum is — 32 y are 3 y and — 35 y. .-. x2 - 32 xy - 105 y'^={x + 3 y)(x - 35 y). Ex.7. 4a-a2 + 21=-(a2-4a-21) = -(a-7)(a + 3) = (7-a)(a + 3). Exercise 54. Factor each of the following expressions : 1. a^ 4- 4 a; + 3. 6. ar + 2a; — 3. 2. a^ — 4 a; -1-3. 7. a:^-fa; — 6. 3. a.-2-h9a;-|-20. 8. a^-|-4a;-5. 4. a.'2_lla;-f-18. 9. a^4-2a;-35. 5. x^-Sx + 15. 10. x'-Sx-lO. 132 ELEMENTS OF ALGEBRA 11. x-x'-i-Q. 35. 4a?2_l2ic-91. 12. a^ + 5a; + 14. 36. oc^-20xy-96y\ 13. ar^ + 18i»4-72. 37. x" - 26 xy + 169 y^ 14. a;-ic2-M32. 38. a^- 23 0^2/ + 132 2/'. 15. a^-5i»-84. 39. 4 a^ + 20 a??/ + 21 2/1 16. a^ + 5aj-150. 40. 9 of - 39 xy -{- 22 y\ 17. a;2 _ 25 a; + 150. 41. ar' + 43 a;2/ + 390 2/^^. 18. a^2 + lla!-180. 42. a" -20 abx-{- 75 b^x\ 19. a;-ar^-f-156. 43. a^ - 29 a& + 54 ftl 20. a;2_3ia; + 240. 44. 130 + 31 a;2/ + a^2/^. 21. a;2_34a;4.288. 45. a^ + 12 a6aj - 28 ?/-a.'l 22. a.-2-35a;-200. 46. .'c^ + 13 aV - 300 «^ 23. ar*- 17 a; -200. 47. x^ - a'x^ - 462 a\ 24. aV-21aa; + 108. 48. a;^ - a^a^ - 132 a*. 25. aV - 21 aa; + 80. 49. 143 - 24 a;a + a^a^. 26. a^^ + 21 aa; + 90. 50. 216 + 35 a? + a^. 27. a^x^ - 19 aa; + 78. 51. e5 + Sxy- xy. 28. aV + 30aa? + 225. 52. 110 - a? -a;^ 29. a^x^ + 54 aa;+ 729. 53. 98-7a?-a^. 30. aV - 38 aa; + 361. 54. 380-aj-a^. 31. a^y^-5xy-24:. 55. 120 - 7 aa; - a^a^. 32. 4a:2^i2aj-55. 56. 105 + 16cy-cy. 33. 9a^ + 6a;-35. 57. (a; + 2/)^ + 6 (a; + 2/) + 8. 34. 16a^ + 8a;-15. 58. (a- 6)2+8 (a- 6) +15. 140. Trinomials of the type-form ax^ -\- bx -\-c. Multiplying and dividing ax^ -{-bx + chj a, we obtain aa^ + &a; + c = \_{axy + h (ax) + ac] -4- a. (1) FACTORS OF EXPEESSIONS 133 By § 139, the trinomial in brackets can be factored by finding the two factors of ac whose sum is b. Ex. 1. 3 ic2 _ 16 X + 5 = [(3 a;)2 - 16(3 x) + 15] -- 3 = (3a;- 16)(3x- l)-3 §139 = (x-5)(3x-l). Ex. 2. 6 x2 4- 32 a; - 21 = [(5 x)^ + 32(5 x) - 105] -- 5 = (5 a; + 35) (5 X - 3) ^ 5 = (a; + 7)(5a;-3). Ex. 3. 3 x2 - 1*7 xy + 10y^ = [(3 x)^ - 17 y(3 x)+ 30 2/2] ^ 3 = {x-6y)(Sx-2y). Exercise 55. Factor each of the following expressions : 1. 2x^-\-3x-{-l. 13. 3a^ + 13a;-30. 2. 3a^ + 5a;-h2. 14. Qa^-{.7x-S. 3. 3a^ + 10a; + 3. 15. 3 a^x^ + 23 ax -\- U. 4. 3a^ + 8x + 4. 16. 3 a^o.-^ + 19 oa; - 14. 5. 2ar^ + 7a;H-6. 17. 6 a'x'' - 31 ax -^ 35. 6. 2a^ + llx + 5. 18. 3a^ + 41a; + 26. 7. 6ar^-{- lla; + 2. 19. 4a^ + 23a; + 15. 8. 2ic2_^3^_2. 20. 3ar_13a; + 14. 9. 4ic2 + lla;-3. 21. 2a^-5xy-3f. 10. 2xr-{-lox-S. 22. 3a^ - 17a;^ + lOyl 11. 3x^ + 7x-6. 23. 12ar'-23x?/ + 10/. 12. 2ar^ + a;-28. 24. 24 ar _ 29 a;?/ - 4 2/2. Factor each of the following miscellaneous expressions : 25. 2a;(7i-l)-2(l-»). 28. 7 ar^ - 15 a?^ - 18 2/'. 26. c}j"' — ay'^+-+ny"'+\ 29. 5x(a—2y)—2(2y—a). 27. 9 «* + 16 6^ - 24 a2^2^ 30. y?/^ -\-f/^^xy/3. 134 ELEMENTS OF ALGEBRA 31. (a;-3)2 + 4(3-a:) + 4. 37. 12^2 + 50aj - 50. ' 32. 1^2x^ + x — l. 38. aa^ + (a — 6)aj — &. 33. aaj^ + (a + 6) a? + 6. 39. x^^2xy—4:XZ—4.yz+A:z\ 34. i»(a; — a)2 — 2/(a; — a). 40. S?/"'-^— 32/*"+^+42/'"+^. 35. x^-^-^xr+^-5x^^\ 41. (a + 6)2 + 5 (a + 6) -24. 36. 121 ar' + 81/ + 198 a;?/. 42. (a; -?/)'- 4 (a; - 1/) - 21. 141. Binomials of the type-form a" — 6", where /i is even. The converse of the identity in § 122 is a^ - 6^ = (a + b) (a - b). (1) That is, the difference of the squares of any tivo numbers is equal to the product of the sumi and the difference of the numbers. Ex. 1. 9 a%^ - 4 c2 = (3 a%^y - (2 c)2 = (3 a353 + 2 c) (3 a%^ -2 c) by (1) The letters a and h in (1) stand for any expressions. Ex. 2. a2-4ay + 4?/2-9c2=(a-2y)2_(3c)2 = (rt-2y + 3c)(a-2?/-3c) by (1) Ex. 3. 9ic2+12a6-9a2-462 = (3a;)2-(3a-26)2 = (3x + 3a-2&)(3x-3a4-26). In factoring a given expression, it may be necessary to use the same principle two or more times in succession as below : Ex. 4. (X2 - ?/ + ^2)2 _ 4 a;2^2 = (X2 - ?/2 + ^2 ^ 2 XZ) (X2 - ?/2 + ^2 _ 2 OJ^) = [(x + 0)2_y2j[;(a;_^)2_y2j = (x + ;s + y) (x + ^ - y) (a: - ^ + 2/) (x - - ?/) . Whenever n is even, a"" — 6" should be factored as the difference of two squares. Ex. 5. X* - «* = (a;2)2 _ (^2)2 = (X2 + a-2) (X2 - ^2) = (x2 + a^) {X 4- a) (x - a). FACTORS OF EXPRESSIONS 135 Exercise 56. Factor each of the following expressions : 1. a'-9, 6. 9a2-1662. 11. Sab^-lSa\ 2. 2oa'-b\ 7. 81/-9a^. 12. 108 ar^ - 3 a.-^. 3. 16-61 8. 36 ar- 49/. 13. 7a:^-28ar'. 4. a^- 9 2/2 9^ Aa-b''-9c\ 14. 32 a^- 8 arY 5. 64a.'2-49&l 10. 4a^-9a^. 15. 7xyz^-7a^f. 16. a^ + 2ab-{-b'--(^. 23. 49 ar' - 1 + 14 a;i/ + ?/2^ 17. a2-2a64-&'-cl 24. a^ _ ig ^^.2 _|_ g ^^ _^ 9 ^2 18. a2_ j2_25c-c2. 25. ar^- 9 ?/'+ 10 ax + 25 a^. 19. a^-b^-{.2bc-c^. 26. 6" - a- - 4ar^ + 4oa;. 20. ar^ + 4a;i/-a=* + 4/. 27. 9 0^- 4 a^- 9 a^^. 12 aa;. 21. ar^-l + 10ca; + 25c2. 28. Actr - y- -9z^ -\-Qyz. 22. l-{-2ab-a'-b\ 29. c^ - 25a2_ 952 ^30a6. 30. a^ + b''-{-2ab-c^-(f-2cd. 31. a2 4-6--2a6-a^-/-2ajy. 32. m^-^ir-2mn-a^-b--{-2ab. 33. a2_,_,i2_2rt,i_^>2_^^2_2 5,,i. 34. 16a2 + 8aa; + ar'-26v-62-2/l 35. 9a'-\-12ab-\-4.b^-(c-{-x-2yy. 36. (a4-2' + c)2-ar-?/" + 2a;?/. 37. (x + Syy-4:f. 39. (5 a^ + 2 y)^ - (3 a; - ?/)2. 38. 9a2-(3a-56)2. 40. (2a;+a-3)2-(3-2a;)2. 41. ixa* — ^\xb^ 45. 5 — 80 a;*. 49. a^" — /". 42. |aV-fa/. 46. aV - 16 6y. 50. 9.T'*-a;"+2. 43. 16 X* — y\ 47. .^•^" — /". 51.4 ar"+3 — x'^+K 44. a* -81. 48. ar''+2_y2n-2 53. x^''+^y^-a^y\ 53. a;2 4- / + «« + 2 a^ — 2 a.-2; + 2 2/2 — 16. 54. a* + 4 62 + 9c2_ 4a6 + 6ac- 126c -a:2_2a:2/_ 2,2. 136 ELEMENTS OF ALGEBRA 142. Binomials of the type-form a" — 6", where n is odd. When n = 3, by § 129 we have Ex. 1. 243-8 cfi = (7)3 - (2 aY = (7 - 2 a) [72 + 7 (2 a) + (2 a)2] = (7-2a)(49+14a + 4a2). Ex. 2. 125 - 8 a^66 = (5)3 _ (2 a262)8 = (5 - 2 a262) [52 + 5(2 a262) + (2 a^lP')'^'] = (5 - 2 a252) (25 + 10 a262 + 4 a^M). Ex.3. (1 -2a;)8-64x3 = (l-2a;)3-(4x)3 = (l-2x-45c)[(l-2a;)2 + (l-2x)(4a:) + (4x)2] = (l-6ic)(l + 12a;2). When n = 5, by § 129 we have a^-b'={a- b) {a' + a'b + a'b' + ab' + b'). Ex. 4. 2 a^ - 64 65 = 2 [a^ - (2 6)5] ^ 2 (a - 2 6) [a* + a3(2 b) + a2(2 5)2 + a(2 6)8 + (2 6)4] =2(a - 2 6) (a* + 2 a35 + 4 ^252 + 8 aft^ + 16 6*). From identity (1) in § 129, we have a** - 6" = (a - b) (a«-i + a^'-^b -\-a^-%^-\ h ab""-^ + &""0, when n is any positive integer. 143. Binomials of the t3rpe-form a" -f 6", where n is odd. When n == 3, by § 130 we have ^3 + 53= (aH-6)(a2-a6 + 62). Ex.1. 8a:3+27?/3 = (2a;)3 + (3?/)3. = (2 a; + 3 y)[(2 a:)2 -(2 x)(3 2/) + (3 ?/)2] = (2 a; + 3 ?/) (4 a;2 - 6 a;?/ + 9 ?/2). FACTORS OF EXPRESSIONS 137 When n = 5, by § 130 we have a!^^b'=(a-\- b)(a* - a^b + a'b^ - ab^ -\- 6*). From identity (2) in § 130 we have, when n is odd, a" -f 6" = (a + b) (a^^^ - a'^-^b H a6"-2 + 6"-!). Exercise 57. Factor each of the following expressions : 1. a^-1. 9. 216-0)3. 17. 32a« + l. 2. 27 -a^. 10. 27 718 + 1. 18. a^&« + 243. 3. a'-Sb^ 11. 8a^-27a«. 19. 1024 ar"^ - 32 2/^. 4. 125-a»6\ 12. a^6*'-ay8. 20. x'-yl 5. x-^ + l. 13. 40a3-13o6^ 21. x^ -1. 6. y3 + 27. 14. 27n3-f 64c». 22. j;^ -f 128. 7. 8ac» + 64. 15. 2/^-1. 23. 1 - (x + yf. 8. 343- 8 al 16. ar^- 32. 24. x^ - f. When n is even, x" — y» should ^rs^ be factored as the difference of two squares (§ 141). a^ - y« = (a;8 - y3)(a;3 + y3) § 141 = (a; - y)(a;2 + xy + y2)(x + y) (x* - xy + y^). 25. a;«-l. 28. x^f-a^b\ 31. 81aV-l. 26. a«-64. 29. a;*-16 6\ 32. a«-729 6«. 27. »«-64/. 30. 16a;*-81a^ 33. 64ic«-729/. 34. (3 + 2a)3-64. 38. aV - (6 - c)*. 35. a^-(x-{-yy. 39. af^f-(xy + iy. 36. a;^_(a_6)«. 40. 16 a^ - (?/ + 2 2)*. 37. a5«_^3._2 5y. 41. 27 aj^^ - (a + &)«. 138 ELEMENTS OF ALGEBRA 144. A trinomial of the type-form a^ + ha^b'^ + 6* can be factored by writing it as the difference of two squares. Note. The two factors of a* + ha'^lP' + 6* are real and equal when A = 2 ; real and unequal when ^ < 2, and complex when ^, > 2. In all the examples given the factors are real and unequal, but as some of them involve surds this article and the next should be omitted until Chapter XVII. has been studied. Ex. 1. Factor m* + m'^n'^ + w*. Adding mhi'^ — mhi'^, we obtain wi* + n* + m^n^ = m'^ + w* + 2 m^n"^ - mH^ = (m2 + n^y - (mw)2 = (wi2 4- n^ + mn) {m^ + n^ — m?i). Ex. 2. m* - 6 ni^n^ -\- n'^ = m* + # - 2 m^n^ -Sm^ri^ = (m2- w2)2 -(mny/S)^ = (m'^-n^ + 17171 ^S){m^-n^-mn V^) . Or, m* - 5 mhi^ + w* = w* + w* + 2 m2n2 - 7 m'^n^ = (m2 + n2 + mw V7)(to2+ ti^ -mn^Jl). Ex. 3. 4 x* + 9 a* - 21 a2ic2 = 4x* + 9 a* - 12 a2x2 - 9a2x2 = (2x2-3a2)2-(3ax)2 EE (2 a;2 - 3 a2 + 3 ax)(2 ic2-3 a2-3 ax). Exercise 58. "Factor each of the following expressions : 1. x^j^x' + l. 9. 25a3*-44i»2/ + 162/'. 2. .T*-3a^4-9. 10. 4x^-4a^/ + 92/^ 3. x* + 9a^ + 25. 11. ^x''-12x'y'' + Uy\ 4. a;* + 9fl^ + 25. 12. 16 x^ - a^/ + 2/'- 5. a;* - 11 aV + a^ 13. 25 a;* - 29 a^2/' + 4 .v'- 6. a;* + (4-c2)a^2/' + 42/'. 14. x^'-s^y^ + y^. 7. (a; + 2/y4-(a5 + 2/)' + l- 15. aj^ + a^?/' + 2/'- 8. 9aj^ + 3a^?/2 + 42/^ 16. a^+a;y + /. FACTORS OF EXPBESSIONS 139 145. Binomials of the type-form a" -f 6", where n is even. (i) a* + b' = a* + 2a-b^-{-b^-2d'b^ = (a' + by - (ab^2y = (a^ + b'-\- ab-y/2) (a' + b'- aby/2). This method can be employed whenever w is a multiple of 4, as when n is 8, 12, 16, etc. (ii) a« + 6«=(ay + W = {a? + b^)[a^-o?b''+b''] = {o? + b^lia' + by - (a6 V3)T = (a2 + b^ (a« + 6^ 4- a6 V^) («' + ?>' - a6 V3). This method can be employed when n is even and one of its two factors is odd, as when n is 10, 12, 14, etc. Exercise 59. Factor each of the following expressions : 1. x^ + 1. 5. 01? -a\ 9. 3?-\-a\ 2. x^ H- cy. 6. a^ + 1. 10. a^ + 1. 3. nx^-\-a\ 7. a^ + 64. 11. a;»« + ai«. 4. a^-1. 8. af-\-cy. 12. a;" + a". 146. Perfect cubes. The converse of identity (1) in § 124 is a^ + 3a'b-\-3 ab^ + b^ = (a-\- bf. Hence, if the four terms of the cube of a binomial are arranged according to the powers of some letter, their extreme terms are the cubes of the terms of the binomial. E.g., if 64 a» - 144 a'^b + 108 ab^ - 27 68 is a perfect cube, it is the cube of 4 a — 3 6 ; for when its four terms are arranged in descending powers of a, the extreme terms are the cubes of 4 a and — 3 6 respectively. The expression is a perfect cube ; for (4 a - 3 6)8 = 64 a^ - 144 a'^b + 108 ab^ - 27 b^ 140 ELEMENTS OF ALGEBRA If a perfect cube which contains only three different powers of some letter is arranged according to the powers of that letter, its factors will often become obvious. E.g., if we arrange the expression, a^+b^ + c^ + S a% + 3 oT-c + 3 a62 + 3 ac^ + 6 ahc + 3 62c + 3 ftc^, according to the three different powers of a, we have a3 + 3 a2 (6 + c) + 3 a (62 + c2 + 2 ah) + (63 + 3 62c + 3 6c2 + c^), or «3_^3rt2(54.c) + 3a(6 + c)2+(6+c)3, which is seen to be (a + 6 -f c)^. Exercise 60. Factor each of the following expressions : 1. a3 + 3a2-f-3a + l. 3. Sm^ - 12m2 -f 6m - 1. 2. a^ + 6a;2 + 12a; + 8. 4. aV - 3 a V^/^ + 3 aa;^^ - /. 5 . 64 a« + 108 ah^ - 144 a% - 27 h\ 6. a;3- 24 0^2/ + 192 a?/ -512 2/3. 7. o? + 6 a^ft - 3 a^c + 12 aV - 12 a6c + 3 ac'-{- 8 6^-12 6^0 ^_3^2 3^_^ 8a^-4'ar^7/2+ 2a.y4_|!. 842 J -r^ y 21 10. 24 62a.-3_36 6V + 18 6=^a;-3 62. 11. a2_|_2a6 + 4c2 + 4ac + 46c + 6». 12. 2 ax^ + 4 aa;22/2 - 4 aifz" -\- 2 az^ -\- 2 ai/ - 4: ax'z^ 13. 3 6aj4 - 6 hxhj + 12 a^^, _ 12 a^^/ + 3 62/' -f 12 a6ar^. 14. a?x'-^d'x^y^-21a'xi/~21aY' 15. a^ + 3a^2/ + 3 V - 3 aa.-' J^f-^a^f-^ axy + 3 a^2/ -{-Sa'x-a^ FACTORS OF EXPRESSIONS 141 147. Summary. To factor any given expression by the foregoing methods, the pupil should first note whether the expression is in any one of the following forms : (i) A sum of terms having a common factor. § 136 (ii) A perfect power. §§ 137, 138, 146 (iii) A difference of squares. § 141 (iv) The type-form ax^-\-bx-\-c or x^ -\- j^x -\- q. §§ 139, 140 (v) The type-form a" - ?/' or a" + 6", ?i odd. §§ 142, 143 (vi) The type-form a* -f /ia-62 + h* or a" + 6", n even. §§ 144, 145 When a factorable expression has no one of these forms, our first aim is to reduce it to one of them. In this reduc- tion much will in the end depend upon the ingenuity of the student. No definite directions which are applicable to all cases can be given. The two following devices will in many cases prove useful : (i) The factors of an expression will frequently become obvious when the exjJi'ession is aii'cmged in ascending or descending powers of one of its letters^ particularly when the expression contains only one power of that letter. Ex. 1. Factor ax + 6y + 6x + ay. Arranging in powers of x, we have ax -\- hy -\- hx + ay = {a -\- b) x ■{- {a + b) y = (a + b)(ix + y). Ex. 2. Factor ax^ — x — a -\- I. Arranging in powers of a, we have ax3 - X - a -f 1 = (x3 - 1) a - (X - 1) = (x-l)[a(x2-Fx + l)-l] = (x - l)(ax2 + ax -I- a - 1). 142 ELEMENTS OF ALGEBRA Ex. 3. Factor «2 (^x-y)-\^ x^ (?/-«)+ y^ (a -x). Arranging in powers of a, we have the given expression =a'^(x — y)— a (x^ — y^) + xy (x — y) = (.^-y) [«^ -(x + y)a-\- xyl = (x-y){a-x)(a-y). (ii) Another device consists in adding to the given, expres- sion some form of zero ; as, y^ — y^, or —1+1. Ex. 1. Factor x'^ - Sy^ - z"^ -2xy + ^ yz. Arranging in descending powers of x and adding y^ _ y2^ y^^e obtain the given expression =x^ — 2 xy + y^ — (4 y"^ -\- z"^ -\- i yz) ~{x-yy-{2y-zy = (x - y + 2 y - z)(x - y - 2 y + z) = (x + y - z)(x - S y i- z). Ex. 2. Factor x^ - 3 ic + 2. Adding — 1 + 1, we obtain x^-Sx-\-2^(x^-l)-S(x-l) = (x-l)(x^-]-x-]-l-S) = ix-l)(x-l)(x + 2). Ex. 3. Factor x^ -Sx^-{-4. Adding x^ — x^, or putting —2x'^ — x^ for — 3 x"^, we obtain a;3 _ 3 a;2 + 4 = a;3 - 2 x2 - x2 4- 4 = (X - 2) x2 - (x2 - 4) = (x-2)(x2-x-2) = (x-2)(x-2)(x + l). Exercise 61. Factor each of the following expressions : 1. a^ 4- a& -h oc -h be. 4. mx — my — nx -\- ny. 2. orc^ 4- acd + ahc -\-hd. 5. Sax — bx — S ay -\- by. 3. a^ H- 3 a + ac 4- 3 c. 6. 6oi^-{-Sxy — 2ax — ay. FACTORS OF EXPRESSIONS 143 7. aaP — 3 bxy — axy + 3 by^. 8. 2 ao/*^ + 3 axy — 2 6x?/ — 3 by\ 9. ama:^ + bma^ — anxy — bny\ 10. ax — bx -{- by -\- cy — ex — ay. 11. a^x -\- abx -\- ac -\- dby -\- b^y -\- be. 12. iB3 + ar^_4a;-4. 23. 2a;3 - 3ar^- 2a; + 3. 13. 57^-x^-^x-\-l. 24. a:3 ^ 5^ _ ^2^, _ ^25^ 14. aa?-\-b3^-\-a + b. 25. a^d^ _ a.2 - 52 4. 1. 15. aar^ + by^ + {a + b) xy. 26. 6a;^ + ay? + bx-\-a. 16. a262 4_rt2_j_ j2^ 1^ 27. x? — f-Yxz — yz. 17. a* 4- a'62 - 6 V - c^ 28. 1 + &^ - (a' + a6)ar'. 18. a'' — a — c^ -}- c. 29. aV + acd + abc + bd. 19. a^ — 52 _ (a — 6)2. 30. ac-{-bd — ad — be. 20. a2-62 + 5c-ca. 31. a 10a^2/^. 4. a^, x- — 3x. 2. 24a^6V, eOa^dV. 5. 21 o^, 7ar> + l). 3. 9 a'b^xy, S x^y^ 6. 6a^- 2ii-, Ga;^- 3 a;. 7. a^ + 2a;, a;2^3^_^2. 8. aj2-5a; + 4, a;2_6a; + 8. 9. a.-^ 4- 4 a; + 4, a;2 + 5 a; 4- 6. 10. a^-a;-6, a^ + i»-2, a;2_4^^3^ 11. a^ + a;-20, ar^-10a;4-24, a;2-aj-30. 12. a;2_^a;-42, ar^-lla; + 30, a^ + 2aj-35. 13. 2a^ + 3.x' + l, 2a^+5a;4-2, a^ + 3a; + 2. 14. 5a;2+lla; + 2, 5a^ + 16a;4-3, a^ + 5aj + 6. 15. x^ — 7 xy -\- 12 y^, x^ — 6 xy -{- S y^, x^ — 5xy -\-6 y^. 16. 2a;2 + 3a;-2, 2a^ + 15a;-8, a^ + 10a;4-16. 17. 8a^-38x?/4-35/, 4 ar^ - a;^/ - 5 /, 2x'-5xy-7f\ 160. L.C. M. by H. C. F. The L. C. M. of two expres- sions can always be obtained by first finding their H. C. F. Ex. 1. Find the L. C. M. ot x^ + x^ - 2 and x^ + 2 a;2 _ 3. The H. C. F. of these expressions is found to be x — 1. By division we find that JC3 + ic2 - 2 = (x - 1) (x2 + 2 X + 2), and x3 + 2x2-3 = (x-l)(x2 + 3x + 3). Since x — 1 is the H. C. F. of the given expressions, their second factors x2 + 2 X + 2 and x2 + 3 x + 2 have no common factor. Hence, the required L. C. M. is (x-l)(x2 + 2x + 2)(x2 + 3x + 3). (1) LOWEST COMMON MULTIPLE 161 16L To find the L. C. M. of three expressions Af B, C, we find 3f, the L. C. M. of A and B; then the L. C. M. of M and C is the L. C. M. required. Exercise 67. Find the H. C. F. and L. C. M. of : 1. 2a^ + 3a;-20, 6ic«-25a^+21a;-|-10. 2. a^-15aj-f36, a^-3a^-2ar-|-6. 3. c)x^-x-2, 3a^-103^-7x-4:. 4. ix^ + x'-Ax-i, a^-{-6a^-\-llx + (i. 5. o^-x^-jx-^lo, x^ + x^-3x-^9. 6. a^-x^-\-x-\-3, x' + a^-Sx^-x-{-2. 7. x*-x^-\-Sx-S, x' + 4:X^-Sx-{-2A. 8. 6x^-{-x^-ox-2, 6x^-\-5a^-3x-2. 9. 4:a^-10x^-{-4:X + 2, 3a^ -2a^-3x-i-2. 10. x^-9xP-{-26x-24, x'-Ua^ + ATx-eO. 11. x^ — ax^ — a^x -\- a% x^+ax^~ a^x — a\ Find the L. C. M. of : 12. x^-(ya^-\-llx-6, ic^ - 9 o^ + 26 - 24, ar^_8a^ + 19a;-12. 13. a^-5x^+9x-9, x^- x^-dx-\-9, x*-Ax'-^12x-9. 162. The L. C. M. of two integral expressions is the product of either expression into the quotient of the other divided by the H. C. F. of the two expressions. Proof Let A and B denote any two integral expressions, H their H. C. F., and L their L. C. M. 162 ELEMENTS OF ALGEBRA Then L by definition contains all the factors of A, and in addition all the factors of B which are not in A ; that is, the factors oi B-i- H. Hence L = Ax(B-^H). (1) 163. From (1) in § 162 we obtain AxB=LxH. (2) That is, the product of two integral expressions is equal to the product of their L. C. M. and their H. C. F. CHAPTER XII FRACTIONS 164. A fraction being an indicated quotient, the fraction a/6 denotes that number which multiplied by the divisor b is equal to the dividend a. Reread § 90. E.g., - 8/4 = - 2, for - 2 x 4 = - 8. 165. Algebraic fractions. The fractions in arithmetic in- volve only arithmetic numbers, and are called arithmetic fractions. In Chapter III. we used arithmetic fractions to denote the arithmetic values of positive and negative numbers, the quality being indicated by the sign ■•" or ~. An algebraic fraction is one whose numerator and denomi- nator are quality-numbers. The sign before an algebraic fraction denotes the quality of its numeral coefficient. Thus, — ^^^ denotes the product of — 1 and the fraction (-4)/(+3). 166. By the law of quality in division it follows that — Changing the quality of both the nuynerator and denominator does not change the quality of the fraction. Changing the quality of either the numerator or denominator changes the quality of the fraction. ^* -8 8' h -b But ^^ and -^^ are opposite in quality. 6 — b 163 - -8 9 8 ~9' «_ — a b or -- a -b abc_ xyz~ ,(z a)(-6)(- xyz ic). — abc xyz = - (- - abc -x)yz 164 ELEMENTS OF ALGEBRA 167. By § 166 and the law of quality in § 48 it follows that — Changing the sign before a fraction and changing the quality of either its nuynerator or denominator does not change the quality of the term. E.g., Again, 168. A fractional literal expression is an expression which has one or more fractional literal terms. E.g., and ax + by -\ are fractional expressions. x-y a+b An integral literal expression, as we have seen, denotes any integral or fractional number; likewise a fractional literal expression denotes any integral or fractional number. E.g., a denotes 2, 5, 3/2, — 2/3, or any other number. Again, when a = 6 and 6 = 2, a/b = 3 ; when a = 12 and 6 = 3, a/b = 4 ; when a = 3 and 6 = 5, a/b = 3/5 ; and so on. 169. A proper literal fraction is a fraction whose numer- ator is of a lower degree than its denominator in a common letter of arrangement. An improper literal fraction is a fraction whose numerator is of the same or of a higher degree than its denominator in a common letter of arrangement. 1 J* 4- 1 E.g., — = — and "-^-^ are wroper literal fractions. While — - — and — ^— t are improper literal fractions. x-2 + 1 x2 + 3x-4 The value of a proper literal fraction may be either less or greater than 1. FRACTIONS 166 REDUCTION OF FRACTIONS. 170. To reduce an expression is to find an identical expres- sion of some required form. 171. To reduce an improper fraction to an expression con- taining no improper fractions, Perform the indicated operation of division. Sometimes the quotient can be obtained by inspection. For examples, see §§ 129 and 133. Exercise 68. Reduce each of the following improper fractions to frac- tional expressions containing no improper fractions : , 5ar^-20a;-15 ^ -^ , ^ o 1. • 0. • y. ox x + 7 x + 2 x + 3* 5x-\-7 x' + a' x-\-a ar^ + a'^ x — a ^-a' X + a a^ + 16 10. 11. 12. x* + a' X — a 2x2 -7x -1 x-3 a^- ■3x X- -2 Sor + 2x 4-1 x-3 x + 2 x-\-4: 4 g^ + 6 a6 + 9 6^ ^^ 60x^-17x^-4.-^ + 1 2a-3h ' ' 5x2 + 9x-2 172. The value of a fraction is not changed by multiplying its numerator and denominator by the same number. That is, a/b = am/(bm). Proof - = - X -, m/m being a form of 1 b b m = am/(bm). § 91 ■^■' 4 4x6 20' x+y (x + y)(x + y) (x-Vy)'^' 166 ELEMENTS OF ALGEBRA 173. The value of a fraction is 7iot changed by dividing its numerator and denominator by the same number. That is, a/b = (a^ m)/(b ^ m). Proof ^^'^^ = (^^^^0^^^ = ± § 172 E.g., b -i-m (b -i- m) m b c -\- ex _ {c + ex) -4- c _ 1 ■\-x G + cy~ (c -\- cy) -^ c~ 1 -^ y 174. A fraction is said to be in its lowest terms Avhen its numetator and denominator have no common factor. 175. To reduce a fraction to its lowest terms, Divide its numerator and denominator by all their common factors, or by their H. C. F. (§ 173). Ex. 1. Reduce ^^ ^ to its lowest terms. 8 a'^xy^ The H. C. F. of the numerator and denominator is 4 axy^ ; and 4 ax^j^ _ 4 ax^y^ -^ 4 axy^ _ x^ 8 a%y* ~ 8 a^xy^ -^ 4 axy^ ~ 2 ay^' Ex. 2. Reduce £JZ^ to its lowest terms. Factoring numerator and denominator, we obtain a'^ — ax _ a(a — x) _ a cfi — x'^~ {a ->t x){a — x)~ a -\- X 173 §173 Ex. 3. X* - 1 _ x^-\ x^ — x^ — x"^ -{- X ic2(a:4 — 1) — x(x* — 1) _ 1 _ 1 ~ x^ — x~x{x — 1) Exercise 69. Reduce to its lowest terms each of the following fractions : ■ -ajy ■ 4aWc FRACTIONS 167 xY^"^ 3- r7-2- — 5 ab^xifz 5 a^b*<^x}f g Sa^bc*x^fz 4 a^b^ca^y^z^ 10. «;-°^ a''* + a6 12. ^-^y". _. a^ + 2a; ,. 4.T-16 15 2ar^-4a;\ ar* — 4 a;'' 17. 20. a-3 9-a- 18. ^'-^'. 0^—1 a;'* — a* x^ — a^ 16. x-2 4.-x^' 15 g^ — 5 aa; g^ - 2 oa; + a^ ar^-g'' g^ + 2g^6^ + 6^ 61 -g* l-5g + 6g» • l_7g + 12g2 jr -\-6x — 55 l-9f-^20y* • 1 + 62/^-552^ a:^ + :r^-2 2g a;'^ + 2a;>' + l (a^+f)(x'+xy+y') - ar^ — ga.*^ + 6^a; — ab^ a^ — aji^ — b^x + g6- 168 ELEMENTS OF ALGEBRA 176. When the factors of the numerator and denominator of a fraction cannot be found by inspection, their H. C. F. can be found by the method given in Chapter XI. Ex. 1. Reduce ^^^ ~ '^^^^^ 23 a; -21 ^^ ^^ lowest terms. 15a;3- 38x^ -2x + 21 The H. C. r. is found to be 3 a; — 7 ; and by division we find 3x3 - 13 x2 + 23x - 21 = (x2 - 2 X + S)(Sx - 7) 15x3 - 38x2 - 2x + 21 = (5x2 - X - 3)(3x - 7). 3 x3 - 13 x2 + 23 X - 21 _ x2 - 2 X + 3 " 15x3 -38x2 -2x + 21" 5x2 -X- 3' Exercise 70. Reduce to the simplest form the following fractions: a3_3a-f-2 ^ 2 0? ^ cux" -\- 4. a^x - 1 a^ !• - — :; :: — 7. r' "■ • a4 _ ^2 _ 12 o. — : r :• o* 2a^-3a2 + 1 a« + 3a2- -20 a>-a^- 12 4 x^ + 3 aa;2 + a 3 x''-^ax'-\- a^x^ a* 4a^-10a; 2 + 4a • + 2 ^x'^-2y? -3aj + 2 6a;^ + a^- -5a;- -2 a^_7aa^4_8a2a'_2a3 2a^ + 3a^ + 4a;-3 6a^ + a^_l a;4 _ a^ _ a; _|_ 1 a,4_2a^-a^-2aj4-l 4a;* + lla^ + 25 4a;*-9a^-f30a;-25 i»4-20a.-2-15x + 4 ^* 6a^-i-5a;2-3x-2* " x'*-f9a^+19ar'-9a;-20 177. Two or more fractions which have the same denomi- nator are said to have a common denominator. The lowest common denominator (L. C. D.) of two or more fractions is the L. C. M. of their denominators. E.g., the L. C. D. of the fractions — -^ and — ^-^— - is (« _ 5)2^(j ^ ft)^ or the L. C. M. of the denominators ol^ — IP- and (a - 6)2. FRACTIONS 169 178. To reduce two or more fractions to identical fractions having the L. C. D., Multiply both the numerator and the denominator of each fraction by the quotient obtained by dividing their L. C. D. by the denominator of thai fraction. Proof The derived fractions have the L. C. D. ; and by § 172, each is identical with its corresponding given fraction. Ex. 1. Reduce , — — -^ , and - — -^ to iden- a^h{x + a) ah-\x-a) ah{x^ - d^) tical fractions having the L. C. D. The L. C. M. of the denominators is a'^b^{x^ - a*). Dividing this L. C. D. by the denominator of each fraction, and multiplying both the numerator and denominator by the quotient, we obtain X _ xxh(x — a) _ bx(x — a) a26(x + a) ~ a-^62(a;-2 _ a^) " a^ft^Cx^ - a^y y - y X a(x + a) _ ay(x + a) ab\x - a) ~ a^b'\x^ - a«) a^b^x:^ - a'^) ' and _ z X ab _ abz abix:'' - a?') a'^b\x^ - ««) a^b^x^ - «^) Ex. 2. Reduce — — ;? -, — — ^ -, — — to iden- «« - 5 a; + 6 x^-ix-\-S a;^ - 3 x + 2 tical fractions having the L. C. D. The denominators equal (x_3)(x-2), (x-3)(x-l), (x-2)(:«-l), respectively. Hence their L.C. M. is (x — 3)(x — 2)(x — 1). 1 _ x-1 {X- -3)(x- -2) (X - -3)(x- -2)(x- -1) 1 X - -2 c^- -3)(x- -l)-(x- -3)(x- -2)(x- -1) 1 X - -3 (x-2)(x-l) (x-3)(x-2)(x- 1) 170 ELEMENTS OF ALGEBRA Exercise 71. Eeduce to identical fractions having the L. C. D. : , 3 4 5 ^234 Ax 6a^' 12a^ * a - b' a + b' a' + b' 5a^ Sbx 1 cy — m ay ax^ xy^ Q>x^y Sfx 10 xz' ' r^' (1-a;/ (1-xf a X O? a '^ '"'^ '*"^^ ^' J • > — -»' *>• —y X — ax — ax- — a^ ab m — n 7. am — bm -\- an — bn 2o? — 2 ab 3 5 2 a^ + 3u; + 2' a;2 + 2a;-3' :f?^hx-^^ ^ 2 a b 3a^ 5 6^ 10. a-b' 2b-2(K 4.(a'-b'')' 6(b'-a') 1 1 1 (x — a){x — by (b — x) (c — ic)' {x — c)(x — a) ADDITION AND SUBTRACTION OF FRACTIONS. 179. The converse of the distributive law for division is a b c _a-\-b — c x x x~ x Hence to add or subtract fractions, Reduce the fractions, if they have not a C. D., to identical fractions having the L. C. D. ; then add or subtract each nu- merator as the sign before the fractiori directs, and write the result over the L. C. D. h c__hy cx_by + cx and X y xy xy xy a b _ax bc_ax — be c x" ex cx~ ex FRACTIONS 171 Note. The student should remember that when either the numer- ator or the denominator is a polynomial, the horizontal line in a frac- tion is a sign of grouping as well as a sign of division. Ex. 1. Combine and simplify 1- The L. C. M. of the denominators is {x — y)(x-\- y); and x-y x-\-y (x-y)(a: + y) (x-y){x + y) _. a; + y-Ka;-y) _ 2x 1 1 Ex. 2. Combine and simplify a;2-5x+ 6 = (a;-2)(x-3), x^~lx^ 12 = (x-3)(x-4); hence, the L. C. M. of the denominators is (x — 2)(x — 3)(a; — 4), and the expression = ^-^^^ ^ ~ (x_2)(x-3)(x-4) (x-2)(x-3)(x-4) _ (x-4)-(x-2) -2 (X - 2)(x - 3)(x - 4) - (X - 2)(x - 3)(x - 4)' Ex. 3. Combine and simplify g^ZL^ _ «£_i^^ _ «fe - c^ he ca ah The L. C. M. of the denominators is ahc, hence the expression =«(«1:^M _ Hac~b') _ c(ab - c^) dbc abc abc _, a(a^ - be)- b(ac - b^)- c(ab - c^) ~ abc _ fl^ + b^ + c^-Sa bc ~ abc Exercise 72. Combine and simplify : , o, — 5b a — 3 b „ a — 36,3a — 6 1. ^' 1 z — 172 ELEMENTS OF ALGEBRA 6 a — 5 b 4 a — 76 2 ^~'^ q ^ + ?/ 3 2 ' '3 4 * _ X x — 4:,x — 5 ^a — b „a — 6 5- T 3 1 — 8. 5 7 4 3b ex 2.T-3 a;4-2 5a; + 8 9 6 "^ 12 ' 10 g — 26 a-5 6« + 7& 2 a 4 a 8 a 11 b -\- c . c-{-a a — b 2a Ab 3c a — a;a4-ic a^ ~ a^ 14. 1- 13. X a 2 ax 2 gg - 62 62 _ g2 ^2 _ ^2 a'-^ 6^ c2 ^^ 2a;-3y 3a;-2g 5 xy XZ X 15 a;-3a;^-9 8-a^ 5a; 10a;2 15a^'^ 16. l-^Mn^ + ^^^hl!. a;?/ xy^ x^y^ Keduce to an improper fraction : .1. a^ + 6^ 17. a + 6 -^— • a — 6 q + 6 a2_^ft2 _ ^2_ 52_(^2^52) 1 a — b ~ a — b ~a — 6~6 — a 18. a-l4--4-T- 20. a;4-2?/4--^^- a 4- 1 X — 2y 19. a + a;+-^. 21. x^ _ 3x - ^^^?^=^. a — a; a; — 2 FRACTIONS 173 Gar* 22. a^ — 2 ax + 4:0^ — 23. x-a + y + '^'-^'y-^y" a-^2x HA x-\- a 24. l + aj + ic' + a^' + r-^- 1 — a; 25. ^;+2a^ + l _ _ 26. l + 2a; + 4a52 + -^-+i. 2ic — 1 ar^ + 13a;-5 27. cc2_2a; + 3- 28. Combine and simplify — ^ + ^^ a — ic a^ — a* Beginners should always see to it that the denominators of the fractions to be added or subtracted are all arranged in descending powers, or all in ascending powers, of some particular letter of arrangement. Arranging the denominators in this example in descending powers of a, we have a ax _ a , — «« a — X x^ - a^ a — X a^ — x^ _ a(a + x) — ax _ a^ - rt2 _ a;'2 " a2 - x^' Combine and simplify : 29. -^ + -*-. 33. ^- + -1-. a — b b — a 3 + x a^—d 30. .-^_ + -«-. 34. l+?-ll^. x — aa — x 1 — ojl-fa; 31. '^ + « . 36. ^- + — ? 32. -^ 2_. 3 a + 2b_a-2b_ 1-x 1-x' a-2b a + 26 174 ELEMENTS OF ALGEBRA x-\-y x — y ' 4y^ x — y x^y xr — y^ 38. 2a ^ 26 ,C' ^h\ a-j-b a—b a? — b^ 39. -^ 1- 1 a — 1 a (a — 1) 4.o?-b^ 2a-[-b 41. -^+ 1 1 a; — 1 ic — 2 ic4-2 x-\-l The character of the denominators in this example suggests that it is simpler first to combine the first and fourth fractions, next the second and third, and then to combine these results, as below : 1 1 _x + l - (r<;-l)_ 2 x-\ x-\-\ a;2 _ 1 a;2 - r ^ 1 _ a; + 2-(a;-2) _ 4 X - 2 a; + 2 " a:2 _ 4 "a;'^ _ 4' and ^ . 4 ^ 2(a;2-4)+4(x2-l) ^ 6x^-12 x2-l x2-4~ (a;2_ i)(a;2_4) "a:*- 6x2 + 4' 42. ^ + ^?_+ 2 ^. 4 1— ic l + a; 1 + a^ l + ic* Here it is simpler first to combine the first and second fractions, next to combine this result and the third fraction, then this last result and the fourth fraction, as below : 2 ' 1-X 1+X 1-X2 1-X 2 2 _ 2(1 +x2)+2(l -x2) _ 4 1-X2 1 + X2~ l-X* ~1-X* 4 4 _ 4(1 + x*)+4(l -x4) ^ 8 1 _ X* 1 + X4~ 1 - X8 1 - XS" 43. FRACTIONS 175 2 1 ic + 6 x-2 x-\-2 ar^ + 4 44. -^ + _iL_+ 2a^ a — ic a + a; a^ + a^ 45 3 -a; 3 + a; l-16a? l-3a; l+3ic 9x2_i' 1 1 a.' + 3 46. x-1 2(a; + l) 2(ar' + l) .^ a , a , 2a2 , 4 a* 47. 1 r-^— — n-T—r-, — i* a — X a-\-x a- -f ar a* -f a;* 48. ^-^+ 3 1 a; — 3 a; — 1 a; + 1 a;-|-3 49. 1 -_±,+6__4_^ 1 a; — 2 x — 1 X x-{-l x-\-2 50. , f ,+ 2 1 aj2_3a; + 2 ar^-a;-2 ar'-l The expression 51. 52. 53. 54. (a;_2)(a;-l) (a;-2)(x+l) (x-l){x + \) 2(x + l)+2(a;-l)-(a;-2) (x-2)(x-l)(a;+l) 3 X + 2 (x-2)(x-l)(x+l)* 1.1 x^-dx-[-20 xr-llx-\-30 1 1 a^- 7x 4-12 a;2- - 5a; + 6 1 1 2ar^ — X -1 2x2^3.. -3 1 3 2aj2 — X -1 6ar^ — X- -2 4 3 55. 4-7a-2a2 3-a-lOa" 176 ELEMENTS OF ALGEBRA 56. ? ? 5 + x-lS^ 2-h5x + 2x' g^ 5x 15(a;-l) 9(x-\-S) 2(x-\-l)(x-3) 16(x-3)(x-2) 16 (a; +1) (a; -2)' 58. ^^^+ 1^ 12 x^-{-5x + 6 ar^ + 9ic + 14 a^H-10a;4-21 69. _^ + -A_+ 40.4-2 a^-1 2a; + l 2a^ + 3ic + l 24 a; 34-2a; 3-2a; • 9_i2a;4-4a;2 3-2a; 3 + 2a;' 61. 1 2 ^ 1 a;^ 4- 5 aa; + 6 a^ ar^ -f 4 aa; + 3 a^ a;^ 4- 3 ax -\-2a^ 62. —^ "'-" + 2 ■ (a; — 2 a)^ a^ — 5 aa; 4- 6 a^ a; — 3 a 63. 1_ 4 ^ 6 _ 4 ^ 1 a a4-l a + 2 a -^3 a 4-4 1 2 1 64. — :--, : — :4- a^_5a; + 6 x'-4,x-\-3 '.x^-3x + 2 65. ,A^-;,-V+ ^ 8_8x 8 4-8a; 4 4-4a.'2 24-2a;* 1 1 1 , 18 66. tt:-:; -tt: - o . .. + 6a-18 6a4-18 a^ 4-9 a^ 4-81 6T. , 1 . + ,. ,^ , + 1 (a — b)(a — c) (b — c) (6 — a) (c — a) (c — 5) In examples of this kind it is best for beginners to arrange all the factors in the denominators of the fractions so that a precedes b or c, and b precedes c. We therefore change b — a into — (« — 6), c — a into — (a — c), and c — b into —(b — c). The expression then becomes 1 1 + 1 (a - &) (« — c) (a - &) (6 - c) (a - c) (6 - c) FRACTIONS 177 The L. C. M, of the denominators is (a — b)(a — c) (6 — c) ; ••• the expressions (a - 6)(a - c)(6 - c) -^- 68. c . a . b 91 (6-c)(6-a) (c-d)(c-b) (a-b)(a-c) 69. ? + - + ^ (y-x)(z-x) (y-z){y-x) (z-x){z-y) 70. y + z I z + x ^ x-\-y {y-x)(z-x) (y-z)(y-x) (z-x){z-y) MULTIPLICATION AND DIVISION OF FRACTIONS. 180. Product of fractions. See § 91. Ex 1 a; + 2 ^ X + 3 ^^ X + 4 _ (a: + 2) (x + 3) (x + 4) _ ^ ■x + 3 x + 4 x + 2 (x + 3)(x + 4)(x + 2) Ex. 2. Simplify ^^-^^ x -M±J/L. x -^M^ The factors common to numerator and denominator can be can- celled before the multiplication is performed, as below : The expression =^(2-ri) x ^(^^J^ ^ W _ 2x ~3(x + 2j/)" 181. To multiply a fraction by any number, , Multiply the numerator^ or divide the denominator, by that number. Proof. ^xm^^X?^^ §91 1 b = a/(b-^m). §173 8 62" 8 62 " 8 62^4' Ex. ^x4=^«iAior ^«' 182. The reciprocal of a fraction is equal to the fraction inverted. That is, 1 -J- {a/b) = b/a. 178 ELEMENTS OF ALGEBRA Proof, b/a multiplied by the divisor a/b is equal to the dividend 1 ; hence b/a is the quotient. 183. To divide by a fraction, Multiply by the reciprocal of the fraction. Proof. Dividing by a number gives the same result as multiplying by its reciprocal (§ 87). Ex 1 3« ■ ^x_Sa ^^7 y_ 2lay * 56 ■ ly~bh 2x~106x' Ex 2 ^ — « . x^ — a^ _ X — a x + a x^ + a^ x + a x^ -^ a^ x^ — a^ \ (x2 - ax + a^) (ofi -\-ax-^ a^) 1 a;4 + a'^x'^ + a* 184. To divide a fraction by any number, Divide the numerator, or multiply the denominator, by that number. Proof «^m = -x- = — §91 b b m bm = (a-^m)/b. §173 Exercise 73. Simplify each of the following expressions : 1. l^x'p- 6. !^X^-^'- 3 c 4a a c a ^ 2a .6 c 5x ^ G? .^ o. "' 3b^ ha^2y ^' b^^ f ' / ^ 2a 5c x'b ^ Sa^ ^ 2c Sa o. X — X * o. — — — X 5b X y 4.b 7ax 7bx^ ^ 2 a' . Sabx ^ 5f^^^21c, . 35 (fy be c'y ' 7 a^ 4: ax 7 a^x 3axy ^ 6aV -^ 2 b , x 9a^ ' 56^ ' lOba^' ' 3a' V 4 6^* 17. x-2'' x — 2 X- X — S ' X - -4 -3' 18. a^-a' x + 2a^ ar-4a ^ x — a 19. a' + a^ Aa-x)\ 9ft Ux"- 7x 2x- 1 FRACTIONS 179 x^ -\-xy xy — y^ _- a^ + 2a; ,, g^-3a; XT — ^y^ X -\- y 14 _a±h_ ab--^ a^-a'b ab + a^ " 12a^ + 24:x' ' x" + 2x ^g ar> + 3a;^ . x + S ^^ 16a^-9a^ . 4a;-3a x + 4 ' x?-\-4:X ' «2-4 ' a;-2 ,. a + 46 . ah-^AW ._ a2/>2-|.3a& aft + 3 (ji-^hab a3 + 5a-6 4a2-l 2a+l ar'-14a;-15 . a^-12a;-45 • a:2_4a._i5 ' ar^_6a;-27* 24 ar^-6a.-^ + 36a; . a^ + 216a; iB2 _ 49 ' Q^-x-^2 y?-x-2^ , x-\-\ . a^H-2a;-8 ar'-25 ' or^-h 5a; ' ar^ - a;- 2 * ar'-lSx + SO . a;^-15a; + 56 a;4-5 • ar*_5a;_50 * ^-^x-1 x-l 27 af^-8a;-9 a;^ - 25 . a;^ + 4a; - 5 ^-Vlx^l2 x^-l ' a^-9x + S 28 a;^-8a; ^ ar^ + 2a; + l . a.-' + 2a; + 4 ^ a;^ — 4a; — 5 ar' — a;- — 2a;* x — 5 29 (g + sy . g-^-ft^ . (a-\-by ' (a - 6)» * (g2 - 62)2 • ^2 _^ ^2 . (a - cy -b'- ' b^-(c- ay 180 ELEMENTS OF ALGEBRA 31. ^^ + / X ^ ~ ^ • ^^ ~ ^''^^ "^ ^* - ' QC^ — y^ x-\-y ' X* -\- x^y^ + y^ First reduce each of the mixed expressions to fractions. 33. (a + ^!^\(b «'' a— bj\ a + 6 34. of_±xy^r_x_ y\ \x — y x-\-y) ^ -\- y^ \^ — y ^ + Vy 35 fci + ba — b '\^/' a-{-b a — b \ \a — b a -j- bj ' \a — b a -\- bj "■(-¥)x(3^-')*J 2 a (a — by \ ctj \ «/ 38 4a^ + a;-14 ^ 4fl;^ ^ a; - 2 _^ 2a^ + 4fl; 6 a;?/ — 14 2/ a^ — 4 4fl; — 7*3ar^ — a; — 14 x^ + a;-2 a^2 + 5a; + 4 . / a;^ + 3a; + 2 .^-{-3 • aj2_^._2o'^ or^-o; * l^a.-' - 2 a^ - 15 a^ J ._ 4a^-16a; + 15 a;^-6a;-7 ^, 4ar'-l )• 2aj2^3a; + l 2a;2_l7a; + 21 4a^-20a; + 25 a^ 4- a6 — ac (a + c)^ — 6^ ab — b^ — be a2-62_c2_26c 62_2?,c + c2-a** 43 a;^-64 ar'-H2a;-64 . a?^-16a; + 64 ic2 4-24a;4-128 aj3_64 • a^^4,x-^16' FRACTIONS 181 185. A complex fraction is a fraction whose numerator and denominator, either or both, are fractional expressions. g + 6 E.g.^ , or /(- + -), is a complex fraction. ^ _L £ c — d/\x yl ^ X y Observe that a heavy line is drawn between the numerator and denominator of the complex fraction. ay bx Ex.1. ^/^=^xy= b/ y b z Ex. 2. + x X x+j,^_x_ y y + x y Sometimes the easiest way to simplify a complex fraction is to multiply its numerator and denominator by the L. C. M. of the denominators of their fractional terms. a-\-x a — x a-\-x a — x\, .. , ; — — ; — (a - x) (a + a — x a -{- X \a—x a + x/ ._.. „^.. .„_.. .^... ^) Ex. 3. _ , a + xa-x (a-\-xa-x\. .. , , h — ; — h — ; — ) (a - x) (a -{- x) a — x a + x \a-x a-\-xJ^ ^^ ^ - (a + x)2 + (a-a:)2 _ 2 ax ~ a2 -1- x2* Here (a-x)(a + x) is the L. C. M. of the denominators of the fractional terms in the numerator and denominator of the complex fraction. ^^4 X _\ X J _x* + a-x^ _ x2 Ex. 5. Simplify x + 2 x + 2 X4- 1 182 ELEMENTS OF ALGEBRA In a fraction of this kind, called a continued fraction^ we first sim- plify the lowest complex fraction as below : X X X4-2 =^ {x + 'l)x §83 ^ _, 2 ^ + 1 (a;+2)x-(a:+l) X x^ + 2x x^-\-x-l xCx"^ -h X- I) X (x'^ + X - 1) - (ic2 + 2 x) Exercise 74. Simplify each of the following fractional expressions: 3a + — - + - 8c ^ b d 4. ^r-r' 7. Q ,76 3c+^ 8a 5. ^. 1. 71 m a 6 m n 2. x-y d 3. 46 "+3 10. 12 3 ^—x X 11. 2x'-x-6 S-' X 8. m X _1 a; l+i a; ar X a;^ X OCT a -\-b a — b a 12. - ^ _ a^ + 6^ 13. (a + 6)^ x^'^a' a^ ax x^ FRACTIONS 183 14. 1 -f l-\-x 2x2 15. 16. a — a + 17. 18. 19. 1- 1 +a; 1 X X x-2- X — x-1 x-2 1 x-\- m 2/4- ic + x + 20. 21. 22. 23. 24. 25. x — y — x-y-\- xy x-y x + y x-\-y xy x + y l-x , l + 2aj2l 1 l-fo: 1 a;-2 1 a;-2 a;-4 X 01? — y^ ^ ~ 9 x--y^ x-\-l 2 0^ + 1* 4 ic — 4 x-2 X — b X—4: — X— 4: a" + h^ 2 ah ^\ aW . 4a6(a-f ?>) ' |i + _^+_-^Ui i_l V ict + by] 1 + a + 6 ^^* U W"^S A2/"^W ^/ 184 27. 28. 29. 30. 6« ELEMENTS OF ALGEBRA a-b a'- a^b' + b* a^-W a' b' ' a' + a'b' + b' x + 2 4aH-5\ /2 a; + 3 3 a; + 4 2aj + 3 5x-{-Q>] V3i» + 4 4.x-{-6 l-\-x l^a?\ fl-\-x^ 1-^x 1 + a? l + xV Kl + x" l-\-x' a - by a + b a + b a — bj a — b J 186. Power of a fraction. The nth power of a fraction is equal to the nth power of its numerator divided by the nth power of its denominator ; and conversely. That is, Proof {a/by = ay b". f«Y^^l.«.«...tonfactors \bj b b b Ex. 1. Ex. 2. (- _ a aa • - • to n factors ~ bbb ••' to n factors = a'yb\ 27 a^^c^ by notation §91 by notation §§ 119, 186 §§ 118, 119 (x'^-7x+ 12)2 {X - 3)-^ ^IM"^^^-^^-^'- 5 x-S Exercise 75 186 Write each of the following powers as a quotient of products : 2a\2 ^ r_2_ax\^ ^ /2a^Y SbfJ' ' \3bYj' 1. 'zay 3 by' f 3 ax'y 3. V 2 62/7 ' V 2 6cVy * FRACTIONS 186 '■(-?)■ -MS)' "-(-sr Simplify each of the following expressions : (»-a/ * (a; + 2/)" ' (a' -by 20. (^^-^)* . 22. ^^-+iYx'^^^. (a^ + a + iy V2/ + V ^' + 1 \2ab' mhjy2ab' mh) Expand each of the following powers : -g.!)-. »-($-^- ...e.^jj. -i^'W -■($-?)■ -(^^o■ V6 .yy \h a) \x a yj Factor each of the following expressions : 34. i-^+i- ' 36. %-"^+i- y y- y by b^ 35. i!i4.1^'4.25. 37. -i^-^4-4. y y 4/ y 186 ELEMENTS OF ALGEBRA 38. 51^ + 52£ + 4. 39. •^-2+-^. 25 9^ 41. -^ + 1^ + 40^ + 8. 27 3 7/^ yiQ 9 2/^ x^ 2ax o? 42. 8ar^-4;.y + |V-A. 48. _--4.___. * 642/3 8/^ ■ ' 25a2"^5a2'^a2 c^ * Eeduce to its lowest terms each of the following fractions : 44. 45. 4aV 9?/' 6^ c^ 46. 47. 40)2 4_^4aj_ a" 51. a;2_ 30^2/ -28^2 64. a.'« + x'- x'-l a.-«- x« + x'-l 55. a^- -a^6 - ab* + b' a'- a3/>- - a'b' + a63 52 (^' - a^) (g 4- a;) gg (x-^y ^zf-jx-y -zf (a^-^x'){a-x) ' 3x(y''-\-2yz-\-z^) gg (x'-f)(x-y) g^ a'-16 {^-f){x^-y^) ' a*-4a3+8a2_l6a+16 CHAPTER XIII FRACTIONAL EQUATIONS 187. A fractional equation is an equation one or both of whose members are fractional with respect to an unknown. E.g., = 4 is a fractional equation in x, 2a; - 1 X while ^ -I- ^ = ^sd: — is an integral equation in x. We cannot speak of the degree of a fractional equation. The term degree as defined in § 101 applies only to an integral equation. 188. If both members of an integral equation are mxdti plied by the same unknown integral expression M, the derived equa- tion has all the roots of the given equation, and, in addition, those of M=0. E.g.., if we multiply both members of the equation 2x+l=a; + 3 (1) by ic — 5 ; the root 5 is introduced in the derived equation. Proof If A and B denote integral expressions in the unknown, and we multiply both members of A = B (1) by any unknown integral expression M, we obtain AM= BM, or {A - B) M= 0. (2) By § 149, (2) is equivalent to the two equations ^-^ = 0and3/=0. 187 188 ELEMENTS OF ALGEBRA That is, the roots of M=0 are introduced in the derived equation (2) by multiplying both members of (1) by M. 189. If both members of a fractional equation in one un- known are multiplied by any integral expression which is necessary to clear the equation of fractions, the derived integral equation will be equivalent to the given fractional equation. 3 Ex. 1. Solve the equation = 5 — x. (1) X —\ Multiplying by ic — 1 to clear (1) of fractions, we obtain 3 = 6 X - x2 - 5. Transpose, x2 - 6 x + 8 = 0. Factor, (x - 2) (x - 4) = 0. (2) No root could be lost, nor could either root of (2) be introduced by multiplying (1) by x — 1 ; hence (2) is equivalent to (1). Therefore, the roots of (1) are 2 and 4. Ex. 2. Solve -^ + -^ = 5. (1) X — 5 X — 3 Multiplying by (x — 5) (x — 3) to clear (1) of fractions we obtain 3 (X - 3) + 2 X (x - 5) = 5 (X - 5) (x - 3). . •. x2 - 11 X + 28 =^ 0. ... (x_4)(x-7)=0. (2) No root could be lost nor could either root of (2) be introduced by multiplying by x — 5 or x — 3 ; hence (2) is equivalent to (1). Therefore, the roots of (1) are 4 and 7. Proof By transposing to the first member all the terms of any fractional equation, adding them, and reducing the resulting fraction to its lowest terms, we derive an equation of the form A/B = 0, (1) where J. and jB have no common factors. FRACTIONAL EQUATIONS 189 By the preceding principles of equivalent equations, the derived equation (1) is equivalent to the given fractional equation. We are to prove that (1) is equivalent to the equation ^ = 0. (2) Any root of (1) reduces A/B to 0. But when A/B is zero, A is zero ; hence any root of (1) is a root of (2). To prove the converse we must first prove that when A = 0, B=^0. If A and B could become for the same value of x, as a ; then by § 132 they would have the factor x — a in common. But by hypothesis A and B have no common factor ; hence when A = 0, B^O. Hence any root of (2) reduces ^ to but not B to 0. Therefore any root of (2) reduces A/B to and is a root of (1). Hence equations (1) and (2) are equivalent. Ex. 3. Solve 1 - -^ = — 6. (1) Transposing and adding the fractions, we have r'2 — 1 1 - + = 0. X — 1 . •. 1 - (x + 1) + G = 0, or a; = 6. (2) By §§ 105 and 100, equation (2) is equivalent to (1) ; hence is the one and only root of (!)• But if, as would be more natural for the beginner, we should clear equation (1) of fractions by multiplying by x — 1, we would obtain x-l-x2 = -l-Gx + e. Transpose, x^ - 7 x + = 0. ...(x-i)(x-e)=0, (3) of which the roots are 1 and 0. 100 ELEMENTS OF ALGEBRA As was shown above, to clear equation (1) of fractions it was not necessary to multiply by x — \ ; hence multiplying (1) by x — 1 is the same as multiplying the equivalent integral equation (2) by ic — 1. In clearing of fractions an equation in one unknown, to avoid introducing roots, the following suggestions should be heeded : (i) Fractions having a common denominator should be combined. (ii) Factors common to the numerator and denominator of any fraction should be cancelled. (iii) When multiplying by a multiple of the denomi- nators, we should always use the L. C. M. Ex. 4. Solve -^— + ^^ = ^i-i + ^^=-^. (1) x — 'l X — 1 x—\ X — iS Transpose so that each member is a difference, x X 4- 1 _ x — 8 a; — 9 05 — 2 X — \ X— 6 X — 7 Combine, ^ = ? : (2) (x _ 2) (X - 1) (x - 7) (X - 6) ' ^ ^ Clear of fractions, x2 - 1.3 x + 42 = x2 - 3 x + 2. . •. 10 X = 40, or X = 4. (3) Since the root 4 could not be introduced in clearing (2) of fractions, 4 is the root of (1). Ex. 5. Solve x-^_^x±^^x±\ ^x + 3, ^^ x+lx+7x+3x+5 Transpose, x-^ _x±l^x±^ _x±b^ ^^ x+lx+3x+5x+7 Combine, ^- = ^ (3) (X + 1) (X + 3) (X + 5) (X + 7) Clear of fractions,.x2 + 12x +35= x^ -f 4 x + 3. .•.8x = -32, or x = -4. FRACTIONAL EQUATIONS 191 Or reducing the improper fractions in (2) to mixed expressions, we have, x+ I x+3 x+5 x+7 1 . I ^ + ^ x-\-l x + 3 x + 5 x + 7 Combining these fractions, we obtain equation (3) above. Since the root — 4 could not be introduced in clearing (3) of fractions, — 4 is the root of (I). Exercise 76. Solve each of the following fractional equations : , 3x-16 5 X 3 5x-5 11. x-\-l a;-1 ^2 x + 1 3 x-2 ^ 2x-5 = 3. 12. X- -5 2x- -2 4a;- -5 13. 14. 5. 2£:z3 = .^_,. 15. 3a;-4 6a;-7 6. -^— = -^-2. 16. x-\-l x -\-2 7. ^^-1-1 = 1. 17. X -\-l X 8. 10^4^ = ._A__7. 18. ^ a — 1 » + 1 1 . 1 4a;+6 6a;+4 2£c+3 19. 12 2 10. — 1 = — = -^ — 20. 3a;-f-9 5a;H-l x+3 Sx-5 4a; + 8 ^ 1 5 7 2a;+3 4a?-f6 6a;+8 4.^ X _3 a; 4-1 a;-2-^' 6x ^ -5 x-7 x-6 ^• 2x 4a; ^ x + 3 x + 7 - 1 + a; + 4 2 3 a;-f-6 a;-h5 3 2 1 x + 1 a; + 2 a;H-3 5 1 2x + 4: _3 1 , 1 3 1 2xH-2 =5 1 +1. 4a; + 3 4a; 2x-5 2a;-7 3a;-7 3a;-5 6a;-2 _3a; + 7 192 ELEMENTS OF ALGEBRA 21. ^±l_^Z:i? = §. 25. 3'^-^-\-2^±l = 5. ic — lie + 3a; ic + 1 x — 1 22. ^+2_£^ = _§_. 26. 5^!^^-2^^ = 3. x-2 x + 2 x + 1 x + 2 x + S x-\-3 ic — 4 ic x^ — 1 aj + l 1 — ic 24. -^ = 3^^ ^. 28. -^+ 1 ^ a;-h2 i«-2 ic + l x--9 x + 3 3 29. 30. 35. 36. 37. 38. 39. 3 g; -f 5 5 ^ 8 4- 3 a; 3a;-l l--9a;*-^~l + 3;»* 1111 a; + 5 x + 6 x + 6 aj + 8 31. 1 +_!=! + 1 x + 2 x + 10 a; + 4 a + S 32. -J_ + _1_ = ^_ + _L.. if — 5 37 + 2 .T — 4 x + 1 33 ^' .a; — 9_a; + l , x — S x—2 x—1 x—1 X— 6 g^ a;4-3 a; — 6 _ a; + 4 . a; — 5 a; + l a; — 4 a; + 2 x — 3 x — 3 a; — 4 _ a^ — 6 x — 7 a; — 4 x — b x — 1 ^ — 8 X a; + l_a; — 8 a; — 9 x — 2 a;— l~a; — 6 x — 1 x-[-iS _ x — Q _ x — 4: _ a; — 15 a; + 4 x — 1 x — 5 x — 16 x—1 _ x — 9 _ x — 13 _ x — 15 _ X — 9 a; — ll~x— 15 x — ll' x-\-3 _ x + 6 ^ x-\-2 _ a;4-5 _ x4-6 x + d~x-{-5 x-i-S FJi ACTIONAL EQUATIONS 193 a? + 2 x — 7 x-^3 __ x — X x — 5 x-\-l if — 4 4a;_17 10a;_13 8aj-30,5a; a;_4 2a;-3 2a;-7 x-l ^„ 5a;-8 , 6x-44 10a;-8 x-S x-2 x-7. x-1 x-6 ^^ S0±6x_^60±8x^^^_^ 48 cc-f-l x-i-3 ic-fl 44 25-^a; 16a^ + 4i ^g 23 a; + l 3a; + 2 aj + l 45. 3_4-...^' =;^+ ' 46. 4-2a; 8(l-x) 2 - a; 2-2a; 60 _ 10^ 8_^ X 4 5aj-30 3a;-12 47. -i 2- = ^ ?i-. a! + 3 K + l 2x + 6 2x + 2 48. (2x-l)(3x + 8) j^Q_ 6a;(a; + 4) In the five following examples first reduce improper fractions to mixed expressions. 4^ 5ar-64 2 ic - 11 4a;-55 aj~6 50. 51. 52. a; -13 X- -6 a;- -14 a; -7 x — H , .T-4 a; -^- a:-7 a;-9 a;- 10 ' a;-G x-\-b _^x • + c 2. a;— c a; — 6 x-\- c x vnx nx 53. ; 1 = m 4- w. m 4- a; n -\- x 194 ELEMENTS OF ALGEBRA 54. 55. x — c _ x — b 2(b — c) x—b X — c X — b — c m -f- r n-\- r _m -\- n -\- 2 r x-\-2 n x-\-2m x-\-m-\-n 190. Problems which lead to fractional equations. Prob. 1. The quotient of a certain number increased by 7 divided by the same number diminished by 5 is 4. Find the number. Let X = the required number. Then by the conditions of the problem, we have ^ + 7^1 x — b Whence a; = 9, the required number. Prob. 2. The value of a fraction is 1/4. If its numerator is dimin- ished by 2 and its denominator is increased by 2, the resulting fraction will be equal to 1/9. Find the fraction. Let X =■ the numerator of the fraction ; then 4 X = the denominator of the fraction ; and, by the conditions of the problem, we have x-2 ^1 4 a; + 2 9' Whence a; = 4, and the required fraction is 4/16. Exercise 77. 1. The value of a fraction is 1/7. If its numerator is increased by 5 and its denominator by 15, the resulting fraction will be equal to 1/5. Find the fraction. 2. The sum of two numbers is 20, and the quotient of the less divided by the greater is 1/3. Find the numbers. 3. What number added to the numerator and denominator of the fraction 3/7 will give a fraction equal to 2/3 ? 4. What number must be added to the numerator and subtracted from the denominator of the fraction 5/11, to give its reciprocal? FRACTIONAL EQUATIONS 195 5. The reciprocal of a number is equal to 7 times the reciprocal of the sum of the number and 5. Find the number. 6. A train ran 240 miles in a certain time. If it had run 6 miles an hour faster, it would have run 48 miles farther in the same time. Find the rate of the train. 7. A number has three digits which increase by 2 from right to left. The quotient of the number divided by the sum of the digits is 48. Find the number. 8. A steamer can run 18 miles an hour in still water. If it can run 96 miles with the current in the same time that it can run 48 miles against the current, what is the rate of the current ? 9. A number of men have $ 80 to divide. If $ 150 were divided among 2 more men, each one would receive $ 5 more. Find the number of men. 10. The circumference of the hind wheel of a wagon exceeds the circumference of the front wheel by 4 feet. In running 200 yards the front wheel makes 10 more revolu- tions than the hind wheel. What is the circumference of each wheel ? 11. A number has two digits which increase by 4 from right to left. If the digits are interchanged and the result- ing number is divided by the first number the quotient will be 4/7. Find the number. 12. A train runs 10 miles farther in an hour than a man rides on a bicycle in the same time. If it takes the man 5 hours longer to ride 352 miles than it takes the train to run the same distance, what is the rate of the train ? 13. A tank can be filled with one pipe in 30 minutes, by a second pipe in 40 minutes, by a third in 50 minutes. How long will it take to nil it with them all running together ? 14. A can do a piece of work in 3^ days, B in 2i days, C in 3| days. If A, B, and C work together, how long will it take to do the work ? 196 ELEMENTS OF ALGEBRA 15. A cistern can be filled in 15 minutes by two pipes, A and B, running together ; after A has been running by itself for 5 minutes, B is also turned on, and the cistern is filled in 13 minutes more. In what time would it be filled by each pipe separately ? 16. A man, woman, and child could reap a field in 30 hours, the man doing half as much again as the woman, and the woman two-thirds as much again as the child. How many, hours would they each take to do it separately ? 17. A and B ride 100 miles from P to Q. They ride together at a uniform rate until they are within 30 miles of Q, when A increases his rate by 1/5 of his previous rate. When B is within 20 miles of Q he increases his rate by 1/2 of his previous rate, and arrives at Q 10 minutes earlier than A. At what rate did A and B first ride ? 18. A and B can reap a field together in 12 hours, A and C in 16 hours, and A by himself in 20 hours. In what time could B and C together reap it ? In what time could A, B, and C together reap it ? 19. The sum of two numbers is n, and the quotient of the less divided by the greater is a/b. Find the numbers. 20. The reciprocal of a number is n times the reciprocal of the sum of the number and a. Find the number. 21. A train ran a miles in a certain time. If it had run b miles an hour faster, it would have run c miles further in the same time. Find the rate of the train. 22. A steamer can run a miles an hour in still water. If it can run b miles with the current in the same time that it can run c miles against the current, what is the rate of the current ? 23. The value of a fraction is 1/a. If its numerator is increased by m and its denominator by n, the resulting fraction will be equal to 1/6. Find the fraction. CHAPTER XIV SYSTEMS OF LINEAR EQUATIONS 191. Equations in two or more unknowns. In the equation y = 3x + 2, (1) where x and y are both unknowns, y has one and only one value for each value of x. E.g., when x = l, y = 5] when x = 2, y = S -, when x = 3, ?/ = 11 ; when a; = 4, y = 14, etc. That is, equation (1) restricts x and y to sets of values. In like manner, any equation in two or more unknowns restricts its unknowns to sets of values. 192. A solution of an equation in two or more unknowns is any set of values of the unknowns which renders the equation an identity. E.g., if in the equation 2/ = 3a; + 2 (1) we put 3 for x aud 11 for y, we obtain the identity 11 = 3 X 3 + 2. Hence 3 and 11, as a set of values of x and y, is one solution of (1) ; 2 and 8 is another solution ; and so on. Note. The word solution denotes either the process of solving or the result obtained by solving. The word is here used in the latter sense. A root of an equation in one unknown is often called a solution. 197 198 ELEMENTS OF ALGEBRA 193. The degree of an integral equation in two or more unknowns is the degree of that term which is of the highest degree in the unknowns. E.g., ax -^ by = 7 is R linear equation in x and y ; while ax^ -\-hy = c or cxy + 3 ic = 2 is a quadratic equation in x and y. 194. Two equations are said to be equivalent when every solution of each equation is a solution of the other. 195. The following principles concerning the equivalence of equations, which have been proved for equations in one unknown, hold true for all equations : (i) If for any exj^ression in an equation an identical expres- sion is substituted, the derived equation will he equivalent to the given one (^ 105). (ii) If identical expressioiis are added to or subtracted from both members of an equation, the derived equation will be equivalent to the given one (§ 106). (iii) If both members of an equation are midtiplied or divided by the same known expression, not denoting zero, the derived equation ivill be equivalent to the given one (§§ 108, 110). (iv) If one member of an equation is zero, and the other member is the jyroduct of two or more integral factors, the equations formed by putting each of these factors equal to zero are together equivalent to the given equation (§ 149). E.g., the equation (ic + 2?/-4)(2a;-32/+ 1) = is equivalent to the two equations x + 2i/-4 = and 2a;-32^ + l=0. (v) If both members of an integral equation are multiplied by the same unknown integral exjyressioyi M, the derived equa- tion has all the solutions of the given equation, and in addition those of M=0 (§188). SYSTEMS OF LINEAR EQUATIONS 199 Proof. If in the proofs of these principles for equations in one unknown we substitute the word "solution" for the word "root," the proofs will apply to equations in any number of unknowns. Exercise 78. Of the following equations state which are equivalent to the equation 2 a; + ?/ = 3, and give the reason : 1. (4 a- -f 2 2/)/2 = 3. 4. 6 a; + 3^ = 9. 2. 'Sx-\-y = x + ^. 5. (2a;4-?/)/3 = l. 3. a; + y = 3 — x. Q. i:X + oy = (S-\-y. State to what two linear equations each of the following quadratic equations is equivalent, and give the reason : 7. (a;-2/)(a-+2y-hl)=0. 8. {■y-x)x-^2y{x-y)=0. Obtain ten solutions of each of the following equations : 9. 2a;-hy = 3. 10. 2a; -f 3?/ = 6. 11. 2 a; — 3y = 4. 12. How many solutions has a single equation in two unknowns ? 13. By (iii) in § 195, show that the two equations ax -\-by = c and a'x -\- b'y = c' are equivalent when a' /a = b'/b = c'/c. 196. Independent equations. Prob. If the sum of two numbers is 10 and their difference is 4, what are the two numbers ? Let X = the less number and y = the greater number. Then by the Jirst condition we have the equation y + x = \0; (1) and by the second condition wc have the equation y - X = 4. (2) 200 ELEMENTS OF ALGEBRA In (1), when x = l, y ^9; when x = 2, y = S ; when x = 3, / = 7, etc. In (2), when x = 1, y = 5 ; when x = 2, ?/ = 6;^when jr = 3, / = 7, etc. Hence, 3 and 7 is a set of values of x and y which will satisfy each of the two different conditions expressed by equations (1) and (2), and are therefore the required numbers. Equations, like (1) and (2), which express different condi- tions are called independent equations. Observe that independent equations express different rela- tions between their unknowns, while equivalent equations express the same relation. Any solution as x = 3, ?/ = 7, can be written briefly 3, 7, it being understood that the value of x is written first. 197. Systems of equations. Two or more equations are said to be simultaneous, when the unknowns are restricted to the set or sets of values which satisfy all the equations. A group of two or more simultaneous equations is called a system of equations. E.g.^ equations (1) and (2) in § 196 are simultaneous^ and form a system of equations. 198. A solution of a system of equations is a set of values of its unknowns which satisfies all its equations. E.g., 3, 7 is a solution of the system of equations, (1) and (2), in § 196. To solve a system of equations is to find all its solutions. 199. Equivalent systems. Two systems of equations are said to be equivalent when every solution of each system is a solution of the other system. E.g.^ the systems (a) and (6) x^2y = b, (1)1 4.x- y = 2, (2)}*^'^' i(5-2y)-y = 2, (4) j x-^2y = b, (1)1 , x = 5-2y, (3)1^^^ [ («) ... ^ . ^ ... r (ft) SYSTEMS OF LINEAR EQUATIONS 201 are equivalent ; for each system has the solution 1, 2 ; and, as will be proved later, neither system has any other solution. Observe that (3) is obtained by solving (1) for x, and (4) by put- ting in (2) the value of x given in (3) ; x therefore does not appear in (4), and is said to have been eliminated. 200. Elimination is the process of deriving from two or more equations a new equation involving one less unknown than the equations from which it is derived. The unknown which does not appear in the derived equa- tion is said to have been eliminated; as a; in § 199. There are in common use two methods of elimination : I. Elimination by substitution or comparison. II. Elimination by addition or subtraction. 201. In this chapter we shall use three principles con- cerning the equivalence of systems of equations. For con- venience of reference we shall number them, (i), (ii), (iii). (i) Equivalent equations. If any equcUion of a system is replaced by an equivalent equation, the derived system will be equivalent to the given system. E.g., since equation (3) is equivalent to (1), and (4) to (2), system (6) is equivalent to system (a). 3a; + 22/ = 8, (1)1 dx 4x-Sy = 5, (2)/^^^ 8x + 6y = -Gy = = 24, = 10. >> The only solution of either system is 2, 1. Ex. Solve the system 3 + 4 a; = 15, Sl<-' 2 + 3y = 8. From (1), a; = 3. From (2), y = 2. S)<'> Since equation (3) is equivalent to (1), and (4) to (2), system (b) is equivalent to system (a) ; hence, the one and only solution of system (a) is 3, 2. 202 ELEMENTS OF ALGEBBA Proof of (i). Let (1) and (2) be a system of equations in two unknowns, ^ = ^' ^^A(a^ "^' = ^'' ^^^(h^ C=D, (2)r^ C' = Z>' (4)]^^ and let (3) be equivalent to (1), and (4) to (2) ; we are to prove that system (h) is equivalent to system {a). Since (1) and (3) have the same solutions, and (2) and (4) also have the same solutions ; it follows that any solution common to (1) and (2) will be common to (3) and (4) also ; and conversely. Hence, systems (a) and (6) are equivalent. 202. The method of elimination by substitution depends upon the following principle of equivalence of systems: (ii) Substitution. If one equation of a system is solved for one of its unknowns, and the value thus obtained is substituted for this unknown in the other equation (or equations) of the system, the derived system will be equivalent to the given one. Ex. 1. Solve the system 2 a:; = 10, (1) y = l2-lSx. (2) j ■(«) From (1), X = 5. 1 Substituting this value of x in (2), we obtain I (?)) ij = 12 -16 = -3. J By (ii), system (6) is equivalent to system (a). Hence, the one and only solution of system (6) or (a) is 5, — 3. Ex. 2. Solve the system Sx + by = 19, (1) 1 bx-4y = 1. (2) J rrom(l), x = il9-by)/S. (3)1 Substituting this value for x in (2), we have [ (6) (5/3)(19-5i/)-42/ = 7. (4) J SYSTEMS OF LINEAR EQUATIONS 203 By (ii), system (6) is equivalent to system (a). From (4), y = 2. (5) Substituting 2 for y in (3), we obtain x=(19-10)/3 = 3. (6) (c) By (ii), system (c) is equivalent to system (6) ; hence, the one and only solution of system (c) or its equivalent system (a) is 3, 2. Ex. 3. Solve the system 2x- 5y=l, (1) 7x + 3i/ = 24. (2)' }(«: From(l), x=(5y+l)/2. (3)1 Substituting this value of x in (2), we have > (/>) (7/2)(5y+l)+3y = 24. (4) J By (ii), system (6) is equivalent to system (a). From (4), y = 1. (5) Substituting this value of y in (3), we obtain (c) x = (5 + l)/2 = 3. (6). By (ii), system (c) is equivalent to (6), and therefore to (a). Hence, the one and only solution of system (a) is 3, 1. The foregoing examples illustrate the following rule for eliminating by substitution. From one of the equations find the value of the unknown to he eliminated, in terms of the others; then substitute this value for that unknown in the other equation or equations. Proof of (ii). Let (1) and (2) be a system in two unknowns, and let (3) be the equation obtained by solving ^ = A (1)K. =^ = P, (3)1 C = D, {2)]^' C' = D', (4)J (b) (1) for X, and (4) the equation obtained by substituting for x in (2) its value F as given in (3); we are to prove that system (6) is equivalent to system (a). 204 ELEMENTS OF ALGEBRA Since x = Fis equivalent to equation (1), system (c) is by (i) equivalent to system (a), X=:F, (5)1 C = D. (6) J («) Any solution of system (c) renders x = F and C=D', hence, any such solution must satisfy (6) after F has been substituted for x (vi, § 32) ; therefore, any solution of system (c) will be a solution of system (b). Conversely, any solution of system (b) renders x = F and C = D' ; hence, any such solution must satisfy (4) after x has been substituted for F (vi, § 32) ; hence, any solution of system (b) is a solution of (c). Hence, system (b) is equivalent to system (c) or (a). In like manner the theorem could be proved if the systems (a) and (6) contained three or more equations. Exercise 70. Solve each of the following systems of equations by the method of substitution : 1. 3 a; = 27, 2 a; 4- 3 2/ = 24. 2. 3 a; + 4 2/ = 58,' 2y = U, 3. 3 a; + 4 2/ = 10,1 4 a; + 2/ = 9. J 4. x-\-2y = lS, Sx + 5. 2/ = 13,| 2/ = 14. J 4x + 72/ = 29,l a; + 3 2/ = 11. J 6. 5 a; + 6 2/ = 17,1 6 a; + 5 2/ = 16. J 7. 8 a; -2/ = 34,1 ic + 8 2/ = 53. 1 8. 6^ -5 a; = 18,1 12 a; - 9 2/ = 0. J 9. 7aj + 42/ = l, I 9a; + 4^ = 3. J 10. x-lly = l, 111 2/ - 9 a; = 99. 11. 3 a; + 5 2/ = 19,1 5x — 4:y = 7. J 12. 8 x- 21 2/ = 5, 1 6 a; + 14 2/ = - 26. J SYSTEMS OF LINEAR EQUATIONS 205 13. 3 a; -112/ = 0, I 16. 1 0^ + 3 2/ + 14 = 0,1 19 a;- 19 2/ = 8. J |a; + 52/ + 4 = 0. 1 14. ^^-^^y = ^'_ ] 17. i(^+x)=i(9+y), I 25 0.-17 2/ = 139. J i(ll+^+2/)=i(9+2/) 15. ^_^_1 3 6~2' 5 10 2 J 18. J(a. + 1)=4(2/ + 2),| i(^ + 2/)=i(2/ + 2).J 19. (a; + l)(2/ + 5) = (.T + 5)(2/ + l),| ] 20. an/-(a;-l)(2/-l)=6(2/-l) a; - 2/ = 1. 203. The following example illustrates a special form of the method of elimination by substitution, which is called elimination by comparison. Ex. Solve the system 2x-3y = \, 6x-^2y = l2Q. ;:;)'•' Solve (1) for a;, Solve (2) for a;, a; = (3y + l)/2. x=026-2y)/5. '> Substituting in (4) the value of x given in (3), or, what amounts to the same thing, putting these two values of x equal to each other, we obtain 3y4-1^126-y ^r y = 13. 2 6 Substituting in (3), x = 40/2 = 20. (c) By principles (i) and (ii), systems (a) and (c) are equivalent ; or, in other words, no solution has been either lost or introduced in pass- ing from system (a) to system (c) ; hence, the one and only solution of system (a) is 20, 13. 206 ELEMENTS OF ALGEBRA Exercise 80. Solve each of the followmg systems by the method of comparison : 1. ^+2/ = 16, 4 2. X — y 6. ^ + ^ = 5, 5 2 • X — y =z4. 7. l-^-TV?y = 3,| 4:X-y = 20. J -? = 2. 8. ^ + -^^ 0, 10, 3. ^ + 2/ 1 i» + 2/ = 50 4. a? = 3 2/, 1 oj + 2/ = 34 5 it* 3 4 Sx-7y = 37. } X -\-l _ 3 y — 5 10 ~~^ a; + 1 X — w 5. 2/ 5 2/ 10. 10 8 a^ + 3 8 - V I 3 (x + ?/) ^ a? + 3 8 ~ 5 ' 204. The method of elimination by addition or subtraction depends upon the following principle : (iii) Addition. If an equation obtained by adding, or sub- tracting, the corresponding members of two or more equation"} of a system is put \n the place of any one of these equations, the derived system will be equivalent to the given system. Ex. 1. Solve the system 3 a; + 7 y = 27, (1) ) bx + 2y = \Q. (2) J To eliminate x, we obtain from (1) and (2) equivalent equations in which the coefficients of x are equal. Multiply (1) by 5, 15 x + 35 «/ = 135, (3) Multiply (2) by 3, Ibx + Qy 48. (4) (&) SYSTEMS OF LINEAR EQUATIONS 207 Subtract (4) from (3), 29 ')x - {5 k -\- 2)y = 2k-\- 10. Putting the coefficient of x in (4) equal to 0, we obtain k = - Substituting — 5/3 for k in (4), we obtain y = 2. (5) Putting the coefficient of y in (4) equal to 0, we obtain A: = — 2/5. Substituting — 2/5 for k in (4), we obtain x = 4. (6) The two equations, y ^2 and x = 4, which result from the two different values of k in (4) form a system which by (i) and (iii) is equivalent to (a). No one method of elimination is preferable for all cases. The learner should aim to select that method which is best suited to the system to be solved. Exercise 82. Solve each of the following systems by that method which is best suited to it : 1. 57 a; + 25^ = 3772, 25 0^4-57 2/ = 1148. 2. 93a;H-15i/ = 123, 15x4- 93 2/ = 201. 3. Wx-hl9y = lS,] 5. + 52/ = -4, 5 SYSTEMS OF LINEAR EQUATIONS 211 6. ''-±l-^iy = 2, y + 11 x + l ^ 11 2 1. 2x — 5y x-^7 _ ^ 8. 3 a:-2 4 2a;-5 ll-2y = 0.J 9. ^-3^ = 0, 2x 7 13 -y 3 16 0. 10. ^-f = 4, 11. 2a;-f2/=:0, i2/-3a; = 8.J 12. \x-\-\y = n, 13. 3. r — 7 2^ = 0, f a; + |2/=7. 14. lx-\y = 0, 3 x + ^y = ll,] 15. aa; -h fty = (a + 6)2 ax — by= a^ — Ir. 16. «x + &y=a24-6V hx-{-ay = 2db. 17. aa; + 6y = a--62, 6a; + 6y = a2-62 , + a?/ = a2_62 J 18. x + y = a-^h, ax — by = b^ — a'. 19. b-x-a'y = 0, bx 4-a?/=:aH-6. J 20. x — y = a~b, ax-by = 2a^-2b-. J 21. ax-by = a' + b^,) x+ y = 2a. j 22. 6a; — a?/ = ft'', I ax — by = al J 23. aa;H-6?/ = l, | 6a; + ay = l. J 24. (a + b)x-(a-b)y = 3ab,] (a + 6)2/ - (a - 6)a; = ab. J a^a; + 6^y = c*. J 26. ^ + a 6 a6' a? _ y^ 1 a' 6' a'6'" 212 ELEMENTS OF ALGEBRA 27. Sx_^2y a b 9x_6y_^ a b 28. qx — rb =p{a — y), " ^+,.=,(1+1). 29. -^-1 =1. m' m 30. a b 3a 66 3 J 31. a a' V J 32. b a b 206. Two conditions are said to be consistent or inconsistent according as they can or cannot be satisfied at the same time. Equations which express consistent conditions and there- fore have one or more solutions in common are called con- sistent equations. Thus the equations in any of the above systems are consistent equations. Equations which express inconsistent conditions and there- fore have no solution in common are called inconsistent equations. E.g.., the equations 3 5c + 3y = 15, (1) (2) express inconsistent conditions and have no solution in common. For if a; -I- 2/ is 4, 3(x + ?/) is 12 and cannot therefore be 15. 207, Each of the foregoing systems of linear equations illustrates the following theorem : If the two equations of a system in two unknowns are linear, independent, and consistent, the system has one, and only one, solution. SYSTEMS OF LINEAR EQUATIONS 213 Proof. By the principles of equivalent equations and (i) in § 201, any system of two linear equations can be reduced to an equivalent system of the form )(«) ax-\-hy = c, (1) a'x + h'y = c'. (2) Multiply (1) by h', ab'x + bb'y = b'c. (3) Multiply (2) by b, a'bx + bb'y = be', (4) Subtract (4) from (3), (ab' - a'b) x = b'c - be'. (5) ^ From (1), ax-\-by = c. (1) J By (i) and (iii), system (6) is equivalent to system (a). When ab' — a'b = and b'c — be' = 0, (5) is an identity, and system (b) or (a) has all the solutions of equation (1) ; hence equations (1) and (2) are equivalent. When ab' — a'b = and b'c — be' ^ 0, no value of x will satisfy (5) ; hence system (6) or (a) has no solution, and equations (1) and (2) are inconsistent. Hence (1) and (2) are not independent and consistent unless ab' - a'b ^ 0. When ah' — a'b ^ 0, x has one, and only one, value in (5), and this value of x will give one, and only one, value for y in (1) ; hence system (6) or (a) has one, and only one, solution. When ab' — a'b ^ 0, from (5) we obtain X = (b'c - be') / (ab' - a'b). (6) Similarly, y = (ae' — a'e) / (ab' — a'b). (7) 208. Systems of three linear equations. Ex. 1. Solve 6x + 2y-5z = lS, (1) 1 3a; + 3y-20=13, (2) i (a) 1x + 5y-Sz = 26. (3) I (ft) (0 214 ELEMENTS OF ALGEBRA To eliminate y, we can proceed as follows : Multiply (1) by 3, \^x + Qy -\^z = 39. Multiply (2) by 2, Qx-^Qy - 4:Z = 2Q. Subtract, 12 x - 11 2; = 13. (4) Multiply (1) by 5, 30 a: + 10 «/ - 25 ^ = 65. Multiply (3) by 2, Ux-{-10y- 6z = 52. Subtract, 16 x - 19 = 13. (5) Solving system (6), i.e., (4) and (5), we obtain z=l, (6) I x = 2. (7) J From (6), (7), and (1), y = 3. (8) The systems (6) and (c) are equivalent. But (b) with (1) forms a system equivalent to (a); hence (c) with (1), or (c) with (8), forms a system equivalent to (a). Hence the solution of system (a) is 2, 3, 1. Ex. 2. Solve 3 X + 2 y + 4 2: = 19, (1)1 2x+ 5^4-30 = 21, (2) K«) Sx-y-\-z = i. (3) J From (3), y = Sx + z-i. (4) Substituting in (1) and (2) the value of y in (4), we obtain 3 X + 2(3 X + - 4) + 4 2: = 19, and 2x4-5(3x + ;3 -4)+32 = 21; or, 9 X -i- 6 2; = 27, | and 17x + 8^ = 41. J Solving system (6), we obtain (5)1 X = 1. (6) J ' ' From (4), (5), (6), y = 2. (7) By (ii), system (6) with (4) forms a system equivalent to (a); hence (c) with (4), or (c) with (7), forms a system equivalent to (a). Hence the solution of (a) is 1, 2, 3. SYSTEMS OF LINEAR EQUATIONS 215 The foregoing examples illustrate the following method of solving a system of three linear equations : From any two of the three equations derive an eq^iation, eliminating an unknown; next from the third equation and one of the other two derive a second equation eliminating the same unknown. Solve for these two unknowns the two equations thus derived, and substitute the values of these two unknowns in the simplest equation which contains the third unknown. 209. From a system of four linear equations we can elimi- nate one of the four unknowns, and thiis obtain a new system with three unknowns. Solving this new system, we can substitute the values thus obtained in the simplest equation which contains the fourth unknown. Solve Exercise 83. 1. .T-f .3?/ -1-42 = 14,] or + 2?/ -1-2 = 7, 2x-\-y-\-2z=2. 2. a; -1-22/ + 22 = 11, 2x + y-\-z = l, 3a; + 42/-h2 = 14. 3. ^x-2y-\-z = 2,] 2aj-j-32/-2 = 5, .T + 2/ -h 2 = 6. 4. x-\-y^z = l, 2a;-|-3 2/ + 2 = 4, 4 a; H- 9 2/ + 2 = 16. 5. 5x + 3y-\-7z = 2y 2a; -42/ + 92 = 7, 3x + 22/ + 62 = 3. 6. x-{-2y-3z = 6, 2.T + 42/-72 = 9, 3a; — ^ — 52 = 8. 7. x-2y + 3z = 2, 2a; — 3 2/ + 2 = 1, 3x-y-\-2z = 9. 8. 3x-\-2y-z = 20, 2x-\-3y-\-6z = 70 a; — 2/ + 62 = 41. J 216 ELEMENTS OF ALGEBRA 9. 2a; + 3?/ + 42 = 20, 3a; + 42/ + 5^ = 26, Sx-{-5y-{-6z = 31. 10. Sx-4:y = 6z-16, 4:X-y = z-\-5, a; = 32/ + 2(2-1). J 11. ax-\-by = l, by -\- cz = 1, cz + ax = 1. 12. cy -\-hz — he, az -\- ex = ca, hx -\- ay = ah. 13. a;-| = 6, y-- ^ 7 = 8. !- = 10. 14. ^(a;+2!-5)=2/-2, i(a;+;2_5)=2a;-ll, 2a;-ll = 9-(a;+22). . 15. 16. 17. 18. 19. 20. a; + 20 = 12/ + 10, a; + 20 = 2 2 + 5, 22 + 5 = 110 -(2/ + 2). ax + &2/ = Ij 1 62/ + C2 = 1, I C2 + aa; = 1. J cy -i-bz = be, az -^ ex = ca, bx -\- ey = ab. X — ay -\- a-z = a^, X —by -\- bh = b^, X — cy + c^2 = c^. J •'c + 2/ + ^ "~ "^ = H> a; + 2/ - 2 + w = 17, a; — 2/ + 2 + w= 9, ■ X -\- y -\- z -\- u = 12. x-\-y -{-z = % x-\-y-{-u = l, a; + 2 + 1< = 8, 2/ + 2 + ?^ = 9. J SYSTEMS OF FRACTIONAL EQUATIONS. 210. In clearing of fractions the equations of a system, no solution will be lost, but new solutions may be introduced even when we clear of fractions in the simplest manner. Ex. 1. Solve the system 4 a; — 2 ?/ =2, 5x + l 3y-l 8 Clearing (2) of fractions and transposing, we have 40 a: -33?/ =-19. 0) (2) (3) (a) SYSTEMS OF LINEAR EQUATIONS 217 The solution of system, (1) and (3), is 2, 3. In clearing (2) of fractions we multiplied by the unknown factor 3 y — 1 ; hence any solution which was introduced will be a solution of the equation 3y — 1=0, or3y + 0x — 1 = 0. Since 2, 3 is not a solution of this equation, it was not introduced in clearing (2) of fractions. Hence 2, 3 is the one and only solution of system (a). Ex. 2. Solve the system 5 « — y = 2, (1) ^ 4-^^ = 0. (2) (a) x-1 y-S Clear (2) of fractions, z-\-y = 4. (3) The solution of system, (1) and (3), is 1, 3. To clear (2) of fractions we multiplied by the unknown factor (a;— l)(y— 3), and 1, 3 is a solution of the equation (x— l)(y— 3) = 0. Hence the solution 1, 3 may have been introduced by clearing (2) of fractions. By trial we find that 1,3 is not a solution of (2) ; hence the solu- tion 1, 3 was introduced, and system (a) has no solution ; that is, its equations are inconsistent. 211. A system of fractional equations which are linear in the reciprocals of their unknowns is readily solved without clearing of fractions, by treating these reciprocals as the unknowns. («) Ex. 1. Solve the system a/x + c/y = :m, (1) b/x + d/y = : n. (2) Multiply (1) by 6, ab(l/x) + cbil/y) = : bm. (3) Multiply (2) by a, ab(\/x) + ad(\/y) = -- an. (4) Subtract (4) from (3) , (6c - ad) (1 /y) = : 6m- - an. .'. y = 6c- bm - -ad -an (5) Multiply (1) by d, ad(\/x) + cd(\/y) = - dm. (6) Multiply (2) by c, 6c(l/x) + cd(l/?/) = = C7l. (7) Subtract (7) from (6), {ad - 6c)(l/x) = - dm - - en. .'. x = ad- -be (S) dm 218 ELEMENTS OF ALGEBRA Multiplying (1) and (2) by xy to clear them of fractions would give us a system of quadratic equations and introduce the new solution 0, 0. Ex. 2. Solve the system 36, 1 + 1 + 1 X tj z 1 + 5-1 = 28, X y z ^y 2z 20. Subtract (1) from (2), 2 (\/y)-2 (l/z) = -S, Subtract (3) from (1), ^ (l/y)+ l(l/z')= W. Solving system (6), y = 1/12, = 1/16. From (1), (6), and (7), x = 1/8. (1) (^) (3) 4) J 5) J (4) ( (6) (7) (8) (a) (&) Solve the system Exercise 84. 1. «-» = !, X y 5. ^ + 1« = 79, X y 15+6^7. X y 15-1 = 44. a; ?/ 2. 5_4=2, X y 6. 6 7 „ 18 + 18 = 10. X y 2 + 11 = 3. ^ y 3.. 5_5 = 9j a; y 7. ^ + 2 -7 1 1-2 = 5. X y 7 1 -3. 6* 10?^ J 4. 5 + ^ = 3, a; y 8. 14-2-3 1 2X + 32,-'' 5-^ = 1 4a; oy 9. SYSTEMS OF LINEAR EQUATIONS 219 9. - +- = a, X y X y 10. m n mr 11. X y n 7i_m_ nr X y m mx ny mx ny 12. - + f = 2, ax by — -- = 7. ax by 13. A_2 ^4, ax by a; 2^ 14. 1 = m + 71, ?ia; ?n.y X y 1 2 15. i_-_j_4 = 0, a; ^ 1-1 + 1=0, 2/ 2 5 + 3 2! a; 14. 16. 1 + 1 + 1 X y z 36, 1 + ^-1 = 28, X y z a; 3y 2z 17. -^ + -^ J_ J J^^l 2x~^4.y 3z 4' 1^ 1^ X Sy i-^ + i = 2A. X by z 18. 3a; + 4^ = 11, 3y + l 7* 19. 2x-\-A = 1, = 2. 3y-l lx-2 .V + 3 20. x±2y±J^^ Sx-\-y-l ' 3x+x±l = i 4a;-2/-2 21. 5a; + 2y 9a; + 4y 2a; ^2 7a; + 2/ 5* 22. 5x — Sy = S, 1 +^- = 0. a; — 3 y — 4: CHAPTER XV PROBLEMS SOLVED BY SYSTEMS 212. A determinate jrroblem is one which has a finite num- ber of solutions. Every determinate problem must contain as many independent consistent conditions, expressed or implied, as unknown numbers. If in any such problem we denote each unknown by a letter, and express each condition by an equation, we shall obtain as many independent con- sistent equations as there are unknowns. The solutions of the system of equations thus obtained will give the solutions of the problem. Prob. 1. Find two numbers such that twice the greater exceeds three times the less by 6, and that twice the less exceeds the greater by 2. Let X = the greater number, and y = the less. Then, by the first condition, we have 2x-Sy = Q, (1). and by the second condition we have \ (a) 2y-x = 2. (2) J From system (a), x = 18, the greater number ; and y = 10, the less number* Prob. 2. A number expressed by two digits is equal to six times the sum of its digits, and the digit in the tens' place is greater by one than the digit in the units' place. Find the number. Let X = the digit in tens' place, and y = the digit in units' place. 220 PROBLEMS 221 Then, from the first condition, we have 10x + i/ = C(x + ?/), (1)- and from the second condition we have ■ (a) x-y = l. (2) . From system (a), a; = 5, the digit in tens' place ; and 2/ = 4, the digit in units' place. That is, the required number is 54. Prob. 3. If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes equal to 5/8, and if the numerator and denominator are each diminished by 1, it becomes equal to 1/2. Find the fraction. Let X = the numerator, and y = the denominator ; then, from the first condition, ^-±-? = -, (1) ] and from the second, - — - = -• (2) J The solution of sy.stem (a) is 8, 15 ; hence the fraction is 8/15. Prob. 4. A man and a boy can do in 15 days a piece of work which would be done in 2 days by 7 men and 9 boys. How long would it take one boy or one man to do it. Let X = the number of days it would take one man to do the whole work, and y = the number of days it would take one boy. Let the whole work be represented by 1. Then in one day a man would do 1/x of the work, and a boy l/y of it. Hence, by the first condition, we have 15/x + Vo/y = 1, (1) ^ and by the second condition we have [ (a) 14/x + 18/?/ = 1. (2) J The solution of system (a) is 20, 60. Hence one man would do the work in 20 days, and one boy in 60 days. 222 ELEMENTS OF ALGEBRA Exercise 85. 1. Six horses and 7 cows can be bought for $1250, and 13 cows and 11 horses can be bought for $2305. Find the value of each animal. 2. Four times B's age exceeds A's age by 20 years, and \ of A's age is less than B's age by 2 years. Find their ages. 3. Find a fraction such that if 1 be added to its denomi- nator it reduces to |-, and if 2 be added to its numerator it reduces to f . 4. A man being asked his age, replied : " If you take 2 years from my present age the result will be double my wife's age, and 3 years ago her age was \ of what mine will be in 12 years." Find their ages. 5. One-eleventh of A's age is greater by 2 years than \ of B's, and twice B's age is equal to what A's age was 13 years ago. Find their ages. 6. In 8 hours A walks 12 miles more than B does in 7 hours ; and in 13 hours B walks 7 miles more than A does in 9 hours. How many miles does each walk per hour ? 7. At an election the majority was 1G2, which was ^\ of the whole number of voters. What was the number of the votes on each side ? 8. A and B have $ 250 between them ; but if A were to lose half his money^ and B | of his, they would then have only $ 100. How much has each ? 9. A man bought 8 cows and 50 sheep for $ 1125. He sold the cows at a profit of 20%, and the sheep at a profit of 10%, and received in all $1287.50. What was the cost of each cow and of each sheep ? 10. Twenty-eight tons of goods are to be carried in carts and wagons, and it is found that this will require 15 carts and 12 wagons, or else 24 carts and 8 wagons. How much can each cart and each wagon carry ? PROBLEMS 223 11. A and B can perform a certain task in 30 days, work- ing together. After 12 days, however, B was called off, and A finished it by himself 24 days after. How long would each take to do the work alone ? 12. Find the fraction such that if you quadruple the numerator and add 3 to the denominator the fraction will be doubled, but if you add 2 to the numerator and quadruple the denominator, the fraction will be halved. 13. The first edition of a book had 600 pages, and was divided into two parts. In the second edition J of the second part was omitted and 30 pages were added to the first part. The change made the two parts of the same length. How many pages were in each part in the first edition ? 14. A marketman bought eggs, some at 3 for 5 cents, and some at 4 for 5 cents, and paid for all $ 5.60; he afterwards sold them at 24 cents a dozen, clearing $ 1.80. How many eggs did he buy at each price ? 15. In a bag containing black and white balls, half the number of white is equal to a third of the number of black ; and twice the whole number of balls exceeds 3 times the number of black balls by 4. How many balls does the bag contain ? 16. A crew that can row 10 miles an hour down a river, finds that it takes twice as long to row up the river as to row down. Find the rate of the current. 17. A certain number between 10 and 100 is 8 times the sum of its digits, and if 45 be subtracted from it the digits will be reversed. Find the number. 18. If A were to receive $ 50 from B, he would then have twice as much as B would have left; but if B were to receive $50 from A, B woukl have 3 times as much as A would have left. How -much has each? 224 ELEMENTS OF ALGEBRA 19. A farmer sold 30 bushels of wheat and 50 bushels of barley for $93.75. He also sold at the same prices 50 bushels of wheat and 30 bushels of barley for $ 96.25. What was the price of the wheat per bushel ? , 20. One rectangle is of the same area as another which is 6 yards longer and 4 yards narrower; it is also of the same area as a third, which is 8 yards longer and 5 yards narrower. What is the area of each ? 21. A boy rows 8 miles with the current in 1 hour 4 min- utes, and returns against the current in 2|- hours. At what rate would he row in still water ? What is the rate of the current ? 22. A, B, C, D have $1450 among them; A has twice as much as C, and B has 3 times as much as D ; also C and D together have $ 250 less than A. Find how much each has. 23. A, B, C, D have $1350 among them ; A has 3 times as much as C, and B 5 times as much as D ; also A and B together have $ 250 less than 8 times what has. Find how much each has. 24. A number consists of 2 digits followed by zero. If the digits be interchanged, the number will be diminished by 180 ; if the left-hand digit be halved, and the other digit be interchanged with zero, the number will be diminished by 454. Find the number. 25. A train travelled a certain distance at a uniform rate ; had the speed been 6 miles an hour more, the journey would have occupied 4 hours less ; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance. Let X = the number of miles the train runs per hour, and y = the number of hours the journey takes. Then xi/ =(x + 6)(?/ - 4), ] and xy=ix-Q)(y + 6).\ PROBLEMS 225 26. A traveller walks a certain distance ; had he gone ^ mile an hour faster, he would have walked it in ^ of the time ; had he gone ^ mile an hour slower, he would have been 2i hours longer on the road. Find the distance. 27. A man walks 35 miles, partly at the rate of 4 miles an hour, and partly at 5 ; if he had walked at 5 miles an hour when he walked at 4, and vice versa, he would have covered 2 miles more in the same time. Find the time he was walking. 28. A fishing-rod consists of two parts ; the length of the upper part is f- that of the lower part ; and 9 times the upper part together with 13 times the lower part exceeds 11 times the whole rod by 36 inches. Find the lengths of the two parts. 29. A man put $12,000 at interest in three sums, the first at 5 per cent, the second at 4 per cent, and the third at 3 per cent, receiving for the whole $ 490 a year. The sum at 5 per cent is half as much as the other two sums. Find each of the three sums. 30. A, B, and C can together do a piece of work in 30 days ; A and B can together do it in 32 days ; B and C can together do it in 120 days. Find the time in which each alone could do the work. 31. A certain company in a hotel found, when they came to pay their bills, that if there had been 3 more per- sons to pay the same bill, they would have paid f 1 each less than they did; and if there had been 2 fewer persons, they would have paid $ 1 each more than they did. Find the number of persons, and the number of dollars each paid. 32. A railway train, after travelling 1 hour, is detained 30 minutes, after which it proceeds at f of its former rate, and arrives 20 minutes late. If the detention had occurred 10 miles farther on, the train would have arrived 5 minutes later than it did. Find the first rate of the train, and the distance travelled. 226 ELEMENTS OF ALGEBRA Let X = the number of miles the train at first ran per hour ; and y = the number of miles in the whole distance travelled. Then y — x = the number of miles to be travelled after the de- tention, = the number of hours required to travel y — x miles at the rate before the detention, and ^^^ — -^^ = the number of hours required to travel y — x miles at the rate after the detention. y — X X 4(2/- -X) -X 4(y -x) , c 5x _io go' Hy-x- 10) 5 5a; 60 10_ X 40. 5x _ 5 60" .-. x = 24 t, y- = 44. Hence ^Lr^ _ li^^^_2 = i!£. (i) X 5x CO ■ ^ ^ Similarly, y - x - 10 _ 4(y -^x - 10) ^ |_^ ^2^ Subtract (2) from (1), Hence the first rate was 24 miles an hour, and the distance travelled was 44 miles. 33. A railway train, after travelling 1 hour, meets with an accident which delays it 1 hour, after which it proceeds at I of its former rate, and arrives at the terminus 3 hours behind time ; had the accident occurred 50 miles farther on, the train would have arrived 1 hour 20 minutes sooner. Find the length of the line, and the original rate of the train. Ans. 100 miles, 25 miles per hour. 34. A jockey has 2 horses and 2 saddles. The saddles are worth $ 15 and $ 10 respectively. The value of the better horse and better saddle is | that of the other horse and saddle ; and the value of the better saddle and poorer horse is |f that of the other horse and saddle. Find the worth of each horse. 35. Five thousand dollars is divided among A, B, C, and D. B gets half as much as A ; the excess of C's share over D's share is equal to ^ of A's share, and if B's share PROBLEMS 227 ■were increased by $ 500 he would have as much as C and D have between them. Find how much each gets. 36. A i^arty was composed of a certain number of men and women, and, when 4 of the women were gone, it was observed that there were left just half as many men again as women ; they came back, however, with their husbands, and now there were only a third as many men again as women. What was the original number of each ? 37. Two vessels contain mixtures of wine and water; in one there is 3 times as much wine as water, in the other 5 times as much water as wine. Find how much must be drawn off from each to fill a third vessel which holds 7 gal- lons, in order that its contents may be half wine and half water. 38. There is a number of 3 digits, the last of which is double the first ; when the number is divided by the sum of the digits, the quotient is 22 ; and when by the product of the last two, 11. Find the number. 39. Some smugglers found a cave which would exactly hold the cargo of their boat ; viz. 13 bales of silk and 33 casks of rum. While unloading, a revenue cutter came in sight, and they were obliged to sail away, having landed only 9 casks and 5 bales, and filled J of the cave. How many bales separately, or how many casks, would it contain ? 40. There are 2 alloys of silver and copper, of which one contains twice as much copper as silver, and the other 3 times as much silver as copper. How much must be taken from each to weigh a kilogram, of which the silver and the copper shall be equal in weight ? 41. A person rows a distance of 20 miles, and back again, in 10 hours, the stream flowing uniformly in the same direction all the time ; and he finds that he can row 2 miles against the stream in the same time that he rows 3 miles with it. Find the time of his going and returning. 228 ELEMENTS OF ALGEBRA 42. A and B can do a piece of work in m days, A and C can do the same piece in n days, and B and C can do it in p days. Find in how many days each can do the work. 43. For $26.25 we can buy either 32 pounds of tea and 15 pounds of coffee, or 36 pounds of tea and 9 pounds of coffee. Find the price of a pound of each. 44. A pound of tea and 3 pounds of sugar cost $1.50; but if sugar were to rise 50 per cent, and tea 10 per cent, they would cost $ 1.75. Find the price of tea and sugar. 45. A person possesses a certain capital which is invested at a certain rate per cent. A second person has $5000 more capital than the first person, and invests it at 1 per cent more; thus his income exceeds that of the first person by $400. A third person has $7500 more capital than the first, and invests it at 2 per cent more; thus his income exceeds that of the first person by $ 750. Find the capital of each person and the rate at which it is invested. 46. Two plugs are opened in the bottom of a cistern con- taining 192 gallons of water ; after 3 hours one of the plugs becomes stopped, and the cistern is emptied by the other in 11 more hours ; had 6 hours occurred before the stoppage, it would have required only 6 hours more to empty the cistern. How many gallons will each plug-hole discharge in an hour, supposing the discharge uniform ? 47. A certain number of persons were divided into 3 classes, such that the majority of the first and second classes together over the third was 10 less than 4 times the majority of the second and third together over the first ; but if the first class had 30 more, and the second and third together 29 less, the first would have outnumbered the last 2 classes by 1. Find the number in each class when the whole num- ber was 34 more than 8 times the majority of the third class over the second. PROBLEMS 229 48. Two persons, A and B, could finish a work in 7n days ; they worked together n days, when A was called off, and B finished it in p days. In what time could each do it ? 49. The fore-wheel of a carriage makes 6 revolutions more than the hind-wheel in going 120 yards ; if the circum- ference of the fore-wheel be increased by \ of its present size, and the circumference of the hind-wheel by \ of its present size, the 6 will be changed to 4. Required the cir- cumference of each wheel. CHAPTER XVI EVOLUTION. IRRATIONAL NUMBERS 213. An nth root of a given number is a number whose nth power is equal to the given number. U.g., one second root of 4 is 2, since 2^ = 4. Another second root of 4 is — 2, since (— 2)2 = 4. A third root of — 8 is — 2, since (— 2)^ = — 8. A second root of a number is usually called a square root ; and a f^irc? root a c«6e root. 214. The radical sign, ^, written before a number, denotes a root of that number. The radicand is the number Avhose root is required. The index is the number which, written before and a little above the radical sign, indicates ivhat root is required. When no index is written, 2 is understood. E.g., -y/lG or ^16 denotes a second, or square, root of 16 ; 16 is the radicand, and 2 is the index. The expression ^u denotes an nth root of u ; u is the radicand, and n the index. 215. Since by definition {-^uY = u, it follows that ^u is one of the n equal factors of u. 216. A rational, or commensurable, number is any whole or fractional number. A rational expression is one which ca7i be written without using an indicated root. All the expressions in the previous chapters are rational expressions. 230 EVOLUTION 231 217. A perfect nth. power is a number or expression whose »ith root is a rational number or expression. E.g., since ^'25 = 5, 25 is 2i perfect square. Since V— Sa;^^ = — 2 icy^, _ 8 ic^?/^ is a perfect cube. Prior to § 238 each radieand will be a perfect power of a degree equal to the index of the root. 218. Two roots are said to be like or unlike according as their indices are equal or unequal. An even root is one whose index is even; as, -^a?. An odd root is one whose index is odd; as, -y/27. 219. Number of roots. (i) An arithmetic niuiiber has one, and only one, nth root. Any odd power of a positive or negative base has the same quality as the base itself; hence, (ii) A positive or a negative number has one odd root of the same quality as the number itself. E.g., one value of \/+ 27 is + S, since (+ 3)3 = + 27. Again, one value of v^— 32 is — 2, since (— 2)^ = — 32. If two numbers, opposite in quality, are arithmetically equal, their like even powers are the same positive number ; hence, (iii) A jyositive number has two even roots, which are arith- metically equal, and opjwsite in quality. E.g., two values of V+81 are + 9 and - 9, since (+ 9)2 or (- 9)'^ is + 81. Again, two values of \/+ 81 are +3 and —3, since (+3)* or (_3)Ms +81. Any even power of a positive or a negative number is positive; hence an even root of a negative number cannot be a positive or a negative number. Even roots of negative numbers give rise to new quality- numbers,, which will be considered in Chapter XVIII. 232 ELEMENTS OF ALGEBRA 220. The principal root of a positive number is its positive root. The principal odd root of a yiegative number is its negative root. E.g., + 4 is the principal square root of 16, and — 3 is the principal cube root of — 27. Unless the contrary is stated, the radical sign will hereafter be understood as denoting only the principal root. 221. The like principal roots of equal numbers are equal ; hence, The like principal roots of identical expressions are identical expressions. 222. Evolution is the operation of finding any required root of a number or expression. In the statement of the following principles of joots, by " the root " is meant " the principal root." 223. The exponent of any base in the root is equal to the exponent of that base in the radicand divided by the index of the root; and coyiversely. That is, ^a"'" = a'". (1) Proof By § 118, {aTf = a"*". Hence, by ? 221, a"* = -^a"*", and conversely (1). E.g., ^a^ = a^^^ = a^ ; ^x^^ = x^. 224. The nth root of a product is equal to the product of the nth roots of its factors ; and conversely. That is, V~^b=ya'^b. (1) Proof By § 119, {^a • ^by =(^ay(-^by=ab. Hence, by § 221, -^a • -y/b = Vab, and conversely (1). Ex. 1. v/- 32 aio = V- 32 • ^a^^ = -2a^. Ex. 2. V- aPh^ = i5 25 Ve 18 -• Vf-i- -• \fl- -• \ 226. 77ie s^7i roo^ of the rth power of a number is equal to the rth power of its sfh root ; and conversely. That is, ^'a^^{-^ay. (1) Proof Let /8r^«c*)3 = (9x«c2)3 = 729x3«c6. 227. The sth root of the qth root of a nmnber is equal to the qsth root of the number ; that is, Proof If a number is resolved into q equal factors, and then each one of these q equal factors is resolved into s equal factors, the number will be resolved into qs equal factors ; that is, , / y^ _ gy^ Ex. 1. ^ v(2^ ^^y^'^) = \/(2® ^y^^) = 2 xy^- § 227 Ex. 2. ^ V(25-^ X 93) = 3 GOa;^ 2/^ 80^3 2/^ 90 ar^ 2/^ + 8a f 108 a; y EVOLUTION 243 If the terms of a perfect fourth power are arranged in descending (or ascending) powers of some letter, from (1) it follows that the first term of the root is the fourth root of the first term of the radicand ; that the second term of the root is the second term of the radicand divided by four times the cube of the fii'st term of the root ; and that the last term of the root is the fourth root of the last term of the radicand. Similarly for any other higher root, E.g.^ the first term of the fourth root of 81a;*+ 108x3+54x2 + 12a; + l (1) is v^81 X*, or 3ic ; the second term is 108 x^ ^ 4(3x)8, or 1, which we know to be the last term of the root. Since (3 a; +1)*= the radicand (1); 3ic+l is the fourth root of (1). Again the first term of the fifth root of 32x6- 80x* + 80x8 -40x2+ lOx-1 ^ (2) is \/32x^, or 2 X ; the second term is - 80x* h- 5(2 x)*, or — 1, which we know to be the last term of the root. Since (2x - 1)5 = the radicand (2); 2x - 1 is the fifth root of (2). The fourth root can also be obtained by finding the square root of the square root; and the sixth root, by finding the cube root of the square root. Similarly for any other root whose index is not a prime number. Exercise 00. By inspection find the fourth root of the expressions : 1. 16 a* - 96 a»aj + 216 aV- 216 aa^ + 81 a;*. 2. x^-%a?a + 24.a?a'-^2xa^ + Ua\ 3. 1 + 4 a 4- 4 a^-f- 10 a«+ a«+ 10 a^-\- 16 a^-f 16 a*+ 19 a\ By inspection find the fifth root of the expressions : 4. 80 a^a^ - 80 ax^ + 32 a.-* - 40 a^x - a^ + 10 a^x. 6. 90aV~16aa^ + a^-270aV + 405a*aj-243a«. 244 ELEMENTS OF ALGEBRA By inspection find the sixth root of the expressions : 6. 192 a; + 64 + 240 a^^ _^ ^.e _^ i2a^ + 60 a;^ + 160 a:^. 7. 1215 a* - 1458 a' - 540 a^ + 135 a^ - 18 a + 1 + 729 a«. 8. 60 a'x'- 16 a^x^-{- 64 a^-^ x^- 12 aa^+ 240 aV- 192 a'x. ROOTS OF DECIMAL NUMBERS. 233. Square roots. ^1 = 1, and ^10^ = 10 ; hence the square root of any number between 1 and 100 lies between 1 and 10 ; that is, if a number contains one or two integral figures, its square root contains one integral figure. Again, ^100 = 10, and V^O^OO = 100 ; hence the square root of any number between 100 and 10000 lies between 10 and 100 ; that is, if a number contains three or four integral figures, its square root contains two figures ; and so on. Hence, in finding the square root of a decimal number, the first step is to divide its integral figures into groups of two figures each, beginning at units' place. We thus determine the number of integral figures in the root, and indicate the part of the number from which each figure of the root is to be obtained. The group to the left may contain only one figure. E.g., in the square root of 5 38 24 there are hundreds, tens, and units ; and the hundreds' figure is the square root of the greatest per- fect square in 5 ; that is, the hundreds' figure is 2. 234. From § 233, we have the following principle : (i) The first figure in the square root of a decimal number is the square root of the greatest perfect square in the first, or left-hand, group of figures. Let A stand for the number denoted by one or more of the first figures of the root, and B stand for the number denoted by the rest ; then the root is A-}- B. Hence the radicand = A^-\-(2A + B)B. (1) Subtract A^ and then divide by 2 ^ + J5, (radicand -A^^{2A-^B)=B. (2) EVOLUTION 245 Let h stand for the number denoted by the first figure in B, i.e., the figure whose order of units is the next lower to the lowest in A ; then from (2) (radicand -A^-i-2A>b. (3) From inequality (3), we have the following principle : (ii) If the square of the first part of the root is subtracted from the radicand, and the remainder is divided by twice this part of the root, the quotient will be greater than the next figure of the root. E.g., by (i), W\Q first, or hundreds, figure in the square root of 5 47 56 is 2 ; since 4 is the greatest perfect square in 5. The radicand less (200)2 is 14756. Hence, by (ii), the tens figure of the root cannot exceed 14756 -2(200), or 3 tens. The radicand less (230)2 is 1856 ; hence the root is greater than 230, and 3 is the tens figure of the root. By (ii), the units figure of the root cannot exceed 1856 H- 2(230), or 4. The radicand less (234)2 jg zero. Hence the square root of 54756 is 234. Instead of finding each square independently, much labor can be saved by using the relation (§ 229) A'-\-{2A-^b)b={A^by, and thus making use of the previous square. The work in the example above is usually written as below : 5 47 56(234 ^2 =(200)2= 4 00 00 2^ + 6 = 2(200) + 30 = 430)14756 (2 ^ + 6)& = 430x30= 129 00 2^ + 6=2(230)4- 4= 464)18 56 (2^ + 6)6 = 464x 4= 18 56 246 ELEMENTS OF ALGEBRA At first A = 200. Subtracting A^, or 200^, from the radicand, and dividing the remainder, 14756, by 2 A, or 2(200), we find that the tens figure of the root cannot exceed 3. Multiply 2 A + b, or 430, by 6, or 30, and subtract the product ; then in all we have subtracted A- + (2 A+b)b, or (A-\-by; that is, 230"'2. Now let A = 230, the part of the root already found, and b = the next figure of the root. Dividing the remainder 1856 by 2 A, or 460, we find that the units figure of the root cannot exceed 4. Multiply 2 ^ + 6, or 464, by b, or 4, and subtract the product ; then in all we have subtracted A^ + (2A + b)b, or (A + 6)2 ; that is, (234)-^. Omitting the ciphers and explanation, and in each remainder writing the next group of figures only, the work will stand as below : 5 47 56(234 4 43)147 129 464)18 56 18 56 235. If a number has decimal places, its square will have twice as many. E.g., O.S^ = 0.64 ; 0.25^ = 0.0625. Hence to determine how many decimal figures there will be in the square root of a number, we divide its decimal figures into groups of two figures each, beginning at the decimal point. If the group to the right does not contain two figures, a cipher must be annexed. Ex. Find the square root of 5727.2976. Formula, A^ +(2 A + b)b = (A + by. 75 27.29 76(86.76 64 166)1127 9 96 1727)13129 120 89 17346)10 40 76 10 40 76 Here at first A = SO, b = 6 ; next ^ = 86, 6 = 0.7 , next A = 86.7, b = 0.06. EVOLUTION 247 Exercise 01. Find the square ! root of the numbers : 1. 2916. 9. 29376400. 17. 0.0022448644. 2. 2601. 10. 52.2729. 18. 0.68112009. 3. 17956. 11. 53.7289. 19. 25/49. 4. 33489. 12. 883.2784. 20. 64/81. 5. 119025. 13. 1.97262025. 21. 121/36. 6. 15129. 14. 3080.25. 22. 144/49. 7. 103041. 15. 41.2164. 23. 169/196. 8. 835396. 16. 384524.01. 24. 225/289. 236. Cube root. Since ^1 = 1, and ^1000 = 10, it fol- lows that the cube root of any number between 1 and 1000 lies between 1 and 10 ; that is, if a number contains one, two, or three integral figures, its cube root contains one integral figure. Again, ^1000 = 10, and ^1000000 = 100; hence, if a number contains four, five, or six integral figures, its cube root contains two integral figures ; and so on. Hence, to determine how many integral figures there are in the cube root of a number, we divide its integral figures into groups of three figures each, beginning at units' place. The last group to the left may contain only one or two figures. When the figures of a number have been divided into groups of three figures each, from what precedes it follows that, (i) The first figure in the cube root of a decimal number is the cube root of the greatest cube in the first, or left-hand^ grouj) of figures. Using a notation analogous to that in § 234, we have radicand - A'' = (3 A' -^ 3 AB + B^ B. '. (radicand - .1'') - (3 A- -\- 3 AB -^ B^ = B. .-. (radicand -yl^) --3. 42 > 6. (1) 248 ELEMENTS OF ALGEBRA From inequality (1) it follows that, (ii) If the cube of the first part of the root is subtracted from the radicand and the remainder is divided by^ three times the square of this part of the root, the quotient will be greater than the next figure of the root. E.g., by (i), the Jirst or tens^ figure in the cube root of 614 125 is 8, since 8^, or 512, is the greatest perfect cube in 614. The radicand less (80)^ is 102125. Hence, by (ii) , the units' figure of the root cannot exceed 102125 -- 3(80)2, or 5. The radicand less (85) ^ is zero. Hence, the required root is 85. Instead of finding each cube independently, much labor can be saved by using the relation A' -{- {S A' + 3 Ab -\- b')b = (A + by, and thus making use of the previous cubes. Thus, the work in the example above is usually written as below, without the explanations to the left : 614 125(85 43 = 512 000 3^2^3(80)2 =19200) 3^6 = 3.80-5= 1200 62 = 52 = 25 20425 102 125 102 125 237. If a number has decimal places, its cube will have three times as many. Thus 0.2^ = 0.008 ; 0.12^ = 0.001728. Hence, to determine how many decimal figures there will be in the cube root of a number, we divide its decimal figures into groups of three figures each, beginning at the decimal point. If the group to the right does not contain three figures, ciphers must be annexed. lEBATIONAL NUMBERS 249 Ex. Find the cube root of 129554.216. Formula, A^ -\-(iS A^ + S Ab + b'^)b = (A + by. 129 554.216(50.6 125 750000 9000 36 759036 4 554 216 4 554 216 Here at first ^ = 50, 6 = 0; next A = 50.0, b = 0.6. Exercise 92. Find the cube root of the numbers : 1. 74088. 2. 15625. 3. 32768. 4. 110592. 5. 262144. 6. 1481544. 7. 103.823. 8. 884.736. 9. 1953125. 10. 7077888. 11. 2.803221. 12. 12.812904. 13. 56.623104. 14. 264.609288. 15. 1076890625. 16. 8/27. 17. 64/125. 18. 343/1728. INCOMMENSURABLE ROOTS, OR IRRATIONAL NUMBERS. 238. The nth power of a whole number is evidently a whole number which is a perfect ?ith power ; and the ?ith power of a fraction (whose numerator and denominator are prime to each other) is a fraction whose numerator and de- nominator are perfect nth. powers prime to each other. Hence, it follows that (i) The nth root of a tvhole mimber which is not the nth poicer of another whole number is not a commensurable number. (ii) The nth root of a fraction whose numerator and ck^ nominator {prime to each other) are not the nth powers oj whole numbers, is not a commensurable number. 250 ELEMENTS OF ALGEBRA E.g., as 2 is not the square of any whole number, y/2 is not a com- mensurable number, and therefore is not as yet included in our number system. The same is true of ^3, y/b, ^1 •••. Again, as the terms of the fraction 2/3 are prime to each other and are not the squares of whole numbers, ^(2/3) is not a commensurable number. 239. To enlarge our number concept so as to give mean- ing to such expressions as ^2, ^5, etc., we assume the identity to hold when the radicand u is not a perfect /ith power. E.g.., yJ2 is the number whose square is 2, i.e. ( \/2)^ = 2. Again, ^5 is the number whose cube is 5, i.e. {^b)'^ = 5. 240. The ?ith root of a number which is not a perfect nth power is called an incommensurable root or an irrational number; as, V2, V3. 241. An irrational number, or any other number which is not a whole or a fractional number, is called an incommen- surable number ; as ^3, -^6, or the ratio of the circum- ference of a circle to its diameter. 242. Approximate values of incommensurable roots. If to 2 we add ciphers and apply the method of finding the square root, we obtain the result below : 2.00000000 )1.4142 ... 1_ 24)100 1st remainder 90 281)400 2d remainder 281 2824)11900 3d remainder 11290 28282)60400 4th remainder 56564 0.00003836 5th remainder IRRATIONAL NUMBERS 251 Each remainder in the above process is the difference be- tween 2 and the square of the corresponding part of the root. This remainder decreases rapidly as we increase the num- ber of figures in the root; hence the square of the root found approaches nearer and continually nearer 2; and therefore the root itself approaches nearer and continually nearer ^2. By continuing the operation indefinitely we obtain a com- mensurable number which approaches indefinitely near and continually nearer ^2, but which, by § 238, can never reach ^2. This increasing commensurable number is said to approach the incommensurable root ^2 as its limit. In like manner we can find a commensurable number which shall differ from any incommensurable root by as little as we please. Exercise 03. Obtain to three places of decimals the value of the roots : 1. V3- 4. V^- 7. V^-3- 10. V^-^^-*- 13. ^2. 2. V^- 5. Vll- 8- V^>-^- 11- V^-^^- 14. ^4. 3. V^. 6. V13. 9. V^-^^- 12- V^-^- 15- v^2.5. 243. The quality-unit + 1 or — 1 multiplied by an arith- metic incommensurable number is a positive or a negative incommensurable number ; as, + ^2, — ^3, — ^5. 244. The fundamental laws which have been proved for commensurable numbers hold also for incommensurable numbers. The proof of these laws for incommensurable numbers will be found in the chapter on the theory of limits. 245. An irrational expression is one which involves the nth root of an expression which is not a perfect nth power ; as, ^^i V(^ + ^)- ^^y i^'^'^lional numeral expression denotes an irrational number. But, just as a fractional literal 252 ELEMENTS OF ALGEBRA expression denotes both integral and fractional numbers, so an irrational literal expression denotes both rational and irrational numbers. E.g., the irrational literal expression ^a denotes a commensurable or rational number, when a = 1, 4, 9, 1/4, 4/9, ••• and an incommen- surable or irrational number when a = 2, 3, 5, •••. Observe that commensurable and incommensurable apply to numbers only ; while rational and irrational apply to either numbers or expressions. Exercise 94. 1. Is the number -^jA commensurable or incommensura- ble? sjQ? V9? V12? V14? V(4/9)? ^(8/27)? -3/3? -^216? 2. Is the expression -^a* rational or irrational ? -^x ? ^a^7 V(^ + ^)^ ^{a + xy? ^(a/x)? 3. Give ten sets of values of a and x for which ^(a/x) denotes a commensurable, or rational, number. CHAPTER XVII SURDS 246. A surd number is an irratioual number in which the radicand is a rational number; in other words, it is an incommensurable root of a commensurable number. E.g., y/b, ^7, ^(2/3) are surd numbers; so also is ^a when a denotes a commensurable number which is not a perfect square. The incommensurable root V(^ + \/2) i^ "^^ ^ surd number, since the radicand 3 + ^^2 is not a commensurable number. 247. A surd expression is an irrational expression in which each radicand is a rational expression ; as -yja^ VS/G, 248. Surds of different orders. A surd of the second order, or a quadratic surd, is a surd with the index 2 ; as ^5, ^a. A surd of the nth order is a surd with the index n ; as -y/a. Observe that -^ ^5 is a surd of the 6th order. 249. A rational number or expression can be written in the form of a surd of any order. E.g., 3 = v^, ^7, {/81, or ^43 and «= V«^ \/«^ v^«S ^^ v^«^- 250. A surd expression is in its simplest form when each radicand is integral, its numeral factor being as small as possible, and its literal factor of as low degree as possible. E.g., the simplest form of the surd y/9> is 2y/2. The simplest form of the surd ^(16 x*y5) is 2 xy^(2 x^). 263 254 ELEMENTS OF ALGEBRA 251. Reduction of surds to their simplest form. The cases which most frequently occur are the three following : I. Radkand integral. Eesolve the radicand into two factors, one of which is a perfect power of a degree equal to the order of the surd, and apply the law, ^{ab) = y/a • ^b. Ex. 1. \/l35 = v'PTs = 3^5. Ex. 2. 7 V50 x'y<' = 7 \/(5 x'^y)^ • 2 y = 7x5 x'^yV2y = 35 x^yV2y, Ex. 3. 5 /75. Ex. 6. X ^x^ = {/7^ X ^a;8 = ^v?. Ex. 6. Multiply 2^3 + 3^2 by 4v3 - 5^2. The work can be arranged as below : 2\/3 + 3V^ 4\/3-5V2 24 + 2V6-30 = 2V6-6. 256. In finding powers of monomial surds we often make use of the law, {-^aj = l? SURDS 259 Obtain the simplest form for each of the following products : 22. 2 V15 X 3 V5. 29. ^168 x ^147. 23. 8V12 X 3V24. 30. 5^128 x 2^'432. 24. Vi2 X V^T X V75- 31. ViO X ^200. 25. ^16 X ^6 X ^9. 32. ^4 x V^'- 26. ^12 +-^75x^30. 33. (V^-V3)(V^ + V3)- 27. ^0x^12x^18. 34. (V^)-V7)(V^' + V)"- 28. v^a; + 2 X -^x - 2. 35. ( V^ - V-*^) ( V^' + V^)- 36. ( - v'c - V«) ( - V^ + V«)- 37. (-V^+V^)(V^+V»')- Find each of the following powers : 38. (y/2f. 42. (^aby. 46. (v'^)«. 39.. (2V3)*. 43. (^ay. 47. (2V^^)*. 40. (Va;)'. 44. (-^'ay. 48. (3^/^2^)*. 41. (-^ay. 45. (-^63)^^ 49. (2^/^::rP)6^ 50. (V3-V5)^ 54. (^2-3. Here the rationalizing factor is ^a*. 8UBDS 261 The simplest rationalizing factor of -s/x"^ is evidently II. When the divisor is a binomial quadratic surd, the simplest rationalizing factor is the conjugate of the divisor. Ex 1 5+V7 ^ (5+V7)(3+v7) ^ 22 + 8v7 ^. ^ • 3-V7 (3-V7)(3 + V7) 9-7 ^' Ex. 2. a/Q^ + a/^ - C a/« + a/^)(V^ + a/^) -- c< + 2 Vo^ + 6 V<^-V^ (V« — \/^)(v'« + V^) a-b III. When the divisor is of the form (^a -f- -y/b) + -y/c, first multiply by the expression ( y/'a + ^b) — ^c. The divisor thus becomes ( V« + V^)' - ( V^)'. or (a + 6 - c) + 2 ^(ab). (1) Next we multiply by the conjugate surd (a 4. 6 _ c) - 2 ^(ab). The divisor thus becomes the rational expression (a-^-b-cf- (2 Vaby. Ex. ^^2 ^ V2(V2-fV3 4-V5) V2+V3-V5 [(V2+V3)-V'5][CV2+V3) + V5] ^ 2+V6 + V/10 ^ 2 + a/6+V10 ^ (2 + yC + ylO) X V6 _ 2^/6 + 6 + 2^15 ^ v/6 + 3 + v/15 12 6 259. When applicable, the identities in Chapter IX. should be used in writing the quotient of two binomial surds. Ex. V^^ + Vy« _ Cv/x)8 + (Vy)'» =x-y/xy -\-y. 262 ELEMENTS OF ALGEBRA Exercise 98. Compute to three places of decimals : 1. 14--V2. 3. 48--V6- 5. 144 --V6. 2. 25 -V^- 4. V2-V^- 6. 4--V243. Rationalize the denominator and simplify : 7. 3V3/(2V2). 11. 12/ V"^- 15- V«/^«- 8. Vl5/V(3/5). 12. 2^6/V2. 16. ^a/^a. 9. V21/V(V3). 10. 10/^5. 19. 2V5 V5 + V3 20. l^+li^^ 15 - 2 V3 21 V5 + 3V3 . ' 2V5-V3 22. 23. 24. 25. 26. V6-3V12 2V<3 + V12 2 V3 + 3 V2 5 + 2V6 V9 + V9 4- x' + 3 r £c -f- Vi^^ — y'^ 1 1 + V2 + V^* 13. 3V2/^9. 17. V x'/^af. 14. -^20/(3^16). 18. ^(ax)/^x. 3 27. 2+V3 4-V5 28. ^+V^+V^. 1+V2-V3 29. 30. 31. 32. 33. 34. V3 V2+v^+V5 3 V2-V6-V'^* ■Vx — 2-\-Vx Va? — 2 — V ic a— Va^ + 3 Va — 6 — Va + b Va — & 4- Va + b 3 + 4V3 V6+V2-V5' SUBDS 268 Write each of the following quotients : 35. (a — x)-i- (y'a — ^x). 36. {ax-ay)/(^x-^^y). 37. (l-l/x)-(l4-l/V'^). 38. (a/b - x/y) h- [ V(« A) + V(V^)]- 39. (v«'-V^')^(V«-V^)- 40. (arV^J- W2/)-^(V^+V2/)• Rationalize the denominator of each fraction : 41. 42. x" 44 Vl + ar^- -VI- X 45. Vl + ar' + Vl- 2 Va + 6 — 3 Va a;- 6 Va^ H- a* H- a 2 Va + 6 — Va — 6 43 V10+V5 4-V3 . 46 (V3+ V5)(Vo + V2) V3+V1<>-V5 * V-+V3 + V5 260. A root of a monomial surd is found by applying the ^^^^ V V« = V«- § 227 Ex. 1. »/ V7 = ^7. Observe that {/{fa= {/{/a, since each member = *^a. Ex.2. Vv^(4x2) = {/V(4a;2)= */(2x). Exercise 09. Simplify each of the following expressions : 1. -^V(27a'). 5. V(^V^)- 9- ^WV^)- 2. V^(9ajO- 6- VA/(25iry/16). 10. ^^--^(x/^x). 3. ^v(«0. 7. V(^V^)- 11- "-V(^/V^O- 4. ^^(A^)- 8. V(2/V2) 12. V(^V/^ = 8V5; §261 or X + y = 18, and xy = 80. (a) By inspection we see that one solution of system (a) is X = 8, y = 10. .'. V18 + 8^6 = V8 + VIO = 2 V2 + VIO. Ex. 2. Extract the square root of 83 — r2y'35. Assume V83 - 12 ^35 =^x- y/y. Square 83 - 12^35 = x + y - 2Vxy. .-. X + y = 83, and 2Vxy = 12 V35, or x + y = 83, and xy = 1260. By inspection, x = 63, and y = 20. .-. V83 - 12V35 = V63 - V20 = 3 V7 - 2y/5. By taking x = 20 and y = 03, we would obtoin the negative root of the given number. Exercise lOO. Find the square root of the binomial surds : 1. 6 + V20. 6. 11-2V30. 11. 4J-IV3. 2. 12-6V3. 7. 7-2V10. 12. 17 - 2 V66. 3. 16 + 6V7. 8. 17-12v'2. 13. 19+8V3. 4. 13-2V42. 9. 47-4V33. 14. 11+4V6. 5. 28-5V12. 10. 19 + 4V22. 15. 15-4V14. CHAPTER XVIII IMAGINARY AND COMPLEX NUMBERS 264. Quality-units V— 1 and — V— 1. A.s we have seen in § 219, an even root of a negative number, as V—2, cannot be a positive or a negative number, and therefore is not as yet included in our number system. To give a meaning to such expressions as V— 1 and V— 2, we assume the identity { 1 / / / / / y B' OA ' V^l = OB. But OA-V^l=(+l).V^^=V'^. From (1), (2), OB = V^^, or i. .'. OB' =- OB = -i. (1) (2) Hence, if the primary quality-unit + 1 is represented by the directed line OA, the quality-units — 1, i, and — i will be represented by the directed lines OA', OB, and OB', respectively. As the lines OB and OB' are just as real as the lines OA and OA', so the quality-units i and — i are just as real as -|- 1 and — 1. Arithmetic multiples of i and — i can be represented by distances along the lines OB and OB or their extensions, just as multiples of -f 1 and — 1 are represented by distances along the lines OA and OA' or their extensions. Again, if in a football game we denote the forces exerted in the direction OA by positive real numbers ; then negative real numbers will denote the forces exerted in the opposite direction OA', positive IMAGINARY NUMBERS 269 imaginary numbers will denote the forces exerted in the direction OB, and negative imaginary numbers will denote the forces exerted in the direction OB'. To express by numbers the magnitudes and directions of the many other forces in the game we need still further to enlarge our concept of quality-numbers, as is done in § 285. 269. Since imaginary numbers are simply arithmetic mul- tiples of the units i and — /, they are added and subtracted the same as real numbers. That is, at ± bi = {a± b)i, (1) which is the converse of the distributive law. Ex. 1. 4 I + C i = (4 4- 6) t = 10 I. Ex. 2. (7/3) i - (0/3) i = (7/3 - 5/3) i = (2/3) i. 270. When the imaginary unit is a factor of a product, the distributive law follows from its converse in § 269, and the associative law follows from the commutative law in § 266 ; that is, (fl ±b)i= ai ± bi, § 269 and ai • bi = i-ab = -ab. § 267 271. Powers of i. From § 264, we have ,^ = -1, i' = ~i, i* = -\-l. (1) Ex. 1. 2-7 = 1-4 . 1-3 = (+ 1) ( _ i) = _ f. l)y (1) Ex. 2. I'lo =(i4)--2i2 =(+ 1)2(- 1) = - 1. by (2) Ex. 3. zi8 =(^iiyi=(^\)Si:^ i^ by ( 1 ) If n is any positive integer including zero, we have ,.4n =(i4y = ^_^iy^_^i. (2) /4«+i_^-4u^- =/. by (2) /4n+2 = ,;4«^-2 = _ 1 ; by (1), (2) /4n+3 = ^•4.^-3 ^ _ / by (1)^ (2) 270 ELEMENTS OF ALGEBRA Hence^ any eve7i power of i is +1 or — 1, and any odd power of i is i or — /. 272. The square root of any yiegative number is an imagi- nary number; that is, V^ = V« • '• (1) Proof By the commutative and associative laws we have ( V« • ^/^l)' = {^a)\-y/^^f = ~a; § 271 hence, y^V^^ = V~a, or conversely (1). § 221 E.g., V- 16 = i . vie = 4 i, or W^l. and y/— ai=i ' Va^ ^ a^, or aV— 1. 273. To add or subtract imaginary numbers given in the form Va, we first reduce them to the typeform, -^a • i. Ex.1. V^l^ -i-y/-Sl-V-3{J = 7i + 9i-ei = (7 + 9_6)i = 10i, orlOV^l. Ex. 2. V-9a2 4. V- 4 &-^ - V- 7 c'^ = 3 a • i + 2 6 • i - c V7 • i = (3a + 26 -Cv/7)i. 274. From the commutative and associative laws we have the following principle : The product of two or more quality-numbers is equal to the product of their quality-units into the product of their arith- metic values. Ex. 1. V- 5(- Vll)= i •-! • V^' \/ll =- ^'V^^i or -V- 55. Ex.2. V^8. \A^ = iV3- V7 =-V21. Ex. 3. V^ . V^ . \/^ = i3y'2 . y'3 . V5= - iy/^^, or - V^M Ex. 4. V^ • V"^ • v^ • v/^ = /I V^ V^^ V^ V' = v^«^- IMAGINARY NUMBERS 271 Exercise 101. Simplify each of the following expressions : 1. V^^+V^=^-V^=loo. 2. V^^-V^^-hV^^^16. 3. v^^^-v^^-v^^n^. 4. V-l)a^-V-4a2_V-l(5a='. 5. V - 36 6^ _ V - 49 6^ _f- V - 81 6^ 6. V — (a; + af.+V— {x — af. 7. 3 V^ + 7 V-c - 11 V-6 + 2 V-(m-?r)- 8. V2V^^. 12. V^^-V^=^. 9. V3-V^^. 13. V^^Vie. 10. V^^ • V^ni. 14. V^=l^ • V^3 . 11. v^^V^. 15. Ve-V^^-V 16. V5 . V^=^ • V^=^ . V^=^. -5. 17. V^=^' • V^=^' • V~^^. 18. V— 3 ax . V— 3 6u; • V— 4 aft. 19. (V:r2 + v^35)(V^^+V^. 20. (2 V"=^ + 3 \/^=^) (5 V^^ - 2 V^=^). 21. (V— a; + V— y)(V— a?— V^^). 24. V^^ 25. (V^)^ 22. V-rt'-^. 23. ( 6)^ 275. Quotient of one quality-unit by another. /^/=l- /-^(+l)^/; /^ (_!)=_/; §84 + 1 ^/=/^--/ = r^ = -/; § 2G4 — 1 -i- / = i^ -^ ij = i. 272 ELEMENTS OF ALGEBRA E.g., 1 ^ i5 = 1 ^ i = _ i ; _ 1 ^ i6 = _ 1 ^ (_ 1) = 1 ; i ^ f = i -^ (— i) = — I. 276. From the commutative and associative laws we "have the following principle : The quotient of two quality-numbers is equal to the quotient of their quality-units into the quotient of their arithmetic values. Ex 1 V^5 _i V5 _ V2. Ex.2. ^^.A.^ = ,^, or^^^^. V7 +1 v/7 7 ' 7 Ex. 3. y/~a/ \/~b = (i/i) ( y/a/y/h) = \/a/b. Ex.4. y/^/V^ = (+l/i)i^a/^b) = -iVa7b, or -\/-(a/6). Exercise 102. Perform the operation of division in 1. i^ -r- i. S. 1 -r- i^ 5. i -J- ^^ 2. ^5^^•. 4. - 1 -m'^. 6. V-14-V-2. 7. V^^ie^V^l. 11. V^'-^V^'. 8. V^=n^^V^^. 12. (V=T2-V-15)-r-V-3. 9. V^^^V~a. 13. (V~a+V--6) ^V~c. 10. V^(^)-^V£c. 14. (Vl6-V8)-=-V^^. 15. i-v-i^l 16. -i-~i''. 17. ^3-^F. 18. i*^ _j. jT-ss^ COMPLEX NUMBERS. 277. The sum of a real number and an imaginary number is called a complex number, as 4: ± 5 i, 7 ± 3 i. The general expression for a complex number is evidently a + bi, where a and b are any real numbers. COMPLEX NUMBERS 273 When 6 = 0, a-\-hi = a, a real number. When a = 0, a-\-hi = hi, an imaginary number. 278. We define addition of complex numbers by assum- ing the commutative, and therefore the associative, law of addition for real and imaginary numbers. Hence, in adding or subtracting complex numbers the real parts can be added or subtracted by themselves and the imroginary parts by themselves. That is, (a + bi) ± (c + di) = (a ± c) -\- (b ± d) L (1) Ex. When is the second member of ( 1 ) a complex number ? When an imaginary number ? When a real number ? 279. If two complex numbers are equal, their reed parts are equal, and their imaginary parts are equal. Proof. Let a-{-bi = c-{-di, (1) where a, b, c, d are all real numbers. Transposing, a — c = di — bi. (2) But, if a real number is equal to an imaginary number, each is zero (§ 267) ; hence, a — c = 0, or a = c, and di — bi = 0, or d = b. An important case of this theorem is the following : If a + bi = 0, then a = and 6 = 0. 280. Two complex numbers which differ only in the signs before their imaginary terms are called conjugate com- plex numbers, as a -f bi and a — bi. Since (a + bi) -\- {a — bi) = 2 a, the sum of two conjugate complex numbers is real. 274 ELEMENTS OF ALGEBRA 281. Multiplication by a complex number is defined by assuming the distributive law ; tliat is, (a H- bi) (c + di) = ac -\- adi -f bci + bdv^ = (ac — bd) 4- (ad -f- be) i. (1) Before multiplying one complex number by another it is convenient to reduce each to the type-form a -f- bi. Ex. (3 + V^^)(4 - V^S) = (3 + v^ • 0(4 - V''^ • = 12 +(4^5- 3^3)1 -fVlS- 282. From (1) in § 281, it follows that the product of two complex numbers is, in general, a complex number. But, the produet of two eonjugate eomplex numbers is real and jjositive. Proof. (a + bi) (a - bi) = a^ - {bif = a^ + b\ E.g., (-3 + V^^)(-3-V^^) = (-3)2-(V32)2:=ll. 283. The quotient of one complex number by another is, in general, a complex number. Proof a + bi _ (a + bi)(c-di) c -f di"~~ (c H- di) (c — di) _ ac + bd be — ad . ^-js ~ c^ + d^ -^ c2 + d=^** ^^ 284. From (1) in § 283, it follows that when the divisor is a complex number, the quotient can be expressed as a complex number by multiplying both the dividend and divi- sor by the conjugate of the divisor. Ex 4 + 3 I ^ (4 + 3 0(3 + 20 S-'2i (3-2 0(3 + 20 ^6 + 17j^A + iIj- 9 + 4 13 13 ' COMPLEX NUMBERS 275 Exercise 103. Find each of the following sums and products : 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. (V^^4-V2) 20. (2 + 3aO'- 2+V-4) + (3-V-l). 3+V-9)-(7-V-16). 3-2 + (6 + 1). 8+V-36)-(o+V-25). 4 + 3 - (3 - 4 1). 2+V^ + (2-V^:r3). 2-hV^^)(3-V^. 4+V-16)(3-V-25). 5 + 2 V^ (2 - 3 V^l). l+V-7)(2-V-T6). 3-hv'^'5)(2-V^. V2 + V^2) ( V3 - V^. 2+V^(2-V^. _4-V^(-4+V^. _7-V^^ll)(-7+V^ll). .rV- X + 2/V^) (xV^^ - 2/V^). V^T 4. 5V^3)(V^^ + 3V^2). 21. (2-3V^2)3. 22. (a -f ci)l 276 ELEMENTS OF ALGEBBA Reduce each of the following expressions to the typical form a -\-hi: 23. 24. 25. 1-V-l 26. 27. 28. a —V— X 1 3 4-4V^5 2_3V^^ 29. 30. 31. 3+2V-1 2 - 3V^=~1 3-4^' 2-V-2 Show that 3)/2]» = + 1 ; also [(- 1 - V=3)/2]-' = + 1 ; and that therefore there are at least three cube roots of +1. 33. In § 281, when is the product a complex number ? When an imaginary number ? When a real number ? 34. In § 283, when is the quotient a complex number ? When an imaginary number ? When a real number ? 285. Geometric representation of a complex number. To find the sum of + 4 and — 1 geometrically, we lay off Oil/ equal to 4 in the positive direction ; then from the end of OM we lay off MB equal to 1 in the nega- tive direction. The straight line OB^ which extends from the origin of the line 03/ to the end B of the line MB, is the sum of + 4 and — 1 ; that is, the sum is the directed line dravm from the origin of the first line to the end of the second. Similarly, to con- struct the sum 4 + 3 1, we lay off OM equal to + 4 ; at il/, the end of OM, we erect a perpendicular and on it, in the COMPLEX NUMBERS 277 direction of positive imaginaries, lay off MP three units long ; then OP, i.e. the directed line extending from the origin O of the first line to the end P of the second line, will represent the complex number 4 + 3 i. From the right-angled triangle OMP we have the length of OP = \AP + 3^ = 5. If we take 011= one unit long and draw HR parallel to MP, then from the similar triangles Olill and OMP we have OB = 4/5 and RH= (3/5)i, that is, the directed unit-line 0^ represents 4/5+ (3/5)2, and OP, which represents 4 + 3 j, is 5 times this unit-line. Hence the arithmetic value of 4 + 3 i is 5, and its quality-unit is 4/5 +(3/5)/, which illustrates § 286. Observe that the quality-unit 4/5 +(3/5)i is obtained by dividing 4 + 3 I by 5, and that 5 = \/42 + 32. In like manner we can represent any other complex number and its quality-unit. 286. The arithmetic value, or modulus, of the complex num- ber a +- hi is the square root of the sum of the squares of a aud 6, or Va^ + ^^; and its quality-unit is (a -f hi)/-\/a--\-h^. That is, a-irU = [(a +- hi)/^a? -f 6^] . Va' + h\ where the complex nnmber a -f hi is written as the product of a quality-unit and an arithme:tic number. E.g. the arithmetic value, or modulus, of 4 + 3i is v42 + 3^, or 5. The modulus of 5 - 3 i is V62+(-3)2, or y/ZA. The modulus of - 1/2 + \/^/2 is V(l/2)2 + (V3/2)2, or 1. CHAPTER XIX QUADRATIC EQUATIONS IN ONE UNKNOWN 287. By the principles of equivalence of equations in Chapter VII. we can derive from any quadratic equation in one unknown, as x, an equivalent equation of the type-form ax' + bx + c = 0. (A) Observe that in (A), ax- is the sum of all the terms in .^^, bx is the sum of all the terms in x, and c is the sum of all the terms free from x. E.g., from the quadratic equation 6 x^ - 3 _ Gx^-S ^ x2- lOx 2 4 8 we derive tlie equivalent equation 13 x2 - 40 X - 9 = 0, (1) which is in the type-form. Comparing (1) with (A) we liave a = 13, 6 = - 40, c = - 9. If a = 0, (A) ceases to be a quadratic equation ; hence in what follows we shall assume that a is not zero. If neither b nor c is zero, (^1) is called a complete quad- ratic equation. If either b or c is zero, or if both are zero, {A) is called an incomplete quadratic equation. When 6 = 0, the incom- plete equation is often called a pure or binomial quadratic equation. E.g., equation (1) is a complete quadratic equation; while 3 x2 + 4 X — 0, 8 ic2 + 9 = 0, and -5 x^ = are incomplete, the last two being pure. 278 QUADRATIC EQUATIONS 279 Ex. 1. In examples 20-30 of exercise 105 reduce each equation to an equivalent equation of the type-form (J.), and state the values of a, 6, and c in each. Ex. 2. Solve the incomplete quadratic equation ax:^-^bx = 0. (1) Factor x (ax + 6) = 0. (2) Equation (2) is equivalent to the two linear equations, X = 0, ax -\- b = 0. Hence the roots of (2) or (1) are and — b/a. Remember that to solve any quadratic or higher equation we must first find its equivalent linear equations. Reread §§ 148 and 149. 288. Since w^ + mu + {m/2f = (u + m/2)2, § 137 TJie expression u^ + mu is made a perfect square by adding {m/2y, or the square of one-half the coefficient of u. The addition of (m/2y is called completing the square. E.g. J x2-7x is made a perfect square by adding (—7/2)"^ or (7/2)-^ that is, x'^-lx-h 49/4 = (x- 7/2)2. 4a;2 + 8x, or (2x)^ + 4(2x), is made a perfect square by adding (4/2)2, or 4 ; that is, (2 x)2 + 4 (2 x) + 4 = (2 x + 2)2. 289. Any quadratic equation can be solved by transposing all its Jtenns to one member, factoring that member by vmting it as the difference of two squares, and then putting each factor equal to zero. The. following examples will illustrate the method. Ex. 1. Solve the pure quadratic equation «X2 + C = 0. (1) Divide by a, a:2 _(- c/a) = 0. (2) Factor, (x-V- c/a)(a; + v -c/a)= 0. By § 149, oj = V— c/a, X = — V— c/a. 280 ELEMENTS OF ALGEBRA Writing these two linear equations together, we have x = ± V- c/a. (3) Ex. 2. Solve the complete quadratic equation x^ + 4x-2 = 0. (1) Add 4-4, a;2 + 4a; + 4-6 = 0; or (x + 2)2-(V6)2 = 0. (2) Factor, (a; + 2 - VC) (x + 2 + V6) = 0. By § 149, x + 2 = V6, x + 2=-V6. Writing these two linear equations together, we have x + 2 = ±V6. (3) ,', x = -2± V6. Observe that in each example the two linear equations in (3) can be obtained from (2) by transposing the known term and then extracting the square root of both members, writing the double sign ± with one member. The principle of equivalence of equations which this illustrates is proved in the next article. 290. Square root. If the square root of both members of an equation is extracted, and the double sign ± is written before one member, the two derived equations (when rational in the unknown) will together be equivalent to the given equation. Proof Let the given equation be A' = B^, (1) Avhere A and B are rational in the unknown. Transpose, A^-B^ = 0. Factor, (A - B) {A + 5) = 0. (2) By §§ 149 and 106, (2) is equivalent to the two equations A = ±B. (3) QUADRATIC EQUATIONS 281 Equations (3) can be obtained from (1) by extracting the square root of both members and writing the double sign ± with the second member.; hence the theorem. The following examples illustrate how this principle, which is proved by factoring, abbreviates the work of finding the two linear equations, which are equivalent to a given quadratic equation. Ex. 1. Solve x2 + 32 - lOx = 0. (1) Transpose 32, a;2 - 10 x = - 32. Add (10/2)-2, x2 - 10 « + 25 = 25 - 32 = - 7. Extract square root, x — 5 = ± V— 7. .-. X = 5 ± V^. (2) By §§ 106 and 290, no root is either introduced or lost in passing from (1) to (2); hence the roots of (1) are 5 + V— 7 and 5— V— 7- Ex.2. Solve xM:_9^xM:J^ ,j^ 4 5 Multiply by 20, 5 x* + 45 = 4 x^ + 4. Transpose, x^ = — 41. Extract square root, x = ± V— 41. (2) By §§ 106, 108, and 290, no root is either introduced or lost in passing from (1) to (2); hence the roots of (1) are + V— 41 and Ex.3. Solve 2x2 = 7x + ll. Transpose 7 x, 2 x^ — 7 x = 11. Multiply by 2, (2 x)2 - 7 (2 x) = 22. Complete square, (2 x)2 - 7 (2 x) + 49/4 = 22 + 49/4 = 137/4. Extract square root, 2 x - 7/2 = ± ^1^7/2. : .-. x=(7±Vl37)/4. Hence, to solve a quadratic equation we can proceed as follows : Reduce the equation to the form aa^ -\-bx = — c. 282 ELEMENTS OF ALGEBRA If the tei'm in xr is not a perfect square, multiply (or divide) both members by a number which will make it a perfect square. Add to both members what is necessary to complete the square of the unknown member. Extract the square root of each member, writing the double sign ± before the known member. Solve the two derived linear equations. Exercise 104. Solve each of the following equations : 1. x'-Jrl = 4.x. 9. 3x'-6x-\-2 = 0. 2. x^-2x = 4:. 10. 5x--6x-\-ll = 0. 3. ^'-^ + 5 = 80;. 11. 3^2 + 4 a; + 7 = 0. 4. 'x''-{-2x = 2. 12. 2x'-6x-{-10 = 0. 5. x'^-\-6x = -S. 13. 5x'-{-Sx + 21 = 0. 6. 4j^ + 4ic = ll. 14. 2x'-5x-\-15 = 0. 7. 9x^-\-6x = 17. 15. 2x'-3ax^2a' = 0. 8. Ax^-4.x-7 = 0. 16. {x-7y = i9(x + 2y. When both members are perfect squares in the unknown, as in example 16 (or can be made so, as in some of the examples which follow), the first step is to extract the square root of both members. 17. {x-{-2y = 4.(x-lf. 21. x''-\-2ax = b^-}-2ab. 18. (x+6y = 16(x-6y. 22. x' -{-2 ab = b^ + 2ax. 19. (ic + 8f = 9i»l 23. 4.x^-\-4:ax = b'-a\ 20. x^-3ax-\-2a^ = 0. 24. x'^ + 3 a^ = 4. ax. 291. To solve the general quadratic equation, ajr^ + 6jr + c = 0, {A) we proceed just as with the particular equations above. Transpose c, ax^ -\-bx = — c. QUADRATIC EQUATIONS 283 Multiply by 4 a instead of a, to avoid fractions in (1) and (2), (2 axf -\-2b(2ax) = -4: ac. Add b', (2 axy -\-2b(2ax)-l-b' = b'-4.ac. (1) Extract square root, 2ax-\-b = ± V6- — 4 ac. (2) Hence, x = (-b± Vb'-^ac) /(2 a). (B) By §§ 106, 108, and 290, no root is either introduced or lost in passing from (A) to (B) ; hence the roots of (A) are given in (B). Let b' and c' denote the values of b and c when a = l. /Ihen when a = 1, equations (^1) and (B) become and x = -b '/2 ± V{b'/2f - c. (B') By § 287, any quadratic equation can be reduced to an equivalent equation of the form (A)-, hence, a quadratic equation in one unknown has ttvo, and only two, roots. 292. Solution by formula. Instead of repeating the pro- cess in § 291 with every quadratic equation, we should here- after find the values of a, b, and c when the equation is reduced to the type-form {A), and substitute these values in the two equations (5), x = {-b± V6^-4ac)/(2 a). {B) Ex. 1. Solve 2 X'^ - 3 X -I- 5 = 0. Here a = 2, 6 = — 3, c = 5. Substituting these values in equations ( B) , we obtain X = (3 ± V9 - 40) /4 = (3 ± V- 31)/4. Ex. 2. Solve - 3 x2 = 3 A: - 2 ax. Here a= — 3, 6 = 2 a, c = — ^k. Substituting these values in equations {B), we obtain X =(- 2 a ± \/4 a^ - 3(5 k) / {- 6) = (aTVa2^^^9lfc)/3. 284 ELEMENTS OF ALGEBRA 293. Equations (JB') of § 291 afford the following simple rule for writing out the two roots of an equation in the form Tlie two roots are equal to minus one-half the coefficient of x plus and minus the square root of the binomial, the square of one-half the coefficient of x minus the known term. Ex. 3. Solve ic2 + 4 a; + 7 = 0, by the rule given above. x = -2± \/22 - 7 = - 2 ± V^. Ex. 4. Solve x2 - 6 a; - 8 = 0, a: = 3 ± V(- 3)2 -(- 8) = 3 ± V17. Exercise 105. Solve each of the following equations by § 293 : 1. cc2-2a; = l. 7. a^ + 31 = 10aj. 2. a^ + 8a; + 5 = 0. 8. «2 + 6a; + ll=0. 3. a;2_^4a; = l. 9. aj^ + 10 a? + 32 = 0. 4. a;2 + 18 = 10a;. 10. x'-\-52 = Ux. 5. x'-{-3 = 2x. 11. x'-\-2x = l. 6. a^ + ll = 4a;. 12. x^ = 4:X-lS. Solve each of the following equations by § 292 : 13. 3a^ + 121 = 44x. 19. 21 + 0^ = 2 a^^ 14. 25x = 6x^ + 21. 20. 9a^-U3 = 6x. 15. 8af-\-x = S0. 21. 12 a;^ = 29 a; - 14. 16. 3a;2 + 35 = 22a;. 22. 20a;2 = 12-a;. 17. x-{-22 = 6x\ 23. 15 aj^ - 2 aa; = a^. 18. 15 = 17 a; + 4a;2. 24. 21 x^ = 2 ax -\- 3 a'. QUADRATIC EQUATIONS 285 Solve each of the following equations by the method best suited to it : 25. 9x'-eax = a~-b'. 26. aix" -{-!) ^x^a" + 1). 27. a(pc^-l)-\-x(a^-l)=0. 28. x'-2(a-b)x-]-h^ = 2ab. 29. (b — c)x^+(c — a)x = b — a. 30. (a 4- 6) a^ 4- c.r = a + & + c. 31. abx^-(a^-{-b^x + ab = 0. 32. (a^ - 62) (a:2_ 1)^4^52.^ 33. {b''-a'^(x' + l) = 2{a'-hb^x, 34. (a-a;)3 + (a;-6)» = («-6)l 35. (a;-a + 2&)3-(a;-2a + 6)' = (a4-&)^ 294. Discussion of the roots, (— b ± VA^ — 4 ac)/(2 a), wlien a, 6, c are real. (i) If 6^ — 4 ac > 0, the two roots will be real and unequal. (ii) If 6^ — 4 oc = 0, the two roots will be real and egzmZ. (iii) If 6^ — 4 ac < 0, the two roots will be imaginary or complex. (iv) If 6 = 0, the two roots will be both real or both imagi- nary, but opposite in quality and arithmetically equal. (v) If c = 0, one root will be zero and the other — b/a. (vi) If 6 = c = 0, both roots will be zero. (vii) Both roots will be real, both imaginary, or both complex. (viii) If br — 4 ftc is a perfect square, the two roots will be rational when a and b are rational. The pupil should give the reasons for each of the above statements. 286 ELEMENTS OF ALGEBRA Ex. 1. What kind of numbers are the roots of the equation Sx'^ -2x^-1? (1) Here a = 3, 6 = - 2, c = 7 ; ... 62 _4(^c =(-2)2-4.3.7<0. Hence the roots of (1) are complex and unequal. 295. Sum and product of roots of jr- -h 6'jr + c' = 0. (A') Representing the two roots of (A') by Xi and x^, we have x,^-b'/2 + V(b'/2y-c, (1) X, = - b'/2 - ■\/(b'/2y - c. (2) Adding (1) and (2) to find the sum, we obtain x, + x, = -b'. (3) Multiplying (1) by (2) to find the product, we obtain ^^ ^ ^gQ _ ^ _ ^4^^_^ ^ _^ 14. 60 ^ ^60 45. A cyclist rode 180 miles at a uniform rate. If he had ridden 3 miles an hour slower than he did, it would have taken him 3 hours longer. How many miles an hour did he ride ? 46. A man drives to a certain place at the rate of 8 miles an hour. Returning by a road 3 miles longer at the rate of 9 miles an hour, he takes 7| minutes longer than in going. How long is each road ? 47. A father's age is equal to the united ages of his 5 children, and 5 years ago his age was double their united ages. How old is the father ? PROBLEMS 299 48. A and B are two stations 300 miles apart. Two trains start simultaneously from A and B, each to the oppo- site station. The train from A reaches B 9 hours, the train from B reaches A 4 hours, after they met. When did they meet, and what was the rate of each train ? 49. If a carriage wheel 14| feet in circumference takes one second more to revolve, the rate of the carriage per hour will be 2J miles less. How fast is the carriage travelling ? Let X = number of miles travelled per hour ; then X 10 X - 2f 50. The number of square inches in the surface of a cubical block exceeds the number of inches in the sum of its edges by 288. Find its edge and volume. 51. A cistern can be filled by 2 pipes running together in 2 hours 55 minutes. The larger pipe by itself will till it sooner than the smaller one by 2 hours. Find the time in which each pipe separately will fill it. 52. My gross income is $ 3000. After paying the income tax, and then deducting from the remainder a percentage less by 1 than that of the income tax, the income is reduced to $ 2736. Find the rate per cent of the income tax. 53. A set out from C toward D at the rate of 5 miles an hour. After he had gone 45 miles, B set out from D toward C, and went every hour -^^ of the entire distance. After travelling as many hours as he went miles in an hour, he met A. Find the distance from C to D. CHAPTER XXI IRRATIONAL EQUATIONS 298. An irrational equation is an equation one or both of whose members is irrational in an unknown. In this chapter, as heretofore, the radical sign will denote only the principal root of a number or expression. E.g.^ Vx^ — 2 = X — 7 is an irrational equation, and Vx^ - 2 de- notes only the principal square root. Note that we cannot speak of the degree of this or any other irrational equation. In solving irrational equations we use the following principle ; 299. If both members of an irrational equation are raised to the same integral power, the derived equation will have all the roots of the given one and often others in addition. Proof Let A = B (1) be the given irrational equation. Squaring, A^ = B\ (2) By §§ 105 and 106, (2) is equivalent to the equation {A-B){A + B) = (>. (3) By § 74, the roots of (3) include those of A — B=0, or (1) ; hence no root is lost by squaring (1). But the roots of (3) include also those of A + B = 0,ov A = — B; hence any root of A = — B which is not a root of ^ = J5 must be introduced by squaring (1). In like manner the principle can be proved for any other positive integral power. 300 IRRATIONAL EQUATIONS 301 Ex. 1. Solve the equation x — 6 = — Vx — 6. (1) Square, x2 - 12 x + 36 = x - 6. (2) Transpose, x^-lSx-{-4t2 = 0. Factor, (x - 6) (x - 7) = 0. (3) Now (3) , or (2) , is satisfied when x = 6 and x = 7 ; but (1) is satisfied only when x = 6. Hence by squaring (1) the root 7 was introduced. By § 299, the roots of (2) include those of (1) and also those of A = - B, or X - 6 = Vx-e. (4) Equation (4) is satisfied both when x = 6 and when x = 7. Hence if it had been required to solve (4), by squaring we would have obtained (2), and no root would have been introduced. Notice that we cannot say that (2) is equivalent to (1) and (4) jointly (as would be the case, by § 290, were (1) and (4) rational equations); for (2) has only two roots, while (1) and (4) together have three roots, 6 being a root of each. Observe that, since we cannot speak of the degree of an irrational equation, we do not know how many roots it has until we have solved it. Ex. 2. Solve 2 - V2 X + 8 + 2 Vx + 6 = 0. (1) Our purpose being to obtain a rational equation, it is better before squaring to put the more complex surd in one member by itself, as below. Transpose, 2 + 2 Vx + 6 = a/2 x + 8. Square, 4 4- 8 Vx + 6 + 4x + 20 = 2x + 8. Transpose, x + 8 = - 4Vx + 5. (2) Square, x2 + 16 x + 64 = 16 x + 80. Transpose, x^ - 16 = 0. (3) Hence, by § 299, if (1) has any root, it is 4 or — 4. But neither X = 4 nor x = — 4 satisfies (1) ; hence (1) has no root, i.e., it is impos- sible, and therefore both roots of (3) were introduced by squaring (1) and (2). If we use both the positive and the negative values of V2x + 8 and vx + 6, we obtain in addition to (1) the three equations. 2-V2x4-8-2Vx + 5 = 0, (4) 2+V2x + 8-2\/x+5 = 0, (5) 2+V2x + 8 + 2Vx-f5=0. (6) 302 ELEMENTS OF ALGEBRA By treating (4), (5), or (6) as we did (1), we would obtain (3). Hence the roots of (3) include the roots of each of the four equations, (1), (4), (5), and (6). By trial we find that — 4 is a root of (4), — 4 and + 4 are both roots of (5), and neither — 4 nor + 4 is a root of (6) ; that is, (6) expresses an impossible condition as well as (1). Hence in rationalizing (1) or (6), i.e., in deriving the rational equation (3) from (1) or (6), two roots are introduced. In rationalizing (4), one root is introduced. In rationalizing (5), no root is introduced. Exercise 109. i^ Solve each of the following irrational equations : 1. Vaj — 5 = 3. 7. Va;4-25 = 1+V^. 2. 7 - Va; - 4 = 3. 8. Vi» + 3 -f ^x = 5. 3. V5a;-l=2Vaj + 3. 9. V8a; + 55 - 3 = 2V2^. 4. 2V3-7a;=3V8fl7-12. 10. 10 -V25 -\-'i) x = 3^x. 5. .V9x'-llx-o = 3x-2. 11. V9ic-8 = 3Va.'-h4-2. 6. V4.x^-7 x-^l = 2x-l^. 12. ■Vx-4.-\-3=Vx-\-ll. 13. In each of the foregoing examples, from what other irrational equation or equations would we have derived the same rational equation ? 14. ■\/Sx-{-17 -V2x = -V2x-\-9. 15. V3a;-ll+V3^=Vl2x-23. 16. -Vi2x-5+VSx-l=^27x--2. 17. Vx + 3 + Va; + 8 = V4 a; 4- 21. 18. Va5 + 2+V4a7 + l=V9aj + 7. 19. ■\/x -i- 4: ab = 2 a -\- ^x. 20. ■^x+V4:a-^x=2Vb + x. 21. Vx — l-\-^x = 2^^x. IRRATIONAL EQUATIONS 303 22. ViC + o + V^ = 10 -=- Y^ic. 23. V^- vaJ-8 = 2H-V.T-8. 24. Vl + a; + V^ =2 -J- VI + a;. 25. 2 VaJ — V4ic— 3 = l^V4a; — 3. 26. V^ - '^ = 1 -^ (V^ + ")• 27. -^Hi- = 3 + ^^ + 1. V« — 1 2 Simplify the first member in example 27. 28. ^ - 2Va; 2 -1 2 = -2 + 1 29. V2 + « + V2 - a; V2 + a;-V2-a! 30. 1 1- + 1 + ^ a; ^x -h 1 V^ ~ - = 0. In the next seven examples, first reduce the improper fractions to lixed expressions : 31. 4- 3 _ S^x - 5 - 2 3 V-x- - 13 32. 9V^ 3v r-23_6V«-17 'X - 8 2 V-» - <3 V^ + - V-'« + '^ 34 6Va?-7 g^ 7Vx-26 V^-l 7V.T-21 35 2Va;-l ^ Vx-2 3^ 12v^-ll ^ 6^^4-5 * Va' + t V^-i ' 4va.'-4f 2V^ + t' 36. 304 ELEMENTS OF ALGEBRA 39. ^(a-x)+^(b-x)=^(a-\-b-2x). 40. ^(ax + b^) — -^(bx i- a^) = a — b. 41. ^(a-{-x)-\--^(b + x)=^(a + b-^2x). 42. ^(a-x)+^(b~-x)=-y/(2a-{-2b). 300. Equations in quadratic form. If an equation has only two unknown terms, and if the unknown factor of one of these terms is the square of the unknown factor of the other, the equation is in quadratic form. E.g., since x^ + Sx is the square of vx^ + 3x, the equation (x2 + 3x)+ 5vxM-3x = 7 is in quadratic form. The following examples illustrate how the principles of quadratic equations can be applied to irrational equations which are, or can be put, in quadratic form. Ex.1. Solve 2x2 + 3x-5\/2x'^ + 3x + 9=-3. (1) Add 9, (2x2 + 3x + 9)-5V2x2 + 3x + 9 = 6. (2) Since 2 x^ + 3 x + 9 is the square of V2 x^ + 3 x + 9, equation (2) is in quadratic form. Transposing 6 and factoring, we have (\/2 x2 + 3 X + 9 - 6) ( V2 x2 + 3 X + 9 + 1) = 0. (3) The roots of (3) include the roots of V2 x2 + 3 X + 9 = 6, (4) and of \/2 x2 + 3 X + 9 = - 1, (5) but no others. The roots of (4) are 3 and — 4J ; while (5) is an impossible equa- tion, since a principal square root cannot be a negative number. What would be the roots of (1), if the sign before the radical were + ? Ex. 2. Solve 3 x2 - 7 + 3 V3 x2 - 16 X + 21 = 16 X. (1) Transposing 16 x and adding 28 — 28, we obtain (3x2 - 16x + 21) + 3V3x2 - 16x + 21 - 28 = 0. (2) IRRATIONAL EQUATIONS 305 Factor, (VSx^- 16 a: + 21 - 4) ( V3 x^ - 16 x + 21 + 7) = 0. (3) The roots of (3) include the roots of V3 x2 - 16 a; + 21 = 4, (4) and of V3 x2 - 16 X + 21 = - 7, (5) but no others. The roots of (4) are 5 and 1/3, and (5) is impossible. What would be the roots of (1), if the sign before the radical were — ? If we could not factor (2) by inspection, by § 293 we would have V3x2-16x + 21 = - f ± Vf + 28 = + 4 or - 7. Exercise 110. Solve each of the following irrational equations : 1. Sx^-4:X-\-V3x'-4:X-6 = lS. 2. a^ _ a; -f 4 4- Vur - a? -h 4 = 2. 3. a^ + 2a; - Var* H- 2a; - 6 = 12. 4 1 1 1 Var* \ X \ 5 — Va;2 + a; -f 5 5. a^ 4. V4ar^ H- 24 a; = 24 - 6 a;. 6. 2a^-\-6x = l-Va^-]-3x + l. 7. 2 (2 a; - 3) (a; - 4) - V2 ar^ - 11 a; + 15 = 60. 8. ^4.0^ -{-2 X -^ 7 = 12 x'-^Gx- 119. 9. 2a;2_2a;_17 + 2V2a;2_3^_l_7^^^ 10. 3a;(3-a;) = ll-4Var'-3a; + 5. 11. 2ar2-4a;-Va;^-2a;-3 = 9. CHAPTER XXII HIGHER EQUATIONS 301. The following examples illustrate how the princi- ples of quadratic equations are applied to higher equations which are, or can be put, in quadratic form. Ex. 1. Solve (x2 + 2a;)2-5(a;2 + 2a;)-14 = 0. (1) Factor, (a:2 + 2 x - 7)(a;2 + 2 a; + 2) = 0. - (2) Equation (2) is equivalent to the two equations x2 + 2x-7 = 0, x2 + 2x + 2 = 0, each of which is readily solved. Ex. 2. Solve x* - 8 x3 + 10 ic2 + 24 X + 5 = 0. (1) Adding C x2 — 6 x^ to the first member, we have (x* - 8 a:3 + 16 a;2) - 6 x2 + 24 X + 5 = 0, or (a:2-4x)2-6(x2-4x) + 5 = 0. (2) Factor, (a;2 _ 4 x - 5) (x2 _ 4 x - 1) = 0. (3) Equation (3) is equivalent to the two equations ic2-4x-5 = 0, x2-4x-l=0, whose roots are 5, — 1, 2 i VS. Ex. 3. Solve -^^ + ^^ = ^. (1) X — 1 x2 4 Here the second term is the reciprocal of the first. Putting y for the first term, and therefore the reciprocal of y for the second, (1) becomes y 4 306 HIGHER EQUATIONS 307 Multiply by 4 y, 4 y^ - 17 ?/ + 4 = 0. Factor, (y - 4) (4 2/ - 1) = 0. .-. 2/ = 4, or 1/4. Hence (1) is equivalent to the two equations ^^ = 4 and -^^ = 1. (2) x-l x-1 4 The roots of equations (2) are 2, 2, (1 ± V— 16)/8. Exercise 111. Solve the following equations : 1. a:^-5ar' + 4 = 0. 3. a;* - 7 ar^ - 18 = 0. 2. a;* -10 0^2.^9^ Q 4^ (oF -l)/9 -i-l/a^ = 1. 5. a^ + lOO/ar^ = 29. 6. (a^ -h a!)2 - 22 (ar' + a-) =-40. • 7. (ar^ - a^)- - 8 (a-2 - .1-) = - 12. 9. 2»2 + 3a;H-l= 30/(2 ar' + 3 x). 10. ar^ + 3a;-20/(ar + 3a;) = 8. 11. a^ + a; + 1 = 42/(a^ + a;). 12. a;^ - 8 ar'^- 12^2 4- 112 a; = 128. 13. aj* + 2ar'-3ar-4.r-96 = 0. 14. a;*-10.t'3-}-30ar'-25a; + 4 = 0. 15. a;^-14ar^-hGla:2_g4^_^20 = 0. 16. ^_ + ^ + l_o 17. a; + l x" X a.-^+1 ^5 X- + 1 a; 2 18 - a r^-f-2 ar^ + 4.r4-1 ^5 ar^ + 4a;-fl ar'4-2 2* 308 ELEMENTS OF ALGEBBA 302. A binomial equation is an equation of the form a;" = a, where n is a positive integer. The binomial quadratic equation x^ = a has already been solved. Certain binomial higher equations are readily solved by previous principles. Ex. 1. Solve the binomial cubic equation x^ — 1 = 0. (1) Factor, (a; - 1) (a;2 + x + 1) = 0. (2) Equation (2) is equivalent to the two equations X - 1 = 0, x2 + X + 1 = 0. (3) The solutions of equations (8) are 1 and (— 1 ± v'— 3)/2. Hence, the cubic equation (1) has one real and two complex solutions. Since by (1), x^ = + 1, tlie cube of each solution of (1) is equal to +1 ; that is, +1 has the three cube roots +1 (— 1+V— 8)/J, and (—1 — V— 3)/2. See example 32, exercise 103. Since + 27 =(+ 1) x 27, the three cube roots of +27, or the three solutions of the cubic equation x^=^21, can be obtained by multiply- ing the three cube roots of +1 by the cube root of the arithmetic num- ber 27. Thus the three solutions of x^ =+27 are +3 and 3(- 1 ± V^^)/2. Ex. 2. Solve the binomial biquadratic equation x^ — 1 = 0. (1) Factor, (x^ - l)(x2 + 1) = 0. • (2) The solutions of (2) are ± 1 and ± V— 1. Hence +1 has four fourth roots, two real and two imaginary. The four solutions of x* = 81, or the four fourth roots of +81, are ±3and±3V^^. Ex. 3. Solve x5 = 1, or x^ - 1 = 0. (1) Factor, (x - 1) (x* + x^ + x^ + x + 1) = 0. (2) One solution of (2) is 1, and the other solutions are those of the equation x4 -f x3 + x2 + X + 1 = 0. (3) Divide by x^, a;2 + x + 1 + - + ^ = 0. HIGHER EQUj^TIONS 309 Addl, a;2 + 2+l+x + - = l, (..i)%(..l)=x. .-. x2+l = K-l±V5)x. (4) Solving the two equations in (4), we obtain four solutions, all of which are complex. Hence 1, or any other positive number, has five fifth roots, one real and four complex. Ex. 4. Solve x6 = 1^ or x5 - 1 = 0. (1) Factor, (x^ - 1) (x^ + 1) = 0, or (a;-l)(a;2 + x+l)(x+l)(x2-x + l) = 0. (2) Equation (2) is equivalent to the four equations X - 1 = 0, x' + X + 1 = 0, X + 1 = 0, x^ - X + 1 = 0. (3) Solving equations (3), we obtain six solutions, two real and four complex. Hence 1, or any oXhQV positive number, has six sixth roots, two real and four complex. Ei. 5. Solve x^ = 1, or x8 - 1 = 0. (1) •Factor, (x* + 1) (x^ + 1 ) (x^ - 1 ) = 0. (2) The roots of (2) are ±1, ± V- 1, and the roots of X* + 1 = 0. (3) Add 2 x2 - 2 x2, x4 + 2 x2 + 1 - 2 x2 = 0. Factor, (^2 + 1 + Xy/'l) (x2 + 1 - Xy/I) = 0. (4) Equation (4) is equivalent to the two equations X2 + 1 + Xy/2 = 0, and x2 + ,l-xV2 = 0, each of which has two complex roots. 310 ELEMENTS OF ALGEBRA Hence, any positive number has eight eighth roots, two real, two imaginary, and four complex. Observe that any root of + 1 or — 1 is a quality-unit. Exercise 112. Solve each of the following binomial equations : 1. ar'' + l = 0. 5. a^ + l = 0. 9. a^^ _ ^4 ^ 0^ 2. a^ + 27 = 0. 6. ar' + 32 = 0. 10. a:* -625 = 0. 3. .T^ + 1 = 0. 7. x^-\-l = 0. 11. af- 243 = 0. 4. x'-\-16 = 0. 8. a.'«H-64 = 0. 12. x«- 729 = 0. CHAPTER XXIII SYSTEMS INVOLVING QUADRATIC AND HIGHER EQUATIONS 303. As in linear systems, so in any other determinate system tliere must be as many independent consistent equa- tions as there are unknowns. In solving systems which involve quadratic or higher equations we have frequent use for the following principle of equivalent systems: 304. If M, N, P, Q denote any integral unknown expres sions, then system (a) PxQ 0,1 (a) is equivalent to the four systems (6), (c), (d), (e). M=0, P=0. M=0,] , ^=0,1 , iV^=0,l ("). Q^oM^^ p=oM'^ e=o:}« Proof Any solution of system (a) must reduce the factor M or N (or both) to 0, and at the same time must reduce P or Q (or both) to 0. Now any solution of system (a) which reduces M to and P to is a solution of system (6) ; any solution of (a) which reduces M to and Q to is a solution of (c) ; and so on. Hence any solution of system (a) is a solution of system (6), (c), (d), or (e). ('onversely, any solution of system (b) reduces M to and P to 0, and therefore reduces M x iV to and P x Q to 0; 311 312 ELEMENTS OF ALGEBRA hence, any solution of system (b) is a solution of system (a), and so on. Hence, any solution of system (6), (c), (d), or (e) is a solution of system (a). Whence system (o) is equivalent to the four systems (6), (c), (d), (e). Ex. 1. Solve thie system a;2 - a:?/ - 2 ?/2 = 0, (1) 3y2_i0y + 8 = 0. (2) By § 201, system (a) is equivalent to (&). Factor (1), (^^ _ 2 ?/) (x + 2/) = 0. (3) Factor (2), (Zy - ^){y -2)=Q. (4) By § 304, (6) is equivalent to the four linear systems (c), a;-2?/ = 0, 1 x-2y = 0, I »: + ?/ = 0, 1 x + ?/ = 0, ' 3 2/-4 = 0. J ?/-2 = 0. J 3?/-4 = 0. J ?/-2 = 0. , (a) (&) (^) The solutions of the four systems (c) are |, f ; 4,2; — -f, | ; — 2, 2 which are therefore the four solutions of (a). Ex. 2. Solve the system x^ + 2xy + y'^ = 36, (1) a:2-2a;i/ = 0. (2) System (a) is equivalent to system (6) . From(l), x^y=±A. ^^H (5) From (2), x{x -2y)=0. (4) J By § 304, (&) is equivalent to the four linear systems (c). x + ?/ = 4, 1 a; + ?/ = 4, 1 a; + !/ = -4, 1 x + y = -4 4, 1 a; + !/ = -4, 1 a; + y = - 0. J x = 0. J ic - 2 ?/ = 0. cc = 0. J a; — 2?/ = In applying the principle of this article to system (6), observe that the two equations in (3) are equivalent to the equation {x-\-y-i)(ix + y + i)=0. The solations of (a) are therefore 0, 4 ; f , f ; 0, — 4 ; — |, — f . SYSTEMS OF QUADRATIC EQUATIONS 313 Whenever one or each of the equations of a system can be resolved into two or more equivalent equations, the Jlrst step in solving the system is to apply the principle of this article. 305. The two examples in § 304 illustrate the theorem : A system of two quadratic equations in two unknowns has, in general, four, and only four, solutions. Exercise 113. Solve each of the following systems of equations : 1. (x-2y)(x-l) = 0,] 6. (x-\-y){x-y + l) = 0,] x-\.y-i = 0. 1 (x-\-2)(y + S) = 0. J 2. (x-3)(y-2) = 0,) 7. (x-hyf = 16,\ a; + 2/ = 7. J (p-yY 3. x'-^xy + Zy'-^OA 8. cr + 2 a^y + 2/' = 144, | X -\-y = \. J Q(? — 2 xy -\- y"^ = A:. J 4. a-?/ -7 2/ + 3 a; = 21,1 9. a?-{-xy = x-\-y. .1 x-\-y = 2. ) y^ — 2xy = Sy — 6x. 5.4:a^ — xy = 0,] 10. a^ — y^ = x -\- y, 1 2a?-32/ = 6. J ix?-Sxy = 5x-15y.) 306. A system of two equations, one linear and the other quadratic, can be solved by first eliminating one unknown by substitution. Ex. 1. Solve the system aj + 2 y = 5, S!<« «2 + 2y2 = 9. Solve (1) for x, x = 5-2y. (3) From (2) and (8), (5 - -2?/)2 + 2?/-^ = 9. (6) Factor, (3 y - -4)(2 2,-4)=0. (4). 314 ELEMENTS OF ALGEBRA By § 201, (a) is equivalent to the system, (3) and (4), or (6). By § 304, (b) is equivalent to the two systems (c) and (d). X = 5 — 2y,] x = 6 — 2y,] ■(c) (d) 3?/ -4 = 0. r ' y-2 = 0. J ' ' The solution of (c) is 7/3, 4/3 ; and that of (d) is 1, 2. After the theory is clearly understood, the work after equation (4) can be abridged as below : From (4), y = 4/3, or 2. When y = 4/3, from (3), x = 6- 8/3 = 7/3 ; When y = 2, from (3), a; = 5 - 4 = 1. This example illustrates the following theorem : A syste7n of one linear and one quadratic equation in two unknowns has, in general, two, and only two, solutions. Exercise 114. Solve each of the following systems : 1. x-\-y = 15,] S. x — y = 3. } xy = 36. J iK^ + 19 + 2/- = 3 xy 2. x-\-y = bl,\ d. 2x — y = 5, xy = 518. J x-\-3y = 2xy 3. Sx-4.y = -12, I 10. 3x-\-2y = 5, 1 3ic2 4- 2 2/2 - 2/ = 48. J x' - ixy -\- 5y' = 2.) 4. x-y = 10, 1 11. 3x'-2xy = 15,] 0^ + 2/2 = 58.1 2x-{-3y = 12. J 5. 3x-\-3y = 10,) 12 xy = 1. J 6. 2x-5y = 0, 1 13. x^ + 3xy-y^ = 23, 3)2-3 2/2 = 13.1 x + 2y = 7. 7. 2x-\-3y = 0, I 14. x'-{-if=lS5, . x + y = 15, I a^2 -f 2/' = 125. J SYSTEMS OF QUADRATIC EQUATIONS 815 15. 2x-7y = 2o, 1 17. x-\-y = 2, 1 5x'-{-4:xy+Sy-=2S.} 2x + 3y = 6xy.} le. 3x — 31 = 5y, 1 18. x-^2y = 7, 1 a^ -\- 5 xy + 2o = y^. } 3y + 6x = 5xy. } 19. x-y = l, I ^-f = (^/6)xy.) 20. x^-2xy=^0, (1)| 4a^ + 92/2 ^225. (2) I Factor (1), a;(x - 2 y) = 0. (3) System, (2) and (3), which by § 201 is equivalent to (a), is equiva- lent to the two systems (6) and (c). 4a;2 4-9?/2 = 225, 1 4^2 + 9y2 = 225, 1 a; = 0. J a;-2y = 0. 21. a:2-3a:?/ = 0, 1 23. Q!^-2xy + o = 0, 5ar' + 3/ = 48.J (a;-2/y = 4. J 22. 2ar-3a;.y = 0,l 24. a^ + 4/ = 4a^ + 16, 1 2/2 + 5an/ = 34. J a^ 4- .v' = 5. J 307. If each of two quadratic equations has onSf and only one J temi below tlie 2d degree^ and these two terms are similar; the system can be solved by first eliminating the term below the second degree by addition or subtra^ion. Ex. 1. Solve the system x"^ + xy + 2 y- = 44, (1) | 2a;2-xy + 2/2 = 16. (2)}^''^ Each equation in system (a) has one, and only one, term below the 2d degree, 44 and 1(3, respectively ; and these terms are similar. We proceed to eliminate the term below the 2d degree. Multiply (1) by 4, 4^2 + 4 a;.v + 8 y2 = ne. (3) Multiply (2) by 11 , 22 x"^ - 11 xy -\- 11 y"- = 176. (4) Subtract (3) from (4), 18 x"^ - Ibxy + Sy^ = 0. (5) Factor, (y - 3 x) (y-2x) = 0. (6) 316 ELEMENTS OF ALGEBRA System, (6) and (1), which is equivalent to (a), is equivalent also to the two systems (&). x2 + ce?/ + 2 y2 = 44, I x2 + a;?/ + 2 2/2 = 44, | y-^x = 0. J y-'2x = 0. J ^^^ The solutions of systems (&) are y/2, Zy/2; —y/2, —3^2; 2, 4; and — 2, — 4 ; which are therefore all the solutions of (a). Ex. 2. Solve the system ?/2 _ 2 a:2 = 4 x, (1) ] 3 ?/2 + xz/ - 2 a;2 = 16 x. (2) J ^^^ The terms below the 2d degree, 4 x and 16 x, are similar. We proceed to eliminate the term in x. Multiply (1) by 4, 4 ?/2 _ 8 a;2 = 16 x. (3) Subtract (2) from (3), y^ - xy - Qx'^ = 0. Factor, (y + 2 x) (y - 3 x) = 0. (4) System, (4) and (1), which is equivalent to system (a), is equiva- lent also to the two systems (6) and (c). ?/2 _ 2 a:2 = 4 X, ] ?/2 - 2 ^2 = 4 x. ] 2/'-2x2 = 4x, 1 ?/ + 2x = 0. ' ^"^ .. o^ A 1 (<^) The two solutions of system (h) are 0, and 2, — 4 ; those of (c) are 0, and 4/7, 12/7 ; which are therefore the four solutions of (a). Observe that by eliminating the term below the second degree in each of the systems above, we obtained a homo- geneous equation in x and y, which we resolved into two equivalent equations. Instead of eliminating the term below the second degree, it is sometimes better to eliminate one of the terms of the second degree. Ex. 3. Solve the system 9 x2 - 8 y2 ^ 28, (1) ] 7x2 + 3 2/2 = 31. (2) J '^^'^ Multiplying (1) by 3 and (2) by 8, and adding, we eliminate y"^ and obtain 83 x2 = 332, or X = ± 2. (3) When X = 2, from (2) we obtain y = ±\. When X = — 2, from (2) we obtain y =±l. SYSTEMS OF QUADRATIC EQUATIONS 317 Hence the four solutions of (a) are 2, 1 ; 2, — 1 ; — 2, 1 ; —2,-1. Ex. 4. Solve the system xy + x = 25, (1) | 2xy-Zy = 2S. (2) J Eliminating the product xy we obtain 2 X + 3 2/ = 22. (3) Solving system, (1) and (3), which is equivalent to system (a), we obtain the two solutions 5, 4 ; 16/2, 7/3. Ex. 5. Solve the system x^ — Sxy = 10, (1) ] 4y2-xy = -l. (2) |(«: Sometimes by adding or subtracting the given equations we obtain an equation which can be resolved into equivalent equations. Add (1) and (2) , x:^ - 4xy + iy^ = 9, or x-2y=±S. (3) System, (1) and (3), which is equivalent to system (a), is equiva- lent to the two systems (6) and (c) . x^-Sxy = lO,] x^-Sxy = lO, x-2y = S. >^'^ - "- '^ ^^'^ ^-Sxy = lO, I x-2y=-3. J Exercise 115. Solve each of the following systems of equations : 1. x^ + xy=12, x!/-if = 2. J 6. ar^ + 52/2 = 84, 3a^ + 17a'?/ + 84 = 2/^ 2. x^-{-xy = 24, 2f-{-3xy = S2. 7. ar-7xy-9y- = 9, x'^5xy-\-lly' = 5 3. ic2 + 3a«/ = 7, f + xy = 6. . 8. x(x-^y) = AO, \ y(x-y)=:6. J 4. 2,x?-hf = 2%\ Sxy-4:f = S. J 9. x^ + xy + f = 7, 6x'-2xy-\-y' = i5.. 5. x^-Sxy-^2f = 3 y 10. x'-^3xy = 2S,\ 2x' + y' = 6. xy-\-4:y' = S. > 818 ELEMENTS OF ALGEBRA In example 10, add the two equations. = 40,1 = 9. J 11. a^ + 3aj?/ = 40, 4 ?/2 + 0^2/ 12. a^ + 3fl7?/ = 54, xy -\-4:y^ = 115. 13. x^-\-xy-\-A4: = 2y^,) xy + 3 / = 80. J 14. 3xy-{-x'^ = 10, 5xy — 2 x'^ = 2.. In example 14, eliminate x^ or the product xy. = 301. J 15. Ax'-Sy^^-ll, 11 a^ + 5 2/' 16. 2x^-\-y^ = 9, 5aP-{-(yy' = 26.} 17. 20 a^- 16/ = 179, 5a;2-336/ = 24. 18. 2a^-2xy-3y^ = lS, 3x'-2y' = 19. 19. i»2 4-3a; — 2?/ = 4 2x'-5x-{-3y u\ 20. (a; + l)(2/ + l) = 10, a?2/ = 3. 21. 4:x'-3xy = 10, y^-xy = 6. 22. x^ — 2xy = 3y, 2x'-9y'' = 9y. 23. 2x^-xy-^y^ = 2y; 2 a^ -]- 4: xy = 5 y. 24. a.'3 + l = 9?/,| a^ 4- a; = 6 2/. J 308. Systems of symmetrical equations. A symmetrical equation is one which is not changed by interchanging its unknowns. E.g., x-\- y = 12, xy = 35, x^ + y^ _ 74^ x^±2 xy + y^ = 16 arc symmetrical equations. The equations x — y = 2, x^ — y^ = 4, x^ — y^ = IQ are symmetrical except for sign. The methods given below for solving systems of sym- metrical equations can usually be employed when the equa- tions are symmetrical except for sign. (a) Ex. 1. Solve the system, X^ + y2 = 74, (1) xy = 35, (2) Multiply (2) by 2, 2xy = 70. (3) SYSTEMS OF QUADRATIC EQUATIONS 319 Add (3) to (1), x^-\-2xy + y^ = 144, x + y = ±l2. (4) Subtract (3) from (1), x-y = ±2. (6) By § 301, system (6) is equivalent to the four systems (c) . a; + y = 12, x-y = 2. X 4- y = 12, x-y=-2. + i/=-12, I x + y = -l2, -y = 2. J x-y = -2. The solutions of systems (c) are 7, 5 ; 5, 7 ; — 5, Ex. 2. Solve the system, x^ - xy -\- ij^ = 49, x + y=lS. or Square (2), Subtract (1) from (3), Subtract (4) from (1), x2 + 2 xy + 1/2 = 169. 3 xy = 120, xy = 40. x-y=±3. System, (2) and (5), is equivalent to the two systems (6). a; + y=13, 1 (1) (2) (3) (&) (c) 5. ) |(« (4) (5) a: + y = 13, 1 x-y = 3. J a; _ y = _ 3. J (6) The solutions of systems (6) are 8, 6, and 5, 8. The four solutions of system, (1) and (5), must include the two solutions of (a), since no solution was lost by squaring (2). Hence the two solutions of (a) must satisfy (2) and also (5). Therefore the solutions of systems (b) are the two solutions of (a). Observe that each of tlie above systems was solved by first finding the values of x +y and x —y. Ex. 3. Solve the system x* + y< = 82, x-y = 2. Let x — v + w^ and y = v — w. From (2), (3), and (4), w=\. From (1), (3), (4), and (6), (v+l)*+ (tj-l)* = 82, or (r2 + i0)(t;2_ 4)3,0. .•.v = ±2, or±^^ 10. (1) (2) (3) (4) (5) (6) 1" (ft) 320 ELEMENTS OF ALGEBRA From (3), (5), and (6), x = S, -1, \ ± V^^TO. From (4), (5), and (6), ^ = i, _ 3, _ i ^ V^TTo. (c) System (a) with (3) and (4) forms a system equivalent to (6), which is equivalent to (c) with (5) and (6). Hence the four solutions of (a) are given in (c). Exercise 116. Solve each of the following systems of equations by first finding the values oi x -]- y and x — y : 1. a^2_^/ = 89, 1 5. x'^l+y' = Zxy, xy = 4.0. J Ss(f — xy-{-3y^ = 13. 2. x'-\-y' = 170y xy = 13. 6. ^ ^ - ^2/ + r = "6, 1 a; + .?/ = 14. J 3. x' + y^=Qo,\ 7. x' + xy + f^^QlA iC2/ = 28. J .'c + ?/ = 9. J 4. a^ + ic2/ + 2/^ = 67, 1 8. x" - 4.xy + y~ = b2,\ a^-^^ + 2/' = 39.J _i^(ar_2/) = l. J 9. Solve the systems in examples 1, 2, 4, 5, 8, 12, and 14 in exercise 113, by first finding the values oi x -\- y and x-y. Solve each of the following systems of equations : 10. x-.y = 3, 1 13. x^ + y^ = 212A a;2 _ 3 ic2/ + / = - 19. J x-y = 2. J 11. x^ — xy-\-y'-=12A 14:. x — y = 2, 1 x-{-y = lA. J x' — jf = 242. J 12. a; + 2/ = 4, 1 15. aj^ 4- ^Z'' = 706, J a;4 _j_ ^4 ^ 82. j a; + 2/ = 8. 16. Solve system (2) in § 263, and observe that x and y are rational only when a^ — 6 is a perfect square. SYSTEMS OF QUADRATIC EQUATIONS 321 309. Division. If the members Qf one equation (1) are divided hy the corresponding memhers of another equation (2), and the derived equation (3) is integral in the unknowns; then the system (a) is equivalent to the two systems (6) and (c). AB = A'B', (1)1 A = A\ (3)1 5 = 0, (5)1 B = B'. (2)r ^ B = B'. (4) J ^ ^ B' = 0. (6) J ^ ^ Observe that (3) is the derived equation, that (4) is the same as (2), and that (5) and (6) are formed by equating to the members of (2). Proof. Substituting B for B' in (1) we obtain the sys- tem (d). B(A-A') = 0, B = B'. )W By § 202, system (d) is equivalent to system (a). By § 304, system (cl) is equivalent to the two systems (6) and (e). ■8 = 0, \ By substitution (§ 202), system (e) is equivalent to (c). Hence (a) is equivalent to the two systems ip) and (c). E.g.^ dividing (1') by (2') we obtain the integral equation (3'); y^ = x{x^y), (1') y^ = X + y. (2 Hence system (a') is equivalent to the two systems (6') and (c') ;;}- V = -, (30 I .^ = 0, (50 1 y^ = ic + y. (40J j; + y = 0. (60 J Whenever the equation B = or JB' = is impossible, sys- tem (c) will be impossible, and system (a) will be equivalent to system (6), (a) 5) J 822 ELEMENTS OF ALGEBRA Ex. 1. Solve the system a;^ -y^ = 21, ^^^ \ X - y = 3, (2) J Dividing (1) by (2) we obtain the integral equation (3); hence, as B' =0, or 3 = 0, is impossible, system (a) is equivalent to (6). x2 + x^ + if = 9, (3) x-y = S. (4) Ex. 2. Solve the system x* + x^ -\. i/ = lii7\, (1) x2 _ a;i/ + 1/^ = 03. (2) ■ Divide (1) by (2), x^ + xy + y^ = 117. (3) Add (2) and (3), x'^ + y- = CO, (4) ] Subtract (2) from (3), •2xy = 54. (5 Since 03 = is impossible, by division (§ 309) and addition (§ 204) system (&) is equivalent to (a). Ex. 3. Solve the system x-y + xy- = 30, (1) x-\-y = 5. (2) Divide (1) by (2), xy = 0. (3) Equations (2) and (3) form a system equivalent to (a). Exercise 117. Solve each of the following systems : 1. a^ + 2/' = 3473, 1 6. x' + x'y' + y' = 2^3 x-y = 4.. i a^-xy-^y' = S7 (a) a; + ?/ = 23. J x'-xy+y' = 9. J 2. a.-^ - 2/' = 218, 1 7. x* + x^ -h y' = 91, ) x-y = 2. J aj2+ a.'^ + 2/' = 13. J 3. x'-7f=^9SS,\ 8. a;^-fa^/ + ^' = 2923,1 ^3 - 7/3 ^ 2197, 1 9. x^-\-a^y^-{-y' = 7371, x-y = 13. J a.-2-j'?/ + 2/2 = 63. a..4 + ^2^2 _^ 2/^ ^ 2128, 1 10. x'-f=56, 1 a;2 4_ 0^2/ + 2/' = "6. J x" -\- xy -{- y' = 2S. J SYSTEMS OF QUADRATIC EQUATIONS 823 11. a^ + f = 126, I 12. x-^y-Vxy = 7, 1 ay'-xy + f = 21.S ^^fj^xy=^ 133. J In the next four systems apply § 304 first. 13. a; + 2/ = 5, 1 15. a; + 2/ = l, 4 icy = 12 - x'y\ J x'y^ + 13 rt^ + 12 = 0. J 14. a?y-\-xf = imA 16. Sa^ - 5/ = a; + ?/, 1 a2/=400. J 3a;2_32^2^3._2, J In the next four systems let the unknowns be the reciprocals of x and y, and let v = \/x and w = \/y. 17. 2/.^ + 1/?/ = 1, ^ + ^ + i=5. y? xy y"^ 18. l/a; + l/7/ = 2, 1/a^ 4- 1// = 20. J 19. 3/ar^-l// = l, ^-1 + 1 = 3. a? xy f 20. 21. l/x2-l/(4y^) i_l + J_ ar* a^ 42/^ x_-j-j x—ji x-y x+y x' + f = 20. 3, = 9. 22. a^A + /A=9/2, a; + 2/ = 3. 310. It should be observed that the methods given in this chapter are applicable only to special systems of quadratic and higher equations, and do not enable us to solve a sys- tem of any two quadratic equations ; for the equation de- rived by eliminating one unknown will, in general, be above the second degree in the other unknown, and we have not yet learned how to solve an equation of a higher degree than the second, except in very special cases. E.g.^ consider the system a;2 + x + y = 3, x2 + y2 = 5. („) Solving the first equation for y and substituting its value in the second, we have a;-2 + (3 - y - x)2 = 5, or x4 + 2a;8-4x-^ -6a; + 4 = 0. (1) Equation (1), which is of the fourth degree, cannot be solved by any methods which have been given in the previous chapters. 324 ELEMENTS OF ALGEBRA Exercise 118. 1. The difference of two numbers is 7, and the sum of their squares is 169. Find the numbers. 2. The sum of the squares of two numbers is 130, and the difference of their squares is 32. Find the numbers. 3. The sum of two numbers is 39, and the sum of their cubes is 17,199. Find the numbers. 4. A person bought some fine sheep for $ 360, and found that if he had bought 6 more for the same money, he would have paid $ 5 less for each. How many did he buy, and what was the i)rice of each ? 5. If the length and breadth of a rectangle were each increased by 1 yard, the area would be 48 square yards ; if they were each diminished by 1 yard, the area would be 24 square yards. Find the length and breadth. 6. The numerator and denominator of one fraction are each greater by 1 than those of another, and the sum of the two is 1 j^ ; if the numerators were interchanged, the sum of the fractions would be 1^. Find the fractions. 7. For a journey of 108 miles, 6 hours less would have sufficed, had the traveller gone 3 miles an hour faster. At what rate did he travel ? 8. The hypotenuse of a right-angled triangle is 20 feet, and its area is 96 square feet. Find the length of the other two sides. 9. A number is divided into two parts such that the sum of the first and the square of the second is twice the sum of the second and the square of the first ; and the sum of the number and the first part is 4 more than twice the second. Find the number. 10. The small wheel of a bicycle makes 135 revolutions more than the large wheel in a distance of 260 yards; if SYSTEMS OF QUADRATIC EQUATIONS 325 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 yards. Find the circumference of each wheel. 11. A man bought 6 ducks and 2 turkeys for $ 15. For $ 14 he could buy 4 more ducks than he could turkeys for $ 9. Find the price of each. 12. The sum of the cubes of two numbers is 407, and the sum of their squares exceeds their product by 37. Find the numbers. 13. A rectangular field contains 160 square rods. If its length be increased by 4 rods, and its breadth by 3 rods, its area will be increased by 100 square rods. Find the length and breadth of the field. 14. A man rows down stream 12 miles in 4 hours' less time than it takes him to return. Should he row at twice his ordinary rate, his rate down stream would be 10 miles an hour. Find his rate in still water, and the rate of the stream. 15. The sum of two numbers is 7, and the sum of their fourth powers is 641. Find the numbers. 16. A gentleman left $ 210 to 3 servants to be divided in continued proportion, so that the first should have $ 90 more than the last. Find the legacy of each. 17. From a sheet of paper 14 inches long, a border of uniform width is cut away 'all round it, and the area is thereby reduced |; but had the sheet been 3 inches nar- rower, and a border of the same width had been cut away, the area would have been reduced ^. Find the breadth of the paper, and the width of the border cut away. 18. A and B set out from the same place, and travel in the same direction at uniform rates. B starts 5 hours after 326 ELEMENTS OF ALGEBRA A, and overtakes him after travelling 100 miles. Had their rates of travelling been a mile per hour less, B would have overtaken A after travelling 60 miles. Find their rates. 19. A man has to travel a certain distance, and, when he has travelled 40 miles, he increases his speed 2 miles per hour. If he had travelled with his increased speed during the whole journey, he would have arrived 40 minutes earlier; but if he had continued at his original speed, he would have arrived 20 minutes later. Find the whole dis- tance he had to travel, and his original speed. 20. A cubical tank contains 512 cubic feet of water. It is required to enlarge the tank, the depth remaining the same, so that it shall contain 7 times as much water as before, subject to the condition that the length added to one side of the base shall be 4 times that added to the other. Find the sides of the new rectangular base. CHAPTER XXIV INEQUALITIES 311. An inequality is the statement that one number is greater or less than another, as 6 > 4, — 3 < — 2. See §§ 7 and 31. 312. When a and h are real, in § 31 we agreed to say that : a > 6, when a — 6 is positive; and a ; and 'a — b is negative ' by a — 6 < 0. In this chapter we shall not consider imaginary or complex numbers. 313. Two inequalities are said to be like or unlike in species according as they do or do not have the same sign of inequality. E.g.y the inequalities 8>4 and a>b are like in species; while 2 < 3 and « > 6 are unlike in species. If a > 6 ; then, conversely, b 6 and its converse b a second, and this second number > a third, then the first number > the third number. That is, if a > 6 and 6 > c, then a'^c. (ii) If the same number is added to both members or sub- tracted from both members of an inequality, the derived in- equality will be like the given one. That is. if a > ft, then a ±7)i>b ± m. 327 328 ELEMENTS OF ALGEBRA (iii) If the corresjyonding members of two or more like inequalities are added, the derived inequality will he like the given ones. That is, if a > 6 and c > 6 + d. (iv) If both members of an inequality are multiplied, or divided, by the same positive number, the derived inequality will be like the given one. That is, if a>b, then a (+n) > b (^w), or a-r-'^n^b-h '^n. (v) If both members of an inequality are multiplied, or divided, by the same negative number, the derived inequality will be unlike the given one. That is, if a > 6, then a (~n) < b (~n), or a-i-~n+&!, ^a2>+bz, ••♦, then +ai • +a2--->+6i • +62"-. (vii) If both members of an inequality are positive, and they are raised to the same positive integral power, the derived inequality ivill be like the given one. That is, if ^a >+6, then ("^a)** > (+6)% where w is a positive integer. (viii) If the same principal roots of both members of an inequality are taken, the derived inequality will be like the given one. That is, if a > 6, then ^a >-^&. Proof of {}). (a — 6) + (6 — c)=a — c. Hence, if a — 6 > and 6 — c> 0, then a — c > ; that is, if a>h and & > c, then a>c. Proof of (it), a — b = (a ± m) — (b ± m). Hence, if a > 6, then a ±m>b ± m. The proof of the other principles is left as an exercise for the pupil. INEQ UA LITIS S 329 316. The following principle is often useful in proving inequalities : Jfa and b are unequal and realy a- + 6- > 2 ab. Proof. (a-by>0, or a^-2ab-\-b'>0. (1) Adding 2 a6 to each member, by (ii) of § 314 we obtain a^ + b'>2ab, when a ^ b. (2) Observe that a^ and b' are both positive. Ex. 1. Prove (x + y)/2>y/xy, if a; > 0, y > 0, and x # y. If in (2) we put x for a* and y for 6*, we obtain X -\-y>2 y/xy. Hence, by (iv), (x + y)/2 >\/xy, where x > 0, y > 0, and x :^ y. Ex. 2. a« + 6' > a^b + ab\ if « + 6 > and a ^ 6. From (1), a^ -ab + 6^ > ah. by (ii) Multiply by a + 6, a^ ■¥b^> a'^b + aft-^. by (iv) Ex. 3. The sum of any positive number, except 1, and its recipro- cal is greater than 2. Let the number be n ; then in (2), putting n for a^ and 1/n for 62 we obtain n + l/«>2. 316. The following examples illustrate some of the uses of the principles of inequalities : Ex. 1. For what values of x is (6x - 7)/3 >(2 - 3 x)/5 ? (1) Multiply by 15, 25 x - 35 > 6 - 9 x. by (iv) Transpose, 34x>41. by (ii) Divide by 34, »> 41/34. by (iv) Hence (1) is satisfied for any value of x gi-eater than 41/34. Ex. 2. For what values of x is x^ - 4 x + 3 > - 1 ? (1) Addl, x2-4x + 4>0, or (x-2)2>0. by (ii) Hence (1) is satisfied when (x — 2)2 >0, i.e.., when x has any real value except 2. 330 ELEMENTS OF ALGEBRA Ex. 3. Find what values of x satisfy the inequalities 4iK-6<2x + 4, (1) \ and 2 X + 4 > 16 - 2 ic. (2) J From (1), 2x<10, or x<5. by (ii), (iv) From (2), 4ic>12, or ic>3. by (ii), (iv) Hence (1) and (2) are satisfied by any value of x between 3 and 5. Ex. 4. Find what values of x satisfy the inequality x^-lx5, (1)| and the equation 6 x -\- 7 y = 12. (2) J Multiply (1) by 5, 15 x + 10 2/ > 25. (3) Multiply (2) by 3, 15 x + 21 y = 36. (4) Subtract (4) from (3), - 11 y >- 11, or ?/ < 1. by (v) Multiply (1) by 7, 1 x + 14 y > 35. (5) Multiply (2) by 2, 10 x+Uy = 24. (6) Subtract (6) from (5), 11 x > 11, or x > 1. Hence any solution of equation (2) in which x > 1 and y < 1 will satisfy both (1) and (2). Exercise 119. If the letters denote unequal positive numbers, prove : 1. a^-\-b'^-\-c^>ah + ac + bc. (1) Use the relation a^ 4. 52 -> 2 ab. 2. a^ + b''>d'b-{-ab\ (2) INEQUALITIES 331 3. ^>^; % + -A+-- <3) 4. a^j^ly' + (?> (a'b + ab^ + a-c + ac" + b^c + 5c2)/2. 5. am + 67i+cr fa; 4- 18. 10. a;^ _^ a; > 12. 7. ^a;-|a;>|a;-3. 11. (a; + 2)/(a; - 3) > 0. 8. -2(a; + 7)>-16. 12. (a; - 7) /(a; + 4) < 0. 9. x^ — ox> — 4:. 13. 3(a;+7)/5>5(a;-3)/7. 14. If 5 a; — 6 < 3 a; + 8 and 2a; + l<3a; — 3, show that the values of x lie between 4 and 7. 15. If 3 a; — 2 > I a; — I and J — fa;<8 — 2 a;, show that the values of x lie between 12/25 and 82/9. Find what values of x and y will satisfy each of the following systems : 16. 2x-\-3y=4:,] 17. Sx-y=Q,) 18. 4:X-2y=e, x—y>2.) 2x+y>4:.) 2x—Sy>5. 19. Show that (1) in example 1 holds, if a, 6, and c are real and either a ^ b, or a ^ c, or b ^ c. 20. Show that (2) and (3) in examples 2 and 3 hold, if a and b are real and unequal and a-\-b>0. CHAPTER XXV RATIO AND PROPORTION 317. The ratio of one number to another is the quotient of the first divided by the second. The dividend is called the first term, or the antecedent, of the ratio ; and the divisor, the second term, or consequent. The ratio of a to ft is written -, a/b, a -^ b, or a: b, each of which b forms can be read ' a is to 6 ' or ' a by &.' The ratio of 8 to 2 is 8/2, or 4 ; the ratio of 7 to 5 is 7/5. It is clear that a ratio is arithmetically greater than, equal to, or less than 1, according as its first term is arithmetically greater than, equal to, or less than, the second. 318. Since a ratio is a fraction, all the properties of frac- tions belong to ratios in whatever form the ratios are written. Thus a:b = am:bm, or a/b = am/(bm) ; §172 and a : b=(a -^ m) : (b -^ m), or a/b={a -^ m)/(b -=- m). § 173 Two ratios can be compared by reducing them as fractions to a common denominator. Ex. 1. Which is the greater, 3:11 or 5 : 19 ? 3:11= 3/11 = 57/209, and 5 : 19 = 5/19 = 55/209 ; hence the ratio 3 : 11 > the ratio 5 : 9. Ex. 2. (a : bf = a^ : b^ ; V^76 = V« - V&- §§ 186, 225. 319. By § 91, (a : &) (c : d) (e :/) = ace : hdf. The ratio ace : hdf is said to be compounded of the ratios a:b, c: d, and e : /. RATIO AND PROPORTION 320. The inverse of a ratio is its reciprocal. Hence the inverse of the ratio a : 6 is the ratio b : a (§ 182). 321. By § 183, a/b :c/d = ad: be. Hence the ratio of any two fractions can be expressed by the ratio of two integers. 322. Two numbers are said to be commensurdble or incoiii- mensurable with each other according as their ratio can or cannot be expressed by the ratio of two integers. E.g., y/2 and 5 are incommensurable with each other, so also are ^3 and ^5. The incommensurable numbers 3^2 and 7v^2 are com- mensurable with each other; for their ratio is 3/7. Compare § 224. 323. Ratio of concrete quantities. If A and B are two con- crete quantities of the same kind, whose numerical measures in terms of the same unit are the numbers a and 6, then the ratio of ^ to J5 is defined to be the ratio of a to b. Exercise 120. Find the simplest expressions for the following ratios : 1. 6 a to 12 a'*. 4. a/x to c/y. 2. 3 a^x/5 to 6 aa^/J. 5. a/{x - 2) to S/(x - 2y. 3. 1/a to 1/6. 6. 9/(a-by to 6/(a-b). 7. Write as a ratio (2x:Syy; (2a: bf ; (« : c)« ; i^oTb. Find the ratio compounded of : 8. The ratio 25 : 8 and the square of the ratio 4 : 3. 9. The ratio 32 : 27 and the cube of the ratio 3 : 2. 10. The ratio 6 : 7 and the square root of the ratio 25 : 36. 11. Arrange the ratios 5 : 6, 7 : 8, 41 : 48, and 31 : 36 in descending order of jnagnitude 334 ELEMENTS OF ALGEBRA 12. For what value of x will the ratio 15 + a;: 17 + a? be equal to 1/2 ? 13. What number must be added to each of the terms of the ratio 3 : 4 to make it equal to the ratio 25 : 32 ? Let X = the number to be added ; then (3 + a;)/(4 + a;)=25/32. 14. Find two numbers in the ratio of 5 to 6, whose sum is 121. 15. Which is the greater ratio, 5:7 or 5 + 2:7 + 2? 16. Which is the greater ratio, 7:5 or 7 + 2:5 + 2? PROPORTION. 324. Four quantities are said to be in proportion when the ratio of the first to the second is equal to the ratio of the third to the fourth. An equality whose members are two equal ratios is called a proportion. Thus, if a:b = c: d, (1) then a, b, c, and d are in proportion, or are proportional, and equation (1) is 2^ proportion. A proportion can be written in the form a/h = c/d, a:b = c: d, or a:b : : c: d, each of which is read ^ a by 6 is equal to c by d,' or ^ a is to 6 as c is to d.' The four numbers in a proportion are called the propor- tionals, or the terms, of the proportion. The first and fourth terms are called the extremes, and the second and third the means. E.g., a and d are the extremes, and b and c are the means in the proportion a :h = c : d. In (1), d is called the fourth proportional to a, b, and c. RATIO AND PROPORTION 335 325. The following theorem aud its converse in the next article are the two fundamental principles in proportion. In any propoHion the product of the extremes is equal to the product of the means. That is, if a\b = c\d, (1) then ad — be. (^2) Proof. Clearing (1) of fractions, we obtain (2). Ex. The first, second, and fourth terms of a proportion are c^, 2 a, and 5 h respectively ; find the third term. Let x = the third term of the proportion ; then c2 : 2 a = X : 5 6. .-. 2 ax = 5 6c2, or X = 5 6cV(2 a). 326. Conversely, if the product of one set of tivo numbers is equal to the product of another set of two numbers, either set can be made the extremes and the other set the means of a proportion. Proof. Let ad = bc. (1) Divide (l)hj db, a:b = c:d, ot c: d = a:b. Divide (1) by dc, a: c = b : d, or b : d = a: c. Divide (1) by a6, d:b = c: a, or c: a — d\b. Divide (1) by ac, d : c = 6 : a, or 6 : a = d : c. From this principle it follows that — (i) A proportion is proved ivhen it is proved that the product of its extremes is equal to the product of its means. (ii) In a given proportion, we can interchange the means, or the extremes, or we can take the means as extremes and the extremes as means. §172 6,91 327. If a: b = c: d, then ma : mb = no : nd, and ma : nb = mo : nd. 336 ' ELEMENTS OF ALGEBRA 328. Any proportion, as a : 6 = c : (/, (1) can he taken by (i) inversion ; that is, b : a = d : c, (2) (ii) alternation ; that is, a: c — b . d, (3) (ill) addition ; that is, a -\- b . a = c -\- d . c, (4) or a -{- b : b = c + d : d, (5) (iv) subtraction; that is, a — b : a = c — d : c, (6) or a - b : b = c - d : d, (7) (v) addition and subtraction ; that is, a + b: a-b = c + d\c-d. (8) Proof. From (1), ad = he. (1') Add hd to (1'), (a -f 6) c/ = (c + d) h. (2') Add - hd to (1'), (a - ?>) d = (c - d) h. (3') By § 326, from (1'), we have (2) and (3); from (2'), (5); and from (3'), (7). Dividing (2') by (1'), we obtain (4). Dividing (3') by (1'), we obtain (6). Dividing (2') by (3'), we obtain (8). Observe that (2) and (3) can be obtained from (1) by (ii) of § 326. 329. The products or the quotients of the corresponding terms of two proportions are x^^^ojyortional. That is, if a: b = c: d, (1) and a':b' = c''. d\ (2) then aa^ : 66' = c& : d&, (3) and ala! : bjb^ = c/c' : d/d'. (4) Proof Multiplying (1) by (2), by §§ 6 and 91 we ob- tain (3). Dividing (1) by (2), since ^ = ^, we obtain (4). a 10 010 330. Like powers or like principal roots of proportionals are proportional. RATIO AND PROPORTION 337 That is, if a:b = c:d, (1) then a" : 6" = c" : d", (2) and {/a:^b=^c:^d. (3) Proof. By §§ 128 and 186, from (1) we obtain (2). Hy §§ 221 and 225, from (1) we obtain (3). 331. In a series of equal ratios the sum of the antecedents is to the sum of the consequents as any one antecedent is to its consequent. That is, if a: b = c d = e .f= ", (1) then a + c + e -{-•' .b + d ^f+ •' = a:b = c:d= •.-. Proof Let a/b = r; then c/d = r, e//= ?', ••• ; hence a = br, c = dr, e =fr, •••. Adding the members of these equations, by § 6, we obtain a + c + e -f- ••• = (& + <« +/+•••) »•• a4- c 4- e + ••• _ _a_c_ ** b-hd-\-f-\--'~~^~b~d • 332. A general and easy method for proving a proportion is to represent the value of one of the equal ratios in the given proportion by a single letter, as was done in the last section. Ex. 1. Given a:b = c:d, prove that a2 + a6 : c2 + cd = 62 - 2 a& : (P _ 2 cd. (1) Let a/b = r ; then c/d = r ; then a = br, and c = dr. Substituting these values of a and c in each ratio of (1), we have a^ + ab ^ b^r^ -h 6^r ^ b^(r^ + r) ^ &« c2 + cd d^r^ + dh' d%r^ + r) d^' and b^-2ab ^ 6 ^-2 fc^r ^ b'^(l-2r) ^ 6^ d2-2cd d^-2d^r d\l-2r) '\/^\ 39. Using negative fractional exponents, write Vi/^; ^17^; -^17^'; ViT^. THEORY OF EXPONENTS 346 344. The meaning of any real commensurable exponent having been determined, it remains to prove that any such exponent obeys all the laws of positive integral exponents. For convenience in stating these laws we shall enlarge the meaning of the word power so as to include whatever is denoted by an exponent. 345. Exponents were so defined as to obey the following fundamental law : Tlie product of the mth and the nth power of any ba^e is equal to the (m + ri)th power of that Inxse. That is, a^^a" = «*"+", I where m and n denote any commensurable real numbers. Ex. 1. x^ . a;^ = x^'^^ = x^. Ex. 2. x~^ . a;"^ = x~^^"^) = x~^. Ex. 3. x^a~^ ' x-hi^ = x-^a~^ = l/(xa^). 346. Tfie quotient of the mth power of any base divided by the nth power of the same base is equal to the (m — n)th power of that base. That is, a'^/a" = a'"-", II where m and n denote any commensurable numbers. Proof a*" H- a" = a"* x a"" = a"-". §§ 342, 345 Ex. 1. x-8 ^ X-* = x-3-(-'«) = X. §§ 342, 345 Ex. 2. J-^ oT^ = a^-(-^) = al Ex. 3. a~^ -r a"^ = a~^~^~^^ = a^. 347. By §251, V^=Va^*; .-. a' = a^. r^p r»^p8 rq+pt Hence a' ' = a*« *« = a '" . Ex. 1. x^ . x^ = x^* = x^2" = x"^. Ex. 2. x^/x^ = x*"^ = x^ = xT^. 846 ELEMENTS OF ALGEBRA Exercise 123. Simplify each of the following expressions : 1. x'x\ 5. x^3fx-\ 9. a^6-2 X a-^h\ 2. a-W. 6. a-^a-V. 10. ^ x~^^y^ x 2 x~^y^ . 3. rt'V. 7. «y. 11. 4 a;~^a~^ X 3 a^' ai 4. a;-2a;-l 8. x'^^-x^. 12. 2A^x5:K"V^r. 13. a/a-\ 15. 4-74-^ 17. a^-yiu-^ 19. x-'^/x-l 14. .'^Va;-^ 6. 7-77-^ 18. x-'/x-^\ 20. t/V^-I 21. (8x--)/(6a.-0. 29. ^-^^"1 22. .^'"-70?-"*. 30. ic^/a; ^. 31. a^^^.aM. 23. x-^/x^-'^. 24. ar'-ya?2-n^ 25. x''^/y?-'\ 32. a'^aj"^^ • a^ici 26. o^-Yr--. 33^ a.V/(^-Vb. 27. a.-^-a^"l 34. a/^^^ v'^^. 28. aWa"^. 35. .J/a?'^ • ^ic'" -j- a^^. 348. If m and n denote any commensurable real numbers, the other three laws of exponents are : The nth poicer of the mth power of any base is equal to the mnth power of that base. That is, (a'")" = a'"". Ill The nth power of the product of any number of factors is equal to the product of the nth powers of the factors. That is, (ab.'.y = a"b" '". IV THEORY OF EXPONENTS 347 TJie nth power of the quotient of one number by another is equal to the quotient of the nth power of the first by the nth power of the second. That is, {a/by = a"/b". V Ex. 1. (a:2)-3 = x2(-3) = x-e = l/x*. By IH Ex. 2. (8-2)-i = 8-2H) = 8^ = 16. By HI Ex. 3. (a"^)~^ = a~^^"^^ = a^. By III Ex. 4. (x-iy-8)-2 = (a;-i)-2(y-8)-2 = x^^. By III, IV Ex. 5. (9"^ x-6)"^ = (9"*)~^(x-6)"^ = 9^ X* = 3 X*. Ex. 6. (x2/a-8)-2 = (a;2a3)-2 = a;-*a-«. Ex. 7. {x^/y^y^ = (x^y"^)-» = x'V. Ex.8. (^_£l\-''^(l^\-'^^:!^ \^y'l \zy^l 3-S2,-t = 27 2/^x8/8. Ex 9 Q^v^^ rT>/R _ aV^ . /a6-2\^ 6\/a^ '^^v^ 6a"^ ' \ba-^l = aV^^(a26-8)^ 349. Proof of laws III, IV, V when m and n are positive fractions. Let p, g, r, and s denote any positive integers. To prove III, (J)' = (V^)' § 337 = (V^y § 227 = [(V«W §226 = (^a)^ = a". Ill 348 ELEMENTS OF ALGEBRA r To prove IV, (aby = - + 1 + 1/ ^a by ^a + 1/ ^a - 1. The terms + 1 and — 1 may be regarded as the coefllcients of a^ ; hence arranging both expressions in descending powers of a, we have a^ + 1 + a~^ g^ - 1 + q~^ a^ + a^ + 1 - a3 - 1 - a 3 + 1 + a~^ + g"^ af +1 + a~^ 350 ELEMENTS OF ALGEBRA Ex. 2. Divide 16 a-^ + 5 a-i - 6 a-2 + 6 by 1 + 2 a-K Arrange in ascending powers of a. 16 a-3 - 6 a-2 4- 5 a-i + 6 [ 2 «-i + 1 16a-3+ 8q-2 8 a'^ - 7 a"! + 6 - 14 a-2 + 5 a-i - 14 a-2 - 7 g-i 12 a-i + 6 12a-i + 6 Ex. 3. Eind the square root of 4 X + 2 x^' - 4 x^ - 4 x^ + x^ + xi Arrange in descending powers of x. x^ - 4 x^ + 2 x^ + 4 X - 4x^ 4- x^ | x^ - 2 x^ + x «" J 2 x^ - 2 x^ xt -4x^ + 2x^ + 4x -4x^ +4x -4 x^ + x^ 2x^ -4x^ + x^ 2x« -4x^ + x^ Exercise 125. Multiply : 1. a^-\-ah^-^b^ by a^-bi- 2. x^ — ic^2/^ + y^ by a?^ + y^. 3. 3aj^ — 5 + 8a;~^ by 4 a;^ + 3 aj~i 4. 3a^-4a*-a~^ by 3 a^ -^ oT^ - 6 oT^ 5. a^ + a^6^ + 6^ by a^ — ah^-{-bK 6. ic^ — a;2 + a?^ — aj by a;^ + x^. 5 3 1 -1 1 3 1 7. a;^ — a?^ + a?^ — a? * by a;^ + ^^. 8. a3 + &^ 4- c^ — bh^ — c^a^ — a^6^ by a^ + b^ -\- ci THEORY OF EXPONENTS 351 n n 9. x'^ + x^-j-l by X-" -\- x~~^ -}- 1. 10. iJ-^i^a^^i-^i^ah-^b^ by ia^ + ^bi 11. (f + 2c-'-7 by 5-3c-'= + 2(f. Divide : 12. 21x + a;^4-a;^4-l by 3a;^4-l. 13. 15a-3a^-2a"* + 8a-i by 5a*4-4. 14. 55^-66^-4fe-^-46~*-5 by 6^-26"^. 15. 21a'^ + 20-27a^-26a2' by 3a*-5. 16. a;^y~^ + 2 H- ic~^y^ by x^2/~^ — 1 + x~^yK 17. a^ + a-6^ — ah^ — ab + ah^ + &^ by a^ 4- &i 18. x^y~^ + y^x~^ by a;'?/"^ + y^a;~». 19. «' — 2 -h a~3 by a^ — a '. 20. 8 c-" - 8 C + 5 c^" - 3 c-3" by 5 c" - 3 c"". Find the square root of the following expressions : 21. 25a* + 16-30a-24a.^ + 49a*. 22. 9x-12x^-\-10-4:X~^-{-x-\ 23. 4 ay^a-' + 12 a;a-^ + 25 + 24 x-^a + 16 aj-^a^ 24. 25 arV' + i y^x-^ - 20 ic^/"^ - ^yx'^ + 9. 25 . x^ — 2 oT^x^ + 2 a^ic^ + a~h^ — 2 a^a;^ + ai 26. 4 .T" + 9 «-" + 28 - 24 aj-i" - 16 a;K 27 . 9 a;-^ - 18 a;- Vy + 15 2/ -^ a^ - 6 Vp ^ a; + ^V^. 352 ELEMENTS OF ALGEBRA 352. The following examples are applications of the for- mulas for products and quotients in Chapter IX., to binomials whose terms involve fractional and negative exponents : (1) (2 a^ - x~i)^ = (2 Jy + 3(2 a^)2(- x~^) +S(2 J)(-x~^y+(-a~^y. (1) = 8 a^ - 12 ax~^ + 6 a^x-i - a"i (2) (2) (x^ + yi) (x^ - yi) = {x^^ - {y^y = x^- yK § 122 (3) (7 x-9 2/-1) (7 x+9 y-i) =49 a;2_8l y-\ (4) (4a;-5a;-i)(4x+3x-i) = 16a:2+(3x-i-5x-i)4ic-15ic-2 §123 = 16 x2 - 8 - 15 a;-2. (5) (pfi - 1) - (X^ - 1) = [(X^)5 - 15] ^ (x^ _ 1) = (x^f + (x^s ^ (a;^-)2 + x^ + 1 § 129 = a;'^ + ic + cc^ + ic^ + 1. (6) (x^ + 27) - (x^ -t- 3) = i(xh^ + 33] - {x^ + 3) = X - 3 a:^ + 9. Exercise 126. Write the value of the following expressions : 1. {ar^-\-h^f. 3. {m^ + Jf. 5. (f^ - s'^f. 2. (x^^y-^f, 4. (c2-6*)l 6. (2a^-a-^f. 7. (r-2 + 6^y. 11. (a^ + &~^/. 8. {i^'-^n-^y. 12. (a^^-2/"^)«. 9. (iVa-i-^6)^ 13. (o?^ + 1) (aJ^ - 1). 10. (a~^-2 62c^)^ 14. (x^-\.y^)(x^-y^), 15. (4 a;^ + 3 a"^) (4 a;^ - 3 a"*). 16. (3 a; - 5 a-^ (3 a; + 2 a-i). THEORY OF EXPONENTS 353 17. (ab-c^(ab-^5x-^). 18. (x-9a)-^{x^ + 3a^). 19. (a-2x_i6)^(a-x_4). 20. (aj-3- + 8) -f- (a;-* 4- 2). 21. (c2* - c-'') -=- (c* - c-i*). 22. (l-8a-3)^(l-2a-^). 23. (a;* + ic* + 1) (a;^ - 1). 24. (a;-'*-l)^(a;-i + l). 25. (ar^+-32)H-(a;'* + 2). 26. (a,-3 - 2/S) ^ (a;^ — 2/^). 27. (ar^ + /)^(a;* + A 28. (a; - 243 2/^) ^(a;^- 3 2/^. CHAPTER XXVII INDETERMINATE EQUATIONS AND SYSTEMS 353. Division by zero. As a quotient, 0/0 denotes the number which multiplied by is equal to (§ 85). By § 74 any number multiplied by is equal to ; hence 0/0 denotes any number whatever, or is indeterminate. That is, when the dividend is zero, division by zero is indeterminate. As a quotient, a/0 denotes the number which multiplied by zero is equal to a. But any number, however large, multiplied by zero, is zero; hence the division of a by 0^ is impossible. That is, ivhen the dividend is not zero, division by zero is impossible in the sense that no number can express the quotient or any part of it. 354. The forms 0/0 and a/0. As an answer to a problem the indeterminate form 0/0 denotes that the problem is indeterminate, i.e., has an unlimited number of answers. As an answer to a problem, a/0 denotes that the problem involves inconsistent conditions, and is therefore impossible, as is illustrated by the following problem : Prob. A and B are travelling in the direction PB at the rates of a and h miles per hour. At 12 o'clock A is at P and B at ^, which is c miles to the right of P. Eind when they are together. P Q B Let distances measured to the right from P, and periods of time after 12 o'clock, be regarded as positive. Let X = the number of hours from 12 o'clock to the time when A and B are together. Then ax = 6x + c. (1) Hence x = — —• (2) • a — 364 INDETERMINATE EQUATIONS AND SYSTEMS 355 Discussion. If c > and a > 6, a; is positive ; that is, A will over- take B at some time after 12 o'clock. If c > and a < &, x is negative ; that is, A and B were together at some time before 12 o'clock. If c = and a =^ b^ x = 0; that is, A and B are together at 12 o'clock, but not before or after that time. If c = and a = b, x = 0/0 ; that is, A and B are always together under the conditions; and the problem is indeterminate, i.e., has an unlimited number of answers. If c ^fc and a = 6, a; = c/0 ; that is, A and B can never be together as they are always at a fixed distance apart ; the problem involves inconsistent conditions, and is therefore impossible. Observe that the fraction — - — assumes the form - by a-b -^ reason of two indej^endent conditions; namely, c = and a = b. In any such case the form 0/0 indicates that the given fraction can have any value under the conditions. 355. An impossible equation is one which expresses a con- dition which cannot be satisfied. E.g., Sx + 5 = Sx — S is an impossible etjuation ; for it expresses the condition • a; = — 13, which no value of x can satisfy. Again, ^a; = — 3 is an impossible equation, when ^x is restricted to its principal value. An impossible system of equations is a system whose equa- tions are inconsistent (§ 206). E.g., the system ax -{- by = c, (1) 3 ax + 3 6y = 5 c, (2) } («) is impossible ; for its equations are evidently inconsistent (§ 206). An impossible equation or system of equations is often a particular case of a more general equation or system, in which the solutions involve the form a/0. Thus, the equation ax = b becomes impossible only when a = 0, and then its root b/a becomes 6/0. It will be seen in § 356 that a system of two linear equations in X and y becomes impossible only for a certain relation between the coefficients of its equations, which makes the values of x and y assume the form a/0. 856 ELEMENTS OF ALGEBBA Again the system x + y = 9, . (1) 2x + y = lS, (2) [ (&) x + 5y = 16, (3). is impossible ; for the only solution common to (1) and (2) is 4, 5, and this reduces (3) to 29 = 16. Equation (3) cannot be obtained from (1) or (2), or by combining (1) and (2); hence it is independent of them separately and jointly. System (b) illustrates the principle that When the number of independent equations in a system ea>- ceeds the number of unknowns, the system is impossible. 356. A defective system is one which, lacks one or more of the full number of solutions which we would expect from the degrees of its equations. E.g.^ the system |(«) a2x2 - 62y2 = c2, (1) ax-(h-\- e)y = c, (2) which has, in general, two solutions (§306), becomes defective when e = 0. For, dividing (1) by (2) when e = 0, we obtain ax + by = c. (3) Equations (2) and (3) form a system equivalent to system (a); hence system (a) has but one solution when e = 0. 357. An indeterminate equation is one which has an un^ limited number of solutions. Thus any equation in two or more unknowns is indeterminate. An indeterminate system of equations is one which has an unlimited number of solutions. E.g. J the system Sx + 4:y-{-5z = 0, x-y-2z = 0, ^^^^ Ida] is an indeterminate system ; for, assigning any value whatever to z, we can find a corresponding set of values of x and y. Hence, system (a) has an unlimited number of solutions, and is indeterminate. INDETERMINATE EQUATIONS AND SYSTEMS 357 Again, the system 2x-\-Sy-z = lo, (1) Sx-y + 2z = 8, (2) 5 X H- 2 y + 2! = 23, (3) is indeterminate. No two of its equations are equivalent, but any one of them can be obtained from the other two ; thus, by adding (1) and (2), we obtain (3). Hence the system contains but two in- dependent equations, and therefore any solution of two of them will be a solution of the third. These examples illustrate the following principle : WJien the number of independent equations in a system is less than the number of unknowns, the system is indeterminate. Ex. By discussing its solution, show that tlie system ax-i- by = c, a'x + b'y = c', |(«) is (i) indeterminate if ^ = _ = 2_ j (1^ a' b' c' and (u') impossible if ^ = ^ r,^ ^. (2) a b' c' By § 207 the values of x and y in system (a) are _ b'c — be' _ ac' — a'c .on ~ ab' - a'b ^ ab' - a'b ^ ^ (i) When condition (1) is satisfied, from (1) we have ab' - a'b = 0, b'c - be' = 0, ac' - a'c = ; hence the values of x and y in (3) each assume the form 0/0 ; that is, the system has an unlimited number of solutions, and is therefore indeterminate. (ii) When condition (2) is satisfied, we have ab' — a'b = 0. But neither b'c — be' nor ac' — a'c is zero. Hence the value of each x and y assumes the form a/0 ; that is, the system has no solution, and is therefore impossible. The equations in (a) are evidently equivalent when (1) is satisfied, and inconsistent when (2) is satisfied. 358 ELEMENTS OF ALGEBEA 368. Sometimes it is required to find the positive integral solutions of an indeterminate equation or system. The following examples will illustrate the simplest general method of finding such solutions. Ex. 1. Solve 7 X + 12 y = 220 in positive integers. Dividing by 7, the smaller coefficient, expressing improper fractions as mixed numbers, and adding the proper fractions, we obtain aj + y + ^pi=:31. (1) Since x and y are integers, 31 — x — y is an integer ; hence the fraction in (1) denotes an integer. Multiplying this fraction by such a number as will make the coeffi- cient of y divisible by the denominator with remainder 1 (which in this case is 3) , we have - — ^ = 2 ?/ — 1 + ^—^ — = an mteger. Hence ^ = an integer = p, suppose. :.y = lp + 1. . (2) From (1) and (2), x = 28-12j9. (3) Since x and y are positive integers, from (2) it follows that^ >— Ij and from (3) it follows that p < 3 ; hence i) = 0,1,2. (4) From (2), (3), and (4), we obtain the three solutions a: = 28, 16, 4; y= 2, 9, 16. Ex. 2. Solve in positive integers the system a; + 2/ + ^ = 43, (1) 10 aj + 5 ?/ + 2 ;s = 229. (2) J INDETERMINATE EQUATIONS AND SYSTEMS 359 Eliminate z, 8x-\-Sy = 143, y + 2x-\-^-^^ = ^7. (3) .-. i^LzA = x-l+ ?-=-i = an integer. o o x—1 .'. = an integer = p, suppose. o .-. x = 3p + l. (4) From (3) and (4), y = io-8p. (6) From (1), (4), and (5), z = 5p- 3. (6) From (6), i) > ; and from (o), /> < 6 ; hence p= 1, 2, 3, 4, 5. . Whence x = i, 7, 10, 13, 16; y = 37, 29, 21, 13, 5 ; z= 2, 7, 12, 17, 22. Thus, the system has five positive integral solutions. Exercise 127. Solve in positive integers : 1. 3a; + 291/ = 151. 8. 12 x -lly + 4:Z = 22, 2. 3.'c + 82/ = 103. -4:X-{-5y-\-z = 17. 3. 7x-^12y = 152. 9. 20a: - 21?/ = 38, 1 4. 13a; + 72/ = 408. S y -\- 4. z = 34.. J 5. 23 a; + 25 2/ = 915. 10. 5ic- 14^ = 11. 6. 13 a; + 11 2/ = 414. H. 13 a; + 11 2; = 103, | 7. 6a;+72/+4.=122, 1 7.-52/ = 4. J lla;+8i/-62=145.J 12. 14 a; - 11 y = 29. 13. A farmer buys horses at $ 111 a head, cows at $ 69, and spends $ 2256. How many of each does he buy ? 360 ELEMENTS OF ALGEBRA 14. A drover buys slieep at $ 4 a head, pigs at $ 2, and oxen at ^17. If 40 animals cost Mm $ 301, how many of each kind does he buy ? 15. I have 27 coins, which are dollars, half-dollars, and dimes, and they amount to $ 9.80. How many of each sort have I ? 16. A drover buys sheep at $ 3.50 a head, turkeys at $ 1.33i, and hens at | 0.50. If 100 animals cost him $ 100, how many of each does he buy ? CHAPTER XXVIII THEORY OF LIMITS 359. A variable is a quantity which is, or is conceived to be, continually changing in value. E.g.^ the time since any past event is a variable ; so also is the height of an ascending or a descending balloon. The amount of water in a cistern which is being filled by a con- tinuous stream is a variable ; and the number which measures this amount is a variable number. Variable numbers are usually represented by the final letters of the alphabet, as x, ?/, z. A constant is a quantity whose value is fixed or invari- able. Constant yiumbers are usually represented by figures or the first letters of the alphabet, as 4, 7, a, 6, c. E.g.^ the time between any two past events is a constant ; and the number which measures this time is a constant number. 360. Limit of a variable. When, according to its law of change, a variable approaches indefinitely near, and con- tinually nearer a constant, but can never reach it, the variable is said to approach the constant as its limit. E.g.^ let A, B, and N be three ^xed points in the straight line AN; then AB and NB will be constant distances. P I I I I I. J A O D E B N Suppose a point P, starting from A, moves 1/2 the distance from ^to 5, or to C, the first second ; 1/2 the remaining distance, or to i), the next second ; 1/2 the remaininl; then in harmony with the meaning of commensurable exponents we assume a"* to denote a number such that a'Ka'^K a'. (1) Let x=:m, and 2; = m ; then It (z-x)^ It (z) - It (x) = 0. Hence by § 385, lt(a^-^) = 1. (2) THEORY OF LIMITS 373 Now a'-a' = a^ (a'-' - 1). .-. It (a* - a') = It (a') [It (a'-=^) - 1] § § 370, 364 = lt(a^)(l-l)=0. by (2) From (1), a' — a' > a"" — a"" ', hence, as a' — a' = 0, a"" — a' = 0, ,'. a"" = It (a') when x = m. That is, any base a icith an incommensurable exponent m denotes the limit of a* when x = m. 387. Proof of laws of exponents I, II, IV, V. Let m and n be any incommensurable constant numbers, and let x and y be commensurable variable numbers such that x^m, y = n. Proof of I. a'a^ = a'^". § 345 Hence It (a'a*) = It (0*+") = ar+". (1) But It (a'a^) = It (a*) • It (a") = a'^a^ (2) From (1) and (2), a'^a'' = a"»+". Law I Proof of 11. By law I we have (i^~^a!* = a"*. Hence by § 32 ci— " = a'^/a\ Law II Proof of IV. a'b' = (aby. § 348 Hence It (a'6*) = It [(«6)*] = (a^)- (o) But It (a'b') = It {a') • It (6^) = a^ • &«. (6) From (5) and (6), (aft)*" = a-^ft"*. Law IV iVoo/ 0/ V. a'/b' = (a /by. § 348 Hence It (a'^/b^) = It [(«/6)^] = (a/ft)"*. (7) But It (a'/b^) = It (a-) / It (6^) = a'"/ 6"*. (8) From (7) and (8), (a/b^ = a'^/b'^. Law V 374 ELEMENTS OF ALGEBRA 388. To prove law III for incommensurable exponents we need the following theorem of limits : Ify and z are commensurable and m and c are incommen- surable; tlieny when y = m and z = c, z^ = c"'. Proof. For all values of z (except and 1), and hence for z variable, by § 386, we have 2;y _ 2;'* = 0, when y = m. (1) By § 375, g'^ _ c"* = 0, when z = c. ' (2) From (1) and (2), (2^ - z^) + {z^ - C') = ; that is, 2^ — c"* = 0, when y = m and 2 = 0. Hence z^ = c"* when y = m and z = c. (3) Proof of III. Using the same notation as in § 387, we have {ay = rt^^. Hence It [(a^)^] = It (a^^ = a^^. (4) But by (3), It [(a^)^] = [It (a^)]"^^) = {ar)\ (5) From (4) and (5), («'")''=«""*. From these laws for incommensurable numbers the other laws follow by the proofs already given for commensurable numbers. VARIATION. 389. Two variables are often so related that the value of one depends upon the value of the other. E.g.^ the distance a train runs at a given speed depends upon the time it runs, and this distance increases when the time increases. The length of an elastic cord depends upon its tension, and this length varies when the tension varies. liy = b x'^, the value of y depends upon the value of x, and y varies when X varies. We shall here consider only the simplest kinds of variation. THEORY OF LIMITS 375 390. Direct variation. When the ratio of two variables is a constant, either variable is said to vary directly as the other. The symbol oc, read ^ varies directly as/ is called the sym- bol of direct variation. When placed between two variables it denotes that their ratio is some constant. The word ' directly ' is sometimes omitted. E.g., 2/ OCX, read ^y varies directly as a:,' denotes that y/x = m^ where m is some constant. Again, if y = 3 x, y/x = 3 ; hence, yocx or xccy. 391. If one vanable varies directly as another, either vari- able is a constant multiple of the other; and conversely. Proof If yc^x,y/x = m] .-. y=mx, or x=(l/m)ys Conversely, if y = mx, y/x = m\ .-.yocx, or xccy. E.g., the area of a rectangle = base into altitude. Hence if the altitude is constant, the area oc the base. And if the base is constant, the area oc the altitude. 392. If y^x, and if x\ y' and «", ?/" are any two sets of corresponding values of x and y, then 2/':a:' = y':a:". (1) Proof If 2/ oca:, y'/x^ = m and y'^/x^^ = m. (2) From the equal ratios in (2), we have the proportion (1). Conversely, if y' : x' = y" : x", y = mx and yocx. 393. Inverse variation. One variable is said to ixiry in- versely as another when the first varies as the reciprocal of the second. That is, / varies inversely as jr, when y oc 1/jr. 394. If one variable varies inversely as another, the product of the two variables is a constant; and conversely. Proof. If yaz 1/x, y = m (1/x) ; .-. yx = m. Conversely, if yx = m, y = m(\/x); .-. yocl/x. E.g., if yx = 3, y varies invereely as x. 376 ELEMENTS OF ALGEBRA 395. Joint variation. One variable is said to vary as two others jointly when it varies as the product of the two. That is, / varies as x and z jointly^ lohen y (jz {xz), or y = m(xz). E.g., it W= the amount of work done by 2lf men in Z> days ; then, if 31 and D both vary, WazM x D ; if M is constant, TT x D ; if D is constant, W^M. One variable is said to vary directly as a second, and in- versely as a third, when it varies as the product of the second into the reciprocal of the third. That is, y varies directly as x, and inversely as z, when yccx(l/z), or y = mx(l/z). E.g., if yz = Sx, y = 3x(l z); hence yccx(\/z'). 396. In each of the preceding cases of variation, the value of the constant, m, can be found when any set of corresponding values of the variables is known. Ex. 1. Given yS = inl2a-{-(n-l)dl. (3) If any three of the five numbers a, d, I, 71, S are given, the other two can be found from equations (1) and (2), or from (3) and (2). Ex. 1. Find the sum of 20 terras of the A. P. -5-1 + 8 + 7+ 11 + -. Here a = - 5, d = 4, n = 20. .'. S=^n{;2a+(n- l)d} = 10{-10 + 19 X 4} = 660. Ex. 2. Find the sum of the first n consecutive odd numbers, 1, 3, 5 .... Here a = 1, d = 2, n = n. .-. S=^n{2a-\-{n-l)d} = iw{2+(>i-l)2} Hence the sum of n consecutive odd numbers, beginning with 1, is w2. Ex. 3. The first term of an A. P. is 6, and the sum of 25 terms is 25. Find the common difference. THE PROGRESSIONS 383 Here a = 6, /S' = 25, w = 25 ; hence from (3) of § 405 we have 25 = ^ X 25(12 + 24 d} .-. d = -6/12. Ex. 4. How many terms must be taken of the series 11, 12, 13, ... to make 410? Here a=ll, (Z = l, S = 410 ; hence from (3) of § 405 we have 410 = in{22+(n-l)}. (1) .-. w = 20, or -41. Since the number of terms must be an arithmetic whole number, the number of terms is 20. See § 297. Ex. 5. How many terms must be taken of the series — 16, — 15, -14, ... to make - 100? Here a = — 16, d = 1, S = — 100 ; hence we have -100 = in{-32+(n-l)}. .-. n = 8, or 25. Hence the number of terms is 8 or 25. The sum of the 17 terms following the first 8 must therefore be zero. These 17 terms are — 8, — 7, — •••, 7, 8, and their sum is evidently zero. Exercise 132. 1. Find the twenty-seventh and forty -first terras in the series 5, 11, 17, •••. 2. Find the seventeenth and fifty-fourth terms in the series 10, 11^, 13, •••. 3. Find the twentieth and thirteenth terms in the series _.S _2 —1 ... 4. If the twelfth term of an A. P. is 15, and the twen- tieth term is 25, what is the common difference ? 5. The seventh term of an A. P. is 5, and the twelfth term is 30. Find the common difference. 6. The first term of an A. P. is 7, and its third term is 13. Find the tenth term. 384 ELEMENTS OF ALGEBRA 7. The first term of an A. P. is 20, and its sixth term is 10. Find the twelfth term. 8. The seventh term of an A. P. is 5, and the fifth term is 7. Find the twelfth term. 9. Which term of the series 5, 8, 11, ••• is 65 ? 10. Which term of the series J, f, f , ♦•• is 18 ? 11. Insert 6 arithmetical means between 8 and 29. 12. Insert 7 arithmetical means between 269 and 295. 13. Insert 15 arithmetical means between 67 and 43. 14. If a, b, c, d are in A. P., prove that a -\- d = b -^ c. 15. The sum of the second and fifth terms of an A. P. is 32, and the sum of the third and eighth is 48. Find the first term. 16. The sum of the third and fourth terms of an A. P. is 187, and the sum of the seventh and eighth terms is 147. Find the second term. Find the sum of each of the following series : 17. 5, 9, 13, •.. to 19 terms. 18. 1, 21,31 ... to 12 terms. 19. — 5, — 1, 3, ••• to 20 terms. 20. i, I, i, ... to 7 terms. 21. 10, -%S ^8-, ... to 7 terms. 22. I, 1, |, ... to 15 terms. How many terms must be taken of : 23. The series 42, 39, 36, ... to make 315? 24. The series 15, 12, 9, ... to make 45 ? 25. The series — 8, — 7, - 6, ... to make 42 ? TUE PROGRESSIONS 385 26. Find the sum of all the numbers between 100 and 500 which are divisible by 3. 27. Find the sum of all the odd numbers between 100 and 200. 28. The sum of 10 terms of an arithmetical series is 145, and the sum of its fourth and ninth terms is 5 times the third term. Determine the series. 29. Divide 80 into 4 parts which are in A. P., and which are such that the product of the first and fourth parts is | of the product of the second and third. 30. Find 4 numbers in A. P., such that the sum of their squares shall be 120, and that the product of the first and last shall be less than the product of the other two by 8. 31. If a body falling to the earth descends a feet the first second, 3 a the second, 5 a the third, and so on ; (1) how far will it fall during the tth. second ? (2) how far will it fall in t seconds ? Ans. (2t — l) a, at-. 32. How many strokes does a common clock make in 12 hours ? 33. A debt can be discharged in a year by paying $ 1 the first week, $ 3 the second week, $ 5 the third, and so on. Find the last payment and the amount of the debt. 34. One hundred apples are placed on the ground at the distance of a yard from one another. How far will a person travel, who shall bring them, one by one, to a basket, placed at a distance of a yard from the first apple ? 35. Two boys A and B set out at the same time, to meet each other, from two places 343 miles apart, their daily journeys being in A. P. ; A's common difference being an increase of two miles, and B's a decrease of 5 miles. On the day at the end of which they met, each travelled exactly 20 miles. Find the duration of each journey. 386 ELEMENTS OF ALGEBRA GEOMETRIC PROGRESSIONS. 406. A geometric progression (G. P.) is a series in which the ratio of any term (after the first) to the preceding term is the same throughout the series. This ratio, which can be either positive or negative, is called the common ratio. •-g-, , the series 2, 6, 18, 54, 162, 8, 4, 2, 1, h 1, -1, h -i ^, or is a geometric progression (G. P.). In the first series, the common ratio is 3 ; in the second series it is 1/2 ; and in the last it is — 2/3. If we multiply any term in either series by the common ratio, the product will be the next term of that series. 407. The nth. term. Let r denote the common ratio, and a the first term of any G. P. ; then by definition the second term = ar^ the third term = ar^, and the nth. term = ar'*"^ (1) E.g.^ if the first term of a G. P. is 8, and the common ratio is 1/2, the fifth term = 8 x (l/2)5-i = 1/2, and the ninth term = 8 x (l/2)9-i = 1/32. Ex. The sixth term of a G. P. is 156, and the eighth term is 7644. Find the seventh term. Here 156 = the sixth term = ar^, (1) and 7644 = eighth term = ar^. (2) Divide (2) by (1), 49 = r2. (3) .-. r = ± 7. (4) But the seventh term = sixth term x r = 156 (±7; = ±1092. THE PROGRESSIONS 387 408. Sum of n terms. Let S denote the sum of n terms ; then S = a-{- ar + a?*^ + ••• -|- ar**-^ + «?•"-* = a (1 -\- r -{- 7^ + r^ ■] h ^'""^ + r**-^) 1 — r" = a- 1 — 7* _ a(l — r") § 129 Hence S = ^^^^^ ^. (1) 1 — r Let Z denote the ?ith term ; then from § 407 I = ar'^-K (2) From (1) and (2) c, a — rl (3) If any three of the five numbers a, I, n, r, s, are known, the other two may be found from equations (1) and (2), or from (2) and (3). Ex. Sum the series 6, — 18, 64, ••• to 6 terms. Here a = C, r = - 18 -=- G = - 3, n = 6. = f{l-36} = - 1092. 409. When three numbers, a, b, c, are in G. P., the middle term b is called the geometric mean of the other two terms a and c. 410. If a, 6, c are in G. P., by § 406, we have c: b = b: a. .-. 6 = Vac. That is, the geometric mean of any two numbers is the mean proportional between them. 388 ELEMENTS OF ALGEBBA 411. All the terms between any two terms of a G. P. may be called the geometric means of the two terms. 412. To insert m geometric means between a and b. Calling a the first term, h will be the (m + 2) th term ; hence by (1) of § 407, we have h = ar^+\ •»+!/ r= ^/bTa. (1) Hence the m means required are ar, ar^, ••• ar"^, in which r has the value given in (1). Ex. Insert 6 geometric means between 56 and — 7/16. Here a = 56, 6 = - 7/16, and w + 1 = 7. ^ IQ > 2* X 23 2 Hence r = — 1/2, and the 6 means required are - 28, 14, - 7, 7/2, - 7/4, 7/8. Exercise 133. Find the last term in the following series : 1. 2, 4, 8, ••• to 9 terms. 2. 2, 3, 4^, ••• to 6 terms. 3. 3, -3'^, 3^ ••• to 2 7i terms. 4. X, 1, 1/x, ••• to 30 terms. 5. The first term of a G. P. is 3, and the third term is 4. Find the fifth term. 6. The third term of a G. P. is 1, and the sixth term is — 1/8. Find the tenth term. 7. The fourth term of a G. P. is 0.016, and the seventh term is 0.000128. Find the first term. 8. The fourth term of a G. P. is 1/18, and the seventh term is — 1/486. Find the sixth term. 9. Insert 3 geometric means between 486 and 6. THE PROGBESSIONS 389 10. Insert 4 geometric means between 1/8 and 128. 11. Insert 5 geometric means between 3 and 0.000192. 12. Insert 4 geometric means between a^b~^ and a~^b^. Find the sum of the following series : 13. 64, 32, 16, ••• to 10 terms. 14. 8.1,2.7,0.9, ..-to? terms. 15. 3, - 1, 1/3, ••• to 6 terms. 16. 1/2, 1/3, 2/9, ... to 7 terms. 17. - 2/5, 1/2, - 5/8, ... to 6 terms. 18. 2, - 4, 8, ... to 2p terms. 413. When r < 1 arithmetically, the successive terms of a G. P. become smaller and smaller arithmetically, and the G. P. is said to be a decreasing progression. 414. TJie limit of the sum of an infinite number of terms of a decreasing G. P. is Proof Prom (1) of § 412, we have 1-r 1 (1) Now if r < 1 arithmetically, and the number of terms, or n, is increased without limit, then 1 — r Hence from (1), by § 364, we obtain The limit of the sum of an infinite number of terms of a series is often called the sum of the series. 390 ELEMENTS OF ALGEBRA E.g., if a = 2 and r = 1/2, we have the decreasing G. P., 5, 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128. .... (1) The sum of an infinite number of terms of this series approaches 4 as its limit. For suppose that we bisect a line four inches long, and take away one of the parts ; then bisect the remainder, and take away- one of the parts ; and continue this process without limit. It is evi- dent that the part remaining will approach zero as its limit, and the sum of the successive parts taken away will approach four inches as its limit. But the numbers of inches in the successive parts taken away will be the terms of series (1). Hence the sum of an infinite number of terms of that series approaches 4 as its limit. Ex. 1. Eind the sum of the series 1, 1/2, 1/4, •••. Here a = 1, r = 1/2. From (2), it (6') = ^— ^ = 2. Ex. 2. Find the sum of the series 9, —3, 1, •••. Here a = 9, r = - 1/3. From (2), ,t (^) = j_^ = f =6i. Ex. 3. Express 0.423 as a common fraction. 0.423 = 0.4232323 ...= — + ^ + H + .". 10 103 105 23 103 H- "l02J 23 103 «s^ 23 990 23 , 23 , 23 , ^^^' 10-3 + 10-5 + 10^ + .-. 0.423 = ^ + ^'^ = IM- Ex. 4. Find the infinite G. P. whose sum is 18, and whose second term is — 8. Here ar = -8, (1) and -^ = 18. (2) 1 — r Divide (1) by (2), r(l - r) = - 4/9. ... ,.2 _ y _ 4/9 = 0. .'. r = - 1/3 or 4/3. THE PROGRESSIONS 391 Only the value — 1/3 is admissible for r, since the series is a decreasing one. From(l), a = -8 -(-1/3) = 24. Hence the series is 24, -8, 8/3, -8/9, .... Exercise 134. Find the sum of each of the following series : 1. 9, 6, 4, .... 4. f, -1, I, .... 2. 1 -J, I, .... 5. 0.9, 0.03, 0.001, .... 3. h h yV •••• 6. 0.8, -0.4, 0.2, .... Express as a common fraction : 7. 0.3. 8. 0.16. 9. 0.24. 10. 0.378. 11. 0.037. 12. Find the infinite G. P. whose sum is 4, and whose second term is f. 13. Find the infinite G. P. whose sum is 9, and whose second term is — 4. 14. If every alternate term of a G. P. is taken away, the remaining terms will be in G. P. 15. If all the terms of a G. P. are multiplied by the same number, the products will be in G. P. 16. Show that the reciprocals of the terms of a G. P. are in G. P. 17. By saving 1 cent the first day, 2 cents the second day, 4 cents the third day, and so on, doubling the amount every day, how much would be saved in a month of 30 days? 18. Suppose a body to move eternally as follows : 20 feet the first minute, 19 feet the second minute, 18 J^ feet the third minute, and so on. Find the limit of the distance passed over. 392 ELEMENTS OF ALGEBRA 19. A ball falling from the height of 100 feet rebounds one-fourth the distance, then falling, it rebounds one-fourth the distance, and so on. Find the distance passed through by the ball before it comes to rest. 20. If in problem 31 of exercise 132, a = 16 ^J^"? ^^^ ^^ng will it be before the ball in problem 19 comes to rest ? To fall 100 feet, it takes Vl00TT6^, or 10\/J^, seconds; to rebound, or to fall, 25 feet, it takes •\/25 -^ 10^^, or 5\/J^2_^ seconds ; to rebound, or to fall, Q\ feet, it takes \/6|~n(3^, or | V^^, seconds ; and so on. Hence the time = 10 vTS + 2(5 VJS + f ^^ + -) = 30 V^ = 1%% V579 = 7.4805 +. HAEMONIC PROGRESSIONS. 415. An harmonic progression is a series of numbers whose reciprocals form an A. P. E.g.^ the series 1, h h 7' •••> or 4, -4, is an harmonic progression ; for the reciprocals of their terms 1, 3, 5, 7,...., or \, -\, -f, ... are in A. P. 416. When three numbers are in harmonic progression (H. P.), the middle term is called the harmonic mean of the other two. 417. Let H be the harmonic mean of a and h ; then by §415, -' 77' 7 ^^® ^^ ^- ^• a H h " H a b h' 2 1,1 rj 2 ah = --{--, or H Hah a + 6 THE PROGRESSIONS 393 418. If A and G denote respectively the arithmetic and the geometric mean of a and 6, then (§§ 403, 409) 2ab 2 a + 6 r.AxH = '^±^X^^==ab = 0\ 2 a + b Hence A:G=G:H. That is, the geometric mean of any two numbers is also the geometric mean of their arithmetic and harmonic means. 419. Problems in H. P. are generally solved by inverting the terms, and making use of the properties of the resulting A. P. Ex. The fifteenth term of an H. P. is 1/25, and the twenty-third term is 1/41. Find the series. Let a be the first term, and d the common difEerence of the corre- sponding A. P. ; then 26 = the fifteenth term = a 4- 14 d, and 41 = the twenty-third term = a + 22 d. .-. (Z = 2, a = - 3. Hence the A. P. is — 3, — 1, 1, 3, 5, •••, and the H. P. is - \, -1, 1, \, \, .... Exercise 135. 1. Find the sixth term of the series 4, 2, IJ, ..•. 2. Find the eighth term of the series IJ, 1||, 2^, •••. Find the series in which 3. The second term is 2, and the thirty-first term is ^\. 4. The thirty-ninth term is ^, and the fifty-fourth term 394 ELEMENTS OF ALGEBRA Find the harmonic mean between 5. 2 and 4. 6. 1 and 13. 7. { and J^. 8. Insert 2 harmonic means between 4 and 12. 9. Insert 3 harmonic means between 2|- and 12. 10. Insert 4 harmonic means between 1 and 6. 11. If a, b, c are in harmonic progression, prove that a — b:b — c = a:c. CHAPTER XXX PERMUTATIONS AND COMBINATIONS 420. Fundamental principle. If one thing can he done in m ivays, and (after it has been done in any one of these ways) a second thing can he done in n ways; then the two things can he done in m x n ways. Ex. 1. If there are 11 steamers plying between New York and Havana, in how many ways can a man go from New York to Havana and return by a different steamer ? He can make the first passage in 11 ways, with each of which he has the choice of 10 ways of returning ; hence he can make the two jourueys in 11 x 10, or 110, ways. Ex. 2. In how many ways can 3 prizes be given to a class of 10 boys, without giving more than one to the same boy ? The first prize can be given in 10 ways, with each of which the second prize can be given in 9 ways ; hence the first two prizes can be given in 10 X 9 ways. With each of these ways of giving the first two prizes, the third prize can be given in 8 ways ; hence the three prizes can be given in 10 x 9 x 8, or 720, ways. Proof. After the first thing has been done in any one of the m ways, the second thing can be done in n different ways ; hence there are n ways of doing the two things for each of the m ways of doing the first ; therefore in all there are mn ways of doing the two things. This principle is readily extended to the case in which there are three or more things, each of which can be done in a given number of ways. 421. The different ways in which r things can be taken from n things, the order of selection or arrangement being 395 396 ELEMENTS OF ALGEBRA considered, are called the permutations of the n things taken r at a time. Thus, two permutations will be different unless they con- tain the same things arranged in the same order. E.g.^ of the four letters a, 6, c, d, taken one at a time, we have the four permutations a, &, c, d. Of these four letters taken two at a time, we have the twelve permu- tations a&, ac^ ad, ha, be, bd, ca, cb, cd, da, db, dc. If after each of these permutations we place in turn each of the letters which it does not contain, we shall obtain 24 permutations of the four letters taken three at a time. The number of permutations of n different things taken r at a time is denoted by the symbol T^. Thus ^P^, ^P^, ^P^ denote respectively the numbers of permutations of 9 things taken 2, 3, 4 at a time. 422. To find the number of permutations of n dissimilar things taken r at a time. The number required is the same as the number of ways of filling r places with n different things. The first place can be filled by any one of the n things, and after this has been filled in* any one of these n ways, the second place can be filled in {ii — 1) ways ; hence with n things two places can be filled in n(n — 1) ways ; that is, -P, = n(n~-1). (1) After the first two places have been filled in any one of these n (n — 1) ways, the third place can be filled in (n — 2) ways ; hence three places can be filled in n(n — l)(n — 2) ways ; that is, -P, = n(n-l)(n-2). (2) PERMUTATIONS AND COMBINATIONS 397 For like reason, we have ^P, = n(n - l)(7i -2)(n- 3); (3) and so on. From (1), (2), (3), •••, we see that in "P^ there are r fac- tors, of which the rth is n — r -\-l; hence "P, = n(n-l) (n _ 2) ... (?i - r -f 1). (A) If all the n things are to be taken at a time, r = n, and (A) becomes «P„ = w(n-l)(n-2)...3.2.1. (B) 423. The continued product ri (?i — 1) (n — 2) . . . 3 . 2 . 1 is denoted by the symbol [n, or w!, either of which is read * factorial ri/ Thus [4 = 4.3.2.1; [9 = 9. 8. 7- 6. 5- [4. With this notation (B) in § 422 can be written ""Pn^ln. (B) That is, the number of permutations of n different things taken all at a time is factorial n. Ex. 1. In how many different ways can 7 boys stand in a row? The number =7P7 = 7.6.5-4.3.2.1= 5040. by (B) Ex. 2. How many different numbers can be formed with the figures 1, 2, 3, 4, 5, 6, taken four at a time ? The number required = ep^ = 6 . 6 • 4 • 3 = 360. by (A) 424, If N denote the number of jyermutations of n things taken all at a time, of which r things are alike, s others alike, and i others alike; then Proof Suppose that in any one of the N permutations the r like things were replaced by r dissimilar things ; then, from this single permutation, without changing in it the 398 ELEMENTS OF ALGEBRA position of any one of the other n — r things, we could form |_r new permutations. Hence from the N original permuta- tions we could obtain N\r permutations, in each of which s things would be alike and t others alike. Similarly, if the s like things were replaced by s dissimilar things, the number of permutations would be N\r\s, each having t things alike. Finally, if the t like things were replaced by t dissimilar things, we should obtain iV^[r|s|^ permutations, in which all the things would be dissimilar. But the number of permutations of n dissimilar things taken all at a time is [ri. Hence iV|r[s[^ = [n. \n Therefore N = r sit Ex. 1. How many different numbers can be formed by the figures 2, 2, 3, 4, 4, 4, 5, 5, 5, 5 ? 110 The number = ,-^==— = 12600. [2 [3 [4 Exercise 136. 1. A cabinet maker has 12 patterns of chairs and 7 pat- terns of tables. In how many ways can he make a chair and a table ? Ans. 84. 2. There are 9 candidates for a classical, 8 for a mathe- matical, and 5 for a natural science scholarship. In how many ways can the scholarships be awarded ? 3. In how many ways can 2 prizes be awarded to a class of 10 boys, if both prizes may be given to the same boy ? 4. Find the number of permutations of the letters in the word numbers. How many of these begin with n and end with s ? PERMUTATIONS AND COMBINATIONS 399 5. If no digit occur more than once in the same number, how many different numbers can be represented by the 9 digits, taken 2 at a time ? 3 at a time ? 4 at a time ? 6. How many changes can be rung with 5 bells out of 8 ? How many with the whole peal ? 7. How many changes can be rung with 6 bells, the same bell always being last ? 8. In how many ways can 15 books be arranged on a shelf, the places of 2 being fixed ? 9. Given "P4 = 12 • "Pg ; find n. 10. Given n : "Pg : : 1 : 20 ; find n. 11. Given "Pg : "+2p^ : : 5 : 12 ; find w. 12. How many different arrangements can be made of the letters of the word commencement ? Of the 12 letters, 2 are c's, 3 are m's, 3 are e's, and 2 are n's ; 112 .-. JVr= . ^— =3326400. [2 [3 [3 [2 13. Find the number of permutations of the letters of the words mammalia, caravansera, 3fississi'p2n. 14. In how many ways can 17 balls be arranged, if 7 of them are black, 6 red, and 4 white ? Prove each of the following relations : 15. n(n — l)(n — 2)"'(n — r -\- l) \n — r = \n. 16. 9-8.7.6/[3 = |9/([3|5). 17. n(n — l)(n-2)"'(n — r-\- l)/\r = \n/(\r \n — r ). 18. [5[5(6/5)=[6|4; . •. [5 [5 < ^6 [4. 19. \a\a(a-^T)-i-a = \a-\-l\a — l }, .: |ft|a<| a + l|a— 1 . 20. |a|a< |a+l|a-l < |a-l-2|a-2 < |a + 3|a-3 <..». 21. |18 — x\x is least when x = 9. 400 ELEMENTS OF ALGEBRA 425. The different ways in which r things can be taken from n things, without regard to the order of selection or arrangement, are called the combinations of the n things taken r at a time. Thus, two combinations will be different unless they both contain precisely the same things. E.g.^ of the four letters a, 6, c, (Z, taken two at a time, there are the six combinations a6, ac^ ad, be, bd, cd. Taken three at a time, there are the four combinations abc, abd, acd, bed. Taken four at a time, there is one combination only. The number of combinations of n things taken r at a time is denoted by the symbol ""C^. 426. To fiyid the number of combinations of n different things taken r at a time. Every combination of r different things has |_r permuta- tions ; hence, ""Cr \r_ will denote "P^; that is, ^Cr\r_=''Pr = n(n — l){n — 2) '" (n — r + 1). Hence „o^^«(n - l)(n- g-(.- r + 1)_ ^^^ In applying this formula, it is useful to note that the suffix r in the symbol "(7^ denotes the number of the factors in both the numerator and denominator of the formula. Ex. How many groups of 4 boys are there in a class of 17 ? The number = i7(74 = iLJ^jiidi = 2380. 4.3.2.1 PERMUTATIONS AND COMBINATIONS 401 427. In (C) of § 426, multiplying the numerator and denominator of the fraction by | n — r, we obtain "Gr=r-T^-- (D) „(„_l)(„_2)...(n-r + l)l!Lz - r \r\n — r fifi — L_ ^-|r|7i-r ) lituting n — r for r in (D), we obtain -0 - ^ . ^-'-|»-r|r (1) From (D) and (1), "O, = "a_,. (E) The relation in (E) follows also from the consideration that for each group of r things which is selected, there is left a corresponding group of n — r things. The relation in (E) often enables us to abridge computation. E.g., i«Ci3 = "Ca = iL2) = «» + 3 a'^b + 3 ab^ + &». 404 BINOMIAL THEOREM 405 431. Binomial theorem. Suppose we have (a -\- b) {a -\- b) (a -\- b) ' • ' to n factors. (1) If we take a letter from each of the ?i binomials, and multiply these letters together, we shall obtain a term of the continued product ; and if we do this in every possible way, we shall obtain all the terms of this product. We can take the a's from all the binomials in one, and only one, way ; hence a'* is one term of the product. We can take b from one binomial and the a's from the remaining (n — 1) binomials, and we can do this in as many ways as one b can be taken from the n binomials, i.e., n, or "Ci, ways ; hence "C\ • a"~^b is a term of the product. Again, we can take the 6's from two binomials, and the a's from the remaining (n — 2) binomials, and we can do this in as many ways as two 6's can be taken from the n binomials, i.e., "(72 ways ; hence "C2 • a"~-6^ is a term of the product. And, in general, we can take the 6's from r binomials (where r is any positive integer not greater than w), and the a's from the remaining (n — r) binomials, and we can do this in as many ways as r 6's can be taken from the n bi- nomials, i.e., "(7, ways ; hence "C^ • a"~''6'' is the (r -f l)th, or general, term of the product. The 6's can be taken from the n binomials in one, and only one, way ; hence we have the term 6", and this is what the general term '•C^a'*~''6'" becomes when r = n. Hence (a -\- b) (a -\- b) (a -\- b) -•• to n factors = a" + ^da^'-'b + " C^"" -2^2 _^ ... 4- ^Cror-'b' + . . . + 6". (2) If we substitute for ^'Ci, "C2, etc., their values as given in § 426, we obtain (?i denoting any positive integer) (a H- by = a" + na'^-^b + '^(^~^) a^-''b^ -f • • • if _^ n(n - 1) (n - 2) ... (n - r + 1) ^n-.^. + ... + 6". (3) \r_ 406 ELEMENTS OF ALGEBRA Identity (2) or (3) is tlie symbolic statement of the bi- nomial theorem. The second member of either is called the expansion of (a -{-by. Observe that (3) states in symbols the laws in § 126, and that therefore (3) can be written out by these laws. Note that the sum of the exponents of a and b in any term is ii. Ex. Expand (x-^ - i/y. = (X-^y + 4 (x-=^)3(- 2/3) + 6 (a;-2)2(_ ^3)2 + 4 (a;-2) ( _ y3)8 = x-s - 4 x--y^ 4- 6 x-^f - 4 x'V + y^^. 432. In the expansion of (a + by, the general term n(n-l)(n-2)..^(n-r + l) ^„_.^. _ ^^^ ^^ _^ ^^^^ ^^^^ \r Observe that there are r factors in both the numerator and denominator of the coefficient of the (r + l)th term. By giving to r the proper value, we can find any term in the expansion of (a + by. When n is a positive integer, the coefficient of the (r+l)th term becomes zero for any value of r greater than n ; hence there are /? + 1 tei^ms in the expansion of (a + by. Thus, when r = n + 1, n(yi-l)(n-2) -'(n-r-\- 1) ^ n(?i-l)(n-2) •-(n-n) _Q^ [r |_r Ex. Find the seventh term of the expansion of (4 x/5 — 5/2 xy. Here a = 4 x/5, b=-5/(2x), w = 9, r = 6. Substituting these values in the formula, we have the seventh term = — ) 1.2.3.4.5.t3V 5 / \2x I = 10500x-3. BINOMIAL THEOREM 407 433. The coefficients of the expansion in (2) of § 431 are i nri nri nri nip nfl nfl . X, L/1, U2, L/3, •••, L/„_2, ^n-\1 Wj hence the (r4-l)th term from the beginning is ''Cra^~''b''j and the (r 4- l)th term from the end is "C^.^a'^^""''. But "O, = "(7„_, for all values of r (§ 427). Hence, in the expansion of (a + by, the coefficients of any two terms equidistant from the beginning and the end are the same, so that the coefficients of the last half of the expansion can be written from those of the first half 434. If, in identity (3) of § 431, we put a = 1 and b = x, we have (l+a;)- = l + nx + ''^V^'^ ^+-+.-7^^— ^"+-+^" 2 \r\n — r This is a convenient form of the binomial theorem^ and one which is often used. Observe that this form includes all cases ; e.gr., if we want to find (a 4- ?>)", we have 435. In (1) of § 434 the coefficients of x, x^, x^, •••, x'^ are the values of "Ci, "Cg, "C3, •••, "(7„; hence (t) can be written (1 + a;)" = 1 + "C,x + ^C^ + ... + "Oa' + ••• + ^'C^a;". (1) Putting x = l, we obtain 2« = 1+ "Ci + '^C^ + - + "C, + - + "a. (2) That is, the sum of the coefficients in the expansion of (1 + xy, or (a + by, is 2\ From (2) it follows also that the sn7n of all the combinor tions that can be made of n things, taken 1, 2, •", n at a time, is 2" - 1. ■408 ELEMENTS OF ALGEBRA Exercise 138. By the laws in § 126 write the expansion of : 1. {3x'-2y)\ Q (r-^_^h)\ n. {x-'^-2c^)\ 2. (2a2-3 6y. 7. {f-^n^)\ 12. (1 - 1 A)«. 3. (c^ + hy. 8. (2 x/^ - 3/xy. 13. (^2/ _ «-f)5. 4. (3a^ + 2/)'- 9. (x-^-afy. 14. (2 a;/3 - a/c)«. 5. (2-3i«y. 10. (a-'--x-y. 15. (a;-^-2?/-^)^ 16. Find the 3d term in the expansion of (a — 3 by^. 17. Find the 7th term in the expansion of (1 — xy^. 18. Find the middle term in the expansion of (1 + xy^. 19. Find the middle term in the expansion of (2 a? — 3 yy. 20. Find the 18th term in the expansion of (1 + x)^. 21. Find the 7th term in the expansion of [4:x/5-5/{2x)J. 22. Find the 17th term in the expansion of (xF — 1/x)^. 436. Binomial theorem, exponent fractional or negative. When the exponent of a binomial is fractional or nega- tive, the laws in § 126, or, what is the same thing, the formula (a + by = a" + na-~^b + <^lz1} a^-^^b' + - _^n(n -l)(n- 2) ... (n - r + 1) ^._,^. ^ _^ ^^^ \l gives an infinite series ; for in this case no one of the factors n, n — 1, n~ 2, etc., in the (r + l)th term can ever be zero. When, however, r increases ivithout limit, the sum of r terms of this series will approach (a -\- by as its limit, provided the first term of the binomial is arithmetically greater than the BINOMIAL THEOREM 409 second term. That is, when 7i is fractional or negative, the infinite series in (1) is the expansion of (a + by provided a>b arithmetically. A proof of this theorem is too difficult to be given here. For a rigorous proof, see Taylor's " Calculus," § 98. Ex. 1. Expand (c-i — (P)~^ and find the general term. Applying the laws in § 126 we obtain (c-i _ cf^)-f = [(c-i) + (- d2)]-t -i¥.(0"^'(-^^)« + - (1) = c^ + f cV + ^ c'^V .+ ^^ c^(P + -. ' (2) The two distinct steps, that of applying the laws to obtain (1) and that of performing the indicated operations in (1) to obtain (2), must be taken separately. In performing the operations indicated in (1), first note the number of negative numeral factors in a term to determine the quality of its numeral coefficient. Thus in the fourth term there are four negative factors, — ^^\ and (— l)^. Substituting in the general term for n, a, and b their values — |, c-i, and — ^2^ ^e obtain the (r+l)th term = (~i)^~^)(~Y^ "• ^~^~^"^^^ (c-0"^""(-; hence (ii). COMMON LOGARITHMS. 447. The logarithms used for abridging arithmetic compu- tations are those to the base 10 ; for this reason logarithms to the base 10 are called common logarithms. Thus the common logarithm of a number answers the question, ^ What power of 10 is the number ? ' Most numbers are incommensurable powers of 10; hence most common logarithms are incommensurable numbers, whose approximate values we express decimally. Hereafter in this chapter when no base is written, the base 10 is to be understood. When a logarithm is negative, for convenience it is ex- pressed as a negative integer plus a positive decimal. E.g., the conmion logarithm of any number between 10 and 100 is +1 + a positive fraction ; between 1 and 10 is + a positive fraction ; between 0.1 and 1 is -1 + a positive fraction; between 0.01 and 0.1 is -2 4- a positive fraction. LOGARITHMS 417 448. The integral part of a logarithm is called the characteristic, and the positive decimal part the mantissa. A negative characteristic is usually written in the form I, or 9 - 10; 2, or 8 - 10; 3, or 7 - 10; etc. E.g., log 434.1 = 2.63759 ; +2 is the characteristic and .+63759 is the mantissa : log 0.0769 = 2.88593, or 8.88593 - 10 ; 2, or 8 - 10, is the negative characteristic, and .+88593 is the mantissa. The sign - is written over the 2 to show that it affects the characteristic alone. 449. The characteristic of the common logarithm of any number is found by the following simple rule: Wlien the number is greater than 1, the characteristic is positive and arithmetically one less than the number of digits to the left of the decimal point ; when the number is less than 1, the characteristic is negative and arithmetically one greater than the number of zeros between the decimal point and the first significant figure. E.g., 785 lies between IO2 and 10' ; hence log 785 = 2 + a mantissa. Again 0.0078 lies between lO-^ and IO-2 ; hence log 0.0078 = — 3 + a mantissa. Proof. Let N denote a number which has m digits to the left of the decimal point ; then N lies between 10*""^ and 10"*; that is, N = 10<'"-^) + * f™^"°°. .-. log -^= (m — 1) + a mantissa. Again let. ^denote a decimal which has m zeros between the decimal point and the first significant figure; then iV lies between lO-^""^^) and lO""* ; that is, N= lO-^'^+i) + * <^'=«°". .-. log N=— (m -h 1) -f a mantissa. 418 ELEMENTS OF ALGEBBA 450. The common logarithms of numbers which differ only in the positio7i of the decimal point have the same mantissa. Proof. When a change is made in the position of the decimal point, the number is multiplied or divided by some integral power of 10 ; that is, an integer is added to, or subtracted from, the logarithm, and therefore its mantissa is not changed. E.g., log 1054.3 = 3.02296, log 1.0543 = 0.02296, log .010543 = 8.02296 - 10, or 2.02296. 451. When a negative logarithm is to be divided by a number, and its negative characteristic is not exactly divis- ible by that number, the logarithm must be so modified in form that the negative integral part will be exactly divisible by the number. Ex. Given log 0.0785 = 2.8949; find log \/0. 0785. Log v/0.0785 = I log 0.0785 = } (2.8949) = }(7. +6.8949)= 1.8421. Adding —5 + 5 to the logarithm does not change its value and makes its negative part divisible by 7. Exercise 142. 1. Log 427.32 = 2.6307. Find log 42732, log 42.732. 2. Log 23.95 = 1.3793. Find log 23950, log 239.5, log 239500, log 0.002395, log 0.0002395, log 2395. 3. Log 4398 = 3.64326. Find log V0.4398, log -^0.4398, log ^439.8, log ^0.04398, log ^0.004398. 4. Log 674.8 = 2.82918. Find log ^0.6748, log ^0.6748, log -^0.06748, log ■{/0.06748, log ^0.006748. 452. Tables of logarithms. Common logarithms have two great practical advantages : (i) Characteristics are known LOGABITHMS 419 by § 449, so that only mantissas are tabulated ; (ii) mantis- sas are determined by the sequence of digits (§ 450), so that the mantissas of integral numbers only are tabulated. At the close of this chapter will be found a table which contains the mantissas of the common logarithms of all numbers from 1 to 999 correct to four decimal places. Note. Tables are published which give the logarithms of all num- bers from 1 to 99999 calculated to seven places of decimals ; these are called ' seven-place ' logarithms. For many purposes, however, the four-place or five-place logarithms are sufficiently accurate. From a table of logarithms we can obtain : (i) The logarithm of a given number ; (ii) The number corresponding to a given logarithm. 453. To find the logarithm of a given number. Ex. 1. Find log 7.85. By § 450, the required mantissa is the mantissa of log 785. Look in column headed "N" for 78. Passing along this line to the column headed 5, we find .8949, the required mantissa. Prefixing the characteristic, we have log 7.85 = 0.8949. Ex. 2. Find log 4273. 2. When the number contains more than three significant figures, we must use the principle that when the difference of two numbers is small compared with either of them, the difference of the numbers is approximately proportional to the difference of their logarithms. By § 450, the required mantissa is that of log 427.32. The mantissa of log 427 = .6304. The mantissa of log 428 = .6314. That is, an increase of 1 in the number causes an increase of .0010 in the mantissa ; hence an increase of .32 in the number will cause an increase of .32 of approximately .0010, or .0003, in the mantissa. Adding ,0003 to the mantissa of log 427, and prefixing the character- istic, we have log4273.2 = 3.6307. 420 ELEMENTS OF ALGEBBA Ex.3. Find log 0.0006049. By § 450, the required mantissa is that of log 604.9. The mantissa of log 604 = .7810. Also, an increase of 1 in the number causes an increase of .0008 in the mantissa; hence .9 of .0008, or .0007, must be added to .7810. .-. log 0.0006049 = 4.7817, or 6.7817 - 10. To find log 30 or log 3, find mantissa of log 300. Exercise 143. Find, from the table, the logarithm of the numbers : 1. 8. 5. 703. 9. 0.05307. 13. 7.4803. 2. 50. 6. 7.89. 10. 78542. 14. 2063.4. 3. 6.3. 7. 0.178. 11. 0.50438. 15. 0.0087741. 4. 374. 8. 3.476. 12. 0.00716. 16. 0.017423. 454. To find a number when its logarithm is given. Ex. 1. Find the number of which the logarithm is 3.8954. Look in the table for the mantissa .8954. It is found in line 78 and in column 6 ; hence .8954 = the mantissa of log 786. .-. 3.8954 = log 7860; or 7860 is the number whose logarithm is 3.8964. Ex. 2. Find the number of which the logarithm is 1.6290. Look in the table for the mantissa .6290. It cannot be found ; but the next less mantissa is .6284, and the next greater is .6294. Also, .6284 = mantissa of log 425, and .6294 = mantissa of log 426. That is, an increase of .0010 in the mantissa causes an increase of 1 in the number ; hence an increase of .0006 in the mantissa will cause an increase of approximately 3% of 1, or .6, in the number; hence I .6290 = the mantissa of log 425.6 ; .-. 1.6290 = log 42.56. LOGARITHMS 421 Ex. 3. Find the number of which the logarithm is 3.8418, Look in the table for the mantissa .8418. It cannot be found ; but the next less mantissa is .8414, and .8414 = mantissa of log 694. Also, an increase of .0006 in the mantissa causes an increase of 1 in the number ; hence an increase of .0004 in the mantissa will cause an increase of f of 1, or .66 in the number ; hence .8418 = the mantissa of log 694.66. .-. 3.8418 = log 0.0069466. Exercise 144. Find the number of which the logarithm is : 1. 1.8797. 6. 8.1648-10. 11. 3.7425. 2. 7.6284-10. 7. 9.3178-10. 12. 7.1342-10. 3. 0.2165. 8. 1.6482. 13. 3.7045. 4. 2.7364. 9. 8.5209-10. 14. 8.7982-10. 5. 4.0095. 10. 3.8016. 15. 3.4793. 455. The cologarithm of a number is the logarithm of its reciprocal. That is, colog N=\og(l -^N) = — log N. To make the fractional part of the cologarithm positive, if log iV > and < 10, colog N is written (lO-logiV)-lO; if log ^ > 10 and < 20, colog N is written (20-logiV^)-20. E.g., cologO.0674 =-(2.7589)= 1.2411; colog 432 = (10 - 2.6263) - 10 = 7.3737 - 10 ; colog 345000000000 = (20 - 11.5378) - 20 = 8.4622 - 20. Instead of subtracting the logarithm of a divisor, we can, by § 87, add its cologarithm. 422 ELEMENTS OF ALGEBRA Ex.1. Find the value of 1M^AM723. 0.0534 X 7.238 log 15.08 = 1.1784 log0.0723 = 8.8591 -10 colog0.0534 = 1.2725 colog 7.238 = 9.1404-10 Add, log (fraction) = 0.4504 = log 2.8213. Hence the fraction = 2.8213. Ex. 2. Find the value of 0.0543 x 6.34 x (- 5.178). Iog0.0543 = 8.7348 -10 log 6.34 = 0.8021 loa: 5.178 = 0.7141 Add, log (product) = 0. 2510 = log 1 . 7824. Hence the product is — 1.7824. By logarithms we obtain simply the arithmetic value of the result ; its quality must be determined by the laws of quality. J 5.42 X 427.2 Ex. 3. Find the value of Ji \3.244 X 0.0231^ log5.42 = 0.7340 =0.7340 2 log 427. = (2.6304) x 2 = 5.2608 4 colog 3.24 = (9.4895 - 10) x 4 = 7.9580 - 10 ^ colog 0.0231 = (1.6364) -2 = 0.8182 5 )4.7710 .-. log (root) = 0.9542 .-. root = 9.00 456. An exponential equation is one in which the unknown appears in an exponent ; as 2* = 5, af = 10. Such equations are solved by the aid of logarithms, Ex. 1. Solve 32* - 14 x 3^ + 45 = 0. (1) Factor (1), (3* - 9) (3^ - 5) = 0. (2) LOGARITHMS 42^ (4) Equation (2) is equivalent to the two equations 3' = 9, (3) 3- = 5. From (3), x = 2. From (4), a; log 3 = log 5. log 5^ 0.6990 _g49 log 3 0.4771 Hence the roots of (1) are 2 and 1.4649. Exercise 145. Find by logarithms the value of : 1. 742.8x0.02374. 7. 4743-5-327.4. 2. 0.3527x0.00572. g 9.345^ (_ 0.0765). 3. 78.42x0.000437. _ 2.476 x(- 0.742) 4. 5234 X (-0.03671). 9. ^3 ^^ ^ (_o.ooi21)* 5. 3.246 X (-0.0746). ^^ 321 x(- 48.1) x (357) 6. - 4.278 X (- 0.357). * 421 x (- 741) x (4.21) 11. 5l 14. («)«. 17. (iiyi 12. 0.02li 15. 714.2i 18. (3|)^27 13. 0.5328. 16. (Iff)^ 19. 4.71=^ /0.035'* X 54.2 X 785^ x 0.0742 206 20. 21. 22. 1 0.035'^ ^ 4.72^ X 7.14^ X 8.47^ 3/ 0.0427^ X 5.27 x 0.875^ V7.42I* X Vl-'^4 X V0.00215 5/ 0.714^x0.1371^x0.0718^ ^' 0.5242x0.742^ x 0.0527^ 424 ELEMENTS OF ALGEBRA Solve each of the following equations : 23. 31^ = 23. 25. 5^ = 800. 27. 5^-3 = 8^*+^ 24. 0.3^ = 0.8. 26. 12^ = 3528. 28. aH^'' = (^, 29. 23*52^-1 = 4^^3^+1. 30. 42=^-15 (4^) +56 = 0. COMPOUND INTEREST AND ANNUITIES. 457. To find the compound interest, $ /, and amount, $ M, of a given principal, $ P, in n years, $ r being the interest on $ 1 for 1 year. Let ^R = the amount of $ 1 in 1 year ; then R = l-\-r, and the amount of f P at the end of the first year is $ PR ; and since this is the principal for the second year, the amount at the end of the second year is $ PR x R, or $ PR^. For like reason the amount at the end of the third year is $PR^, and so on; hence the amount in n years is ^PR""', that is, M=PR'', or P(l + r)^ (1) Hence I=P(R--1). (2) If the interest is payable semi-annually, the amount of $P in 1/2 a year will be $P(l-\-r/2); hence, as n years equals 2 n half-years, M=P(l + r/2y\ (3) Similarly, if the interest is payable quarterly, M=P{l+r/4:y\ (4) Ex. Find the time in which a sum of money will double itself at ten per cent compound interest, interest to be "converted into prin- cipal" semi-annually. Here 1 + r/2 = 1.05. Let P = 1 ; then M=2. Substituting these values in (3), we obtain 2 = (1.05)2« .-. log2 = 2w .log 1.05. 2 log ^^gl- = ^^^ = 7.1 years, ^ns. 02 1.05 0.0424 ^ LOGARITHMS 425 458. Present value and discount. Let f P denote the pres- ent value of the sum $M due in n years, at the rate ?-; then evidently in n years at the rate r, $P will amount to $M-j hence M=PB^, or P = MR-\ Let $ D be the discount ; then D = M- P= 3/(1 - Z2-**). 459. An annuity is a fixed sum of money that is payable once a year, or at more frequent regular intervals, under certain stated conditions. An Annuity Certain is one pay- able for a fixed number of years. A Life Annuity is one payable during the lifetime of a person. A Perpetual An- nuity, or Pei'petuity, is one that is to continue forever, as, for instance, the rent of a freehold estate. 460. To find the amount of an annuity left unpaid for a given number of years, allowing compound interest. Let ^ J. be the annuity, n the number of years, $ R the amount of one dollar in one year, $ JHf the required amount. Then evidently the number of dollars due at the end of the First year = A ; Second year = AR + A ; Third year = AR' + AR -{- A -, nth year = AR'-^ + AR" ---\ [-AR-\-A ^ A(R^-l) R-1 That is, M=-(R''-1). • (1) Ex. 1. Find the amount of an annuity of 1 100 in 20 years, allow- ing compound interest at 4 J per cent. r ^ 0.046 426 ELEMENTS OF ALGEBRA By logarithms, 1.04520 = 2.4117. ... 3/ ^l^lill^ 3137.11. 0.045 Hence the amount of the annuity is $3137.11. Ex. 2. What sum nmst be set aside annually that it may amount to $ 50,000 in 10 years at 6 per cent compound interest ? rrom(l), .4 = -^!^=.^M00Aa06^ 3793.37. Hence the required sum is $3793.37. 461. To find the preseiit value of an annuity of ^A pay- able at the end of each of n successive years. Let $P denote the present value; then the amount of ^P in n years will equal the amount of the annuity in the same time : that is, PR^ = A{R'-l)r-\ (1) .-. P=A{l-R--)r-\ (2) If the annuity is perpetual, then n = oo, i2~" = 0, and (2) becomes P = Ar-\ Exercise 146. 1. Write out the logarithmic equations for finding each of the four numbers, M, R, P, n. 2. In what time, at 5 per cent compound interest, will $100 amount to f 1000? 3. Find the time in which a sum will double itself at 4 per cent compound interest. 4. Find in how many years f 1000 will become f 2500 at 10 per cent compound interest. 5. Find the present value of $10,000 due 8 years hence at 5 per cent compound interest. LOGARITHMS 427 6. Find the amount of $1 at 5 per cent compound in- terest in a century. 7. Show that money will increase more than thirteen- thousand-fold in a century at 10 per cent compound interest. 8. If A leaves B $1000 a year to accumulate for 3 years at 4 per cent compound interest, find what amount B should receive. 9. Find the present value of the legacy in example 8. 10. Find the present value, at 5 per cent, of an estate of $ 1000 a year to be entered on immediately. 11. A freehold estate worth f 120 a year is sold for $ 4000 ; find the rate of interest. 12. A man has a capital of $20,000, for which he re- ceives interest at 5 per cent; if he spends $1800 every year, show that he will be ruined before the end of the 17th year. 428 ELEMENTS OF ALGEBRA N 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0263 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1761 1790 1818 1847 1875 1903 1931 1969 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2465 2480 2604 2529 18 2563 2577 2601 2625 2648 2672 2695 2718 2742 2766 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3679 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4266 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4466 28 4472 4487 4502 4518 4633 4548 4664 4679 4694 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4767 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 6011 6024 5038 32 5051 5065 5079 5092 5105 6119 6132 6145 6169 5172 33 5185 5198 5211 5224 5237 6250 5263 5276 6289 6302 34 5315 5328 5340 5353 6366 5378 5391 6403 6416 5428 35 5441 5453 5465 5478 5490 5602 6514 6627 5639 5561 36 5563 5575 5587 5599 5611 6623 6636 6647 6658 6670 37 5682 5694 5705 5717 6729 5740 5752 5763 5776 5786 38 5798 5809 5821 5832 6843 6865 6866 5877 6888 5899 39 5911 5922 5933 5944 5955 5966 6977 5988 6999 6010 40 6021 6031 6042 6053 6064 6075 6086 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6376 6385 6395 6406 6416 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6671 6580 6590 6699 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6876 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7060 7069 7067 51 7076 7084 7093 7101 7110 7118 7126 7136 7143 7162 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7236 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 TABLE OF MANTISSAS 429 N 1 2 3 4 5 6 7 8 9 55 7404 7412 7419 7427 7436 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7613 7520 7528 7536 7543 7661 67 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7762 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7826 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8036 8041 8048 8056 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8166 8162 8169 8176 8182 8189 66 8196 8202 8209 8216 8222 8228 8236 8241 8248 8264 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8388 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8446 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8637 8643 8549 8566 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8616 8621 8627 73 8633 8639 8645 8651 8667 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8746 75 8751 8756 8762 8768 8774 8779 8786 8791 8797 8802 76 8808 8814 8820 8826 8831 8837 8842 8848 886^4 8869 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8964 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9176 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9316 9320 9326 9330 9335 9340 86 9345 9360 9355 9360 9365 9370 9375 9380 9386 9390 87 9395 9400 9405 9410 9416 9420 9426 9430 9436 9440 88 9445 9450 9465 9460 9466 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9613 9518 9523 9528 9633 9638 90 9542 9547 9552 9567 9562 9566 9571 9576 9681 9686 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9676 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9764 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9806 9809 9814 9818 96 9823 9827 9832 9836 9841 9846 9860 9864 9869 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 CHAPTER XXXIII GRAPHIC SOLUTION OF EQUATIONS AND SYSTEMS M' 4--4- M P 462. Let XX' and YY' be any two fixed straight lines at right angles to each other at 0. Let the directions OX and OF be positive directions; then the directions ^Y . OX' and 0Y> will be negative direc- tions. The lines XX and YY' are called axes of reference, __ and their intersec- tion 0, the origin. From P, any point in the plane of the axes, draw PM parallel to IP M Fig. 1 YY' ; then the po- sition of P will be determined when we know both the lengths and the directions of the lines OMsindMP. The line OM, or its numerical measure, is called the abscissa of the point P; and MP, or its numerical measure, is called the ordinate of P. The abscissa and ordinate together are called the coordinates of P. E.g., OM' and M'P' are the coordinates of P' ; the abscissa, OW, is negative, and the ordinate, M'P', is positive. OM'", the abscissa of P"', is positive, and M'"P"', its ordinate, is negative. 430 SOLUTION OF EQUATIONS AND SYSTEMS 431 An abscissa is usually deaoted by the letter x, and an ordinate by y. Observe that the numerical measure of OM or MP is a positive number, if it extends in the direction OX or OF; and a negative number if it extends in the direction OX' or or. The axis XX' is called the axis of abscissas, or the jr-axis ; and YY\ the axis of ordinates, or the /-axis. The point whose coordinates are x and y is denoted by {^y y)' E.g., (2, — 3) denotes the point of which the abscissa is 2, and the ordinate — 3. We use a system of coordinates analogous to that explained above whenever we locate a city by giving its latitude and longitude ; the equator ie one axis, and the assumed meridian the other. Ex. Plot the point (- 2, 3) ; (-3,-4). In the figure lay off OJf = - 2, and on MT* parallel to YT lay off JiPP' = + 3 ; then P' is the point (-2, 3). To plot (-3, - 4), lay off OM" = - 3, and on M"P" parallel to YY' lay ofE M"P" = - 4 ; then P" is the point (-3, - 4). The lines XX' and YY divide the plane into four equal parts called quadrants, which are numbered as follows : XOY is the Jirst quadrant j YOX' the secondy X'OY the third, and YOX the fouHh. Exercise 147. 1. Plot the point (2, 3); (4, 7); (3, -5); (-2, +3); (-3, +5); (4, -2); (-2, -3); (-5, -3); (-2, 4); (-4,-1); (0,0). 2. In which quadrant is (+a, +b)? (+a, "6)? ('a, +b)? (-a,-b)? 3. What is the quality of x and of y, when the point (Xy y) is in the first quadrant ? Second quadrant ? Third quadrant? Fourth quadrant? 432 ELEMENTS OF ALGEBRA 4. In which quadrants can the point (x, y) be, when x is positive ? X negative ? y positive ? y negative ? 5. In what line is the point (a;, 0) ? (0, y) ? 6. Where is the point (0, 0) ? (4, 0) ? (- 3, 0) ? (0, 2) ? (0,-5)? 463. Graphic solution of equations in x and /. The locus, or graph, of an equation in a; and y is the line or lines which include all the points, and only those, whose coordinates satisfy the equation. Ex. 1. Draw the locus of y = x"^ — x — 6. (1) If in (1) we put a; = - 3, - 2, - 1, ..., we obtain when X = - 3, - 2, - 1, 0, 1/2, 1, 2, 3, 4, ..., y = 6, 0, - 4, - 6, - 6|, - 6, - 4, 0, 6, .... Drawing the axes XX' and TY' in fig. 2, and assuming 01 as the linear unit, we plot the points (-3,6), (-2, 0), (-1, -4), (0, -6), .... The relative positions of these points indicate the form of a curve through them. Whenever there is any doubt about the form of this curve between any two plotted points, as between (0, —6) and (1, —6), one or more intermediate points should be found and plotted. As X increases indefinitely from 3, y (^or x^ — x — 6) continues posi- tive and increases indefinitely ; hence the locus has an infinite branch in the first quadrant. As x decreases indefinitely from — 2, y con- tinues positive and increases indefinitely ; hence the locus has an infinite branch in the second quadrant. Drawing a smooth curve through the plotted points we obtain the curve ABC in fig. 2, which, with its infinite branches, is the locus of equation (1). This curve is called the locus of the equation because each and every real solution of equation (1) is the coordinates of some point on the curve. SOLUTION OF EQUATIONS AND SYSTE3IS 433 Imaginary or complex solutions of an equation cannot be represented by the coordinates of any points in the plane XO Y, since by definition the coordinates of every point in this plane are real. Note. The pupil should use coordinate or cross-section paper, and with a hard pencil draw the loci of equations neatly and accurately. X X- t—^ Ex. 2. Draw the locus ot y = 3? — 2x. (1) When x = -2, -y^, -1, -0.8, 0, 0.8, 1, y/2, 2, ..., y=-4, 0, 1, 1.1, 0, -1.1, -1, 0, 4, .... As X increases indefinitely from 2, y (or x^ — 2x) continues positive and increases indefinitely ; hence the locus has an infinite branch in the first quadrant. As X decreases indefinitely from —2,y continues negative and arithmetically increases without limit ; hence there is an infinite branch in the third quadrant. Plotting these points, as in fig. 3, and tracing a smooth curve through them, we obtain the curve ABCD, which, with its infinite branches, is the locus of (1). 434 ELEMENTS OF ALGEBRA Ex. 3. Draw the locus oi y = x^ — S x"^ + i. When x = -3/2, - 1, - ^, 0, 1/2, 1, 2, 3, ..., y=- 6.1, 0, 3.1, 4, 3.4, 2, 0, 4, .... As X increases from 2, y increases indefinitely from ; and, as X decreases from —!,' 2 or <— 2, the values of y in (1) are real ; when x lies between — 2 and +2, y is imaginary ; hence there is an infinite branch in each of the four quadrants, but no point of the locus lies between the lines x = — 2 and x = 2. 16. Draw the locus of 2/^ = 4 x. 17. Using one set of axes, draw the loci of X — y = — 3 and a; -f 1/ = 1. Observe that these equations are independent and consistent, and that the one and only point common to their loci is (— 1, 2). 18. Using one set of axes, draw the loci of 2 ic — 2/ = — 1 and 2x — y = — 3. Observe that these equations are inconsistent, and that their loci are parallel and hence have no point in common. SOLUTION OF EQUATIONS AND SYSTEMS 437 19. Using one set of axes, draw the loci of 2 a; + 2/ = 1 and 6x-\-3y = 3. Observe that these equations are equivalent, and that their loci coincide and hence have all points in common. 20. What is the greatest number of points in which a straight line can cut the locus in fig. 2 ? In fig. 3 ? In fig. 4 ? In fig. 5 ? Compare each answer with the degree of the equation of each locus. 464. Graphic solution of systems of equations. Ex. 1. By the aid of loci discuss the system ax + by = c, (1) ax + b'y = c'. (2) (a) Let the locus of (1) be the straight line MN, and that of (2) the line liP. Then the coor- dinates of the point P, which is common to both loci, will be the solution common to (1) and (2), or the solution of the system (a). By measuring the coordinates OA and AP, the numerical solution of the system could be obtained. This example illustrates graphically the theorem in §207. The loci will have one, and only one, point in common, if a/a' ^ b/h>, i.e., if (1) and (2) are independent and consistent (§ 207). The loci will coincide throvghotit their whole extent, if a/a' ^ b/V = c/c', i.e., if (1) and (2) are equivalent (§§ 207, 357). The loci will be parallel and have no point in common, if a/a' = b/b' and a/a^ =^ c/c', i.e., if (1) and (2) are inconsistent (§§ 207, 357). 438 ELEMENTS OF ALGEBBA Ex. 2. By the aid of loci, discuss the system for different values of c. x2 + ?/2 = 25, (1) (2) (a) The locus of (1) is the circle PP'BP" ; and, if c = 1, the graph of (2) is the straight line MN; hence the coordinates of the two points P and B are the two solutions of system (a). By measurement we find the two solutions to be 3, 4 and — 4, — 3. As c increases, the locus MN moves upward parallel to itself, and P and B approach P'. When c = 5\/2, the locus of (2) is the tangent N'3I', and the two solutions of the system are equal. Similarly, when c = — 5\/2, the locus of (2) is M"N". When c < 5\/2 and > - 5V2, the locus of (2) lies between N'M' and N"M", and the two solutions of the system are real and unequal. When c > 5\/2 or < — 5V2, the locus of (2) does not cut the circle, and both solutions of the system are imaginary or complex. SOLUTION OF EQUATIONS AND SYSTEMS 489 Ex. 3. By aid of loci discuss the system x^+ y^ = r^ (1) }(«) C2 + ^2 _ ^2 (2) for different values of r. The locus of (1) is the ellipse ABBS, in which OA=S and 0B=2. If r = 5/2, the locus of (2) is the circle PP'P"P"', and the four solutions of the system are the coordinates of the four points P, P', P", P"', and thus are real and unequal. If r = 3, the circle will be tangent to the ellipse at A and B ; hence two solutions of the system will be 3, 0, and the other two — 3, 0. If r = 2, the circle will be tangent to the ellipse at B and S. If r < 2 or > 3, the two loci will have no common points, and all four solutions of the system will be imaginary or complex. When r = 5/2, by clearing (2) of fractions and then subtracting it from (1) we obtain 5]/^ = 11, the locus of which is the parallel lines PP and P"'P'. These lines cut either the ellipse or the circle in all the points which are common to these curves, and only in these points. This illustrates the equivalency of system (a) to the system 4x2 + 9y2 = 36, 6y2 = 36, I x'^-hy^ = ^,] = 11.1 °^ 5?/2=ll. J 440 ELEMENTS OF ALGEBRA (a) Ex. 4. By aid of loci discuss the system xy = 12, (1) y = mx + n, (2) , for different values of m and n. The locus of (1) is the curves AB and 02), whose infinite branches approach the axes. When n = and m = 3/4, the locus of (2) is the line PP', and the two solutions of system (a) are the coordinates of the points P and P'. Let m = ; then P will move out along the infinite branch PB, and P' along the infinite branch P'C ; that is, y = and x = + go or — oo. y lA Again, when n = 0, the two solutions of system (a) are 2V:i/m, 2\/3hi, and -2\/^i/m, - 2V8to. For m = 0, the solutions in (.3) assume the forms a/0, and - a/0, ; (3) hence equation (1) and y = are inconsistent, and system (a) is then impossible. This agrees with the figure, for the locus oiy = coincides with X'OX, and does not intersect the locus of (1). When m is negative, the solutions in (3) become imaginary. This agrees with the figure ; for when m is negative, x and y in y = mx are SOLUTION OF EQUATIONS AND SYSTEMS 441 opposite in quality, and hence the locus of y = mx will lie in the second and fourth quadrants, and will not cut the locus of (1). If »i = and n ^ 0, the locus of (2) will be parallel to XOX', and will cut the locus of (1) in only one point ; hence system (a) will be defective in one solution. Exercise 149. 1. By aid of loci show that system (a) is equivalent to the four systems in (6). a^ -h y- = 25 xy r = 25 x?/ = 12 J x + y = 7] x + y = 7 ) x-^y = -7) x-\-y = -7] a; — l = lJ x — y = — lj x — y = \ i x — y = — l^ 2. By aid of loci show that the following six systems are equivalent : 0^ + 2/2 = 25 ar'=16 0^ + 2/^ = 25 7?-f = 7 J 2/2 = 9 x'-f = 7 1 ^-f=7 a^ = 16 a^ = 16 1 / = 9 i / = 9 I 1- GRAPHIC SOLUTION OF EQUATIONS IN X. 465. A variable whose value depends upon one or more other variables is called a dependent variable, or a function of those variables. A variable which does not depend upon any other variable for its value is called an independent variable. E.g., x^, 2 a;"2 — .3 X + 7, or x"* — 7 x^ + 9, is a function of the inde- pendent variable x. Again, y in each of the equations in this chapter is a function of the independent variable x. The symbol f{x), read ^function x,' is used to denote any function of x. 442 ELEMENTS OF ALGEBRA The symbols f(a), f(2), /(I) represent the values of f(x) when x = a, 2, 1, respectively. E.g., iif(x) = ^3 + X, then /(a) =a^^a, /(2) = 2^ + 2 = 10, /(I) = 2. Since /(a;) denotes any function of x, y =f{x) denotes any equation in x and y, when the equation is solved for y. Thus, any one of the equations in the first ten exsrcsples in exercise 146 is a particular case of y =f(x). 466. A continuous variable is a variable which in passing from one value to another passes successively through all intermediate values. A function, as /(a;), is said to be continuous between x = a and ic = 6, if when x changes continuously from a to h, f(x) varies continuously from /(a) to f{b). In other words, /(a:) is continuous between x = a and x=b, when the locus of y =f(x) is an unbroken curve between the lines x = a and x = b. E.g., the time since any past event varies continuously. The veloc- ity acquired by a falling body and the distance fallen are continuous functions of the time of falling. In each of the four examples in § 463, y is a continuous function of X for all real values of x. In example 2 of § 464, y in equation (1) is real and a continuous function of x between x = — 6 and x = + 5. The examples in § 463 illustrate the fact that Any rational integral function of x is a continuous function. In what follows we shall use f(x) to denote a rational integral function of x. 467. The ordinates of the points in the locus ot y = a^ — x — 6 in fig. 1, of § 463, are the successive values of a^ — x — 6 corresponding to successive values of x ; hence, the locus of y =f{x) is often caUed the graph of f{x). SOLUTION OF EQUATIONS AND SYSTEMS 443 E.g., in fig. 2, while x increases continuously from —3 to zero, the function x^ — x — 6 decreases continuously from + 6 though zero to — 6 ; and while x increases from zero to +4, x^ — x — Q first de- creases from — 6 and then increases to -f 6. Again, in fig. 3, while x increases continuously from — 2 to — 0.8, the function x^ — 2x increases continuously from — 4 to 4- 1.1 ; while X increases from —0.8 to +0.8, x^ — 2x decreases from +1.1 to — 1.1 ; while x increases from + 0.8 to + 2, x' — 2 x increases from -1.1 to 4. In like manner, in the other figures, the pupil should follow the changes in f{x) as x increases. 468. The abscissas of the points in which the graph of f(x) cuts or touches the axis of x are the real vahies of x for which f(x) is zero ; that is, they are the real roots of the equation f{x) = 0. At a point of tangency the graph is properly said to touch the axis of x in two coincident j)oints. E.g., from the graph in fig. 2, we learn that one root of the equa- tion z^ — x — Q = is — 2 and the other is 3. From the graph in fig. 3, we learn that the three roots of the equa- tion ar^ — 2 X = are — ^^2, 0, and y/2. In fig. 4, the graph cuts the axis of x at (— 1, 0) and touches it at (2, 0) ; hence, one root ofx'' — 3x2 + 4 = 0is — 1 and the other two roots are 2 each. Hence, to find the real roots of f(x) = 0, we can draw the graph of f{x), or the locus of y =f(x), and measure the abscissas of the points of intersection and tangency with the ar-axis. Exercise 150. Construct the graph of f(x), and find approximately the real roots of each of the following equations : 1. 3^^x-2 = 0. 4. x^-Sar-Ax-\-ll = 0. 2. ar'-f 2a;-5 = 0. 5. a^ - 4.0(^ - 6x - S = 0. 3. .r^-3a;+-4 = 0. 6. a;^- 4 a^-3a;+- 2 = 0. CHAPTER XXXIV THEORY OF EQUATIONS 489. Horner's method of synthetic division. Let it be required to divide Ax^ + Bx^+ Cx-^ D hj X- a. If for convenience we write the divisor to the right of the dividend and the quotient below it, by the usual method we have : Ax^ + Bx"^ + Cx -\- D Ax^ — Aax^ Ax' + (Aa + B)x + (Aa^ + Ba+ O) (Aa + B)x:^ (Aa + B)x^ - (Aa^ + Ba)x (Aa^ + Ba + C)x {Acfi + Ba+ C)x - {Aa^ + ga^ + Ca) Aa^ + Ba^ + Ca + D In the shorter or synthetic method, we write only the coefficients of the dividend and place a at their right, as below : A B G D\a _ Aa Aa^ + Ba Aa^ + Bii^ + Ca A Aa-\- B Aa'^-h Ba+C Aa^ + Ba"^ + Ca -\- D Multiplying A by a, writing the product under J?, and adding, we obtain Aa + B. Multiplying this sum by a, writing the product under O, and adding, we obtain Aa'^ + Ba + C. In like manner the last sum is obtained. Now A and the first two sums are respectively the coefficients of x"^, X, and x'' in the quotient obtained above by the ordinary method, and the last sum is the remainder. In like manner any rational integral function of x can be divided by X — a. If any power of x is missing, its coefficient is zero, and must be written in its place with the other coefficients. Observe that the shorter or synthetic method of division includes only that part of the usual method given above which is in black-faced type. 444 THEORY OF EQUATIONS 445 Since we omit the sign — before the second term of the divisor, we must omit also that sign before the second term of each product, and then add that term to the dividend, as in the shorter method above. Here the remainder Aa^ + Bar -{■ Ca -\- D is the value of the divi- dend Ax^ + Bx:^ + Cx-t D for x = A, which affords a second proof of § 131. Ex. 1. Divide 2 x* + a;^ - 29 x2 - 9 x + 180 by x - 4. "Write the coefiBcients with 4 at their right and proceed as below : 2 +1 -29 - 9 +180|4 4-8 +36 +28 +76 2 +9 +7 +19 +256 Hence the quotient = 2 x** + 9 x^ + 7 x + 19, and the remainder, or /(4),= 256. Ex. 2. Divide 2 x* + x^ - 29 x^ - 9x + 180 by x + 5. 2 +1 -29 - 9 + 180 1 - 5 _ 10 +45 - 80 +445 2 - 9 +16 -89 +626 Hence the quotient = 2 x^ - 9 x^ + 10 x - 89, and the remainder, or /( - 5), = 625. Ex. 3. Divide x^ + 21 x + 342 by x + 6. 1 +0 +21 +342 _ +36 - 342 1 _6 +57 Hence the quotient = x^ — 6 x + 57, and the remainder, or /(— 6),= 0. Hence the division is exact, and x + 6 is a factor of /(x). Exercise 151. By Horner's method 1. Divide x^ -2x^ - 4.X + 8 by x-3; by x - 2. 2. Divide 2a;^ H-4a^-ar^- 16 a; - 12 bya;-|-4; by x + 3. 446 ELEMENTS OF ALGEBRA 3. Divide 3 a;* - 27 aj^ + 14 a? + 120 by x-%; by a; + 5. 4. Find the value of 2 a;* — 3 a^^ + 3 a? — 1 when a; = 4 ; when a; = — 3 ; when a; = 3 ; when x = 5. 5. Show that one factor of a;^ -f 8 x-^ + 20 a; + 16 is x -{-2, and from the quotient find the others. 6. Show that two factors of a^* + a^ - 29 a;^ - 9 a; + 180 are x — 3 and a; -f 3, and find the others. 7. Show that two factors of a?^— 4 ar^— 8 a; + 32 are x — 2 and a; — 4, and find the others. INTEGRAL RATIONAL EQUATIONS IN ONE UNKNOWN. 470. If all the terms of an integral rational equation in x are transposed to the first member and arranged in descend- ing powers of x, we shall obtain an equivalent equation of the form Aa^" + ^1^"-' + ^l2^"-' + • • • + ^n-l^ + -4n = 0, (B) where Aq, Ai, A2, •••, A^-i, A^ denote any known numbers, real, imaginary, or complex, and n denotes the degree of the equation. Denoting the first member of (B) by f(x), (B) can be written 471. To solve equation (B), or f(x) = 0, by § 149 we need to factor its first member, equate each factor to zero, and solve the resulting equations. But when (JB) is above the second degree in x, the first member cannot be factored by inspection except in certain special cases. The methods which follow should be used when, and only when, f(x) cannot be factored by inspection. 472. If a is a root of the equation f(x) = 0, that is, if f{a) = 0, theii f{x) is divisible by x—a (§ 131). THEORY OF EQUATIONS 447 Conversely, if f(x) is divisible by x — a, then f(a) = ; tJiat is, a is a root of the equation f{x) — 0. E.g.^ if 2 is a root of the equation a;3_2x-2-4a; + 8 = 0, (1) then its first member is divisible by x — 2 (§ 132). Conversely, if the first member of (1) is divisible by a; — 2, then 2 is a root of this equation. 473. It was proved in § 148 that n linear equations in x are jointly equivalent to an equation of the ?ith degree in x. In proving the converse of this theorem in the next article we assume the following theorem : Any integral rational equation in one unknown has at least one root, real, imaginary, or complex. Note. The proof of this theorem is too long and difficult to be given here. 474. Any equation of the nth degree in one unknown has n, and only n, roots. Pi'oof By § 473, the equation f{x) — has a root. Let rtj denote this root ; then, by § 472, f{x) is divisible by a; — Oj, so that f{x) = {:x-a,)f{x), (1) in which, by the laws of division, f (x) has the form of f{x), and is of the {n — l)th degree. Now the equation f (x) = has a root. Denote this root by a^ ; then f{x) = (x-a.;)f(x), (2) in which f2(x) is of the (?i — 2)th degree. Repeating this process n — 1 times, we finally obtain fn-l(x) = (^-Ctn)A^, (n) where A^ is the coefficient of x'' in f(x). 448 ELEMENTS OF ALGEBBA From (1), (2), •••, {n), we obtain f(x) = {x-a,)f^{x) = (x-a,)(x-ao)f2(x) = (x~ ai) (x — as) (^* — ttg) • • • (a? — a„) A- (3) Hence the equation f(x) = is equivalent to the n linear equations X — ai = Oy a; — 0-2 = 0, • • •, x — «„ = 0, and therefore has n and only n roots. From (3), it follows that any expression of the 71th. degree in X can be resolved into n linear factors in x. 475. Equal roots. If two or more of the factors x — a^, X — a2, '•', X — a^ are equal, the equation /(x) = has two or more equal roots. E. g. , of the equation (x-4)3(a; + 5)2(a;-7) = 0, three roots are 4 each, and two are — 5 each. Ex. One root of 2 ic^ - 5 x^ - 37 x + 60 = is 5. Find the others. One root being 5, one factor of /(oj) is x — 5 (§ 472). By division the other factor is found to be 2 x^ + 5 x — 12. Hence the two roots required are those of the equation 2x2 + 5x-12^0. (1) The roots of (1) are evidently — 4 and 3/2. Exercise 152. Solve each of the following equations : 1 . cc^ — 6 £c^ + 10 a; — 8 = 0, one root being 4. 2. 3 a^ — 25 .^•2 + 42 ic 4- 40 = 0, one root being 5. 3. 2 ic^ 4- a;^ — 15 a; — 18 = 0, one root being — 2. 4. 3 a^ - 8 a;2 - 31 a; + 60 = 0, one root being - 3. THEORY OF EQUATIONS 449 5. 4 a^ - 9 x^ — 3 a; -h 10 = 0, one root being — 1. 6. a;^ + a^ - 29 x2 - 9 X 4- 180 = 0, two roots being 3 and -3. 7. a;'' — 4a;^ — 8x4-32 = 0, two roots being 2 and 4. 8. 2 x^ — 15 x^ 4- 35 x^ — 30 a; + 8 = 0, two roots being 1 and 2. 9. 3 a;^ — 5 a;'' - 17 ^2 4- 13 X 4- 6 = 0, two roots being —2 and 3. By § 148, form the equation whose roots are : 10. The two numbers, ± V— 2. 11. The four numbers, ±V— 3, ± V— 5. 12. The four numbers, 3 ± V^^, 5, — 2/3. 13. 3/4, liV^Ts, l-t-V^=^. 14. 2, ±V^^, 3±V^^. 15. 3, -4, V^^. In each of the last six examples, observe that the coefficients of the equation obtained are all real when, and only when, the imaginary or complex roots occur in conjugate pairs. This illustrates the converse of the next article. 476. In any integral rational equation having only real coefficients^ imaginary or complex roots occur in conjugate pairs; that is, if a 4- bi is a root, then a — bi is also a root. Proof. If the coefficients in f(x) are all real, then all the terms of the expression obtained by substituting a 4- hi for X in fix) will be real except those containing odd powers of hi, which will be imaginary. Representing the sum of all the real terms by A, and the sum of all the imaginary terms by Bi, we have f{a + hi) = AJrBi. (1) 450 ELEMENTS OF ALGEBRA Now f(a — bi) will evidently differ from f(a + bi) only in the signs before the terms containing the odd powers of bi ; that is, in the sign before Bi ; hence f(a - bi) = A- Bi. (2) Since a + bi is a root of f(x) — 0, from (1) we have A + Bi = 0. Therefore ^ = and ^ = 0. § 279 Hence by (2), f(a - bi) = 0. That is, when a -\- bi is a root of f(x) = 0, a — bi is also a root. Ex. One root oix^ - 4x^ + 4x - S = (1) is (1 + V^3)/2 ; find the others. Since 1/2 + aA^/2 is a root, 1/2 - V^^/2 is also a root (§ 476). Hence two factors of the first member of (1) are X - 1/2 - V^/2 and X-1/2+ V^^/2, whose product is (x — 1/2)^ + 3/4, or x^ — x+1. But cc3 - 4ic-2 + 4rB - 3 = (x2 - ic + 1) (a; - 3); (2) hence the third root of (1) is 3. Identity (2) illustrates the following principle : 477. Any rational integral function of x wliose coefficients are real can be resolved into real factors, linear or quadratic in X. Proof. If the coefficients of f{x) are real, the imaginary or complex roots of f{x) = occur in conjugate pairs, as a 4- bi and a — bi-^ hence the complex factors of f{x) occur in conjugate pairs, as x — a — bi and x — a-{- bi, whose product is a real quadratic expression in x-, that is (x — a—bi) (x - a H- bi) = {x - ay -|- b^. THEORY OF EQUATIONS 451 Exercise 153. Solve each of the following equations, and find the real factors of the first member : 1. a^ - 6 ar' + 57 a; - 196 = 0, one root being 1 - 4 V^3. 2. a:^ — 6 a; + 9 = 0, one root being (3 + V^^)/2. 3. ar' - 2 a^ + 2 a; - 1 = 0, one root being (1 + V^r3)/2. 4. a;*-f4aj^ + 5ar^-|-2a; — 2 = 0, one root being — 1 + ^. 5. a;*4-4a^4-6a^ + 4a; + 5 = 0, one root being i. 6. a^ — ar* 4- a;'' — a.*^ + a; — 1 = 0, two roots being — i and (l + V^/2. 7. Show that in an equation with commensurable real coefficients, surd roots occur in conjugate pairs; that is, if a + -y/b is a root of f(x) = 0, a — ->/6 is a root also, -y/b being a surd number. All the terms in /(a + ^/b) will be rational except those containing odd powers of -^6, whiph are surd. Denote the sum of all the rational terms by A and the sum of all the surd terms hy B^b ; then /(a+ y/b) = A + By/b. Hence /(a — y/b) = A — By/b; and so on as in § 476. 8. Solve 6 a^ — 13a;^ — 35 ar^ — a; + 3 = 0, one root being 2-V3. 9. Solve X*- 36 a^+ 72 a; - 36 = 0, one root being 3 - V3. 478. The graph of /(») illustrates the fact that equal real roots form the connecting link between unequal real roots and imaginary or complex roots, and that imaginary or com- plex roots occur in pairs. E.g., by slightly diminishing the term 4 of the function x^—Sx^-\-i, its graph in fig. 4 of § 46.3 would be moved downward, and would then cut the axis of x in three points ; by slightly increasing the term 452 ELEMENTS OF ALGEBRA 4, the graph would be moved upward, and would then cut the axis of x in hut one point. That is, the two equal real roots of the equation x3 _ 3 x2 + 4 = would become unequal real roots or complex roots according as the known term 4 were diminished or increased. From fig. 5 in § 463 the pupil should follow the changes in the roots of the equation ic4 + x3 - 3 x2 - a; + 2 = 0, (i) when the term 2 is decreased continuously to - 1 ; (ii) when the term 2 is increased continuously to 4. 479. An equation of the form (B) in § 470, is said to be in the type-form when the coefficient of ic" is 1. E.g., ic4 _ I a;3 + 3 ^.2 _^ 4 _ is in the type-form. 480. If an equation of the nth degree is in the type form, then — the coefficient of x""'^ = the sum of the roots; the coefficient of x""'^ = the sum of the products of the roots taken two at a time; — the coefficient of jr"~^ = the sum of the products of the roots taken three at a time. (— 1)** (the coefficient of x^) = the product of the n roots. Proof. Let a^, a2, a^, ••• a„ denote the n roots; then, by § 148, the equation can be written in the form (x — tti) (x — ag) (x — as)"'{x— a„) = 0. (1) When n = 2, by multiplication (1) becomes a^ — (ai + ttg) ^ + cti<^2 = 0, which proves the theorem when n = 2. When 71 = 3, by multiplication (1) becomes ^ — («i + «2 4- cis)^ + (<^i«2 + «i0t3 + a2a3)x — aia^^ = 0, (2) which proves the theorem when ?i = 3. THEORY OF EQUATIONS 45B From the laws of multiplication it is evident that the same relation holds when w = 4, 5, 6, •••. Observe that, if the term in a;""^ is wanting, the sum of the roots is 0, and ?.f the kno^n term is wanting, at least one root is 0. E.g., in the equation x4 + 6«2_iix-6 = 0, the sum of the roots is ; the sum of their products taken two at a time is 6 ; the sum of their products taken three at a time is 11 ; and their product is — 6. Note. The coeflBcients in any equation are functions of the roots ; and conversely, the roots are functions of the coefficients. The roots of a literal quadratic equation have been expressed in terms of the coefficients (§ 291). The roots of a literal cubic or biquadratic equa- tion can also be expressed in terms of the coefficients, as is shown in college algebra. But the roots of a literal equation of the fifth or higher degree cannot be so expressed, as was proved by Abel in 1825. Ex. Its roots being in arithmetic progression, solve 4 x3 - 24 x=« + 23 X + 18 = 0. (1) Let a denote the second term in the A. P. and b the difference ; then the three roots are a — b, a, a + b. Hence their sum is 3 a ; the sum of their products taken two at a time is 3 a^ — b'^ ; and their product is a{a^ — b^). Divide (1) by 4 to reduce it to the type-form; then, by §480, we have 3 a = 6, 3 a2 - 62 = 23/4, a(a^ -b^) = - 9/2. (2) Solving the first two equations in (2), we obtain a = 2, b = ± 5/2 j and these values are found to satisfy the third equation in (2). Hence the roots are — 1/2, 2, and 9/2. Exercise 154. 1. The sum of two of its roots being zero, solve The sum of the three roots is — 4 ; hence the third root is — 4. 2. Its roots being in arithmetic progression, solve 454 ELEMENTS OF ALGEBRA 3. Its roots being in geometric progression, solve 3 0^3 _ 26 a^ + 52 a: - 24 = 0. 4. One root being 1 —V— 3, solve aj3-4«2^g^_3^Q^ One root being 1 — V— 3, a second root is 1 + V— 3. The sum of these two roots is 2, and the sum of the three roots is 4 ; hence the third root is 2. 5. By § 480, solve each of the first five examples in exercise 153. 481. If the coefficients of /(a?) are all +, /(a;)> when a;>0; hence, if the coefficients off{x) are all positive, f(x) = has no positive real root. If the coefficients of f(x) are alternately + and — ; then, when a;<0, /(aj)>0 or <0 according as n is even or odd; hence, if the coefficients of f{x) are alternately 4- a7id — , f (jr) = has no negative real root. If the sum of the coefficients of f(x) is zero, /(I) = ; hence, when the sum of the coefficients of f(x) is zero, one root of f(x) = Ois -\- 1. E.g., x^ + 6 x^ -\- llx -\- 6 = has no positive root, since /(x)> when OS > 0. o;^ — 6 a;2 _|- 10 X — 8 = has no negative root ; since /(x)< when aj<0. /k4 + 2 ic3 - 13 ic2 - 14 X + 24 = has + 1 as a root ; since /(I) = 0. 482. If all the coefficients of an equation in the typeform are whole numbers, any commensurable real root of the equation is an integral factor of its known term. E.g., any commensurable real root of the equation a;3 _6a;2 + i0a;-8 = is an integral factor of its known term — 8 ; that is, any such root is ±1, ±2, ±4, or i8. THEORY OF EQUATIONS 455 Proof. Let all the coefficients of the equation a-" + Aix""-^ + A^x""-^ -i \-A^ = (1) be whole numbers, and suppose that s/t, a fractional num- ber in its lowest terms, is one of its roots. Substituting s/t for x, we obtain oft ©n-l ©n— 2 Multiplying by r~^, and transposing, we obtain s^^/t = - (Ais^-^ + A^s""-^ + • • • 4- A„r-^) (2) Now (2) is impossible, for its first member is a fractional number in its lowest terms, and its second member is a whole number. Hence a fractional number cannot be a root, and there fore any commensurable root must be a whole number. Next, let a be an integral root of (1). Substituting a for ic, transposing A^, and dividing by a, we have a"-i +A,a--'' + ^,a-« + • • • + A,,_, = - AJa. (3) The first member of (3) is a whole number; hence the quotient AJa is a whole number, ^.e., a is an integral factor of A^. Ex.1. Solve a;8 - 6x2 + 10x- 8 = 0. (1) By § 481, (1) has no negative root ; hence, by § 482, any commen- surable real root of (1) is + 1, + 2, + 4, or + 8, i.e. it is one of the positive integral factors of 8. The work of determining whether -)- 4 is a root can be arranged as below : 1 _6 +10 -S[± +4 - 8 +8 1-2+2 The division is exact, and the quotient is x^ — 2 a; + 2. Hence the roots of (1) are 4 and the roots of x2-2x + 2 = 0. (2) 456 ELEMENTS OF ALGEBRA Solving (2), x = l ± V^. Hence the roots of (1) are 4 and 1 ± V— 1. Ex.2. Solve a;* + 2 ic3 - 13x2- 14^ + 24 = 0. (1) By § 481, one root of (1) is + 1, and by § 482 any other commen- surable real root is ±1, ±2, ±3, ±4, ±6, ±8, ± 12, or ± 24, i.e. it is one of the integral factors of 24. 1 + 2 + 1 -13 + 3 -14 -10 + 24|J_ -24 1 + 3 -2 -10 - 2 -24L + 24 -2 1 +1 - 12 Hence the roots of (1) are 1, — 2, and the roots of a;2 + a; - 12 = 0. Hence the roots of (1) are 1, — 2, 3, and — 4. Usually it is better to try the smaller factors of An first. Exercise 155. Solve each of the following equations : 1. i^ + 2x' + 9x-{-lS = 0. 2. a^-6a^-{-llx-6 = 0. 3. a^-4.x^-6x-{-9 = 0. 4. x^-3a^-{-x'-^2x = 0. 5. x^-Sa^-\-13x-6 = 0. 6. aj3 + 6a;2-f-9aj + 2 = 0. 7. a^ + 5a^-9a;-45 = 0. 8. x^-Aa^-Sx-^32 = 0. 9. x*-6x^ + 2Ax-16 = 0. 10. x^-SiK^-Ux'-^4Sx-32 = 0. 11. a^-3x*-9a^-\-21x'-10x-\-24. = 0. 12. a^-\-2x^-23x-60 = 0. THEOBY OF EQUATIONS 467 483. Limits of real roots. Superior limit. In evaluating /(4) in example 1 of § 469, the sums are all positive, and they evidently would all be greater for x > 4. Hence f{x) can vanish only for x < 4 ; and therefore all the roots of f(x) = are less tlian 4. Hence, if in computing the value of f('^c) all the sums are positive, the real roots off(x) = are all less than +(?. The least integral value of +c which fulfils this condition is called the superior limit of the real roots of f(x) = 0. Inferior limit. In evaluating /(— 5) in example 2 of § 469, the sums are ^ternately — and -f, and they evidently would all be greater arithmetically for x< — 5. Therefore all the real roots of f(x) = are greater than — 5. Hence, if in computing the value of f{~b) the sums are alter- nately — and -}-, all the real roots of f{x) = are greater than ~b. The greatest integral value of ~h which fulfils this condi- tion is called the inferior limit of the real roots of f(x) — 0. Observe that the above reasoning holds when we regard a zero sum as either positive or negative, and that when the last sum is zero, the limit obtained is itself a root. E.g., if /(x) = x* + 2x8- 13a:2-14a; + 24 = 0; (1) then in evahiating/(4), the sums are all + ; and in evaluating/(— 5), the sums are alternately — and + ; hence the real roots of f(x) = lie between — 5 and 4. Hence, by § 482, any commensurable roots of (1) must be ±1, ±2, ±3, or -4. Compare this result with example 2 in § 482. Exercise 156. 1. Show that any commensurable real root of a:3_2a:_50 = lies between — 2 and 4 ; and hence is ± 1 or 2. 458 ELEMENTS OF ALGEBRA 2. Show that any commensurable real root of is ± 1, ± 2, ± 4, ± 5, ± 8, or 10. 3. Show that any commensurable real root of a;4_i5a^-|.10a; + 24 = (1) is ±1, ±2, ±3, or -4. 4. Eind the roots of equation (1) in example 3. Solve each of the following equations: 5. x*-9ic^ + lTa;'^+27a;-60 = 0. 6. «^-45a^-40x + 84 = 0. 7. a^_4a;*-16i»« + 112a;2-208a; + 128 = 0. 8. ic4-ar^~39a;2 + 24ic + 180 = 0. 9. a^ + 5ar^ - 81 a;^ - 85 ar^ + 964a^ -f 780aj - 1584 = 0. 10 . ic^ + aj« - 14 a^ - 14 a;^ 4- 49 aj^ + 49 a^ - 36 a; = 36. 11. aj6-10x4-3a.-2 + 108 = 0. 12. a;«-2ar^-7a;* + 20aj3-21a^-18a;-f-27 = 0. 484. To transform an equation into another whose roots shall be some multiple of those of the given one. Proof. If in the equation of" + ^ix"-i + A^x^-^ + ^3X"-3 H h ^„ = 0, (B) we put X = Xi/a, and multiply by a", we obtain a;i« + ^laa^i"-^ + ^gOt V~' + ^sO^V"^ + • • • + -4^ = 0. (2) Since Xi = aa;, the roots of (2) are a times those of (1). Hence, to effect the required transformation, multiply the second term of (B) by the given factor, the third term by its square, and so on. Observe that before the rule is applied the equation must THEORY OF EQUATIONS 459 be put in the type-form, and any missing power of x must be written with zero as its coefficient. This theorem becomes evident also when we observe that if in equation (2) in § 480 each root is multiplied by a, the second term will be multiplied by a, the third term by a^, and the fourth term by a\ The chief use of this transformation is to clear an equa- tion of fractional coefficients. Ex. Solve the equation x3-.i^x2 + fx-x^^ = 0, (1) first transforming it into another with integral coefficients. Multiplying the second term by a, the third by a-, the fourth by a^, we obtain x3 _ J^ ax2 -f f a^x - i^« a' = 0. (2) By inspection we discover that 4 is the least value of a which will render the coefficients of (2) integral. Putting a = 4, we obtain a;8 _ 11 x2 + 36 X - 36 = 0. (3) The roots of (3) are found to be 2, 3, and 6. But the roots of (3) are four times the roots of (1); hence the roots of (1) are 1/2, 3/4, and 3/2. Exercise 157. Solve the following equations by transforming them into others whose commensurable real roots are whole numbers : 1. a^-^x^-^^ix + ^\ = (). 2. a^-x'/4:-x/2-\-l/S = 0. 3. 8a^-26a;2-hllx-hl0 = 0. 4. a^- x^/3 - x/m + 1/108 = 0. 5. 24:3^ -520^ + 26x-S=-0. 6. 9x*-9a^ + 5x'-3x-^2/S = 0. 7. x' - ^76 - ar^/12 - 13 a^/24 + 1/4 = 0. 8. 2a;^-12ar^ + 19a:'-6a;4-9 = 0. 460 ELEMENTS OF ALGEBRA 485. If f(a) and f(b) are opposite in quality, an odd number of real roots off(x) = lies between a and b. Iffip) and fib) are like in quality, no real root, or an even number of real roots of f(x) = lies between a and b. Proof If the ordinates of two points in the graph of fix) are opposite in quality, the points are on opposite sides of the a^axis, and the part of the graph between these points must cross that axis an odd number of times (§ 466) ; that is, f{x) is zero for an odd number of values of x between a and 6. If the ordinates of two points are like in quality, the points are on the same side of the avaxis, and the part of the graph between these points either does not cross that axis or crosses it an even number of times, touching it being regarded as crossing it twice. E.g., in fig. 3 of § 463, the graph cuts XJT an odd number of times between A and B or A and Z>, and an even number of times between A and C or B and D. In fig. 5 of § 463, the graph cuts XXf an odd number of times between A and B or B and E, and an even number of times between A and (7, C and E, or A and E. Ex. Find the first figure of eacli real root of the equation ic3 _ 4 a;2 _ 6 ic + 8 = 0. (1) By §§ 474 and 476, (1) has either three or only one real root. By Horner's method we find that : - when x = -2, -1, 0, 1, 2, 3, 4, 5, f(x)=-4, +9, +8, -1, -12, -19, -16, +3. Since /(— 2) and /(— 1) are opposite in quality, at least one root of (1) lies between — 2 and — 1. For like reason a second root lies between and 1, and a third between 4 and 6. Hence two roots are -(!• + ) and 4- + , and, since /(0.9) is + and /(I) is -, the third root is 0.9 +. THEORY OF EQUATIONS 461 486. Any equation of an odd degree in which Aq is positive has at least one real root whose quality is opposite to that of its known term A^. Proof If J-o > and f{x) is of an odd degree, then /(-^) is -, /(O) = A, /(+^) is +. Hence if A^ is positive, one root of f(x) = lies between —00 and (§ 485) ; and if A^ is negative, one root lies between and +qo. 487. Any equation of an even degree in which A^ is positive and the known term A^ is negative has at least one positive and one negative real root. Proof If A > and f{x) is of an even degree, then /(-oo) is +, /(O) is -, X+^) is +. Hence one root of f{x) = lies between — oo and 0, and another between and +qo. Exercise 158. Find the first figure of each real root of the equations : 1. a^-3x2-4a:4-ll = 0. 5. a^-2aj-5 = 0. 2. a^ + a^-2x-l = 0. 6. a:^ ^_ a; _ 50O = 0. 3. x*-4:2i^-3x-\-2S = 0. 7. x^ -\- 10 x" + 5 x = 260. 4. x^-4:X--6x = -S. 8. a;*-12a.-2+12a:-3 = a 48 FRENCH. The Qa Ira Series of French Plays. Edited by Professor B. W. Wells, of the University of the South. 6 volumes, i6mo, cloth. Each, 36 cents. THE plays selected have not heretofore been edited for use in American schools. The series contains works acfapted to the most varied needs, but these works are so treated as to ex- clude all expressions or suggestions which could stand in the way of their use in mixed classes. The introductions give brief biographies of the authors and such comment on their work as may seem helpful. The Notes explain peculiarities of idiom that would not naturally be sought in a dictionary. Allusions to social and political customs, as well as to literature and history, receive such comment as will aid the pupil to put himself in the place of the original audience. In this way it is hoped that the reading of these plays will help the student not only in the study of French, but also in the development of a literary taste. The following works are contained in the series : — Moi, par Labiche et Martin. Gringoire, par Theodore de Banville, et L'Ete de la Saint Martin, par Meiihac et Hel6vy. La Question d' Argent, par Alexandre Dumas,//?. La Camaraderie, par Eugene Scribe. Le Luthier de Cremone, et Le Tr6sor, par Frangois Copp6e. Le Fils de Giboyer, par Emile Augier. Professor A. G. Cameron, Yale University : The volumes are as admi- rable in editing as they are dainty in form. Professor H. A. Rennert, University of Pennsylvania : It (Moi) is an excellent book in every way, and I shall use it. Professor George D. Fairfield, University of Illinois : I heartily commend both the editorial and typographical excellence apparent all through it (Moi). Professor W. A. Cooper, Marietta College, Ohio : I have already used two of the Qa Ira Series with classes and shall use another this term. The books are not only very delightful to look at, but the editor's work gives the student just what help he needs. SCIENCE. 51 Physics for Uniuersity Students. By Professor Henry S. Carhart, University of Michigan. Parti. Mechanics, Sound, and Light. With 154 Illustrations. i2mo, cloth, 330 pages. Price, ;^i.5o. Part II. Heat, Electricity, and Magnetism. With 224 Illustrations. i2mo, cloth, 446 pages. Price, ^1.50. THESE volumes, the outgrowth of long experience in teach- ing, offer a full course in University Physics. In preparing the work, the author has kept constantly in view the actual needs of the class-room. The result is a fresh, practical text-book, and not a cyclopaedia of physics. Particular attention has been given to the arrangement of topics, so as to secure a natural and logical sequence. In many demonstrations the method of the Calculus is used without its formal symbols ; and, in general, mathematics is called into ser- vice, not for its own sake, but wholly for the purpose of establish- ing the relations of physical quantities. At the same time the course in Physics represented by this book is supposed to pre- cede the study of calculus, and its methods will in a general way prepare the student for the study of higher mathematics. Professor W. LeConte Stevens, Rensselaer Polytechnic Institute, Troy, N. Y. : After an examination of Carhart's University Physics, I have unhesitat- ingly decided to use it with my next class. The book is admirably arranged, clearly expressed, and bears the unmistakable mark of the work of a successful teacher. Professor Florian Cajori, Colorado College : The strong features of his Uni- versity Physics appear to me to be conciseness and accuracy of statement, the emphasis laid on the more important topics by the exclusion of minor details, the embodiment of recent researches whenever possible. Professor A. A. Atkinson, Ohio University, Athens, O. : I am very much pleased with the book. The important principles of physics and the essentials of energy are so well set forth for the student for which the book is designed, that it at once commends itself to the teacher. Professor A. E. Frost, Western University, Allegheny, Pa. : I think that it comes nearer meeting my special needs than any book I have examined, being far enough above the High School book to justify its name, and yet not so far above it as to be a discouragement to the average student. 56 SCIENCE. Herbarium and Plant Descriptions. Designed by Professor EDWARD T. NELSON, late of Ohio Wesleyan University. Portfolio, 7% X 10 inches. Price, 75 cents. THIS is an herbarium and plant record combined, enabling the student to preserve the specimens together with a record of their characteristics. A sheet of four pages is devoted to each specimen. The first page contains a blank form, with ample space for a full descrip- tion of the plant, and for notes of the circumstances under which it was collected. The pressed specimen is to be mounted on the third page, and the entire sheet then serves as a species- cover. Each portfolio contains fifty sheets, which are separate, so as to permit of scientific rearrangement after mounting the specimens. The preliminary matter gives full directions for collecting, pressing, and mounting plants, as well as a synopsis of botanical terms. The portfolio is strong, durable, and attractive in appearance. In the class-room and in the field this work has been found helpful and stimulating. It encourages observation and research, and leads to an exact knowledge of classification. Professor D. P. Penhallow, McGill University, Montreal, Can.: The idea is a good one, and well carried out. I am sure it will prove most useful in the botanical work of schools and academies, for which I would strongly recommend it. Professor G. H. Perkins, University of Vermont, Burlington, Vt. : It is the best thing of the sort I have seen ; very attractive and very helpful to beginners in calling attention to points that would be overlooked. Professor B. P. Colton, Normal University, III. : It is a very ingenious ar- -rangement, and neatly gotten up. It speaks well for the publishers, as well as the designer. It is the neatest scheme of the kind I have seen. 0. D. Robinson, Principal of High School, Albany, N. Y. : It appears to me to be a very complete arrangement, admirable in every respect, and very moderate in price. F. S. Hotaling, Formerly Principal of High School, Framingham, Mass. : Last year's work in botany was made so much more interesting and valua- ble by the use of the Herbarium that we find it now a necessity. SCIENCE. hi Lessons in Elementary Botany for Seco^idary Schools. By Professor Thomas H. MacBRIDE, Iowa State University. i6mo, cloth, 244 pages. Price, 60 cents. THIS book is designed to make the laboratory study of botany possible in every High School in the country. While call- ing for the use of none but the simplest appliances, and for no material but such as is easily accessible to every teacher, its method is nevertheless in perfect harmony with the best scien- tific instruction of the time. The student is not asked to study illustrations or text ; he is sent directly to the plants themselves, and shown how to study these and observe for himself the vari- ous problems of vegetable life. The plants passed in review are those which are more or less familiar to every one, and the life history of which every child should know. Austin C. Apgar, State Normal School, Trenton, N.J. : There are many points in the book which please me. Not the least of these is that the author has not been misled by the craze for that " logical order " which begins with protoplasm and, some time or other, if the subject be pursued long enough, reaches such things as can easily be found and examined. He begins with the known and gradually advances to the unknown, — the only order in which successful teacliing can be accomplished. A New Flora. To accompany Macbride's Elementary Botany. IN resjDonse to a desire expressed by many teachers, Professor Macbride has prepared a Key to the More Common Species of Native and Cultivated Plants occurring in the Northern United States. The material thus furnished has never before been offered in such convenient form for class-room work. Its use will in no way change the original plan of the Elementary Botany, but it offers many advantages for additional study. The Key will be furnished without charge to those ordering the author's Botany. Copies ordered without the Botany will be furnished at twenty-five cents each. 58 MA THEM A TICS. Elements of Algebra. By Professor JAMES M. TAYLOR, Colgate University, Hamilton, N.Y. At Press. IN this book Professor Taylor aims primarily at simplicity in method and statement, and at a natural and logical sequence in the series of steps which lead the pupil from his arithmetic through his algebra. An introductory chapter explains the meaning and object of literal notation, and illustrates the use of the equation in solving arithmetical problems. This is followed by a drill on particular numbers before the pupil is introduced to the use of letters to represent general algebraic numbers. General principles are brought out by induction from particular cases, and proofs are given in their natural places where the pupil will be unlikely to memorize without comprehending them. Nomenclature has been looked to carefully. Many of the misleading terms of the older text-books have been discarded and others more useful and help- ful have been applied. The methods of working examples have been chosen for their simplicity and the scope of their application. Suggestions as to method of attack are given, but formal rules are stated but rarely. Positive and negative numbers are so explained and defined as to give clear and true concepts, such as lead naturally to still broader views of numbers. Factoring is made a fundamental principle in the solution of quadratic and higher equations. Particular atten- tion is given to the theory of equivalent equations. In illustrating the meaning of numbers, equations, and systems of equations, the graphic method is used. In general the aim is to render as clear as possible to the pupil all fundamental processes and to simplify the statement of rules. The book is particularly adapted to beginners, and is intended at the same time to prepare for any college or scientific school, as each subject is so treated that the pupil will have nothing to unlearn as he advances in mathematics. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewals only: Tel. No. 642-3405 Renewals may be made 4 days prior to date due. Renewed books are subject to immediate recall. RY ;low. >70 ' Jt Ni 78 LI LD21A-50m-2,'71 (P2001810 ) 476 — A-32 General Library University of California Berkeley P306047 T3 THE UNIVERSITY OF CALIFORNIA LIBRARY