Cl 
 
 5iV^Bk.S 5 S°-E 
 
 TRINITY COLLEGE 
 LIBRARY 
 
 DURHAM, NORTH CAROLINA 
 
 Rec’d 
 

 
 
Digitized by the Internet Archive 
 in 2016 with funding from 
 Duke University Libraries 
 
 https://archive.org/details/elementsofalgebr01shou 
 
THE 
 
 Elements of Algebra. 
 
 BY 
 
 F. A. SHOUP, 
 
 PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF THE SOUTH; LATE PROFESSOR OF LOGIC, 
 METAPHYSICS, ETC., IN THE UNIVERSITY OF MISSISSIPPI. 
 
 5 f l 8 
 
 NEW YORK: 
 
 E. J. HALE & SON, PUBLISHERS, 
 
 1874. 
 
 IP 
 
Entered according to Act of Congress, in the year 1S74, by 
 E. J. HALE & SON, 
 
 In the Office of the Librarian of Congress, at W ashington, D. C. 
 
 Lange, Little & Co., 
 
 PRINTERS, ELECTROTVPERS AND STEREOTYPERS, 
 
 108 to 114 Wooster Street, N. Y. 
 
i '/ 1 . o z. 
 PREFACE. S ; 
 
 This book was begun with the hope of preparing something for 
 beginners in Algebra which would give them, in a narrow compass, a 
 philosophical and, therefore, a thorough start in their analytical 
 studies. By adopting a somewhat new arrangement of the subject, 
 it was found that the matter the author had in mind could be pre- 
 sented in such a small space, that he determined to extend the scope 
 of the work far enough to give all that is really necessary to enable 
 the student to prosecute with profit the higher mathematics. This 
 he did the more readily, since he is of the opinion that the further 
 course in Algebra presents rather more difficulties than any other 
 branch of mathematics, and that, consecpiently, the student at the 
 beginning of his analytical course can derive little real benefit from 
 his efforts to master it. In truth, so far as his information goes, it 
 is generally either omitted altogether, or so little appreciated by the 
 young student, as to be almost lost labor. It is, withal, so impor- 
 tant, that it ought to be insisted upon, and so should be taken up at 
 a later stage of the student’s course, and then presented in connection 
 with the general philosophy of analytical investigations. Accordingly, 
 if this little work meets with approval, the author will be encouraged 
 to attempt the preparation of a supplementary work to meet this 
 end. 
 
 It is not necessary to say much with regard to this essay. It will 
 be found in great part novel in its treatment of the subject. The 
 discussions are considerably extended, and at all points the rationale 
 
4 
 
 PEEFACE. 
 
 of operations fully brought out. An effort has been made to pre- 
 serve the continuity of the subject, so as to present a harmonious 
 whole. The author hopes that it may, in some measure, prove the 
 means of stimulating the reasoning powers of such lads as may 
 chance to use it, and that the mere mechanism of operations will not 
 content them. 
 
 University of the South, 
 Sewanee, Tenn., March , 1874. 
 
TABLE OF CONTENTS 
 
 SECTION I. 
 
 ARTICLE 
 
 Definitions and Explanations 1-17 
 
 SECTION II. 
 
 ALGEBRAIC TERMINOLOGY. 
 
 Introductory Remarks 18 
 
 Translation from English into Algebraic Language 10 
 
 Translation from Algebraic into English Language 20 
 
 Translation from English into Algebraic Language 21 
 
 Translation from Algebraic into English Language 22 
 
 Translation from English into Algebraic Language 23 
 
 Numerical Value 24 
 
 SECTION III. 
 
 TREATMENT OF THE + AND — SIGNS. 
 
 Introductory Remarks 25 
 
 The Nature of the Signs + and — 26 
 
 Algebraic Sum 27 
 
 Addition of Quantities 28 
 
 Subtraction of Quantities 29 
 
 Subtraction of Polynomials 30 
 
 Terms Sum and Difference 31 
 
 SECTION IV. 
 
 MONOMIALS, EXPONENTS AND THE SIGNS X AND -f-. 
 
 Exponents in General 32 
 
 Exponents Entire and Positive. — Multiplication 33 
 
 Law of Signs in Multiplication and Division 34 
 
 Operations on Fractions 35 
 
 Powers 36 
 
 Roots • 37 
 
 Division of Monomials 38 
 
 Zero Power 39 
 
 Entire and Negative Exponents 40 
 
6 
 
 CONTENTS. 
 
 ARTICLE 
 
 Fractional Exponents. — Radicals 41 
 
 Fractional Exponents. — Radicals 42 
 
 To Change the Index of a Radical 43 
 
 To Bring Radicals to a Common Index 44 
 
 Product of the nth Roots 45 
 
 Quotients of the nth Roots 45 
 
 Division of Radicals 46 
 
 Simplification of Radicals 47 
 
 To Pass a Factor under the Radical 48 
 
 To Transform the Sum of Radicals 49 
 
 General Principles of Exponents 50 
 
 SECTION V. 
 
 TRANSFORMATION OF POLYNOMIALS. 
 
 Introductory 51 
 
 Multiplication of a Polynomial by a Monomial 52 
 
 Multiplication of Polynomials 53 
 
 Two Terms Unchanged by Simplifying 54 
 
 Division of Polynomials 55 
 
 Formulas for Transformations 56 
 
 Formulas for Transformations 57 
 
 Formulas for Transformations 58 
 
 Factoring 59 
 
 Factoring 50 
 
 Division of the Difference of Like Powers 61 
 
 Division of Like Powers 02 
 
 Binomial Formula 03 
 
 Powers of Polynomials 04 
 
 Common Multiple 05 
 
 Operations upon Algebraic Fractions 06 
 
 Essential Sign 07 
 
 Imaginary Quantities 08 
 
 SECTION VI. 
 
 ROOTS OF NUMBERS. 
 
 Extraction of the Square Root of Numbers 09 
 
 Extraction of the nth Root of Numbers ‘0 
 
 No Exact Root of an Imperfect Power 1 
 
 SECTION VII. 
 
 EQUATIONS. 
 
 Definitions 
 
 Axioms N 
 
 First Transformation - 
 
CONTENTS. 
 
 7 
 
 ARTICLE 
 
 Second Transformation 75 
 
 Third Transformation 79 
 
 Fourth Transformation 77 
 
 The Change of Signs 78 
 
 Solution of Simple Equations 79 
 
 Degree of Equations 80 
 
 Complete and Incomplete Equations 81 
 
 Complete Equations of the Second Degree 82 
 
 Solution of Incomplete Equations S3 
 
 Solution of Complete Equations 84 
 
 Trinomial Equations 85 
 
 Simultaneous Equations 86 
 
 Elimination by Addition 87 
 
 Elimination by Substitution 88 
 
 Elimination by Comparison S9 
 
 Elimination in general 90 
 
 Equations of Condition 91 
 
 Simultaneous Equations of a Higher Degree 92 
 
 Simultaneous Equations of the First and Second Degree 93 
 
 Homogeneous Equations of the Second Degree 94 
 
 Solution of Simultaneous Equations of any Degree 95 
 
 Radical Equations 96 
 
 Inequalities 97 
 
 Problems...., 98 
 
 SECTION VIII. 
 
 SYMBOLS 0 AND 00 . 
 
 Meaning of the Symbols 0 and 00 99 
 
 Combinations of the Symbols 0 and co 100 
 
 Vanishing Fractions 101 
 
 Problem of the Couriers 102 
 
 Problem of the Lights 103 
 
 Properties of Equations of the Second Degree 104 
 
 Properties of Equations of the Second Degree 105 
 
 Properties of Equations of the Second Degree 106 
 
 Properties of Equations of the Second Degree 107 
 
 Discussion of the Four Forms 108 
 
 SECTION IX. 
 
 ARITHMETICAL PROGRESSION". — RATIO AND PROPORTION. — GEOMET- 
 RICAL PROGRESSION. 
 
 Definition of a Series 109 
 
 Arithmetical Progression 110 
 
 Formula for Last Term Ill 
 
 Sum of Means at Equal Distances from Extremes 112 
 
8 
 
 CONTENTS. 
 
 ARTICLE 
 
 Formula for Sum of Terms 113 
 
 Eatio 114 
 
 Proportion 115 
 
 Product of the Extremes 116 
 
 Equi-multiples, Powers, Roots 117 
 
 Inversion 118 
 
 Composition 119 
 
 Like Parts 120 
 
 To Multiply Proportions together 121 
 
 Sum of Antecedents and Consequents 122 
 
 Continued Proportion 123 
 
 Mean Proportional 124 
 
 Reciprocally Proportional 125 
 
 Geometrical Progression 126 
 
 Formula for Last Term 127 
 
 Formula for Sum of Terms 128 
 
 To Find Formula for any Element 129 
 
 To Insert Geometrical Means • 130 
 
 Sum of an Infinite Progression 131 
 
 SECTION X. 
 
 LOGARITHMS. 
 
 Definitions 132 
 
 Characteristic 133 
 
 Table 134 
 
 General Principles 135 
 
 Solution of Equations 136 
 
 Logarithms of Decimals 13 1 
 
 General Properties 138 
 
 Modulus 133 
 
 Indeterminate and Identical Equations 1-40 
 
 Indeterminate Co-eificients 141 
 
 Principle of Indeterminate Co-efficients 142 
 
 Developments 143 
 
 Partial Fractions 144 
 
The Elements oe Algebra. 
 
 SECTION I. 
 
 DEFINITIONS AND EXPLANATIONS. 
 
 1. Algebra is a branch of Mathematics in which letters are com- 
 monly used to represent quantities, while the operations to be per- 
 formed upon them are merely indicated. 
 
 2 . The leading letters of the alphabet (a, b, c, etc.) are used to 
 denote quantities whose values are given, or may be assumed at 
 pleasure. They are called known or arbitrary quantities. 
 
 Any numeral, as 5 or 25, is made up of a fixed number of units, and admits 
 of no change whatever in this regard : 5 cannot represent 25, or any other num- 
 ber than itself. But if we say that the letter a shall stand for any fixed number 
 whatever, it may have all possible values at pleasure. It cannot, however, 
 have more than one value at a time. Its several values must be taken in suc- 
 cession. 
 
 In like manner a may represent any kind of quantity, as pounds, dollars, 
 men, etc., but always in succession. The moment we attribute a specific nu- 
 merical value to it, it ceases to be an algebraic quantity for that particular value, 
 and becomes an arithmetical quantity. Whatever is here said of a, is equally 
 true of any other quantities, b, c, d, etc. 
 
 3. The final letters of the alphabet ( x , y, z, etc.) are used to denote 
 quantities to be determined. They are called unknown quantities ; 
 or, when they admit of an indefinite number of values, in succession, 
 they are called variables. 
 
 Unknown quantities always depend for their values upon certain other quan- 
 tities ; for example, if we say x shall be a quantity whose third part shall 
 always be equal to 5, the value of x will depend upon 3 and 5, and, although 
 unknown for a moment, readily becomes known. 
 
 If we should take two quantities, x and y, and say that their product shall 
 always be equal to a fixed number, as 100, x may be 5 and y 20 ; or x 4 and 
 y 25, etc. Here x and y may have any number of relative, but never any abso- 
 lute, values. In such a case they are called variables. The difference between 
 an arbitrary quantity and a variable is, that all arbitrary quantities may be 
 assumed at the same time and independently of each other ; while variables 
 may be assumed only in relation to each other. It is manifest that x and y, iu 
 
10 
 
 ELEMENTS OP ALGEBRA. 
 
 tlie above example, could not both be assumed at the same time, and without 
 regard to their relative values. 
 
 4. Since quantities, as long as they remain purely algebraic, have 
 no fixed numerical value, it is impossible to perform any arithmetical 
 operation upon them ; such as addition, multiplication, etc. All 
 that can be done, therefore, is to indicate such combinations as it 
 may be desired to make. For this purpose certain signs are em- 
 ployed. The sign + indicates addition ; — , subtraction ; x, multi- 
 plication y -r-, division; equality. 
 
 If a represents the number of days one man works, and b the number another 
 works, since a may have any value, and b any value, we manifestly cannot tell 
 liow great their sum must be ; but we can write them so as to show that their 
 sum must be taken, if their values ever become fixed ; thus, a + b (read, a plus b) 
 indicates that these quantities must be added together, when their numerical 
 values are determined upon. In algebraic language they are said to be already 
 added ; and algebra knows nothing of any other kind of addition. All actual 
 additions must be made by the laws of arithmetic. The same may be said of 
 all other operations upon numbers. 
 
 To show that a quantity is to be subtracted from another, we write the quan- 
 tity to be subtracted after that from which it is to be taken with the sign — 
 between ; thus a— b (read, a minus b) shows that b is to be taken from a, or, as 
 is commonly said, b is subtracted from a. 
 
 To multiply quantities algebraically, we simply connect them together by the 
 sign x ; thus, axbxc (read, a multiplied by b, multiplied by c) shows that 
 their product is required. 
 
 It is far more usual, however, to write the quantities together, thus, abc, with- 
 out any sign between. Sometimes dots are used ; thus, a - b ■ c. When num- 
 bers are to be multiplied together the sign x must be used ; thus, 4xo. 
 
 To show that one quantity is to be divided by another, we may write them 
 thus, a-i-b (read, a divided by b). It is much more common, however, to write 
 
 the divisor under the dividend, thus, j . 
 
 b 
 
 The sign = is placed between quantities to show that they are equal ; thus, 
 a=b (read, a equal to b) shows that an equality subsists between these quanti- 
 ties. The quantity to the left of the sign = is called the First Member; that to 
 the right, the Second Member. 
 
 5. When it is desired to show that two or more quantities, already 
 connected by the signs plus or minus, are to be considered as a single 
 quantity, the sign ( ), called a parenthesis, is used; thus, (a + b— c) 
 shows that the several quantities within are not to be separated, and 
 are to be considered as a single quantity so long as the parenthesis 
 remains. 
 
 This sign often takes the form [ ], or j j, called brackets. A bar over 
 
DEFINITIONS AND EXPLANATIONS. 
 
 11 
 
 several quantities, tlius, a + b— c, means tlie same tiling 1 . Also tlie quantities 
 + ax 
 
 may be written thus — b ; here the quantities on the left of the bar are to be 
 + c 
 
 severally multiplied by the quantity, x, on the right. 
 
 6. The sign >, indicates inequality. 
 
 The opening is turned towards the greater quantity ; a> b (read, a greater 
 than b), shows that a is greater than b ; a <b, shows that a is less than b. 
 
 7. The sign oc, shows that quantities connected by it vary to- 
 gether ; thus, a oc b means that a and b increase or decrease to- 
 gether. 
 
 s. The sign is an abbreviation for therefore, hence, or conse- 
 quently ; v means since or because. 
 
 9. An Algebraic Expression is any quantity or combination of 
 quantities written in algebraic language. 
 
 Thus, | is an expression for the quotient of any two quantities ; ( a+b)c is an 
 expression for the product of the sum of two quantities by a third quantity. 
 
 10 . A Factor is any quantity which enters an expression as a 
 multiplier. 
 
 Thus, in the product abe, a, b, and e are each factors ; so also ab, be, and ac 
 are factors of the same expression. Unity enters every expression as a factor. 
 A factor may be made up of several quantities connected by the sign + or — , 
 thus, in ( a + b)c and (a+b)(a—b), the quantities (a+b) and (a—b) we factors. In 
 fractional expressions unity divided by the denominator, or by any factor of the 
 
 denominator, is a factor of the expression; thus, in - , is a factor of in 
 
 11. A Co-efficient (co-factor) is any factor of an expression. 
 
 If a numerical factor enters an expression, that factor is usually written first, 
 and is especially spoken of as the co-efficient. Thus, in the expressions 3 ab, 
 
 fed, and — — ; 3, §, and £ would be called the co-efficients, respectively. 
 
 Literal factors, however, are often called co-efficients ; thus, in ax, - • x , and 
 
 b 
 
 bey ; a, - , and be, would be called co-efficients. 
 
 So, also, numerical and literal factors taken together may be spoken of as 
 
 co-efficients ; thus, in the expressions 3<z.r, , and 3 (a — b)z ; 3 a, — , and 
 
 3(a— 6) would be called the co-efficients of the unknown quantities, respectively. 
 Since multiplication is but abbreviated addition, each factor of an expression 
 
12 
 
 ELEMENTS OF ALGEBRA. 
 
 shows how many times all the other factors enter it additively ; thus, in the ex- 
 pression 3 ab, the 3 shows that db enters the quantity three times by addition, so 
 that it may be written ab+ab+ab. 
 
 So if we had a+a+a+a we could evidently write 4 a instead. r + % + t 
 
 0 0 0 
 
 , n a 3a a a a a , , x 2a 
 
 may he written 3 r or — . r T - must be equal to — . 
 
 b b b b b b b 
 
 If we had (a+b)+(a+b)+ written c times, we could write c(a+b). 
 
 Here c or ( a+b ) may, either of them, he regarded as the co-efficient. 
 
 12 . An expression composed altogether of simple factors is called a 
 monomial. 
 
 By simple factors is meant those which are composed without the aid of the 
 signs + or — . 
 
 ab, — , 3ab, are monomials. A monomial is spoken of as a term, especially 
 c 
 
 when found connected with other quantities hy the sign plus or minus ; thus, 
 
 in the expression 7 ax——^+g, any one of the three monomials which enter it is 
 
 a term. Sometimes complex expressions tied together hy the parenthesis, or 
 
 otherwise, are called terms; thus, in a—(b+c) and — , the expres- 
 
 ’ c b—c 
 
 sions (a + c), and may he called terms. 
 
 13 . A Binomial is an expression composed of two terms ; a Trino- 
 
 mial has three terms. a + b, 
 
 dab— 3c, ^ 
 
 4 b 
 ac ' 
 
 are binomials. 
 
 cl 
 
 ac + fi—S, 
 
 and 
 
 3d 
 
 4« — 5c + , are trinomials 
 
 A Polynomial is 
 
 an expression composed of two or more terms; thus, a + b, 5 ab 
 
 2 
 
 + ~ 
 a 
 
 - + etc., are polynomials. 
 
 d, a + b + c 
 
 14. When the same factor enters an expression more than once, as 
 in 3 aaciabbccc, the expression can be greatly shortened by writing 
 any such factor but once, with a small figure to the right, and a little 
 above, to show the number of times the quantity enters as a factor; 
 thus the above expression would take the form, 3 a i b~c z , read, three a 
 to the fourth power, b to the second power, c to the th ird power. The 
 small figure so used is called au Exponent or Index, because it shows 
 the number of times the quantity, to which it is affixed, enters as a 
 
 factor. 
 
 It is important to distinguish clearly between exponents and num- 
 bers which enter an expression as factors, called co-efficients. In the 
 above example, the 3 written first is a multiplier, whereas the small 
 
DEFINITIONS AND EXPLANATIONS. 
 
 13 
 
 figures do not enter the expression ns quantities at all ; they are 
 merely signs, to show how many times other quantities, namely a, b, 
 and c, enter as factors. 
 
 An exponent, therefore, is any quantity used to show how many 
 times another quantity must be taken as a factor. 
 
 Letters are often used as exponents; thus, a, 5°, {a + bf, a/, 6", (read, a 
 to the x power, etc., a to the one-half power, b to the m divided by n power.) An y 
 symbol of quantity may be used for the same purpose. 
 
 15. Terms are said to be Homogeneous, when they contain the same 
 number of literal factors; thus, 9 a 2 b z , 2 cx i , 25x 3 y 2 , abcxij, are all 
 homogeneous. They are said to be homogeneous with respect to a 
 certain quantity or class of quantities, when there are the same num- 
 ber of such quantities in each ; thus, ax 2 , b 2 cxy, Hiy 2 , are homogene- 
 ous with respect to the unknown quantities which enter them. 
 
 A polynomial or an equation is homogeneous when all of its terms 
 are homogeneous. 
 
 16 . The Reciprocal of any quantity is unity divided by that quan- 
 
 Thus, -, — r, , are the reciprocals of a, a—b, and a 2 b. 
 
 17. The Square Root of a quantity is indicated by placing over it 
 the sign ; when any other root is to be taken, a small number is 
 written to the left and above the sign, to show the degree of the root 
 required; thus, ^/, indicates the third root ; ty, the fourth root ; 
 %/, the with root, and the sign is used thus, Va, *\/a 2 b 3 , Va 2 b + d, 
 
 . The small figure used to show the degree of the root re- 
 
 quired, is commonly called the index. 
 
 This sign grew out of the custom of the older algebraists of writ- 
 ing r, signifying root, before the quantity whose root was required. 
 When used before an expression tied together by a bar, thus, 
 r • a + b, it would take very nearly the form of the present radical 
 sign. 
 
 tity. 
 
 Ill 
 
14 
 
 ELEMENTS OF ALGEBRA. 
 
 SECTION H. 
 
 ALGEBRAIC TERMINOLOGY. 
 
 18 . Algebra has a language of its own, which must be thoroughly 
 mastered before the subject can be at all comprehended. In the fore- 
 going Section we have, so to speak, learned the alphabet. Yfe may 
 now proceed to use it. Let it be borne well in mind that it is all pure 
 conventionality. Any other signs or symbols might have been used; 
 but now that we have agreed to use those already explained, we must 
 adhere to them strictly. 
 
 19. First let us translate from English into Algebraic symbols. 
 
 1. Add together any tivo quantities. 
 
 Explanation . — Since they must be any two quantities, we cannot take partic- 
 ular numbers ; and since numbers from their nature are always definite, we 
 cannot take numbers at all. a and b are two such quantities, and since their 
 values must remain undetermined, we can only write, a + b. Any other letters 
 would have done as well. 
 
 2. Write the difference of any two quantities. Ans. a — b. 
 
 3. Write the product of any two quantities. Ans. ab. 
 
 4. Write the quotient of any two quantities. Ans. 
 
 5. Write the product of the sum and difference of two quantities. 
 
 Ans. ( a + b ) ( a—b ). 
 
 G. Write the quotient of the sum and difference. Ans. 
 
 Remark. — When a quantity has 2 for an exponent it is often read square, and 
 though, perhaps not strictly correct, it is convenient to retain the custom ; 
 thus a 2 may be read a square. When a polynomial has 2 for an exponent, as 
 (a + b) 2 , it must be read a plus b squarED. In like manner a 3 may be read a 
 cube; (a + b) 3 , (a + b) cub ed . 
 
 7. Write the sum of the squares of two quantities. 
 
 8. The square of the sum. 
 
 9. The square of the difference. 
 
 10. The square of the quotient. 
 
 Ans. a 2 + b 2 . 
 Ans. (a + b) 2 . 
 Ans. (a—b) 2 . 
 
 11. The sum of the square roots. 
 
 12. The square root of the sum. 
 
 Ans. \Za + Vb. 
 Ans. y'a + b. 
 
 13. The product of the square roots. 
 
 Ans. \> r a x or \/ a. Vb. 
 
ALGEBRAIC TERMINOLOGY^ 
 
 14. The square root of the product. 
 
 15. The quotient of the square roots. 
 
 16. The square root of the quotient. 
 
 17. The square root of the product of the squares. 
 
 Ans. V a 2 b 2 . 
 
 18. The square root of the product of the sum and difference. 
 
 Ans. V(a + b) ( a—b .) 
 
 19. The cube root of the product of the cubes. Ans . V a 3 b 3 . 
 
 15 
 
 Ans. A Jab. 
 
 A A fa 
 
 Ans. — — . 
 
 Vb 
 
 Ans. 
 
 20. Translate the following expressions into English : 
 1. a—b, a + b, a 2 +b 2 , ab, a 2 b 2 , \. 
 
 2 . 
 
 3. 
 
 4. 
 
 5. 
 
 a 2 —b 2 , (a—b) 2 , a 2 +b 2 , (a + b) 2 , (a + b) (a—b). 
 a 2 a + b a—b a+b (a + b) 2 a — b 
 b 2 ' a ’ b ’ a—b’’ a 2 +b’ (a—b) 2 ’ 
 
 Va + Vb, 's/a , Vb, Vab, 
 
 Vb 
 
 V a—b, V a 2 +b 2 • 
 
 V (a + b)(a—b). 
 
 6 . 
 
 V 7 a 2 +b 2 , Va 2 // 
 
 a 2 /~a/ 
 
 T ’ y lA’ 
 
 V(a + b) 2 , Va 2 -b 2 , Va 2 -b 2 , 
 
 21. Translate the following into algebraic language : 
 
 1. The square of the sum of two quantities is equal to the square 
 
 of the first, plus twice the product of the first by the second, plus the 
 square of the second. Ans. (a + b) 2 = a 2 + 2ab + b 2 . 
 
 2. The square of the difference of two quantities is equal to the 
 
 squai'e of the first, minus twice the product of the first by the second, 
 plus the square of the second. Ans. (a—b) 2 =a 2 —2ab + b 2 . 
 
 3. The product of the sum and difference of two quantities is 
 equal to the difference of their squares. Ans. (a + b) (a—b) = a 2 — b 2 . 
 
16 
 
 ELEMENTS OF ALGEBRA. 
 
 4. The quotient of the difference of the squares of two quantities 
 by the difference of the quantities, is equal to the sum of the quanti- 
 
 ties. 
 
 Ans. 
 
 « 2 — b 2 
 
 — Oj -f b . 
 
 5. The quotient of the difference of the squares of two quantities 
 by the sum of the quantities is equal to the difference of the 
 quantities. a 2 — b 2 
 
 a +b 
 
 - — a—b 
 
 G. The product of the sum of two quantities by the first is equal 
 to the square of the first, plus the product of the first by the second. 
 
 Ans. ( a + b ) a = a 2 + ab. 
 
 Let the algebraic expressions in this article be translated back into 
 English. 
 
 22 . Translate the following expressions into English : 
 
 1. \fa x Vb = V ab. 
 
 Ans. The product of the square roots of two quantities is equal to the square 
 root of the product. 
 
 3. (Va+ Vb){Va — Vb) = a — b . 
 
 4. (Va + Vb)~ = a + 2Vab +b. 
 
 5. {Vet— Vb)~ = a — 2V~ctb + b. 
 
 6 . V ) Va=a 3 , V a ~ ~ 
 
 l » -i 
 
 L y a — a n , V am = a ”> V a ~ m — — . 
 
 n m * 
 
 9. (« + 3)£ = V a Ab, [(a + 5) 2 ]i = a + b. 
 
 10. (ai+bi)(ai—bi) = a—b. 
 
 23. Express in algebraic symbols the following : 
 
 1. Divide a certain quantity into three equal parts. 
 
 a a a 
 
 o 5 q J Q * 
 
 o o o 
 
ALGEBRAIC TERMINOLOGY. 17 
 
 2. Divide a given quantity into two parts, one of winch shall be 
 
 three times the other. , a 3 a 
 
 Jins. — . 
 
 4 4 
 
 3. What two numbers are those which differ from each other by a ? 
 
 Ans. x and x—a; or, x and x + a. 
 
 4. The a part of a quantity added to its b part is equal to m. 
 
 Ans. — + r = m. 
 a b 
 
 5. Three times a certain number minus that number is equal to 
 
 f the number less a. Ans. 3x—x — f- x—a. 
 
 6. The sum of two numbers is to the difference of those numbers 
 
 as m is to n. Ans. a + b : a — b :: m : n. 
 
 7. The ratio of the square of the sum of two numbers is to the 
 sum of the squares of those numbers as m is to n. 
 
 Ans. 
 
 (i a + b) 2 m 
 
 24. If at any time particular values are given to the quantities in 
 an expression, the indicated operations may then be performed. A 
 result so found is called the numerical value of the expression. For 
 each new set of values for the several quantities which enter an ex- 
 pression, there will be a new result as the numerical value of such 
 expression. 
 
 The numerical value of the expression ab — ~ , when a = 4, and b = 2, gives 
 
 4 x 2 — 1 = 6, which is the value of the expression in this case. 
 
 If a = 7, and 6 = 5, we should have 7 x 5 — \ = i §- a . , 
 
 Find the numerical values of the following, letting a = 1, b = 2, 
 c — 3, and d — 4 : 
 
 1 . 
 
 cd ab 
 
 3 a> -j’j + 
 
 a— b 
 
 (a + b) 2 2(Z 
 cd a * 
 
 2. (a + b)(a + b), *t*+*^±, [( a + b)(c-a)]d . 
 
 \/ b + a Vd-C abed V c — b 
 
 3. Vd [rt(5+c)]V«, x ^Ta ’ Vb II a X abcd ' 
 
 55—3 ^ 1 (a+b—2)(c — a + d) 
 
 4 6 * — a b ’ c\5 d)\c a) / a c\ la a\ 
 
 \b + d) \b~~~c) 
 
 B 
 
18 
 
 ELEMENTS OP ALGEBRA. 
 
 SECTION III. 
 
 TREATMENT OF THE + AND - SIGNS. 
 
 25. All elementary operations of Algebra fall under one of the 
 two following heads, namely : 
 
 I. Indications. 
 
 II. Transformations. 
 
 As we have seen in the previous section, Indications are the opera- 
 tions of expressing in algebraic language whatever may be stated in 
 spoken language. 
 
 Transformations comprise all changes which may be lawfully per- 
 formed upon expressions after they are written in algebraic symbols. 
 
 The simplest transformations are those made upon quantities 
 connected by the signs phis or minus. 
 
 Thus, a + a + a-\-a, mav evidentlv be written 4 a. This is a change 
 of form without a change in the numerical value. 
 
 a + a—a + a = 2a,\s another such change. 
 
 Algebraists have commonly called these transformations addition; 
 but manifestly the algebraic sum is altogether accomplished in the 
 operation of Indication. The quantities are no more added after 
 the form is changed than when they are simply connected together 
 by their proper signs. 
 
 26. The Nature of the Signs + and — . 
 
 Let us now try to understand the nature of the signs + and — . 
 
 Their first and most obvious use is to connect quantities together 
 so as to show that they are to be added or subtracted. But, then, 
 certain quantities may, from their nature, be additive, while others 
 are subtractive, before any combination takes place. For example, 
 a man, casting up his accounts, would consider all amounts due him 
 to be augmentative, while all amounts which he owed would be 
 diminutive of his capital. The first class would take the sign + ; 
 the second the sign — . It is, however, a mere matter of agreement 
 as to which shall be called plus and which minus ; hut they are al- 
 ways contrary the one to the other, so that the establishment of 
 ' either determines the other. Quantities affected with the sign + are 
 called Positive ; those with the sign — are called Negative. 
 
 To illustrate this, suppose we take a right line, ab, and agree to 
 reckon all distances 
 
 a 
 
 o 
 
 b 
 
TREATMENT OF THE + AND — SIGNS. 
 
 l'J 
 
 to the right of the point o as positive ; then, of necessity, those to 
 the left would be negative. If we had determined to reckon dis- 
 tances to the left of o positive, those to the right would have been 
 negative, and so in general. 
 
 The sign of a quantity must be known before we can use it in com- 
 bination with other quantities. Plus is always understood where no 
 sign is written. A quantity with the signs + and — , thus ±a, is 
 to be used first positively and then negatively. 
 
 27. To find the algebraic sum of several quantities. 
 
 When quantities are given with their respective signs, to find their 
 aggregate, or, as it is called, their algebraic sum, we have this simple 
 principle : 
 
 Write the several quantities one after the other in any order, con- 
 nected by their respective signs. 
 
 Examples. 
 
 1. Add, dab, 4 a 2 b, 2 ab, —ab, and —5 a 2 b. 
 
 A ns. 3ab + 4jrb + 2ab—ab—5a 2 b. 
 
 J __ Q I 
 
 2. Add, —la, —j- , —be, cAtP, and 'fa. 
 
 Ans. — — j— — 1 
 d 
 
 %a—bc + a ~b\ + V a. 
 
 Remark. — It is usual to place a positive quantity at the beginning. 
 
 3. Add, + 24, -\fWUi, -iy, and (a + b)(a-b). 
 
 Ans. + 24— \/67a + j^- + (« + £)(«— &). 
 
 , , , , a 2a 5a , , /a\ m , /a\ m+n 
 
 4 Add , — V ’ and U) ' 
 
 , 2a a 5a , , /a\ m [ a\ m+n 
 
 Am - TS- J + l + i + {b) + (s) 
 
 5. Add, 7 a 2 — 2 aW, 3« 2 + 4 ab-, and 2 a—a m + 5 a T 
 
 Ans. 7a 2 — 2ab 2 + 3 a 2 + 4 crfF + 2 a—a m + 5 oh 
 
 28. To reduce a polynomial to the least number of terms. 
 
 Terms are said to be like or similar when they contain the same 
 letters, and the several letters have the same exponents, respectively. 
 2 x^b m c 3 and 5aAb m <? are like terms. The algebraic sum of several 
 
20 
 
 ELEMENTS OF ALGEBBA. 
 
 quantities can often be much shortened by gathering into one all 
 the terms which are alike. 
 
 For example, 3«# + 4« 2 J + 2aZ>— ab— 5a 2 b may be written 4«Z> — 
 a 2 b ; for, 3 ab and 2 ab, both being +, give 5 ab; but — ab (that is, one 
 ab to be taken away) leaves 4 ab', so 4 a~b less 5a 2 b, leaves one a 2 b 
 minus. 
 
 So in genera], to reduce a polynomial to the smallest number of 
 terms we may say: 
 
 Gather like terms together, giving the results their proper signs, 
 respectively. 
 
 Great care must be taken to combine only terms whose literal parts are en- 
 tirely the same, and to add and subtract the numerical parts according to their 
 signs. 
 
 Examples. 
 
 Reduce the following polynomials to the smallest number of 
 terms : 
 
 1. 2«— 4« 2 J 2 +3« + 5a 2 5 2 — 4m Ans. a + a-b 2 . 
 
 2. 3b%c— 2bic + a m — b m+ '+5a m — 2b m+ \ Ans. bic + Ga m —3b m+ \ 
 
 n o, cd n cd „ . , _ 
 
 3. ■= + g + 2- - 3— + g + 24 -12. 
 
 b a J b a 
 
 . 3 a j:d 
 
 Ans. — 2 1- 12. 
 
 b a 
 
 4. 3 a 2 + 5 a m b n — a' 2 d + Gai b + 2a' 2 d - 2 a m b n — aib. 
 
 A ns. 4 a 2 c * + 3a m b n + oaib. 
 
 5. a 2 +2ab-\-b 2 +a 2 — 2ab + b 2 = 2a 2 + 2b 2 . 
 
 6. a + 2\/ab + b + a — 2\Zab + b—2a + 2b. 
 
 7. V« + 2 \/ r ab-{- Vb+ V«— 2\fab + \Gb=2\fa + 2*Jb. 
 
 8. 2(a+5)i-3(a-S)i + 5(a+5)i-2(ffl-5)i 
 
 Ans. 7(a + &)^— 5(a— b)i. 
 
 9 ^ a + h 
 
 ~7c 
 
 /a — cV A /a + b /a — cV a m ~ x 
 
 \bd) 7/T + \bd) ' b- 1 ' 
 
 V c 
 
 . 2 VaA-b a 1 ” -1 
 
 Ans. 1- y— r. 
 
 V c b 
 
 10. 25a 2 b* c + 12a m b’ 1 — 5a 2 b 3 c + a m b-— 3a 2 b 3 c + 5a m b’. 
 
 Ans. 17« 2 Z> 3 c+18«’”Z> 3 . 
 
TREATMENT OP THE + AND — SIGNS. 
 
 21 
 
 29. Subtraction of quantities. 
 
 To subtract a positive quantity, as + b from a, we should, strictly, 
 write a — ( + £), using the parenthesis to prevent the confusion of 
 signs; but since + b is but another way of writing b itself, we may 
 drop the + sign, and the parenthesis with it, and write a — b. When 
 "the quantity to be subtracted is already negative, as —b, the indica- 
 tion is made in the same way, a— (—6); but we cannot, as before, 
 drop the — sign of the quantity, since, in that ease, we should have 
 for a result, a—b as before, and should thus have taken away, not 
 —b, but + b. To take away a negative quantity is really to augment 
 the quantity from which it is said to be taken ; thus, (25—5) is 20; 
 now from (25—5) remove the —5, that is, strike it out, and, of course, 
 the 25 will be left. By taking away —5 we have really added 5 to 
 the quantity (25—5), from which it was taken; thus, 25 — 5 — (—5) 
 =25—5 + 5=25, or in general, a — b — {—b)=a — b + b=a. 
 
 A man who is insolvent owes, say, a dollars. It would be the same 
 thing to say, he has —a dollars due him. To subtract or take away 
 any part of his debts, is really to augment his fortune by just that 
 amount ; and so in general, 
 
 To subtract a negative quantity, is to aclcl the numerical value of 
 the quantity ; or practically, 
 
 To subtract any quantity, write it after the quantity from which 
 it is to be taken with its sign changed. 
 
 Examples. 
 
 1. From a take b, ab, —cl, 3b, —2ab, +cl, and then simplify the 
 resulting expression. 
 
 Ans. a—b—ab + cl—3b + 2ab—cl = a—Ab + ab. 
 
 2. From ci + b take b, —c 2 , 3c, —2b, and simplify. 
 
 A ns. a + b—b + c 2 —3c + 2b — a + 2b + c 2 —3c. 
 
 3. From 3*^ —cd take ^ —cel and . 
 
 b b o 
 
 A ns. 3 T -cd- 
 o 
 
 a 7 a m _ a a” 
 'l +ci ~W~ 2 i~¥ 
 
 4. From a + yfab + b take a — 'fob and —b. 
 
 A ns. a A V ab + b—a+ V ab + b = 2 *f~ab + 2b. 
 
 11 A 1 1 1 
 
 5. From a 2 — b? take 2ci 2 , —3 b 2 , —5 a 2 , and b 2 . 
 
 A ns. a? — b?—2a 2 -f-3bi + 5ai — b~=-la^ +bi. 
 
22 
 
 ELEMENTS OF ALGEBRA. 
 
 6. Prom 2 V a + 3Vb take — Va, 2 Vb, i\ fa, and — Vb. 
 
 A ns. 2 Va + 3V b -\- V a — 2 Vb — i Va + Vb — | V a + 2 Vb. 
 
 7. From take -3- ^ 4 SZ+Q. and -J- a ~ b 
 
 Am. ^ + 3- “ 
 
 c c c 
 
 _i a ~b 
 *' c -*“■ 
 
 8. From V ab— c take — 2V~ab, 5 c, ^ ^ and —6c. 
 
 M«s. V ab— c + 2\/ab — 5c— ^-^ + 6c = \V~ab- 
 
 9. From v^— take 9y / a, — 2aVb, and — Sy«. 
 
 ^4«s. -\/a— 2aV~b— 9Vff + 2-v / £ + 8v / ft = 0. 
 
 30. Subtraction of polynomials. 
 
 To indicate the subtraction of any polynomial, we have but to 
 write such polynomial within a parenthesis, and connect it by the 
 — sign with the quantity from which it is to be taken ; thus, to take 
 a + b—c from cl, we have cl — (« + b— c). 
 
 Now the minus sign here means that we are first to get the alge- 
 braic sum of all the quantities within the parenthesis, and then take 
 this sum from d. Manifestly the same thing can be accomplished by 
 taking, first a, and then b, and then — c, away from cl; thus, cl— a— 
 b + c. The parenthesis has disappeared, and the several quantities 
 have changed their signs. We may then say, that, 
 
 To subtract a polynomial from any quantity, we have but to write 
 the severed terms in succession after the quantity from which it is to 
 be taken with their signs changed. 
 
 Examples. 
 
 1. From 3a — 2b take a+Aib and simplify. 
 
 Ans. 3a — 2b — a—Tb=2a — 6b. 
 
 2. From a n — a take a m + a n — a 2 . 
 
 Ans. a n —a—a m —a n + a-=a-—a m —a. 
 
 m m 
 
 3 . From a » 5 -5 - (-1 take — 3««£ -2 + l . 
 
 Ans. a » b~* + 1 + 3anb~' J — 1 = du f b~\ 
 
TREATMENT OF THE + AND — SIGNS. 
 
 23 
 
 4. From 8 a p- V take oof l c q + a » c' !+ - p_1 — j/«n. 
 
 1 c q oa p \f—a *c ?fp =. 5a p ~ 1 c q — a~fic q+7l ~ 1 + ^/~a^. 
 
 5. Simplify a~—b m —(2a'~—b m — 1). 
 
 Ans. a~ — b m — 2a^ + b m + l = l—df i . 
 
 6. Simplify a 2 — 7a 2 b 3 c— J— (a 2 — 7ci 2 b 3 c + i). 
 
 Ans. a 2 — 7a 2 b 3 c—\—a 2 + 7 a 2 b 3 c—\— —1. 
 
 7. Simplify x 2 — x — 1 — (— x 2 + .T + 1). 
 
 Ans. x 2 —x—l + x 2 —x— 1 =2x 2 —2x—2. 
 
 Remark. — It will be observed that in removing a parenthesis from a poly- 
 nomial when it has the minus sign before it, we change the signs of all the 
 terms within. When the sign before the parenthesis is plus, the parenthesis 
 may be removed without any other change. 
 
 Eemove the parentheses and simplify the following: 
 
 8. a—(—b + c—a)=a + b—c + a=2a + b—c. 
 
 9. l + (a—b—l)=l+a—b—l=a—b. 
 
 10. a — (—ct)=2a; a — (—l) = a + l. 
 
 11. 3 a 2 b- ~ -(-3 a 2 b+~]=6a 2 b~ j. 
 
 12. or — 1 + ( — cir + 1) = 0. 
 
 13. -* + l-(-* + l + V3) = -V& 
 
 14. a m ~ l —bff—(a m -' — b~k + l) = — 1. 
 
 15. 1— %a+(l— 1-«)=2— a. 
 
 16. a—(—a) — (—a) — (2a)=a. 
 
 Remark. — We may put a parenthesis upon the algebraic sum of several quan- 
 tities without any other change, provided that the positive sign, stands before the 
 parenthesis ; but if the negative sign occupies that place, the signs of all the 
 terms within must he changed. 
 
 Enclose the following in parentheses, first with a + sign, and 
 then with a — sign. 
 
 1. a— b + c=(a— b + c) 
 
 = —(—a + b—c). 
 
24 
 
 ELEMENTS OF ALGEBRA. 
 
 3. a=(a) ; 1— rt=(l— «); a m — ( a m ) 
 
 = — (—1), l—a= — ( — l + a), = — (— a m ). 
 
 Remark . — Where there are several parentheses or brackets, one within another, 
 in removing them begin by removing the outer one, and then the next, and so 
 on. We may, however, begin with the inner one. 
 
 Kemove the brackets from the following : 
 
 1. — [a— {a + b — 1)] = — a + {a + b— 1)=£— 1. 
 
 2. a— [ — { — {a + 5)}] —a + { — (a + b)} —a — (a + b) =a—a—b= — b 
 
 3. 1 — [«—(« + b) + {a—b)+b\ — 1 —a + (a + b) — (a—b) — b = l— a 
 + « + b— a + b— b=l— a + b. 
 
 The management of fractional quantities in Algebra and Arith- 
 metic is entirely the same. 
 
 Now, since tve can separate a fraction, such as ]U t° as many 
 partial fractions as tve please to break the numerator into parts, thus 
 ^ + 1 y + t 5 t , we may do the same thing with an algebraic fraction. 
 
 Now, if such a fraction as this, having the algebraic sum of two 
 or more quantities for its numerator, has the — sign before it, and it 
 should be thus broken into parts, the removal of the division bar 
 will act altogether like the removal of a parenthesis. The place of 
 the bar must, thei’efore, be supplied by a parenthesis ; or the signs 
 of the partial fractions must all be changed. 
 
 Examples. 
 
 4. l-{ + [-(-l)]}=0. 
 
 5. -(! + (— 1)) = — 2. 
 
 For example, — , may oe written, ^ ^ ^ 
 
 Examples. 
 
 Separate the following into partial fractions : 
 
 a b c 
 
 -7 — i 
 
 1 . 
 
 d 
 
TREATMENT OF THE + AND — SIGNS. 
 
 25 
 
 2 . 
 
 a + b—c 
 
 I 
 
 _a b 
 ~d + d~ 
 
 c 
 
 d' 
 
 2a 2 b—c m _ /2 a 2 b c” \_ 2a 2 5 c m 
 
 3a"c 2 \3a n c 2 3a n c 2 ) 3a n c 2 ^ 3a n c 2 ' 
 
 4. 
 
 —ai+b 
 a + b 
 
 i 
 
 aJ 
 
 a + b 
 
 + 
 
 b 
 
 a + b ’ 
 
 Remark . — Tlie sign — standing before the division bar shows that the whole 
 fraction is to be subtracted ; or, in other words, that, after all the operations 
 indicated are performed, the result is to be subtracted ; but when the sign 
 stands before the first term of the numerator, as in the last example, it affects 
 only that term. 
 
 5. 
 
 — 5 + 4a— 2? 
 
 -5_ 4a_ _2+ 
 
 + 5“ 5” ' 
 
 6 . 
 
 —5 + 4a— 2 3 
 5“ 
 
 (_5_ 4a _2 2 \_ 5 
 \ 5" + 5 m 5 m J~5‘ 
 
 4 a 2^ 
 
 5" + 5" 
 
 7. 
 
 a + b 
 a—b 
 
 c\_ fa b 
 d)~ \a—b a—b 
 
 c\_ a b c 
 d) a—b a—b d‘ 
 
 8 . 
 
 a + b 
 (a + b) 2 
 
 a—b 
 
 (a + b ) m - J — 
 
 r a b / a b \1_ 
 
 L (a + b) 2 + (a + b) 2 \(a + b) m (a + b) m ) 
 
 a b a b 
 
 (a + b) 2 (a + b) 2 + (a + b) m (a + b) m ’ 
 
 31. Meaning- of the terms SUM and DIFFERENCE. 
 
 The terms Sum and Difference must be understood in Algebra in 
 their largest sense. To add does not necessarily mean to augment, 
 nor does subtraction always mean a diminution; thus, —b added to 
 + b gives 0 ; while —b subtracted from +b gives 2b. 
 
 By the sum we are to understand the result obtained from con- 
 necting the quantities by their own signs ; a difference results when 
 certain terms are connected with their signs changed. 
 
26 
 
 ELEMENTS OF ALGEBRA. 
 
 SECTION IV. 
 
 MONOMIALS— EXPONENTS AND THE SIGNS x AND h-. 
 
 32. An Exponent is, by definition, a sign which shows how many 
 times the quantity to which it is affixed is to he reckoned as a 
 factor. 
 
 Exponents may be entire or fractional, positive or negative. 
 
 33. Entire and Positive Exponents. 
 
 5 3 is the same as 5 x 5 x 5, or 125 : a 3 is but another way of writing 
 a • a • «; a 2 b 3 — a-a-b-b-b; (a + b) 2 = (a + b)(a + b ) ; {\fa) 3 = 
 V a ■ ''fa • fcr, a m —a • a • a • a written m times. 
 
 Quantities united by the sign x or — are simply made co-factors; 
 thus, 3 a 2 b x5ab 3 shows that the continued product of all the factors 
 is required. The sign x may, however, be dropped altogether and 
 the quantities be written together, where no confusion is like to re- 
 sult; thus, 3a 2 boab 3 ; but in such a case as this it is usual to retain 
 the x or use the • 
 
 Now the order in which factors are taken makes no difference in 
 the product; so that we may write the above expression thus, 
 3 x 5 a 2 abb 3 . It will be observed that the sign x must now be used 
 between the numerals to prevent mistake. 
 
 But here we have the indicated multiplication of two factors which 
 can be actually performed ; so that 15 can be written in the place of 
 3x5. 
 
 We see, further, that a enters three times as a factor, so that a 2 a 
 may be written a 3 . In like manner bb 3 is equal to b 4 . It is thus 
 manifest that 3 a 2 b x 5ab 3 = l5a 3 b i . 
 
 We have simply multiplied the numerical factors together and 
 written each letter once with an exponent equal to the sum of the 
 several exponents of that letter. Transformations of this kind are 
 called multiplication. 
 
 \<le may say, then, in general, that : 
 
 To multiply monomials together, multiply the numerical factors 
 together, and after this product write the several letters ivliich enter 
 the monomials, giving to each an exponent equal to the sum of all the 
 exponents of that letter in the several terms. 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -P. 
 
 27 
 
 Examples. 
 
 1. Multiply 4 a 2 bc by 5 ale 2 . 
 
 2. Multiply a : d 2 f by 2 atl 3 f 2 . 
 
 3. Multiply x m y by x m y. 
 
 4. Multiply x m ~ l y by xy m+ \ 
 
 5. Multiply 5 ax m by 3 a n x n . 
 
 6. Multiply 25 a r lfc by 3 a 3 b a c. 
 
 7. Multiply l~a 2 b 3 by \ab 2 . 
 
 8. Multiply | a x y n by f a v yz*. 
 
 9. Multiply a m b n c p by abc. 
 
 10. Multiply %a m+n b by -§ a m+n b. 
 
 11. Multiply x m+n b n+l by l z x m ~ n b ] ~ n . 
 
 12. Multiply x m ~ n+l b r ~ q by x n ~ m ~' i b q ~ p . 
 
 Ans. 20 a 3 b 2 c 3 
 Ans. 2«°c/ s / 3 
 Ans. x^y 2 
 Ans. x m y m+ 
 Ans. 15 a n+ ' i x n+n 
 Ans. 75a p+3 b a+2 c !! 
 
 Ans. %a 3 b 5 
 Ans. %a x+v y n+l z 2 
 Ans. a m+1 b r,+i c F+l 
 Ans. -i ^a im+Sn b 3 
 Ans. \x lm V 1 ' 
 Ans. x°b° 
 
 Remark . — We shall find that the same laws govern the management of 
 fractional and negative exponents as when they are positive and entire, so that 
 we may employ them in our examples at once. In finding the sum of fractional 
 exponents, let the student remember to apply the rules for the addition or sub- 
 traction of fractions in arithmetic. 
 
 , _ l 1 
 
 13. a-i x a# = a. 
 
 14. abc x aibzd=aibici. 
 
 15. ixzy~3 x \x m y a —\:C^y. 
 
 11 5 3 
 
 16. 5x 3 y 2 z x axy~azn = 5ax i y'Zz2. 
 
 17. ar-bc% xa 3 b~ i d=ab~ 1 c. 
 
 m p m p 
 
 18. 25 a m b^ x 5«"&?==125a m+M Z>» ?. 
 
 19. a m b n c p x cr m b~ n c=a°b°c p+ '. 
 
 nn aAb c x~ l a F b a c aA +p b a+c cx~ x 
 
 20. — - — x— = yw 
 
 34. The law of signs in Multiplication and Division. 
 
 So far no mention has been made of tlie signs of the terms multi- 
 
28 
 
 ELEMENTS OP ALGEBRA. 
 
 plied together. Let us now inquire what effect these signs will have 
 upon the product. 
 
 In the multiplication of two quantities w T e simply repeat one of 
 them additively as many times as there are units in the other. Let 
 us, then, multiply two quantities, a and b, together, and first let them 
 both be positive. + a must be added to itself until there are alto- 
 gether as many times a as there are units in b, so that we shall have 
 + ab. 
 
 Thus we see that, 
 
 The •product of two positive quantities is positive. 
 
 Now, let a be negative and b be positive. — a must now be added 
 to — a until — a has been repeated b times. But — a and — a gives 
 — 2a; —a and — a and — a gives — 3 a; and when taken b times 
 we must have — ba. 
 
 If a had been positive and b negative, — b would have been taken 
 additively a times, and the result would have been again — ab. Thus 
 we see that, 
 
 The product of a positive and negative quantity is always 
 negative. 
 
 Now let us take a and b both negative. We shall have (— a ) 
 (— b). But (—a) = — (+«); and so we may write — (+ a) (— b) 
 = — (— ab) — + ab. Thus we see that, 
 
 The product of a negative quantity by a negative quantity is 
 positive. 
 
 Since the product of the divisor and quotient must always produce 
 the dividend, the same principles hold good in division. 
 
 The laAV of signs may then be summed up as follows: 
 
 In multiplication and division like signs give plus, and unlike 
 minus. 
 
 Examples. 
 
 1. Multiply 5 a~b by — 
 
 2. Multiply — l-a m b n by a m b. 
 
 3. Multiply — \a m b n by — a m b. 
 
 4. Multiply \a m b n by — 5 a m b. 
 
 5. Multiply Urb" by ■§ a m b. 
 
 6. Multiply —9 a m + l b~ 2 by 5 ab m c. 
 
 5 
 
 Ans. — 35rtV& 3 . 
 
 A ns. — ia° m b" + l . 
 Ans. $a*"b n + 1 . 
 
 Ans. — | a-”b' H . 
 Ans. \ar m b n + x . 
 
 Ans. — 45«”' + 5 & m ~c. 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND = . 
 
 29 
 
 IX- _1 —1 - 
 
 7. Multiply a&bsc* by — 3a ~b 3 c ». 
 
 Ans. — 3 a°b°c°. 
 
 8. Multiply— a°b~ l c by— 1. 
 
 9. Multiply ab°c~ l by a~"bc. 
 
 10. Multiply ab°c~ x by ab~°c. 
 
 Ans. a°b~'c. 
 Atis. a}~"bc°. 
 
 Ans. a 2 . 
 
 Remark . — When a quantity has an exponent zero, such exponent shows that 
 the quantity does not enter as a factor at all. It may therefore be omitted alto- 
 gether. Remember, however, that the quantity itself is not zero. Any quantity 
 to the zero power is thus 1 ; but we shall see further upon this point. 
 
 11. abxabx — ab = —a z b z . 
 
 i i Li 
 
 12. —ax% x —ax 3 x ax- = a z x e . 
 
 It has already been said that an algebraic fraction does not differ 
 from a numerical fraction in principle. 
 
 In the treatment, therefore, of algebraic fractions, we have but to 
 apply the laws of arithmetic to those already established for the 
 management of algebraic symbols. 
 
 15. 10 a m - n b~ p x — a° = -10 a m ~ n b~ p . 
 
 16. «(— «)(—«)«(— a) = — « 5 . 
 
 17. ( — 1)(— 1)(-1)(— «°) = 1. 
 
 18. (— 0(- 6°)(— o°) = - 1- 
 
 is. (hmri})°v~a = vs. 
 
 35. Operation upon Fractions. 
 
 Examples. 
 
 1. Multiply | by |. 
 
 2. Multiply — | by ~. 
 
 3. Multiply -|by - C -. 
 
 4. Multiply — | by ^ . 
 
30 
 
 ELEMENTS OP ALGEBRA. 
 
 5. Multiply — — bv — . 
 
 — b " c 
 
 6. Multiply - p by . 
 
 7. Multiply^ by ^ . 
 
 ^\ r r u ^ 7/P 
 
 8. Multiply by -1. 
 
 _ « — « « — a « 4 
 
 9. T X - 7 — X r X t = 7 — 
 
 
 abc 
 
 — 1 
 
 10. 
 
 
 X -7— = 1. 
 
 
 ■^1 
 
 abc 
 
 
 a n 
 
 lr m 
 
 11. 
 
 y- X 
 
 — x — 2 —- 
 
 
 b m 
 
 a 
 
 
 a° 
 
 b a 
 
 12, 
 
 — X 
 
 77 X 7 = 1, 
 
 
 a 
 
 b° b 
 
 
 -A 
 
 —be 
 
 a 
 
 3 
 
 2 
 
 ,4 MS. 
 
 cCV 
 
 —a 2 b 3 
 A ns. 
 
 a 2 b ' 
 15 2?* 
 
 w 
 
 A ns. 
 
 OX” 'y p 
 
 Sab-' ' 
 
 36. The Powers of Quantities. 
 
 The Poiver of a quantity is the product obtained by using the 
 quantity a certain number of times as a factor ; thus, the second 
 power of 3 is 9, the third power of 2 is 8. a 2 is the second power of 
 a ; a 3 , the third power ; a m , the mth power of a. 
 
 To form the poiver of any quantity multiply the quantity by itself 
 as many times as there are units in the exponent of the poiver less 
 one. 
 
 A practical rule for finding the power of a monomial is : 
 
 Raise the numerical factor, if there be one, to the required poiver 
 and multiply the exponent of each letter by the exponent of the power. 
 
 This process is called Involution. 
 
 Examples. 
 
 1. (a 2 ) 2 , (2a) 5 , (—3 z&) 2 , (-£.^) 3 , (-cr 1 ) 5 , (a m ) n , {a»Y. 
 
 3 P m 
 
 Ans . a 4 . 32a 5 , 9 x, — — a“ 5 , a*”, a » . 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -L. 
 
 31 
 
 Remark . — It will be observed that tlie sign of an ecen power must always he 
 positive ; and 
 
 Tliat the sign of an odd power must always be the same as the sign of the 
 term itself. 
 
 Ans. 
 
 144 aW 
 49a 3 
 
 iri y 3 
 
 Cl ® 
 
 V* ’ 
 
 bT 
 
 a 
 
 a 
 
 a » 
 
 Ans. 1, v— tjtt-j- — • — , j 
 ’ b-i’b 2m ’ A\ 
 
 a n b z b 
 
 ar aA 2 7a~ 3 b 3 
 
 4a 4 b 
 
 37. The Boots of Quantities. 
 
 The operation of extracting the root of a monomial is just the re- 
 verse of that of raising it to a power. Practically, 
 
 Extract the required root of the numerical factor, if there beany, 
 and divide the exponents of the titered factors by the index of the root. 
 
 This process is called Evolution. 
 
 If a quantity be raised to an even power, the sign of the power 
 must be plus, whether the quantity (that is the root) is positive or 
 negative; thus, 
 
 a 3 — a x a or —ax —a, 
 
 a 4 =a x ax ax a or —a x —a x —a x —a. 
 
 Whence it follows, that when the index of the required root is 
 even, the root may have either the plus or minus sign ; it is there- 
 fore given both ; thus, 
 
32 
 
 ELEMENTS OF ALGEBRA. 
 
 Hence we may say that, 
 
 The even root of a quantity has the double sign ±. 
 
 When the root is odd, the sign of the power must always he the 
 same as that of the root; thus, 
 
 —ax—ax—a=—a 3 ; ( — a ) 5 = —a 5 ; (— a ) 2 "^ 1 — — a 2m+r 
 
 axaxaxa — +a 3 ; { + a) 5 —+a & -, ( + «) 2 " 1+1 = +a 2m+1 
 
 It follows, therefore, that, 
 
 An odd root always has the same sign as that of the quantity 
 itself. 
 
 Examples. 
 
 Extract the roots of the following, as indicated: 
 
 1. VaS Via 2 , V^a 2 , V 9 a 4 6 2 > VSa 3 , V — 8a 3 , V\a 6 b 3 . 
 
 Ans. ±a 2 , ± 2a, ± \a, ±3a 2 b, 2 a, —2a, l-a 2 b. 
 
 Ans - V” ±8 “‘ s "’ a ’ a>3 - 
 
 d C a 3 c 1 d, fiabdh, f 36« ^ A 
 
 ^ 3 jj 1112 rn 0 ^ 
 
 Ans . , 2 ,n a m b‘ m c m , ±6a ~b 2 , — j. 
 
 ])n 
 
 4. V — 27a 3, ‘V 
 
 /u\i 
 
 Hns. —3a n l 2 , 3a m b 2 , y-> ’ 
 
 5 . VT, ^ • 
 
 -I "I "1 ^ ^ 
 
 21^5. i 1? 1? 1? t) ? 4 ) * 
 
 There is no such thing, so far as the human understanding can 
 reach, as the even root of a negative quantity; for we cannot conceive 
 of a quantity which is not either + or — : but the product of any 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND 
 
 33 
 
 quantity, + or — , taken an even number of times as a factor, is +; 
 therefore, since an even root must enter the power an even num- 
 ber of times, we cannot conceive of such a negative power; thus, 
 V— a 2 can be neither + a nor — a ; for -fax + a=+a 2 and — ax 
 —a=+a 2 : so that — a 2 cannot be produced by the multiplication 
 of any thing whatever by itself. Such indicated roots are called 
 Imaginary. In general, 
 
 An imaginary quantity is the indicated even - root of a negative 
 quantity ; thus, 
 
 V — 4, a/— a, V— 3, a 3 b 2 , and V — 
 
 are all imaginary. 
 
 38. The Division of Monomials. 
 
 Division and Multiplication are reciprocal operations. Any quan- 
 tity united to another by the sign x may be replaced by its recipro- 
 cal (unity divided by it) with the signs-; thus, 5x2 = 5 ---J- or 
 
 a x l— a -j- ^ . Whatever has been said, therefore, of multiplication, 
 
 taken in the converse sense, will apply to division. 
 
 The signs x and always indicate operations upon factors. The 
 sign x means to add as a factor ; and means to subtract as a fac- 
 tor. a 3 xa shows that the quantity a 3 is to be further augmented 
 by the introduction of another a as a factor, giving a 4 ; while 
 « 3 -i- a shows that a factor a is to be withdrawn from a 3 , giving a 2 . 
 
 To divide one quantity by another, is to withdraw the divisor as a 
 factor from the dividend. Where a factor of the divisor, or the di- 
 visor itself, enters the dividend exactly, it maybe taken out at once; 
 thus. 
 
 15 a 2 b i cd 
 3ab 3 c 
 
 = bahd. 
 
 Numerals are disposed of as in arithmetic, and where the same 
 letter is found in dividend and divisor, the difference of their expo- 
 nents gives the new exponent of that letter in the quotient. 
 
 Let it be remembered that like signs give 4- and unlike — . 
 
 Examples. 
 
 1. Simplify 2 §<Tb(?d = 5 o?c. 
 
 2. Simplify 9 w’lAdx* -f- 3a 3 cx. 
 
 c 
 
 A ns. baled. 
 Ans. aa'lf'cx. 
 
34 
 
 ELEMENTS OF ALGEBRA. 
 
 3. Simplify 
 
 4. Simplify 
 
 5. Simplify 
 
 a m + x bhf 
 -aTbhj ' 
 
 — a s b*x n 
 t ibx 2 
 
 —a m b n y r 
 -aby ’ 
 
 Ans. — aby. 
 A ns. —a?bz n ~-. 
 Ans. a n ~ 1 b n ~ l y p ~ 
 
 39. The Zero Power of a Quantity. 
 
 Iii dividing one quantity by another it often happens that the ex- 
 ponents of the same letter in the two quantities are the same, in which 
 
 case that letter disappears entirely ; thus, ( ~ = a. 
 
 We may, however, leave the letter in the quotient by giving it the 
 
 exponent zero ; thus, "^-—ab°. Here, manifestly, b°= 1. The zero 
 
 simply shows that b is not a factor of the quotient at all. To com- 
 prehend how any quantity to the zero power must be equal to unity, 
 we have but to remember that unity enters every expression as a 
 factor, and that thus a° = 1 .a°. 
 
 Now, removing a, since the exponent ° shows that it is not a 
 factor, the co-efficient 1 remains. 
 
 1. 
 
 Examples. 
 
 oa^bci 25 a m bic, — xAifq 
 
 ci 2 b 
 
 -5 a'"b 
 
 1 —a 
 xy ’ — 1 ’ a 
 
 1 _ L “ _ , r _j 
 
 Ans. 3c-, — 5 b * c , — x* y* , 1, — 1. 
 
 ©V 1 ) 
 
 0 a° —a 0 3 ax° \x°{y°) 
 
 ' ~Za ’ -(4)° ’ (25)° • 
 
 Ans. 1, — 1, 1, 4, — 1. 
 
 (5 ) m xny 3 ( a + by (a — b) m (a + b) n — [n + (« s 6 4 ) ; ] 
 
 ( o) n x l y 2 ’ ( a + b)' ! ’ ( a—b)a ' — [ + ( — !)] 
 
 Ans. b m ~ n x* + y, (a + b), 
 
 (. a-b) m -\a + by 
 
 ■, —a—a*b s . 
 
 4. — = a m ~ n =a°. But = 1 a° = 1. 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -r- . 
 
 35 
 
 40. Entire and Negative Exponents. 
 
 When the exponent of a factor in the divisor is greater than the 
 exponent of the same letter in the dividend, the subtraction of the 
 greater from the less must give a negative exponent; thus, 
 
 3« 2 5 
 
 = 3 a-'b. 
 
 Now, in this case, a enters the numerator but twice as a factor, 
 so that when it becomes necessary to take out four such factors, we 
 shall have a deficiency of two. This fact the — sign shows : that 
 is to say, we must still withdraw the factor a twice. 
 
 This meaning of the negative exponent follows from the general 
 definition. A negative exponent, then, shows that not only is there 
 no factor such as that over which it is written in the quantity, but 
 that such quantity must still be diminished by that factor a certain 
 number of times. 
 
 or 1 =- : 
 a' 
 
 for 
 
 a i ~ s =cr\ 
 
 But 
 
 a 6 a 
 
 -i 1 
 
 , a = -. 
 a 
 
 In general, let p and q be two quantities whose difference is m, q 
 being the greater ; then, 
 
 a p 
 
 But - = — . 
 
 fi 7 a m 
 
 = — . That is, 
 cC ’ 
 
 Any quantity raised to a negative power is equal to unity divided 
 by the quantity raised to the corresponding positive poioer. 
 
 Again, 
 
 cf_ 1 _ 1 . , , 
 
 d 1 « 2_s a -1 ’ U 
 
 a z 1 
 
 — = a:, a — — = 
 
 a; a 
 
 whose difference is m, q being the greater. 
 
 cf _ JL_ _ JL 
 a p ~~ a p_7 ~ a~ 
 
 or, in general, let p and q be two quantities 
 
 but 
 
 a 7 
 a f 
 
 — a" 
 
 a = 
 
 That is, 
 
36 
 
 ELEMENTS OF ALGEBRA. 
 
 Any quantity raised to a positive power is equal to unity divided 
 by that quantity raised to a corresponding negative poiver. 
 
 It follows from the foregoing that, 
 
 Any factor may be transferred from the numerator to the denom- 
 inator ; or from the denominator to the numerator by changing the 
 sign of its exponent. 
 
 Examples. 
 
 Convert the following expressions into fractional forms with unity 
 for numerators : 
 
 1 1 m 
 
 1. a, ab, a s , a m , a~b 3 , 5 ab», 20 z~ y y~. 
 
 1 1 1 I 1 1 1 
 
 a a b a a~nb ~ 3 o-'a-'b— ^ y 
 
 2. ( a + b)~\ \ , a~'b m , x » if . —a, —(—a) K 
 
 Ans. 
 
 (a + o)’ (a\ ctb m ’ a 
 
 1 -1 -1 _ 1 
 ’ ~ -1 ’ —a a' 
 
 Convert the following into expressions containing no negative 
 exponents. 
 
 „ 0 _! a~ m b o-'xy~? (a + b)- 1 (a—b)(a + b)- 1 
 
 6 • 2-'ab~ m ’ ( a-b )->’ {a-b)-\a+b)‘ 
 
 8 b 2 2 xb m a—b (a—b)(a—b) 
 
 AnS ‘ a’ tfV (a + J)(a+J) # 
 
 Convert numerators into denominators, and the converse. 
 
 4. 
 
 a~^b m ^a h cx~ 3 9 ~'xy 
 2c~ l d * cd~ p ’ — (ZV - ’ -1 
 
 2- , cf?- 1 d~ m c~ q (-I)” 1 
 
 Ans. 
 
 7 ~' a ' b ' ah~ m (f ’ 9 ‘ r ^ ' 
 
 Remark . — It must be carefully borne iu mind that only factors can be tlius 
 transferred. Thus, in the expression tlie a or 6 canuot *> e written in 
 
 the denominator. The a + b must be taken together, thus, ^ 7 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -4-. 
 
 37 
 
 We could write, however, 
 
 1 
 
 or 
 
 c 
 
 c 
 
 1 1 
 
 a a 
 
 41. Fractional Exponents.- Radicals. 
 
 From the general definition of an exponent, it follows that a frac- 
 tional exponent must show how many times the quantity to which 
 it is affixed is to be reckoned as a factor : that is to say, that the 
 quantity is to be resolved into as many equal factors as there are 
 units in the denominator of the exponent, and that as many of these 
 factors are to be taken, as there are units in the numerator ; thus 
 
 Now, the Root of a quantity is that factor of it which, taken a 
 certain number of times, will produce the quantity itself. 
 
 If there are but two equal factors, either of them is the square 
 root ; if there are three, any one of them is the cube root ; if there 
 are m equal factors, any one of them is the mth root ; thus, 
 
 25—5 x 5 .*. 5 = V25; 27=3x3x3 .-.3 = ^27. In like man- 
 
 ner, if a contains b m times as a factor, b—’\/a. 
 
 It follows therefore, that, 
 
 The denominator of a fractional exponent shows the root to be ex- 
 tracted ; thus, 
 
 am = y a. 
 
38 
 
 ELEMENTS OF ALGEBRA. 
 
 It is thus manifest, that, 
 
 The radical sign is identical in signification with a fractional ex- 
 ponent, having unity for its numerator and the index of the radical 
 for its denominator. Any radical may, therefore, be removed by en- 
 closing the quantity in a parenthesis, and writing instead of it such 
 a fractional exponent. 
 
 There is thus no necessity for the use of the radical, and it is only 
 retained here because it is so universally found in mathematical 
 works. 
 
 Examples. 
 
 Transform the following into expressions with fractional expo- 
 nents : 
 
 1. a/5, V / c, ■y/d, Va + b. 
 
 1 1 l_ J_ 
 
 Ans. 5 2 , c 3 , d«, ( a + b )-. 
 
 Remark . — When the radical sign is removed from a polynomial, as in the 
 last example, the bar must be retained, thus, a + b- , or replaced by a parenthesis, 
 
 thus, (a + &) 2 
 
 ; so, also, where any mistake might result, as in or y , the 
 
 parenthesis must be used ; thus, (§) 2 and J . 
 
 /25x m _ (25) - (x m ) 2 3 / a-b~ _ a 3 b' c 
 
 3 'V *r'~ sV')*’ 
 
 b) 3 ' 
 
 vV“ = 
 
 V ]/ 
 
 (O ’ 1 
 
 - = X ri . 
 
 5 . Vax Vb=a^b^= (ab)^, \Za-Vb=anb<* . 
 
 6 . j / Va = ( y/a^j - = (V 2 ) 2 = a T , |/ V n — ( V 7 ^) ”*= («" ) " = ■ 
 
 7 . a~i= \/a , a 3 !)* — ^/~aA • y/b 3 , cfrbdc' 1 =Va. / y/b(V\ t 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -P. 
 
 39 
 
 43. Fractional Exponents.— Radicals. 
 
 Now, while the denominator of a fractional exponent shows 
 the degree of the root to be extracted, the numerator shows the 
 number of times such root is to be taken; that is, it shows the power 
 to which this root is to be raised ; thus, 
 
 S3=2 2 =4. 
 
 But, (v / a)”= -y/cT, for we may write (f/a) m =an = i a ”') n = \/a m \ 
 hence, we can say in general, that 
 
 The numerator of a fractional exponent shows the power to which 
 the quantity affected by it is to be raised, while the denominator slioius 
 the root to be extracted. 
 
 Examples. 
 
 Transform the following into equivalent expressions : 
 
 f «"*, yV' ? , ’vV 
 
 m, P _» lm 5 ? 1 
 
 1. v a n , ai , a*, a 2 , an, ««+». 
 
 n 9 U / 1 
 
 Ans. a™, V " /if — , 
 
 2. j/ ffa 2 , ^a 2 ‘Va z ‘ l \/a i ,\/ 
 
 Ans. (^)-== («•)", a^aTav, (^)K^) " =-^V r - 
 
 3. a/28- V~5- v'a. 
 
 An, SW-* = & 
 
 4 . 
 
 a*Jb 5 \/ a 2 3 a 2 b(cd)™ cd{ 25) ! 
 cV~d 4 5 6ffa 3 
 
 4 ax 
 
 ab ¥ 5 a» 3 a 2 bycd bed 
 Ans. j , 3" . ^ 
 
 cd~ 6«4 4 
 
 ’ 4 a.r' 
 
40 
 
 ELEMENTS OF ALGEBRA. 
 
 43. To change the Index of a Radical. 
 
 It is obvious that we cau multiply or divide the numerator and 
 denominator of a fractional exponent by the same thing, without 
 
 2 772 2m yni m 
 
 changing its signification ; thus, a 2 —aA, oA—om, aT^—a ». But the 
 numerator is the exponent of the power and the denominator is the 
 index of the root; whence we may say, in general, that 
 
 We can multiply or divide the index of a radical by any quantity, 
 provided that ioe,at the same time, multiply or divide the exponent of 
 the quantity under the radical sign. 
 
 For example, take \/a 2 +b—c. If we should multiply the index 
 of the root by 2, we must raise a 2 +b— c to the second power; thus, 
 vV 2 +b— c = V' (a 2 +b—c) 2 . The quantity under the radical must 
 be considered as one quantity. We may, however, multiply or divide 
 the exponent of each factor under the sign; thus, a/3« 2 6 = 
 i \/9a i b 2 , or \/‘66a i b 6 = \/6a 2 b 3 . 
 
 Examples. 
 
 • Change the indices of the following by multiplying by 2. 
 
 1. *fa, Vo, ‘s/x, \/‘da 2 b, f/?>a 2 b, \/‘ia"‘b". 
 
 Ans. V^ v^ 2 , VWtfb 2 , a/9 a^b 2 , f / 9 
 
 Multiply by 3. 
 
 \/a \/a + b Vl5 a 2 x~ 1 i/ a 2 V a—b 
 
 ‘ V ’ ~ v'j ’ * r 
 
 ty{a + b) 2 v / (15) 3 « 6 .t -3 ,{/ a ~^ 
 
 W’ * ’ ^3 ’ v ’ 
 
 Ans. 
 
 \/ (a—b) 3 
 \/ (a — b) 3 
 
 Multiply by m. 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND 
 
 41 
 
 Divide by 2. 
 
 4. y/tf, V64a:*r 4 , 
 
 ^4 ms. V a, \/%a z b, 8 xy~ s , 
 
 6 a~ 1 b'h 
 5 :% 2 
 
 44. To bring 1 Radicals to a common Index. 
 
 Radical quantities with different indices may be changed into 
 equivalent radicals having the same indices. For let 'sfaP and \ / b q 
 be two such quantities. They may be written, 
 
 p J? 
 
 ffm and b n . 
 
 Causing these fractional exponents to have a common denominator, 
 we have, 
 
 pn qm 
 
 a mn and b mn ; 
 
 and thus, ’^/aT and "f/W. We may, therefore, say that, 
 
 To bring radicals to a common index, find a common multiple of 
 the several indices, and take this for the common index ; then divide 
 this neiv index by each of the old ones in succession, and raise each 
 quantity under the radicals, respectively, to the power indicated by 
 these quotients; or 
 
 Use f ractional exponents and bring them to a common f ractional 
 unit. 
 
 For example, take -|-v / 3 b 2 and a\/c 3 . The least common 
 
 multiple of the indices is 12. We may then write 4 V" ", iV , 
 aV ; now raising the quantities under the radicals, respectively, to 
 the powers found by dividing 12 by 2, by 3 and by 4, we have, 
 
 4v^(2a) # , 4^(3 b 2 )* and aV(c 3 ) 3 ; or 
 4v /, 64 a 6 , Vv' 81 5 8 and a'\/ c 9 . 
 
 It will be observed that the quantities outside of the radicals re- 
 main unchanged. 
 
 Examples. 
 
 Transform the following radicals into equivalent ones having 
 common indices: 
 
42 
 
 ELEMENTS OF ALGEBRA. 
 
 1. l-Voa 2 , 
 
 y%bc, 
 
 Ans. l-\/l 25« 6 , 
 
 VWb 2 , 
 
 2. ao/b, ^v / 3«, 
 
 A 2 3. A 
 
 4. a 2 , & 3 . A«s. « 6 , 6 b . 
 
 5. (a + 6) 2 , («-5)t 
 
 Am. f/ 
 
 7/i p qvn pn 
 
 a* , bT. Ans. as* , h* . 
 Ans. ( a + iy , (a—b) c . 
 
 45. Product of the «th Roots. 
 
 The product of the nth roots of two quantities is equal to the nth 
 root of the product of the quantities, and the converse. 
 
 For, 
 
 i i i 
 
 ci n b n = {ah) n ; whence, 
 f/ a • f/b = f/ ah. 
 
 Again, 
 
 The quotient of the nth roots is equal to the nth root of the quo- 
 tient, and the converse, 
 
 For, 
 
 i i 
 
 whence, 
 
 yA 
 
 f/h 
 
 These principles enable us to combine radicals by multiplication 
 or division. Thus, 
 
 a'fT x c\/7i=acV~b • ‘\/d=-ac\ / 'hd ; 
 
 hence, 
 
 To multiply radical quantities together, multiply the factors out- 
 side of the radical signs together, and also the factors under the radi- 
 cal signs, retaining the radical over the latter product. 
 
 For example, 
 
 3 a 2 h\f hcd m x 2ab 3 VAd—Sa^b x 2ab 3 V ocd m x cd=6a 3 b i \/bc 3 d '"~ 1 . 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -4-. 
 
 43 
 
 The indices of radical expressions so combined must be the same ; 
 if not so already, they must be changed into corresponding radicals 
 with a common index. 
 
 Examples. 
 
 1. 3a\fb* xab\f7. 
 
 2. 3a(b 2 )? x5b(7)?. 
 
 3. %Vax\Vbx 
 
 4. x x - 
 
 3 a 2 b 3 r-7- irr 
 
 5. v 4c x \ V 
 
 X 
 
 3 a 2 b 
 
 (4c)‘^ x UiY*. 
 
 7. — V'2a x \/2a x 4 Y/2a. 
 
 8. Y/ a n x — v'g” x — Yfa. 
 
 n m 1 
 
 9. a™ x —an x —a7. 
 
 10 - zj/Y^f/fr 
 
 11. 2(~V x — c(-^ 
 
 12. 5aV —3x x —2b\/2x 2 . 
 
 13. 5a(-3x)i x -2b(2x 2 y*. 
 
 Ans. 15 abV”b 2 . 
 Ans. 15ab(7b 2 ) 2 . 
 
 Ans. —\Vb. 
 Ans. — -jt(J) 2 . 
 
 Ans. 
 
 3 aH 
 2x 
 
 j/f- 
 
 - my- 
 
 Ans. -iy8192a 13 . 
 Ans. m \/ a n -p+ m ' 2 p+ mn . 
 
 n2p m2p mn 
 
 Ans. a mn p^~ mn p Jr mnp' 
 
 Ans. 
 
 . / ac 
 
 ic ym 
 
 Ans. —2c 
 
 Ans. — lOab V — 108a; 7 . 
 
 Ans. — 10«5(— 27« 3 ) 6 (4a; 4 )' 3 — — 10«5( — lOS-r 1 ) 6, 
 
 46. Division of Radicals. 
 
 To show how to divide one radical expression by another, we have, 
 
 aYfb a Y/h a , m /b 
 
 -s-= =-•—?=— -\/ ^5 whence, 
 
 cYjd c Y/d c\ d 
 
 To divide one radical quantity by another , divide the quantities 
 
44 
 
 ELEMENTS OF ALGEBRA. 
 
 ■without the signs, and the quantities within, respectively, retaining 
 the radical sign over the latter quotient. 
 
 The indices must be made common, if not so already. 
 
 Examples. 
 
 3Va _ 3 , /a 
 
 3(a)* 
 
 2# 
 
 3/a\i 
 
 2W ’ 
 
 3u 2 (k)~ 
 
 5. 
 
 3.r(2 a*Y _ 3g/ 2 10 g >0 \^y 
 
 
 2yV 
 
 2 7 J 
 
 8 « 2 5 ( 5 « 5 2 )» 
 
 2a6(7c6~ 1 )^ 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -f-. 
 
 45 
 
 47. Simplification of Radicals. 
 
 Any factor may be removed from within a parenthesis and made a 
 factor without, by multiplying its exponent by the exponent of the 
 parenthesis; thus, 
 
 2 a(b*c M dy= 2ab 4 (c 3m dy. 
 
 2 a(lfc Zm d )2 = 2 ab(c 3m d)^. 
 
 2a(b 2 c 3m d)i = 2«c m (5VZ)x. 
 
 1 11 
 
 2 a(b‘ 2 c 3m cl) m = 2ab n c 3 d"‘. 
 
 Numerical factors may be treated in the same way ; thus, 
 2a(12b*c 3 d)s =2a(2* x 3b*c*cd)i 
 —2 x 2abc(3cd)^ 
 
 =iabc(3cd)i. 
 
 The radical sign means that the root of the quantity under it is to 
 be extracted ; which may be done by extracting the root of the several 
 factors successively: whence it follows that we may extract the root 
 of any factor under the radical, and write the root so found as a 
 factor without; thus, 
 
 2aV b' 2 c 3m d = 2aV b*<? m d“d, 
 
 2 «V 12hVW = 2a x 3bVcd, 
 
 = 4 abcy 3cd. 
 
 This operation .is called simplifying a radical. A radical is said 
 to have its simplest form when there is no factor under the radical 
 of as great a power as the degree of the radical. Practically, 
 
 To simplify a radical expression, look out all the factors under the 
 radical sign vjlnch are exact powers of the degree of the root required, 
 or may be made so j divide the exponents of such factors by the 
 index, and write the factors, with their new exponents, without the 
 radical. 
 
 Or if the parenthesis is used instead of the radical sign, 
 
46 ELEMENTS OF ALGEBEA. 
 
 Multiply the exponent of the factor by the exponent of the paren 
 thesis. 
 
 Examples. 
 
 1. 2 a(WcY~ =2 ax 2b(c) 2 = 4ab(c)i. 
 
 2. 2aV Wc — 2 a x 2 bVc = 4a5 \/ c. 
 
 5. 5x(x i (a+b)'j 3 — 5x 3 (a + bf . 
 
 6. 5xVx i {a + b) — 5 x^'s/a + b. 
 
 7. 3a (d j A y {a + b) - ~ = 9 az'(a + b)y^. 
 
 8. 7a V 1 6xy\a + b ) 4 = 28a 2 y(a + J) } v x. 
 
 9. c(18« 3 & 5 )“ = c( 2 x 9 a 2 ab 4 b)~ — 3ab' 2 c(2ab)*~. 
 
 10. |V48 « m+2 = ?,V3 x 16«"‘ft s = 2«V3«”. 
 
 11. [32ft 7 (2 + J) 4 p= [4 x 8« 6 fl(2 + b)\2 + b)^=2a\2 + b) [4 a (2 + 6)] 
 
 12. \/ll2a 3 —4aV7a, V ^°-.a m+3 =l-aV a m+l , 
 
 / 324 _ 6 /3 
 
 27a 3 “3«y a 
 
 13. 'V / f^« 7 5 = ?pr \Tab, = fr' Mar', = \fa-\Ta f ‘. 
 
 14. 
 
 2 7ft 4 5 
 Scd 3 
 
 b b 
 
 _5 
 
 b 3 ‘ 
 
 15. vf=i\/3, V&=Wb W&=Wb 
 
 1 6 . VW= Vw= Vfi x io=t ^10. 
 
 17. V-V = ^11= VoV x 63 — y a/63. 
 
 is. vn=vj^M=wi 
 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND -f-. 
 
 47 
 
 48. To pass a Factor under the Radical. 
 
 It is evident that we may pass a factor from without to within a 
 parenthesis, or the radical sign by the reverse process ; that is, by 
 dividing the exponent of the factor by the exponent of the paren- 
 thesis, in the one case ; and by multiplying the exponent of the 
 factor by the index of the radical, in the other. 
 
 Examples. 
 
 1. 3ff(6ff)^=(54« 3 )2. 
 
 2. 2aV ab—2V a 3 b, 3c\koa 4: d= 's / 135 a i c 3 d. 
 
 3. 2ff(rt + Z>) 3 = 2(«3(ff + £)) 3 , 8a 3 (a— b) 3 = (2a(a — l?)) 3 . 
 
 4. 5a 2 \ / cc + b—5Va 4: (a + b), 5aVa—b=aV25(a—b). 
 
 K + /27« 3 (rt + £)\L 
 
 b \d) ~ \b 2 ‘ d) ’2 o\ c ) ~\ 8c 4 ) • 
 
 6. 3« 2 (fl 2 5 m c- 1 ) 2 = (3%(«25’"c- , )) 2 , 2arb"(j)™= (— 
 
 mm m vi p qm p 
 
 8. cin(a)n =(a 2 )T, «n(«)? = (ffps +I )7. 
 
 9. cu k/ a T — 
 
 p / pm 
 
 ~ +T , a m+ ' m ya= V« (m+I)(m_1)+l . 
 
 10. b m -\b)^i=(b^ +1 )^ i, a r(cr$L=(ar i7 '*- 1 )^. 
 
 49. To Transform the sum of two or more Radicals. 
 
 Eadicals are said to be Like or Similar when they have the same 
 indices and the same quantities under the radical signs ; thus, 
 aVb and kVb are like; (a + b)^/a m b and da 2 Va^b are like; so also 
 . 3 / a— b , „ . / a—b 
 
 ° re y — * ni °j/ ■ 
 
 The algebraic sum of two or more like radical expressions may 
 readily be transformed into a single term; thus, 
 
 3 a*\/ be a 2 b *Jbc — 2\^bc={3a — a 2 b—2)'\/bc. 
 
48 
 
 ELEMENTS OE ALGEBRA. 
 
 We have simply to take the algebraic sum of the quantities 
 without the radicals, and after this write the common radical. 
 
 If the quantities without are numerals entirely, they are actually 
 added or subtracted according to their signs ; thus, 
 
 3 Va^+l) + 5 */a + b = 8 v / «~+£- 
 
 yiVH/i- 
 
 Radicals which are not like as given, may sometimes be made so ; 
 thus, 
 
 a^/^c—bVbcd 2 = abVbc—bd\/bc 
 — (ab—bd)Vbc. 
 
 This operation of reducing radical expressions to a single term is 
 generally called addition or subtraction of radicals, as the case may be. 
 
 Examples. 
 
 1. 3Va + 2Va = 5V~a, aVa + bVci=(a + b)\/a. 
 
 2. 8a 2 X'V / 3c 2 + 4a 9 z*/3c* = 12a 2 x\/3c 2 , oa^/b — hb^/b = 
 {2>a—5b)Vb. 
 
 3. 5a{b)^-2a(b)' 2 = 3 a(b)^, 3(a + S) 2 - (a + b)$= 2 (a +3)*. 
 
 6 . 2aVb^c—3b i \/bc^ =2ab'\/bc—3bc / \/bc=(2ab—3bc)^bcI 
 
 7. 4(« 3 ^ 3 )'- !(«&)*= iab(ab)^-i((tb)i=(^ab-^){ab)k 
 
 8. 9^l6^-10^54rt- 1 =18^^ 1 -30^2^ = -12'V / 2« i: i. 
 
MONOMIALS — EXPONENTS AND THE SIGNS X AND 
 
 49 
 
 10 . 
 
 a 2 b la nr 5a n- 31a rr 
 
 T=8-. V » + 3 V? =1T VS: 
 
 11. Vi+VA=iV3 + iV3=fV3. 
 
 13 ' \/ 1 +i^=i^ a - 
 
 50. General Principles of Exponents. 
 
 From the nature of exponents we have the following general re- 
 sults : 
 
 a m Xa n —a m+n 
 
 — =a m_n 
 a n 
 
 (a m ) n =a mn 
 
 f/a m =a n . 
 
 Thus, we may say that, 
 
 I. The addition of exponents takes place in the multiplication of 
 quantities. 
 
 II. The subtraction of exponents takes place in the division of 
 quantities. 
 
 III. The multiplication of exponents takes place in the formation 
 of powers. 
 
 IV. The division of exponents takes place in the extraction of 
 roots. 
 
50 
 
 ELEMENTS OF ALGEBRA. 
 
 SECTION V. 
 
 TRANSFORMATIONS OF POLYNOMIALS. 
 
 51. We have now pretty well disposed of tlie operations upon mo- 
 nomials or single terms ; let us proceed to investigate the transform- 
 ations which may be performed upon polynomials. If the prin- 
 ciples already established are carefully borne in mind, we shall find 
 little difficulty. 
 
 52. Multiplication of a Polynomial By a Monomial. 
 
 First, let a be the sum of all the positive terms in any polynomial, 
 and b be the sum of all the negative terms in the same ; the polyno- 
 mial itself will be a— b. 
 
 The product of this polynomial by any single term, as c, will be 
 
 (a — b)c. 
 
 Let us now convert this product into an algebraic sum. The prod- 
 uct of a and c is ac ; but this is greater than the true value of the 
 given expression, ( a—b)c , since b should have been taken from a be- 
 fore it was multiplied by c. If, then, we subtract be from ac, we 
 shall have the true product ; hence, 
 
 (a—a)c—ac—bc. 
 
 We should have obtained the same result by simply multiplying 
 each term of the polynomial by c, observing the law of signs, and 
 uniting the partial products by their respective signs. 
 
 If c had been negative we should have found the same principle 
 to apply. 
 
 We may say, then, in general, that, 
 
 To multiply a polynomial by a monomial, multiply each term of 
 the polynomial by the monomial, remembering that like signs give 
 plus, and unlike minus. 
 
 1. 5« 2 (3«5 + 2e— 1). 
 
 2 . \a^{fa^b~ x + $). 
 
 3* 3\/a(5 \/ 6 — V«). 
 
 Examples. 
 
 Ans. 15« 3 Z> + lOrt’c— Sab 
 x 
 
 Ans. ab +£a • 
 Ans. 15^/6# — 3a. 
 
TRANSFORMATIONS OF POLYNOMIALS. 51 
 
 KI-v^h) 
 
 . a° a , a 
 
 Ans ' i’—b^ + Tv 
 
 Vc(j / c-20 + 2\/c). 
 
 Ans. fyd— 20y/c + 2c. 
 
 —25a-x i (l — 3abx~ 1 — a 2 ). 
 
 Ans. — 2 Wa?x* + r tha?bod + 25 aJx*. 
 
 2 a 2 [ a + b 3 «- s ). 
 
 Ans. 2a3 + 2a * b 6 . 
 c 
 
 Va/Va \/a\ 
 
 1r ~ a vV 
 
 Vb\Vb Vd 
 
 Ax /be, • — £ 
 
 h Vbd J/b 
 
 We are said to factor an expression when we transform it so that 
 the several factors which enter it are made visible to the eye ; thus, 
 ab + ac—ad—a{b+c—d). When we have an algebraic sum in which 
 a monomial and a polynomial factor enters, the operation of factor- 
 ing is the reverse of that in the above examples. 
 
 Let the student be required to factor the answers given in the 
 above examples. 
 
 53. The Multiplication of Polynomials. 
 
 Resuming the polynomial, a—b, let c—d be any other polynomial. 
 Their product will be 
 
 (a—b){c—d). 
 
 To transform this product into an algebraic sum, let us begin by 
 multiplying a— b by c; we shall have 
 
 ac—bc. 
 
 This result is d times greater than the product of the two poly- 
 nomials, since d should have been subtracted from c before we used it 
 as a multiplier. If we, then, subtract from ac—bc the quantity d{a — b), 
 we shall have the true product of the polynomials ; that is 
 
 (a—b)(c—d)=ac—bc—(ad—bd)=ac—bc—ad + bd. 
 
 It will be observed that we have simply multiplied each term of one 
 polynomial by every term of the other. No mention was made of the 
 signs in deducing this result, so that by inspection we see again, that 
 like signs give plus, and unlike minus. 
 
 We may, then, say, that 
 
52 
 
 ELEMENTS OF ALGEBRA. 
 
 To multiply one polynomial by another, multiply each term of the 
 one by every term of the other, and simplify the result. 
 
 It is sometimes more convenient to set one polynomial under the 
 other, and write like terms under each other as the operation pro- 
 ceeds; thus, 
 
 a 2 — ab + b 2 
 a 2 + ab + b 2 
 
 a l —a 3 b + a 2 b 2 
 
 a 3 b—a 2 b 2 +ab 3 
 
 a 2 b 2 —ab 3 + b i 
 
 « 4 +a 2 b 2 +b 4 
 
 Examples. 
 
 I. (3a-2b 2 c-l)(ha 2 c + b-of). 
 
 Ans. 15a 3 c— 10a 2 b 2 c 2 — oa 2 c+oab—2b 3 c—b—oax+2b 2 cx + z. 
 
 3. (Va + }Vb— i)(Va— \Vb + \). Ans. a—\b + \Vb— 
 
 r . r ,2 1 2 . 2 . 7 
 
 — a 2 b— \ + c 2 + a 2 bc 3 c*. 
 
 b- 
 
 Ans. JS+^l+^Z t«_l. 
 
 c 
 
 6. (a + b)(a— b). 
 
 7. (« + £)(« + 5)- 
 
 8. (a—b)(a—b). 
 
 Ans. a 2 —b 2 . 
 
 Ans. a 2 + 2ab + b 2 . 
 
 Ans. a 2 —2ab + b 2 . 
 
 b 
 
 Vbd d ' 
 
TEANSFOBMATIONS OF POLYNOMIALS. 
 
 53 
 
 i 
 
 i 
 
 (V« + %/ab— V~b)(^Va— / \/~ab + Vb')- 
 
 Ans. a-'^H 2 +2^/a 2 b 3 +b. 
 
 14. (— p + Vq+p z ){ — p— Vq+i> 2 )- 
 
 Ans. —q. 
 
 54. Two Terms unchanged by Simplifying’. 
 
 A polynomial is said to be arranged with respect to the descend- 
 ing powers of a particular letter, when the first term contains that 
 letter with its highest exponent, and the next contains it with its 
 next highest exponent, and so on. It is arranged according to the 
 ascending powers of the letter Avlien the order of terms is reversed. 
 
 Now, when two polynomials are multiplied together, there are 
 always two terms in the result which cannot disappear in the pro- 
 cess of simplification. These are, first, the term which results from 
 combining the two terms with the highest exponents of a particular 
 letter ; and, second, the term which results from combining the two 
 terms with lowest exponents of the same letter; thus, 
 
 In this case, x 5 and — 2x 2 come immediately from the combina- 
 tion of certain terms; while the others are the results of simplifying. 
 
 55. The Division of Polynomials. 
 
 The dividend must always be produced by multiplying the quotient 
 by the divisor ; and when these are polynomials, we know, from the 
 last article, that there must always be two terms at least in the divi- 
 dend which will undergo no change from the process of simplifica- 
 tion. 
 
 Let us take the polynomial x 5 — 5.r 4 + 7x 3 — 2x 2 , and let the divisor 
 x 2 —2x be given to find the quotient. 
 
 Now since x 5 must have resulted from the multiplication of x 2 in 
 the divisor by that term in the quotient which contains the highest 
 power of x, we can tell at once, by dividing x 5 by x 2 , what that term 
 must have been. One term of the quotient must then be x 3 . 
 
 (x 3 —3x 2 +x)(x 2 —2x) — 
 x 3 — 5x 4 +7x 3 —2x 2 . 
 
51 
 
 ELEMENTS OE ALGEBRA. 
 
 Let us for convenience arrange both polynomials, and write the 
 dividend first and the divisor after it ; thus, 
 
 x 5 —ox 4 + 7a; 3 —2a; 2 x 2 —2x 
 x 5 — 2a; 4 a; 3 — 3x 2 +x 
 
 —3x 4 + Hx 3 —2x 2 
 — 3a; 4 + 6a; 3 
 
 x 3 — 2x 2 
 x 3 — 2a: 2 
 
 0 
 
 writing the first term of the quotient under the divisor. Now, as 
 this term of the quotient must be multiplied into every term of the 
 divisor, in the process of forming the dividend by multiplying the 
 divisor and quotient together, if we perform this multiplication and 
 subtract the result from the dividend, we shall free it from all partial 
 products in which this term, x 2 , of the quotient enters. Multiplying 
 and subtracting, as shown above, we have a remainder, — 3a; 4 + 7a; 3 
 -2a; 2 . 
 
 Now, the term —3a; 4 of the remainder must have resulted from 
 multiplying x 2 by that term of the quotient which has the next 
 highest power of x in it. We may thus find another term of the 
 quotient by dividing the first term of the remainder by the first term 
 of the divisor; thus we have, — 3.r 2 . Writing this after the term of 
 the quotient already found, and multiplying the divisor by it, as be- 
 fore, and subtracting the result from the remainder, we have a second 
 remainder, x 3 — 2x 2 . By continuing this process, we shall find all 
 the terms of the quotient. If there should prove to be a final re- 
 mainder, the division cannot be exactly performed. In this case, we 
 may write the remainder in the form of a fraction, having the divisor 
 for the denominator, and unite it with the quotient by its proper 
 sign. 
 
 We may then say, practically, that 
 
 To divide one polynomial by another', arrange the dividend and 
 divisor according to the powers of the same letter ; divide the first 
 term of the dividend by the first term of the divisor. This will be the 
 first term of the quotient. 
 
 Multiply the divisor by th is term of the quotient and subtract the 
 result from the dividend. 
 
 Divide the first term of the remainder ( arranged ) by the first term 
 of the divisor ; multiply as before , and subtract from the remainder. 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 55 
 
 Continue the operation until there is no remainder, or until the first 
 term of the remainder will not contain the first term of the divisor. 
 
 Examples. 
 
 1. (a 2 A2abAb 2 )-A(aAb). 
 
 2. (a 2 — 2abAb 2 )A-(a— l). 
 
 3. (a 2 — b 2 )A-{aAb). 
 
 4. (4« 3 4-4« 2 — 29«4-21)-4-(2a— 3). 
 
 5. {a^-b^Afi-b). 
 
 6. (x s Ay s )A-(xAy). 
 
 7. (x 3 Ay 3 )A(x 2 + y 2 ). 
 a 3 -b 5 
 
 8 . 
 
 A ns. aAb. 
 Ans. a—b. 
 Ans. a—b. 
 Ans. 2a 2 A 5a — 7. 
 Ans. a 3 +a 2 b + ab 2 +b 3 . 
 Ans. x i —x 3 yAx 2 y 2 — xy 3 Ay i . 
 
 Ans. x i —x 2 y 2 +y i . 
 
 Ans. a 4 +a 3 b + a 2 b 2 Aab 3 + b i . 
 
 a m —b m 
 a — b 
 
 10 12 ^~ 192 
 • 3 y-5 ' 
 
 11. l-4-(l-«). 
 
 Ans. a m ~' + a m ~ s b + a m - 3 b' 3 + ■ • • b m 
 
 Ans. 4?/ 3 4- 8?/ 2 4-16^/4-32. 
 
 Ans. \ A a A a 2 A a 3 A a^ A 
 
 12 . 
 
 -fia 2 +4 V a 3 +8'V / a + 8v / « 
 
 \/a + 2 'fia 
 
 13. 
 
 14. 
 
 15. 
 
 a + a 2 b 2 + b 
 
 ~x i i I ' 
 a 2 4- ft 4 # 4 + b 2 
 
 fa e\ f's/a \fic\ 
 \h d) ' ~ rfj)' 
 
 A A 'AA tA A 1C 
 
 a 4 + b^ 
 
 Va — fib 
 
 ' 'fia- 'fib 
 
 1 —a 
 
 Ajis. Va + 2^+4. 
 
 1 Li i 
 
 Ans. a 2 —a‘ L b i Ab 2 . 
 
 'fia 'fic 
 
 Ans. —n 4 — , 
 
 'fib fi d 
 
 L i 
 
 Ans. « 4 A 'A. 
 
 Ans. fia A ‘fib- 
 
56 
 
 ELEMENTS OF ALGEBRA. 
 
 56. Formulas used in Transformations. 
 
 A Formula is any general truth expressed by means of symbols. 
 
 The transformation of polynomials can often be much shortened 
 by the use of certain simple formulas. We shall now give a few of 
 these. 
 
 (a + b) 2 — {a + b){a + b)=a 2 + 2 ab + b 2 ; that is. 
 
 The square of the sum of tivo quantities is equal to the square of 
 the first, joins twice the product of the first and second, plus the square 
 of the second ; and the converse. 
 
 (a—b) 2 = (a—b)(a—b) — a 2 —2cib + b 2 ; that is, 
 
 The square of the difference of tivo quantities is equal to the square 
 of the first, minus twice the product of the first and second, joins the 
 square of the second ; and the converse. 
 
 (a + b)(a — b)=a 2 — b 2 , 
 
 The product of the sum and difference of tivo quantities is equal to 
 the difference of their squares, and the converse. 
 
 When the second member of an equation is the algebraic sum re- 
 sulting from operations indicated in the first, such sum is called the 
 development of the first member; thus, in 
 
 (a + b) 2 =a 2 +2 ab + b 2 , 
 
 the second member is the development of the first. 
 
 Examples. 
 
 Develop the following expressions, 
 formulas. 
 
 by the aid of the foregoim 
 
 1. {x + y) 2 . 
 
 Ans. x 2 +2 xy+y 2 . 
 
 2. (2a + 2b) 2 . 
 
 Ans. 4r< 2 +8u5 + 45 2 . 
 
 3. (a — l) 2 . 
 
 Ans. a 2 — 2a + 1. 
 
 4. (i ~a) 2 . 
 
 Ans. \—a + a 2 . 
 
 5. (\a+\b) 2 . 
 
 Ans. }a+^ab+£b 2 . 
 
 „ fVa ,\ 2 
 
 . a 2 \/a , 
 
 6. [ A— — 1 ). 
 
 Ans. y 1^+1. 
 
 ' V b o 
 
 t Vb 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 57 
 
 7. 
 
 \(a-b) s J 
 
 8. (p— Vq+p 2 ) 2 - 
 
 9. (1 -a~ i y. 
 
 10. (x + y)(x-y). 
 
 11. (£a— b)(^a + b). 
 
 12. (3a-i-b)(3a + ib). 
 
 _ (a c\(a c\ 
 
 13 -(i + S 
 
 14. (V« + 5)(a/« — b). 
 
 15. (1— a)(l + a). 
 
 16. (l-i«)(l +*«). 
 
 17. (« 2 “ + 5 4m ) (a 2 " 1 — b im ). 
 
 18. (‘\fa— v'i)(v / a + y7>). 
 
 ^4 as. 
 
 a + Z> 
 a— Z> 
 
 2a^(a + Z>)^ 
 (a— b) 2 
 
 +a. 
 
 Ans. 2 )‘ z — 2p‘\/q+p) 2 + {q+p 2 ) 
 Ans. 1—2 a _I + a~ 2 
 -d?2s. a: 2 — y 2 
 
 JhiS. Jft 2 — 5 2 
 ^4?is. 9a 2 — JJ 2 
 
 -4^5. TIT 7T 
 
 6 2 d 2 
 
 Ans. a—b 
 Ans. 1— a 2 
 Ans. 1— ia 2 
 Ans. a im —b 8m 
 
 Ans. 'i/a 2 — 
 
 57. Formulas used in Transformations— Continued. 
 
 By developing the following, we have 
 
 (i a + b)(a + c)—a 2 + (b + c)a + bc 
 (a—b){ci—c) — a^ — (b + c)a + bc. 
 
 "Whence we see that when two binomials are to be multiplied to- 
 gether, having their first terms the same and the second terms differ- 
 ent, with the same middle signs, we may write out the result at once: 
 
 1. The first term of the result must be the square of the common 
 term ; 
 
 2. The second term must be the numerical sum of the two last 
 terms into the first term with the sign of the last terms; and 
 
 3. The last term must be the product of the last terms with the 
 plus sign. 
 
 Examples. 
 
 1. (a + 2)(a + 3)=a 2 + 5a + 6. 
 
 2. (x— 5)(z— 4)=.-r 2 — 9£ + 20. 
 
 3. (1 +a) (1+ b )— 1 + (a -|- V) + cib. 
 
58 
 
 ELEMENTS OF ALGEBRA. 
 
 4. (a 2 5 + 5)(« 2 6 + 7) = « 4 5 2 + 12a 2 & + 35. 
 
 5. (3a-J-)(3a-i)=9a 2 -fffl + ^. 
 
 / Va 
 
 wf 
 
 'V a 
 
 Wl 
 
 a 2 
 
 11 a 
 
 ~ ¥ 
 
 + ~b 
 
 —20j 
 
 i-f : 
 
 b 
 
 18. 
 
 21 '/a 
 
 Vb 
 
 + 20 . 
 
 8. (a -2 — l)(a -2 — o)=a~ Ji — 6« _2 + 5. 
 
 58. Formulas used in Transformations — Continued . 
 
 Now let the last terms be different with different signs ; thus, 
 
 {a + b){a — c) = a 2 + (b—c)a—bc. 
 
 Whence we see that in this case the middle term will be the differ- 
 ence of the last terms into the common term, and the last term will 
 have the — sign. 
 
 Examples. 
 
 1. (« + 3)(ffl— 5)=a 2 — 2a— 15. 
 
 2. (a*-9)(a*+3)=a-6o*-27. 
 
 3. (3a 2 Z> 4 — 2)(3a 2 5 4 + 8) = 9a 4 £ 8 — 18« 2 5 4 — 16. 
 
 4. (\/a— l)(V«+i)=«— \Va— h 
 
 5. ( / \/~x—5)(\/ x + V)= \/x 2 +2\/ x—3o. 
 
 6. (a _1 + 9)(a _1 — 11)=« -2 — 2a~ l — 99. 
 
 7. (a m - 4) (a m + 12) = d im + 8 a m — 48. 
 
 m m 2to to 
 
 8. (xn — 9)(^n + 2)=a:+ — 7%n — 18. 
 
 9. (ai-l)(«f + 25)=a^ + 24al-25. 
 
 59. Factoring:. 
 
 By the aid of the simple formulas already explained, we may often 
 resolve trinomials into their factors by mere inspection. Let us first 
 take some examples under the three following formulas: 
 
 a~ +2 ab + b 2 — (a + b)(a + b). 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 59 
 
 a 2 —2ab + b 2 — (a—b) (a—b). 
 a 2 — b 2 — (a + b)(a—b). 
 
 Examples. 
 
 1. a 2 b 2 —c 2 = (ab + c){ab—c). 
 
 2. 4a 2 — 4a + l = (2a— 1)(2«— 1). 
 
 3 . z 2 +z+l=(z+%)(z+l). 
 
 5. a— 5=(V«+ Vb)(Va— Vb. 
 
 6. 4« 2 — i=(2a— + 
 
 7. 25 a 2 b i +20ab 2 c + 4c 2 — (5ab 2 + 2c){5ab + 2c). 
 
 0 25a -2 x_/'oa ~ 1 f oer 1 
 
 ^b^~y-{jr + ^j/^b 
 
 9. J4 = (ai+ 
 
 10. V .r— j=(v / .t+J)( v^— 
 
 60. Factoring— Continued. 
 
 Let us now show the operation of factoring under the formulas, 
 
 (i a + b){a + c)—a 2 + ( b + c)a + bc , 
 
 (a—b)(a—c)=a 2 — (b+c)a + bc. 
 
 For example, take 
 
 a 2 - f- 5a -f- 6. 
 
 If there are any binomial factors in this expression, the product of their first 
 terms must produce a' 2 , and, since a enters the middle term, they must both be 
 a : since the last term is positive the last terms must have like signs. Looking 
 at the middle term we see that they must be + ; we may then write so much of 
 the required factors ; thus, 
 
 a+ , a+ . 
 
 Now we must have two such quantities for the second terms, that when mul- 
 tiplied together they will produce 6, the last term of the trinomial, and when 
 added will give 5. These terms, then, can only be +3 and +3, and hence the 
 factors are a + 2 and a + 3, and we have 
 
 a 2 4- 5a + 6 = (a + 2)(a + 3). 
 
60 
 
 ELEMENTS OF ALGEBRA. 
 
 Again, take 
 
 a 2 — 5a + 6. 
 
 Here the middle term being — , the last terms of the factors must be — also ; 
 thus. 
 
 a 2 — 5a+6=(a— 2)(a— 3). 
 
 When the last sign in the trinomial is — , thus ; 
 
 a 2 +2a— 8, 
 
 the last terms must be unlike to give a — quantity, and the + must be the 
 greater to give +2 in the middle term. Those terms must thus be, +4 and 
 —2, and we have, 
 
 a 2 +2a— 8=(a+4)(a— 2). 
 
 Examples. 
 
 1. x 2 + 3x + 2=(x+2)(x-\-l). 
 
 2 . x 2 + 11x + 72=(cc + S)(x + 9). 
 
 3. x 2 — 12x + ?>o — (x— 7)(x— 5). 
 
 4. x 2 +2ix—2o—{x + 2b){x—l). 
 
 5. x 2 — f£ + i=(.r— 4)(a— J). 
 
 6. a 2 x 2 — 74ax— 75 = (a.T + l)(«.r— 75). 
 
 7. 9rt 2 +12a— 5 = (3« — 1)(3« + 5). 
 
 10. \/a— 3-v / « + 2 = (\/a— 1)( v'ff— 2). 
 
 11. « _2m + 5a~ m + 6 = (ar® + 3)(a~ m + 2). 
 
 12. 25x 2 — 60j: + 35 = (5x 2 —7)(5x 2 — 5). 
 
 Remark . — When there is a monomial factor present, remove it first and then 
 factor. 
 
 13. a 2 x 2 -l2a 2 x + Zba 2 =a 2 {x-7){x-o). 
 
 14. 5« 2 +15a + 10 = 5(rt + l)(« + 2). 
 
 15. 2x 2 y 2 -^^- + ^|~ = 2t/ 2 (.r-i)(.T-i). 
 
 16. a 2 \To— ZaVb + 2^/b= Vb(a — l)(fi — 2). 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 61 
 
 61. The Division of the Difference of like Powers. 
 
 The difference of the like powers of any he o quantities is always 
 divisible by the difference of the quantities themselves. 
 
 For, let a and b be any two quantities, and m be any positive whole 
 number. The difference of the like powers will be a m —b m . 
 
 Beginning the division, we have, 
 
 a m —b m a—b 
 a m —a m ~'b a" 1-1 
 
 a m ~ x b—b m , or 
 b{a m ~ x -b^). 
 
 Now, if this remainder is exactly divisible by a—b, then, oT—b m is 
 itself exactly divisible by it. But the remainder is b times a m ~ l — 
 b m ~\ so that if the factor a m ~ x —b m -' is exactly divisible by a—b, the 
 entire remainder is divisible by this difference, and hence the divi- 
 dend a m —b m is likewise so divisible. We see, thus, that if a m ~ x —b m ~ x 
 is divisible by a— b, a m — b m is also divisible by the same quantity; 
 that is to say,. 
 
 If the difference of the like powers of two quantities is exactly di- 
 visible by the difference of the quantities themselves, then, the differ- 
 ence of such powers greater by unity, is also exactly divisible by the 
 difference of the quantities. 
 
 But we know that a~—b~ is exactly divisible by a—b; hence, 
 from the hypothetical proposition just established, a 3 —b 3 must like- 
 wise be so divisible, and hence « 4 — d 4 must be, and so on to in- 
 finity ; which was to be proved. 
 
 This method of proof is called Mathematical Induction. 
 
 The form of the quotient will be, 
 
 ft 711 Iff 1 
 
 — — a r-^J r a m - i b + a m - 3 b z + + 
 
 a—b 
 
 62. The Division of like Powers. 
 
 The following propositions may also be readily demonstrated: 
 
 1. The differejice of the like even powers oftivo quantities, is always 
 divisible by the sum of the quantities. The form of the quotient 
 will be, 
 
 /,2m l-2m 
 
 - 'b + a? m ~*b + • • • • -J*— *. 
 
 a + b 
 
 /y4 7i4 
 
 f - =« 3 -a*b + ab* -b 3 . 
 
 a + b 
 
62 
 
 ELEMENTS OP ALGEBRA. 
 
 2. The sum of the like odd powers of two quantities is always di- 
 visible by the sum of the quantities themselves. The form of the 
 quotient will be, 
 
 =a !m — a 2 "* _1 £ + a 2m_2 Z> 2 — • • • + J 2 ™. 
 
 a + b 
 
 a \ =a i — a 3 b + a 2 b 2 — ab 3 + b i . 
 a + b 
 
 Examples. 
 
 Write out the developments of the quotients indicated. 
 
 x 2 —y 2 x 3 — y 3 
 
 1. — —x + y, — =x 2 +xy+ y 2 . 
 
 x—y J x ~y 
 
 n x 3 + y 3 „ „ x i — y i 
 
 2. : — =x 2 — xy + y 2 , — 
 
 x + y J J x + y 
 
 Q/y37.3 ^3 
 
 3. — = 4a 2 £ 2 + 2abc + c 2 . 
 
 2ab—c 
 
 4. -~ + '’ 2 = a 4 -2a 3 + 4a 2 - 8a + 16. 
 
 a + 2 
 
 5. Factor a G —b c , x 3 — l, 1— a’ 3 , 8a 3 — 27. 
 
 6. a 5 b—9b 3 , 12x 4 -192, 
 
 -x 3 —x 2 y + xy 2 —y 3 . 
 
 7. a~ 3 -b- 3 , 
 
 8 . 
 
 a i — b i (a 2 —b 2 )(a 2 +b 2 ) (a— b)(a + b)(a 2 +b 2 )' 
 
 a 8 —b 3 _(a 4 — b i )(a i +b i ) (a 2 + b 2 )(a 2 —b 2 )(a 4 + b i ) _ 
 
 '' a 2 + 2ab + b 2 ( a + U)(a + b ) 
 
 ( a 2 + b 2 )(a + b)(a—b(a 4 + 1 > i ) 
 
 (a + b) (a + b ) 
 
 (a + b)(a + b) 
 
 63. The Binomial Formula. 
 
 The process of raising a binomial to any power may be greatly 
 shortened by using what is called the Binomial Formula. By actual 
 multiplication we have the following developments: 
 
 1st Power (a + b) =a + b 
 
 2d “ (a + b) 2 =a 2 +2ab + b 2 
 
 3d “ (a + b) 3 =a 3 +Za 2 b + 9ab 2 +b 3 
 
63 
 
 TRANSFORMATIONS OF POLYNOMIALS. 
 
 4th. Power (a + Z») 4 =« 4 +4a 3 5 + 6a 2 5 3 +4«5 3 
 
 5th “ (a + b) 5 —a 5 + 5a i b + 10a 3 b 2 + 10a 2 Z» 3 +5«5 4 + b 5 . 
 
 Examining any one of these developments, as the last, we see that 
 a (called the leading letter) appears in the first term with an expo- 
 nent equal to the exponent of the power of the binomial in that case, 
 and that this exponent goes on diminishing by unity to the last term, 
 in which it enters to the zero power, that is, not at all; thus, 
 
 a 5 o 4 a 3 a 2 a 1 a°. 
 
 With b the order is just reversed ; thus, 
 
 b° b 1 b 2 b 3 5 4 b 5 . 
 
 It will be observed, further, that any numerical co-efficient may be 
 found from the preceding term by multiplying the co-efficient of 
 that term by the exponent of a, the leading letter in that term, and 
 dividing this product by the number of terms preceding the re- 
 quired term ; thus, to find the co-efficient of the fourth term, in the 
 development of the fifth power above, multiply 10, the co-efficient of 
 the third term, by 3, the exponent of a in that term, and divide the 
 product 30 by 3, the number of terms preceding the fourth, and we 
 have 10, the co-efficient required. In like manner, the co-efficient 
 of any other term may be found, remembering that the co-efficient 
 of the first term is always unity. Generalizing, we may write, 
 
 (a+b) m —a m + ma m ~ l b + m ^ - ^ -a' n ~-b 2 + 
 
 — — pjTg , -a m ~ 3 b 3 + +b m . 
 
 This is The Binomial Formula, and it may be rigidly demon- 
 strated to be true, whether m be entire or fractional, positive or 
 negative. When m is fractional or negative, the number of terms 
 will be infinite. 
 
 The power of any binomial may be developed by means of this for- 
 mula. For example, let us take, 
 
 (3rc 2 — 2«5 2 ) 4 . 
 
 Let ox 2 — a and —2 ab 2 —b. 
 
 We have at once, 
 
 (a + b) i =a i +4« 3 5 + 6rf 2 Z> 2 +4 ab 3 + b l . 
 
64 
 
 ELEMENTS OF ALGEBRA. 
 
 Now in this, writing for a and b their values as above, we have, 
 (3a; 2 — 2a& 2 ) 4 = (3a: 2 ) 4 + 4(3a; 2 ) 3 (— 2a5 2 ) + 6(3a; 2 ) 2 (— 2a£ 2 ) 2 + i(3z 2 ) 
 (~2a& 2 ) 3 + (-2aS 2 ) 4 . 
 
 (3x 2 —2ab 2 ) i =81x 8 —216x e ab 2 +216x±a 2 b i —96z 2 a 3 b s + 16a i b s . 
 
 Examples. 
 
 1. Develop (a 2 +1) 3 , («6 2 -2) 4 , (|+|) 3 . 
 
 2. Develop (V- 1 ^ , (y/a-l) 3 , (Va + V^) 4 . 
 
 3. Develop ( a + b ) 8 , (a + ^^a^ + ^a^ -1 #— ^a^~ 2 b 2 +etc. 
 
 4. Develop (a + &) _1 , (a + &) -1 =a -1 — a~-b+a~ 3 b 2 — etc. 
 
 5. Develop (a 2 '— l) 3 , (V^+^J , (x—y) x . 
 
 6. Develop (2-3a;)- 2 , (V^-V^) 3 . 
 
 7. Develop (1— a)^, (a + 1) 3 , (a + b)”. 
 
 64. The Powers of Polynomials. 
 
 The power of any polynomial may he developed by the use of the 
 binomial formula. For example, let it be required to find the third 
 power of 2a 2 — 4aZ>4-3c 2 . 
 
 Let 2a 2 — ^ab—a and 3c 2 — b ; then 
 
 (a + b) 3 =a 3 + 3a 2 b + 3ab 2 + b' s . 
 
 Replacing the values of a and b, we have 
 (2a 2 - 4ni + 3c 2 ) ■ 3 = (2a 2 - 4 ab ) 3 + 3 (2a 2 - 4 ab ) 2 (3c 2 ) + 3(2a 2 - 4a£) 
 (3c 2 ) 2 + (3c 2 ) 3 . 
 
 Examples. 
 
 1. Develop (a + 5 + c) 3 . 
 
 2. Develop (1— 2a;— 3» 2 ) 2 . 
 
 3. Develop (a’" + 5’ > ) 3 . 
 
 4. Develop (a 2 + l+a~ 2 ) 2 . 
 
 5. Develop (VTi+l — V^) 3 - 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 65 
 
 65. Common Multiple. 
 
 The Common Multiple, or a Common Dividend of two or more 
 quantities, is any quantity which is exactly divisible by each of 
 them: such a dividend may always be found by multiplying the sev- 
 eral quantities together ; thus, 4« 2 , 'dab and 3b 2 c would give 3 6a 3 b z c 
 for a Common Dividend. 
 
 The Least Common Multiple is the least quantity which is so di- 
 visible; thus, V2a 2 b 2 c, is the least Common Multiple, or Dividend, 
 of the quantities above given. 
 
 A common multiple must obviously contain every factor which 
 enters any of the quantities by which it is to be divisible, as many 
 times as such factor enters any one of the quantities ; hence, if all 
 the factors are made to appear, we have but to take each factor the 
 greatest number of times it enters any one of the quantities, and 
 multiply the results together ; thus, a 2 + 2ab + b 2 , a 2 —b 2 and a 2 — 
 2 ab + b 2 , resolved into their simplest factors give, 
 
 (a + b) {a + b), (a + b) {a—b), {a — b) {a—b). 
 
 Here ( a + b ) enters the first expression twice, and ( a—b ) enters the 
 third twice : the second expression containing no other factors than 
 these, we have, 
 
 (a + b) 2 (a-b) 2 =2a 2 +2b 2 . 
 
 It is commonly better, however, to retain the indicated product as 
 in the first member of this equation, than to develop it as in the 
 second member. Thus, w r e may say that, 
 
 To find the Least Common Multiple of two or more quantities, re- 
 solve the quantities into their simple factors, and take each factor the 
 greatest number of times it is found in any one of the expressions. 
 The product of these factors will be the multiple required. 
 
 Examples. 
 
 Find the Least Common Multiple of the following : 
 
 1. a 2 —b 2 , ab + b 2 , a 2 —ab, { a + b ) 2 , {a + b){a— b), b{ci + b), a{a—b), 
 
 {a + b) 2 . Ans. ab{a-'-b) 2 {a—b). 
 
 2. x 2 +x—2, x 2 +2x— 3. Ans. (x—l)(x+2)(x + 3). 
 
 3. {x 2 — a) 2 , x 2 —a, 5. Ans. 5{x 2 —a) 2 . 
 
 4. x~ +Qx+20, x 2 + ox — 4, x~ T 4a: 5. 
 
 Ans. (.r + 4)(a: + 5)(.T— 1). 
 
6G 
 
 ELEMENTS OF ALGEBEA. 
 
 5. 6a 2 — 23a + 7, Ga 2 —25a + 14. Ans. (. 2a-7)(3a-l)(3a-2 ). 
 G. 3a, (a + £) 2 , a 2 —2ab + b 2 . Ans. 3a(a + b) 2 (a—b). 
 
 66. Operations upon Algebraic Fractions. 
 
 The student is supposed to be already familiar -with the manage- 
 ment of fractions in arithmetic. Their treatment in algebra is al- 
 together the same. The transformation, however, of the sum or 
 difference of two fractions into a single expression, requires some 
 care, and we shall now give a number of examples for practice. 
 
 Let the student remember to first convert the fractions into equiv- 
 alent fractions with a common fractional unit, and then to add or 
 subtract the numerators, placing the result over the common denom- 
 inator. 
 
 Examples. 
 
 ^ 1 1 _ a — l) a + 1) _a—b+a + b _ 2a 
 
 a + b a — b a 2 — b 2 a 2 — b 2 a 2 — b 2 a 2 — b 2 ' 
 
 „ 1 2 _ x + 2 2(.r + l) _x + 2— (2(a; + l)) 
 
 x + 1 x+2 (a; + l)(rc + 2) (.r + l)(a; + 2) (x + l)(a; + 2) 
 
 —x 
 
 (x + l)(x + :>) ' 
 
 1— a 1 + a _(1— a){a + b) — (l + a)(rc— b) 
 +rZ-J+~la + b) 2 ~ (a-b){a + b)~ 
 
 a + b— a 2 —ab — a + b— a 2 +ab _ 2b— 2a 2 
 
 (a — b){a + b) 2 (a—b)(a + b) 2 ' 
 
 . 1 — a 1 + a 1 
 
 4 . 1 1 : 
 
 1 + a 1 —a (1 + a) 2 
 
 1 — 5 a — 4a 2 
 
 (1— a 2 )(l + a) ' 
 
 _ a 2b b 
 
 5. — -i j • 
 
 a+b a—b a+b 
 
 „ a + 3 2a— 5 
 
 6 . — 1 — - 
 
 o 3a 
 
 1— a)(l— a 2 ) — (l+a)(l + a) 2 + 1— a 
 ( 1 + a) (1 + a) (1 — a) 
 
 A ns. 
 
 a + b 
 
 A ns. 
 
 a—b 
 
 3a 2 + 19a— 25 
 15a 
 
 a+b a—b 
 a—b a + b' 
 
 _ I ns. 
 
 4 ab 
 
 a 2 — b 2 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 67 
 
 x I x—a\ 
 
 V \ X -—} 
 
 8. 3x + y 
 
 9. * + am ~ l 
 
 10 . 
 
 a m + b m ' a im —¥ m 
 
 1 1 _ 
 
 a*— I)* a+ + b s 
 
 u Va Vb 
 
 \/a—Vb y/a + Vb 
 
 12 - — h + l fr- 
 
 a— b a+b 
 
 Ans. 2 x+ 
 
 Ans. 
 
 cx + bx—ab 
 be 
 
 2a m —2b m 
 
 a im _b im 
 
 Ans. 
 
 Ans. 
 
 2b~ 
 a—b ' 
 
 a + b 
 a—b ' 
 
 Ans. r . 
 
 ab 
 
 67. Essential Sign. 
 
 The sign by which one expression is connected with another is 
 called the Sign of Operation ; the sign resulting from the combina- 
 tion of this sign with the sign of the quantity itself is called the 
 Essential Sign of the expression: thus, in the expression 
 
 a—(—b), 
 
 the second term is to be subtracted from the first, and the sign of 
 operation is — ; but removing the parenthesis the two — signs com- 
 bine and give +b. This resulting sign is the essential sign of the 
 term. 
 
 We sometimes cannot tell the essential sign of an expression abso- 
 lutely ; thus, in 
 
 a 
 
 X — “7 , 
 
 b—c 
 
 the essential sign of the fraction, and so the sign of x, will depend 
 upon the relative values of b and c. If b>c it is + ; if 5<c it is — . 
 
 Examples. 
 
 1. What is the essential sign of x in x — 
 tive and b>c? when 5<c? 
 
 a 
 
 b-c 
 
 a—b 
 
 , when a is nega- 
 
 2. What is the essential sign of x in x— " — - , when a>b and 
 
 c — ct 
 
 c<d? when a<b and c<cl? when a<b and c>d? when a>b and 
 c>d? 
 
68 
 
 ELEMENTS OF ALGEBKA. 
 
 68. Imaginary* Quantities. 
 
 We have seen that the even root of a negative quantity is imagi- 
 nary. Let us now take an imaginary quantity of the second degree, 
 as V — and multiply it by itself, applying the general law of signs ; 
 we shall have, 
 
 V —a x a/— a— */af= ±a. 
 
 But the square of the square root of a quantity is, from the defi- 
 nition, the quantity itself : so that, 
 
 {V— a) 2 = — a. 
 
 We should thus have —a=±a, or, taking the upper sign, — a— 
 + a, which is impossible. 
 
 To avoid this difficulty we must modify the law of signs in the 
 multiplication of imaginary expressions. 
 
 Now, every imaginary expression of the second degree may be put 
 under the form of, 
 
 a's/ — 1 : 
 
 in which a represents any real quantity, whether its exact value may 
 be found or not; and a/— 1, an imaginary quantity, which is called 
 the imaginary factor. 
 
 For example, let us have, 
 
 V — ab. 
 
 This may be written, 
 
 V —ab=Vcib x ( — 1 ) — ^/ab . a/— T. 
 
 Again. 
 
 V— (a + b) 2 = V(a + b) 2 . V"— 1 = (a + b) a/— 1. 
 
 Let us now multiply V — 1 by itself any number of times, until 
 we discover the law of such combinations. 
 
 Since the square of the square root is the quantity itself, 
 a/— 1 x V — 1 = (a/ — l) s = — 1. 
 
 The third power will be found by multiplying the second by the 
 first; hence, 
 
 (a /^) 2 x = = 
 
 In like manner, 
 
 ( a/ 11 !) 4 = ( 2 ( 2 = ( - 1) ( - 1) = + 1. 
 
TRANSFORMATIONS OF POLYNOMIALS. 
 
 69 
 
 We may in the same way continue the operation, and, for the sev- 
 
 (y=I) 3 =(V^ 2 (y=I)= = --iy=I= -V=i. 
 
 (v^) 4 =(y^) 2 (v~i) 2 =(-i)(-i)= +1. 
 
 ( v ^ ) 5 = ( v ) 2 ( v ^1 ) 3 - ( - 1 ) ( - v “y ) - v ^ • 
 
 (y=I) 6 =(V^i) 2 (V^i) 4 -(-i)( + i)= -i. 
 
 (V :z I) 7 =(V = i) 6 (a/ :i i)=(-i)(V^i)= -V^l. 
 (v^iy=(v~i) i (v^ ; i) i ={+i){+i)= +i. 
 
 It will be observed that the last four results are but a repetition 
 of the preceding four, and so they would continue to repeat them- 
 selves in sets of fours. 
 
 The multiplication of imaginary quantities is effected by the use 
 of this imaginary factor ; that is, we first resolve each expression 
 into two factors, one real and the other the imaginary factor, V — 1 ; 
 we then combine the real factors by the ordinary laws, and the 
 imaginary factors according to the laws just deduced for the forma- 
 tion of its several powers. 
 
 If, however, only one of the quantities is imaginary, the ordinary 
 rules apply. 
 
 eral powers of V—l, we shall have, 
 
 (V— l) 1 = 
 
 (V^1) 2 = 
 
 V=L 
 
 -1. 
 
 Examples. 
 
 1. Multiply 5 a/— 4 by 2 a /— 4. 
 
 oa / iV” — 1 x 2 \/4 a/ — 1 
 10a/=Tx4a/=I 
 40(a/ — l) 2 = — 40. Ans. 
 
 2. Multiply a\/ — b by c^/ — cl. 
 
 a*fb -V -1 x cV cl *a/— 1 
 
 acVbd-(V — l)' 
 —ac*fbd. Ans. 
 
 3. Multiply (a + b)V—{ci + b) by (a + b)V—(a + b). 
 
 Ans. — ( a + b ) 3 . 
 
70 
 
 ELEMENTS OF ALGEBRA. 
 
 5. V-2-V-2 V-2 = -2vW-l. 
 
 6. («— V — b)(ju + a/~ b)=a z + b. 
 
 7. («+ V— #)(« + V— b)=a s +2av^— 6 — 
 
 8. 3a/ 5 x 5 a/— 2=15^/— 10. 
 
 9. (3 + a/— 5)1-2— a/— 5) = 11 — V — 5. 
 
 10. a/ — « 3 -a/ — £ 2 -V —c 2 -a / d 2 -^/ —f 2 =abcdf. 
 
 11. (a/ 3 3 + V 3 i)(a/ Z: 12-a/^4:)=-1. 
 
 12. (a/—2 — a/3)(\/2 — V — ^) = a/^— 9 + a/— 4- 
 
 The division of imaginary quantities of the second degree is man- 
 aged in a corresponding way, being careful to introduce the im- 
 aginary factor a/ — 1. 
 
 Examples. 
 
 1. Divide 10 a/ — 1 by 5 a/ — 1. 
 
 2. Divide a a/ — by ca/ — <7. 
 
 3. Divide 4 a/ —a 2 b 2 , by 2 a/ — i 3 . 
 
 4. Divide 9 a/ — 10 by 3 a/ —2. 
 
 Nns. 2. 
 
 , «a/£ 
 
 cVd 
 Ans. 2a. 
 Ans. 3 a/s'. 
 
 5. Divide « 2 +5 by u— v 7 — A A«s. a + V— A 
 
 Imaginary expressions of a higher degree than the second, may 
 be treated in a corresponding manner. 
 
 SECTION VI. 
 
 THE ROOTS OF NUMBERS. 
 
 69. The extraction of the roots of numbers properly belongs to 
 Arithmetic ; but it may be well to show here the rationale of the 
 process. 
 
 Since the square root of 100 is 10, we know that the square root 
 of any number which contains but two digits, must be expressed by 
 
THE ROOTS OF NUMBERS. 71 
 
 a single figure. We can readily find the root of such a number by 
 trial, when it is a perfect power. 
 
 AYhen the number contains three or more figures, there will be at 
 least two figures in its root ; that is, the root will contain a certain 
 number of tens and a certain number of units. 
 
 Then, let n be any number, and a the number of tens, and b the 
 number of units in its root. The square root of n will be (a + 5), and 
 we may write, 
 
 n = {a + b)~ =a~ + 2ab + b 2 • • • • (1). 
 
 We see from this that the number contains a 2 , that is, the square 
 of the tens, and 2 ah, that is, twice the product of the tens and units j 
 and b 2 , the square of the units. 
 
 Let it now be required to find the square root of 1441. We shall 
 begin by finding the number of tens in the root; 
 and since the square of a single ten ' gives units 1444i38 
 followed by two 0’s; thus, (10) 2 = 100, the square I 
 
 of any number of tens can give only 0’s in the last 60|544 
 
 two places. We may then place a point over the 480 
 
 third figure from the right, as here seen, to show 04 
 
 that the last two places, 44, may be regarded as oc- _G4= (8) 2 
 cupied by 0's, in our search for the tens of the 0 
 
 root. 
 
 Noav, since the square of the tens of the root must be found ex- 
 clusively in 14, the square root of the greatest perfect power in 14, 
 which is 9, will be the tens of the root. 3, then, is the first figure of 
 the root, and 30 will correspond to a in formula (1). Write the 3 to 
 the right, as shown. Now, squaring it, subtract the result from 14, 
 and bring down the 44. We have really subtracted (30) 2 = 900, and 
 the remainder is what is left after taking away a 2 from the formula. 
 It must then correspond to the 2ab + b 2 o‘L the formula. We should 
 be able to find b at once from this remainder, if it were 2ah alone, by 
 dividing by 2a, (2 x 30), a being now known. But, at any rate, since 
 b 2 is quite small compared with 2 ab, we shall not come far from b by 
 dividing the remainder, as it stands, by 2a. Then, doubling 30, and 
 dividing the remainder by the result, we get 9. Now, multiplying 60 
 by 9 (giving 2 ab), and subtracting the result from the remainder, we 
 shall have 4 left. But this second remainder must be equal to the 
 square of the units, b 2 ; in this case (9) 2 = 81. If our original number, 
 then, is a perfect power, 9 is too great. 
 
72 
 
 ELEMENTS OF ALGEBRA. 
 
 Trying 8 we find 64 fora remainder, and tins is just equal to (8) 2 : 
 38, then, is the root required. 
 
 To shorten the last part of the operation we may double the 3, and 
 
 divide the remainder, exclusive of the last figure, by it ; 1444;38 
 
 then write the result in the unit’s place, after 6. Now, 
 
 when we multiply 68 by 8, we form at once (2 a + b)b 
 
 =2 ab + b 2 . 0 
 
 When the number requires more than two places in the root, we 
 might apply the same reasoning to the discovery of the two figures 
 of the root of the highest denomination, and then, regarding all 
 these as forming one denomination, proceed as before. This is ac- 
 
 complished by simply pointing off from the right, . . . 
 
 in places of two figures each, and continuing the 15129 123 
 
 operation as above. If we had two more places of 1 
 
 figures in the above example, we should bring them 22; 51 
 down, and then double 38, and divide the remain- 44 
 
 der, exclusive of the right hand figures, by the re- 243] 729 
 suit, writing it in the root, and also in the new trial 729 
 
 division, and so proceed as before. 0 
 
 70. Extraction of the ?fth Hoot of Numbers. 
 
 Let it be required to find any root of a number, JV. The root will 
 consist of a tens and b units. Thus we shall have, 
 
 N—(a + l)"=a n + na n -'b + , etc. 
 
 Now the wtli power of the tens must have O's in the last n places, 
 so that in looking for the tens of the root we may point off the last 
 n places of the number. We may now find the tens of the r jot by 
 finding the highest perfect root in the left hand period. Subtracting 
 the «th power of the tens so found from the number, we shall have 
 na"~ l b + , etc., left. Now, by forming n times the «■ — 1 power of 
 the tens, and dividing the remainder from it, we shall have the units 
 of the root, or something too great. By raising the trial root so 
 found to the «th power, we shall find whether it produces the num- 
 ber or not. If too great, the figure in the unit’s place must be 
 diminished. 
 
 We shall now give an example in cube root. Any other root may 
 be found m like manner. 
 
THE BOOTS OF NUMBERS. 
 
 73 
 
 Let us find the cube root of 12977875. 
 
 It will be observed 12977875 [235 
 
 8 
 
 that only the first fig- o 22 _ , 0175 
 
 ure of the second period 12977 = 1st two periods. 
 
 (23) 3 = 12167 
 
 is brought down for a 3 x ( 23 ) * = ISBTjsIoS 
 
 first remainder. This 12977875 = The three periods. 
 
 . , (235)3 = 12977875 
 
 is because 12, the first q 
 
 divisor, is really 1200, so that the remaining two figures would be 
 cut off in dividing, if brought down. 
 
 Examples. 
 
 1. Find the square root of 651219. Ans. 807. 
 
 2. Find the cube root of 12167. Ans. 23. 
 
 3. Find the cube root of 421875. Ans. 75. 
 
 4. Find the square root of 1058841. Ans. 1029. 
 
 5. Find the cube root of 256047875. Ans. 708. 
 
 6. Find the fifth root of 248832. Ans. 12. 
 
 Remark. — Let it be remembered that 
 
 \ / V / « = v / «, and that, therefore. 
 
 when the index is a multiple of two or more factors, we may find the roots in- 
 dicated by those factors, successively, instead of extracting the entire root at 
 once. 
 
 7. Find the sixth root of 244141625. 
 
 Ans. 25. 
 
 8. Find the eighth root of 214358881. Ans. 11. 
 
 Remark . — If the number is partly or altogether decimal, begin to point off 
 from the decimal point going to the left for the entire part and to the right for 
 the decimal. Any number of 0’s may be added to the decimal and the opera- 
 tion thus continued to any degree of approximation. 
 
 9. Find the square root of 657.4096. Ans. 25.64. 
 
 10. Find the square root of .140625. Ans. .375. 
 
 Remark . — A common fraction may be converted into a decimal and the root 
 extracted to any desired degree of accuracy. 
 
 11. Find the square root of 4j- 9 -. 
 
 Ans. 2.3604. 
 
74 
 
 ELEMENTS OE ALGEBBA. 
 
 12. Find the square root of J- to five places of decimals. 
 
 Ans. .74535. 
 
 13. Find the square root of 7, to four places of decimals. 
 
 Ans. 2.6457. 
 
 71. No Exact Boot of an Imperfect Power. 
 
 The a/ 5, or or the root of any imperfect power of the degree 
 indicated, cauuot be found exactly, either in entire or fractional quan- 
 tities; hut we may approximate such a root as nearly as we please: 
 
 for example, -v/7^2-64575131 + and so on indefinitely. 
 
 Such quantities are said to be incommeasurdble, and are commonly 
 called surds. 
 
 The distinction between an imaginary quantity and a surd is, that 
 we may obtain the root of a surd as nearly as we please without being 
 able to find it exactly, while we cannot make the first movement 
 towards finding the root of an imaginary expression. 
 
 Let us now prove that we cannot find the exact root of a surd. 
 
 Let p be any whole number whose nth root is | , a fraction having 
 
 no common factor in the numerator and denominator. ^ cannot, then, 
 be reduced to an entire quantity. TVe shall have, 
 
 Raising both numbers to the nth power, since they must still be 
 equal, we shall have 
 
 a" 
 
 Now, in raising any number to a power, we but repeat the factors 
 composing it, a certain number of times, introducing no new ones, 
 so that a n and b n are still prime, with respect to each other ; that is, 
 
 (l n 
 
 they have no common factor, and thus, ^ is an irreducible fraction ; 
 
 so that we have p, a whole number, equal to an irreducible fraction, 
 which is impossible. Then, since y ' p cannot be a whole number 
 nor a fraction, it cannot be found exactly at all. 
 
EQUATIONS. 
 
 75 
 
 SECTION VII. 
 
 EQUATIONS. 
 
 72 . An Equation is the indicated equality of two algebraic ex- 
 pressions ; thus, 
 
 ax+by=c+d 
 
 is an equation. 
 
 The sign = divides the equation into two parts, called Members : 
 the part written first being the First Member, and that following 
 the sign of equality, the Second Member. 
 
 73. The essence of an equation (its life, so to speak) lies in the 
 equality of its members. Any operation may be performed upon it, 
 so long as the equality is preserved intact. 
 
 The following are self-evident truths, called Axioms : 
 
 1. If equal quantities be added to both members of an equation, 
 the equality will still subsist. 
 
 2. If equal quantities be subtracted from both members, the 
 equality will still subsist. 
 
 3. If both members be multiplied by the same quantity, the 
 equality will still subsist. 
 
 4. If both members be divided by the same quantity, the equality 
 will still subsist. 
 
 5. If both members be raised to the same power, the equality 
 will still subsist. 
 
 6. If the same root of both members be extracted, the equality 
 will still subsist. 
 
 All operations upon equations are founded upon these axioms. 
 
 74. The First Transformation. , 
 
 There are four principal transformations to which equations are 
 submitted. We shall consider them m their order. 
 
 I. The object of the First Transformation of equations is to clear 
 an equation of fractional quantities. 
 
 Take the equation, 
 
 x 
 
 a 
 
 Now, we may multiply both members by any quantity, and still 
 
76 
 
 ELEMENTS OF ALGEBRA. 
 
 preserve the equality. Let us, then, multiply each term by aid, the 
 common multiple of all the denominators. We shall have, 
 
 There is now a common factor in the numerator and denominator 
 of each fraction. Stinking them out, we have, 
 
 an equation in which all the terms are entire. 
 
 It is generally more convenient to strike out the common factor 
 from the multiple before multiplying by it. We may thus say, that 
 to clear an equation of fractions, or, 
 
 To make the first transformation of an equation, form the least 
 common multiple of all the denominators; divide this by each denom- 
 inator in succession and multiply each term respectively by the re- 
 sult. Entire terms are to he multiplied by the common multiple as it 
 stands. 
 
 It is sometimes better to multiply each numerator by all the de- 
 nominators except its own, and the entire terms by all the denom- 
 inators. 
 
 Clear the following equations of fractions. 
 
 Explanation— Here tlie least common multiple is 4b. Dividing it by 2 and 
 multiplying the numerator of the first term by 2b, we have 2 bx for the first term 
 of the result. The other terms are found in like manner. 
 
 abd.x aJid.u alula 
 
 bdx—ady=abc+ abdfi 
 
 Examples. 
 
 Ans. 2 to — 4a + 3 be— by — 4ib. 
 
 Ans. 2a-x + c—\Day + 2ax. 
 
 Ans. 3x—Za—x-+x=l2ax. 
 
 1 _« + £ 
 
 Ans. a—x—a s +x 2 = (a+x) 2 . 
 
 a 
 
 a 2 
 
 c 
 
 Ans. acVz—cV%—c=5a~b. 
 
EQUATIONS. 
 
 77 
 
 V x + y 
 
 1 
 
 a 
 
 / Vx+y 
 
 Ans. x + y—aVx+y—a. 
 
 Ans. 2b 2 x 2 — a 2 y~ 1 =12ab 2 . 
 
 75. The Second Transformation. 
 
 IT. The object of the Second Transformation of equations is to trans- 
 pose terms from one member of an equation to the other. 
 
 Take the equation, 
 
 How, as we may add the same quantity to both members without 
 affecting the equality, let us add — 2 by to both members. We shall 
 have, 
 
 In the second member we now have two quantities equal with 
 contrary signs, which we may cancel, and we shall thus have, 
 
 It will be observed that 2 by has disappeared from the second 
 member, and appeared in the first with its sign changed. 
 
 How suppose we desired to move the term — cd from the first 
 member. Adding cd to both members, and cancelling as before, we 
 have, 
 
 The term has disappeared from the first and appeared in the 
 second with a contrary sign. Thus, 
 
 We may transpose any term from one member of an equation to the 
 other by changing its sign. 
 
 Examples. 
 
 Transpose all terms containing an unknown quantity to the first 
 member and all others to the second. 
 
 ax —cd—2by + a. 
 
 ax—cd—2by—2by + a—2by. 
 
 ax—cd—Tby—a. 
 
 ax—2by=a + cd. 
 
 1. ax—b—Zx 2 — y. 
 
 2. a + 25 = — x 2 +y. 
 
 Ans. ax—3x 2 +y=b. 
 Ans. x 2 — y—— a— 25. 
 
ELEMENTS OF ALGEBRA. 
 
 3. 21 a 2 b — xy—l—x. 
 
 . a ~ , x 
 
 4. — r 1 = ---4. 
 
 o 2 
 
 —xy—x=l—21a 2 b. 
 
 A X A a ~ i 
 
 Ans. - = 1 — - — 4. 
 2 <y 
 
 Re mark. — Equations are usually cleared of fractions before transposing, but 
 transpositions may be made at any time. 
 
 Transpose all the terms of the following to the first member: 
 
 5. ax 2 — 1 = bx— .cx z . Ans. cx z + ax 2 — bx— 1=0. 
 
 6. 0—a+jy—\y 2 + y z —y /k . Ans. y i — y 3 +\y 2 — jy—a= 0. 
 
 1 ) /v /v y 
 
 7. *Ja + b— — {ci—b)x 2 + V a-b-y. 
 
 Ans. ( a — b)x 2 — Va—b-y + Va + b=0. 
 
 8 . 1 : 
 
 a + b 
 
 a—b 
 
 , a+b x _. 
 
 Ans. r +1=0. 
 
 x a — b 
 
 Remark.— If there are indicated products in an equation, it is generally better 
 to develop such expressions before transposing. 
 
 In general, transpose unknown terms to the first member, and known terms to 
 the second. 
 
 9 (a + b)' 2 x c __ (x—1) x 3 . 
 
 c 5 
 
 5(a + b) 2 x— 5r 2 = (x— 1) x 3c. 
 
 5a 2 x + 10abx + 5b 2 x—5c 2 =3cx—3c. 
 Ans. 5a 2 x + 10abx + 5b 2 x— 3cx=5c 2 — 3c. 
 
 76. The Third Transformation. 
 
 III. The object of the Third Transformation of equations is to 
 so change an equation that the several powers of the same unknown 
 quantity shall enter it but once. 
 
 Let us take an equation with several powers of the same unknown 
 quantity ; thus, 
 
 3ax 2 — bx + x z — 2x + x 2 —ax z — 1. 
 
 Writing the terms containing the same power of the unknown quan- 
 tity together, and factoring with respect to these several powers, we 
 have, 
 
 (l—a)x z + (3a+l)x 2 —{b + l)x 2 =l. 
 
 Or taking a numerical equation, 
 
 5x 2 — 2x 2 +10.r— x=5 ; 
 
EQUATIONS. 
 
 79 
 
 Simplifying, we have, 
 
 3x 2 +9x=5. 
 
 If an equation has any number of unknown quantities, the same 
 course may be pursued. 
 
 Practically to effect the third transformation, 
 
 Gather together the terms containing the like 'powers of the same 
 unknown quantity or quantities, and factor with respect to those 
 several powers. 
 
 Examples. 
 
 Submit the following to the Third Transformation, transposing 
 unknown terms to the first member, and known to the second. 
 
 1. ax— l + cx 2 — x + l=2x s . Ans. (c—2)x 2 — (l—a)x=h—l. 
 
 Caution . — Be careful in putting on or taking off a parenthesis with the — 
 sign before it. 
 
 2. l = a — V^-x 2 — ax + dx 2 — x 2 + ^fa-x. 
 
 Ans. (l + V% — d)x 2 — (Va— a)x= a— 1. 
 
 3. 3x 2 — 2x + l=a 2 x 2 —h 3 x + cx+f. 
 
 Ans. (3—a 2 )x 2 + (b 3 —c—2)x=f=l. 
 
 t it i i 
 
 4. d 2 x 2 —cx—ax 2 — 2x + 25. Ans. (2— c)x+ (a 2 — a)x 2 —25. 
 
 5. 3a; 2 — 2x + l — 7x 2 — 4x. Ans. —4x 2 +2x— — l. 
 
 77. Fourth Transformation. 
 
 IV. The object of the Fourth Transformation of equations is to 
 make the co-efficient of the highest power of the unknown quantity 
 unity. 
 
 Let us take an equation upon which the first three transforma- 
 tions have been already performed ; thus, 
 
 (3a— b)x 2 — (4 + c)a;=l. 
 
 We may divide both members, that is to say every term, by the co- 
 efficient of x 2 , and we shall have, 
 
 „ 4 + c 1 
 
 — o i x= o I* 
 
 3a — o 3 a — o 
 
80 
 
 ELEMENTS OF ALGEBRA. 
 
 The same course may be pursued in any case; hence, having per- 
 formed the previous transformations, to make the Fourth, 
 
 J Divide each term by the co-efficient of the highest power of the un- 
 known quantity. 
 
 Examples. 
 
 Apply the Fourth Transformation to the following: 
 1. (a + b)x 2 —x=b. Ans. x 2 
 
 a + b 
 
 2. {3 — c 2 )x 3 — {a 2 +b)x 2 +2x—4:d 3 —c. 
 
 a 2 +b 
 
 Ans. x 3 - 
 
 3-c 2 
 
 x 2 + 
 
 3 — c 2 
 
 x = 
 
 b 
 
 a + b’ 
 
 Ad 3 — c 
 3-c 2 ’ 
 
 3. (2 —d 2 )x—a^—b 2 . 
 
 4. 25x=40. 
 
 5. (a — b) n x=c n . 
 
 Ans. x : 
 
 
 2—d 2 
 
 Ans. x = . 
 
 2o 
 
 Ans. x ■ 
 
 (a— b) m 
 
 Perform all four of the transformations in succession on the fol- 
 lowing; transposing unknown terms to the first member and known 
 to the second. 
 
 7. 
 
 1 . 
 
 Ans. x — 
 
 be + bed 
 ac—b 2 
 
 „ a 2 + b 2 c—d , a 
 
 8. 5- X — X + T , 
 
 c c~ b 
 
 Ans. x — 
 
 ac 2 +a 2 bc + b 3 c 
 
 bcl—bc—bc 2 
 
 10 . 
 
 25 
 
 3 
 
 5x _ lO.r 
 
 IT ~ 1 ~ 12 
 
 x. 
 
 Ans. x—— 102. 
 
 Ans. x — — 
 
 88 
 
 12 * 
 
 78. The Change of Signs. 
 
 Let us have the equation, 
 
 — ax+by— — c. 
 
 Multiplying both members by —1, we have, 
 
 ax—by=c, 
 
EQUATIONS. 
 
 81 
 
 an equation in which all the signs have undergone a change. Hence, 
 we may change the signs of every term of an equation, and the equality 
 will still subsist. 
 
 79. Solution of Simple Equations. 
 
 An equation containing but the first poiver of the unknown quan- 
 tity or quantities is called a simple equation, or an equation of the 
 first degree. 
 
 It will be observed, that when we have submitted a simple equa- 
 tion containing a single unknown quantity to the four transforma- 
 tions in succession, the unknown quantity is made to stand alone in 
 the first member, equal to known quantities in the second. We 
 have thus found the value of the unknown quantity, and are said to 
 have solved the equation* 
 
 Then, to solve a simple equation containing but one unknown 
 quantity, we have but to submit it to all four of the transformations 
 in succession. These four steps, in few words, are, 
 
 1. Clear the equation of fractions ; 
 
 2. Transpose unlcnoion terms to the first member and biown to 
 the second ; 
 
 3. Factor the first member with respect to the unlcnoion quan tity ; 
 
 4. Divide both members by the co-efficient of the unknown quan- 
 
 The value of the unknown quantity so found is called the Root of 
 the equation. The root of an equation, then, is such a value of the 
 unknown quantity as will verify the equation ; that is, when substi- 
 tuted for that quantity in the equation it will show that the two 
 members are identical ; thus, the root of 
 
 tity. 
 
 is 
 
 2 b 
 
 2 a + b 2 ’ 
 
 Writing this in the equation, we have. 
 
 Simplifying and clearing of fractions, we have, 
 4 ab + 2b 3 = iab + 2b 3 . 
 
 F 
 
82 
 
 ELEMENTS OF ALGEBRA. 
 
 Thus, the two members are entirely the same ; and the equation 
 must be true. 
 
 An equation may be made to have any two equal quantities in the 
 two members ; thus, cancelling equal terms in both members, we 
 have, 
 
 The verification of a numerical equation is more simple ; thus, 
 solving 
 
 we have x =50. 
 
 This for x in the equation gives, 
 
 0 = 0 . 
 
 By dividing each member by itself, any equation will be 
 
 1 = 1 . 
 
 300-50=250 
 
 250=250. 
 
 Examples. 
 
 Ans. x=3. 
 
 — 3x—12. 
 
 Ans. x=6. 
 
 3 * 4 6 2 3 
 
 Ans. x=3. 
 
 Ans. x = — 
 a 
 
 a + b 
 
 5. a”{x— 1) +ab[x— 2)=b 2 . 
 
EQUATIONS. 
 
 83 
 
 ai ~ x ~ x + 1 = 2 . 
 
 8 . 
 
 x 
 c 2 
 
 7 + 9a; 
 
 Ans. x- 
 
 5c$-2a$ 
 
 
 Aiis. x= 
 
 35 
 
 167' 
 
 80. Degree of Equations. 
 
 The degree of an equation is determined by the greatest number 
 of times the unknown quantity or quantities enter any one term as a 
 factor; thus, 
 
 5x 1x—o ) are 0 £ £jj e g rs £ d e g ree> 
 
 a z x + oi/—4:Z—c ) ° 
 
 \ are of the second degree. 
 ci 2 xy + ox=4:Z—c ) ° 
 
 5x 3 -7x=3 l 
 
 „ , 7 . ■ are ot the third degree. 
 
 a 2 xyz + oxy=iz— c ) ° 
 
 81. Complete and Incomplete Equations. 
 
 A complete equation of any degree is one which contains all the 
 several powers of the unknown quantity from that which determines 
 its degree down to the zero power; thus, 
 
 ax 3 —bx 2 + cx + dx° = 0 
 
 is a complete equation of the third degree of one unknown quantity. 
 x° being iinity, we may write it or not as we please. 
 
 When one or more of the intermediate powers are wanting, the 
 equation is said to be incomplete ; thus, 
 
 x 3 + bx 2 =d 'j 
 
 x 3 —5x=d > are incomplete. 
 x 3 — d / 
 
 82. Complete Equations of the Second Degree. 
 
 When an equation has undergone the four transformations already 
 given, it is said to have its Simplest Form. 
 
 Let us take a complete equation of the second degree containing 
 but one unknown quantity ; thus, 
 
 x 2 
 
 a 
 
 x , 2x , x 2 
 
 -2 + C = d ~- 3 + T 
 
84 
 
 ELEMENTS OF ALGEBRA. 
 
 Submitting it to the four transformations, we hare, 
 
 . ab ab(d-c) 
 
 x +771 V x =—7 "• 
 
 6 (b—a) b—a 
 
 Now, the co-efficient of a; is a known quantity, and we may let 2 \p rep- 
 resent it. So we may let q be equal to the second member. Using 
 these values in the equation, we have, 
 
 x 2 +2 px—q - - (1). 
 
 This is called the Reduced Form of a complete equation of the 
 second degree containing but one unknown quantity. The terms, 
 however, may have different signs. 
 
 In any case, after making the sign of the first term plus, if not 
 so already, the other two terms, 2px and q, must present one of the 
 four following phases : 
 
 Both plus; 
 
 2px minus and q plus; 
 
 2px plus and q minus; 
 
 Both minus. 
 
 The equation (1) written in every possible form gives the follow- 
 ing: 
 
 x 2 + 2px= q. 
 x 2 —2px~ q. 
 x 2 +2 px= —q. 
 x 2 —2 px——q. 
 
 The application of the four transformations will manifestly bring 
 any equation of this character to the form of one of these equations. 
 They are thus called the Four Forms of a complete equation of the 
 second degree. 
 
 S3. Solution of Incomplete Equations. 
 
 If the equation is incomplete, there will be no term containing the 
 first power of the unknown quantity; that is, 2p— 0, and the four 
 forms will become, 
 
 x 2 — +q. 
 
 We may extract the square root of both members of these equa- 
 tions, and shall have, 
 
 x—±Vq. 
 
EQUATIONS. 
 
 85 
 
 It will be observed that the values of x in the last case are ima°f- 
 inary. 
 
 We have thus found the roots of the incomplete equations; and so 
 in any case, 
 
 To solve an incomplete equation of the second degree, reduce it to the 
 form of x 2 =q, and then extract the square root of both members. 
 
 Reduce the following equations to the form of x 2 + 2 px—q. 
 
 1. 
 
 a 2 x 2 
 
 ~W 
 
 2 ax b 2 
 
 c c 2 
 
 : 0 . 
 
 A ns. x 2 
 
 2 ab 2 c _ 5 4 
 
 a 2 c 2 1 a 2 c 2 
 
 2 . 
 
 ce-l-12 
 
 x 
 
 X 
 
 £ + 12 
 
 26 
 
 5 
 
 Ans. £ 2 + 12£— 45. 
 
 3. 
 
 ax 3x 2 , 1 + 5 
 
 T + -T + 1== — i — 
 o 4 b 
 
 x 2 x 
 
 4 + a' 
 
 Ans. x 2 + 
 
 a 2 — b 1 
 ~aT X ~b' 
 
 Reduce the following equations to the form of x 2 =q, and solve. 
 
 7i 
 
 Ans. x— 
 
 4. ax 2 -\ — — — cx 2 + 1. 
 c 
 
 c—b 
 
 3x~ 2 
 
 5. — =4.r 3 — 14. 
 
 ac +c 2 
 Ans. x=±2. 
 
 7. (a + x)(a—x) — ^~ 2x 2 . 
 
 9. ax(b— x)^=a[l — 4 3 ) 2 . 
 
 Ans. x— ±3. 
 
 Ans. x— ± 
 
 Ans. x— ±6. 
 
 Ans. x= 
 
 84. Solution of Complete Equations. 
 
 Resuming the equation, 
 
 x 2 +2px=q • • • • (1), 
 
 we see that the first member may be made a perfect square by add- 
 ing p 2 to it. This we are at liberty to do, provided that the same 
 quantity be added to the second member also. We may thus have, 
 
 x 2 +2 px+p 2 =q+p 2 . 
 
86 
 
 ELEMENTS OF ALGEBRA. 
 
 Extracting the square root of both members, we have, 
 
 x+p=±Vq+p 2 - 
 
 Transposing p to the second member, 
 
 x=~P±^qTp 2 (2). 
 
 These, then, are the values of x in equation (1). There are always 
 two roots; written separately, they are, 
 
 x=— p + ^q+p) 2 and x— —p— V q +JJ 2 - 
 
 It will be observed that the entire part in each root, —p, is alto- 
 gether the same, and is half the co-efficient of the first power of x in 
 the equation (1) from which they came, taken with a contrary sign ; 
 the radical parts are also the same, but one of them has the plus, 
 and the other the minus sign. 
 
 It is obvious that we may at once write out the roots of such an 
 equation, after it has been reduced to this trinomial form, by substi- 
 tuting the values of p and q, taken from the equation itself, in these 
 roots. 
 
 For example, let us have, 
 
 ax 2 —-x=7—x~. 
 
 k A 
 
 The application of the four transformations gives us, 
 
 c _ 14 
 
 r '" 2a + 2‘ X ~2a+2' 
 
 Q 
 
 Now, in this equation, p, the half co-efficient of x, is ^ - j - ; and 
 14 
 
 * 13 2«T2- 
 
 These, in the formulas for the roots (2) give us 
 
 c 
 
 + 4/ 14 +1 
 
 f c y 
 
 4a + 4 
 
 +/ ]/ 2a + 2 + 
 
 
 c 
 
 < / 14 1 1 
 
 ( c V 3 . 
 
 4a + 4 
 
 \ 2a + 2 + 
 
 Ua + 4/ ' 
 
 We may solve any such equation in the same manner; so that we 
 can say, in general, 
 
 To solve a complete equation of the second degree, containing one 
 unknown quantity, 
 
EQUATIONS. 
 
 87 
 
 1. Apply the four transformations in succession, thus reducing it 
 to the form x 2 -\-2pz=q. 
 
 2. Write the unknown quantity equal to half the co-efficient of the 
 first power of that quantity, with a contrary sign, plus and minus 
 the square root of the second member, augmented by the square of 
 this half co-efficient. 
 
 Examples. 
 
 1. Solve 
 
 10 14- 2x 22 
 
 x x “ 
 
 By the four transformations, 
 
 . 108 126 
 ^ 22 X ~ 22 ' 
 
 Writing out the roots, 
 
 x =^±i/- 
 
 22 ± y 22 + \ 22 / ' 
 
 Performing the operations under the radical sign, as indicated, 
 
 54 /Iir 
 
 22 ± y (22) 3 
 
 Extracting the root. 
 
 Whence, 
 
 54 , 12 
 ^”“22 ± 22 ' 
 
 21 
 
 x=3 and z=— . A ns • 
 
 „ „ . 2 x 2 bx x x 2 1 
 
 2. Solve \ — - 2 = z -r + -~ + -■ 
 
 a a 2 b a a 
 
 a 2 —b 2 
 
 ab 
 
 2 = 1 . 
 
 2 ab ± V \ %ab ) 
 
 a 2 -b 2 , 
 x— - ^ - ± 
 
 2ab 
 
 V 
 
 ka 2 b 2 + a 2 — 2a 2 b 2 + b i 
 
 4 a 2 b 2 
 
 ka 2 b 2 
 
88 
 
 ELEMENTS OE ALGEBRA. 
 
 I A / u i + Za*b 2 +tA 
 2ab y Aci 2 b 2 
 
 a 2 — b 2 a 2 +b 2 
 
 /y — 4- ; 
 
 2a£ 2ab 
 
 2=7 and 2= — . Ans. 
 b a 
 
 3. 3x 2 — 22=65. 
 
 A 
 
 s 
 
 II 
 
 Oi 
 
 « 
 
 ii 
 
 i 
 
 «£ 
 
 4. ax 2 —b=bx—^~. 
 
 b 2 
 
 . b V4«3 3 — 4a+J 4 
 
 Ans. x= — ± — — 
 
 2 a 2ab 
 
 5 * _ 7 
 2 + 60 32—5 ' 
 
 Ans. 2=14, 2=— 10. 
 
 _ a 2 x 2 2 ax b 2 
 
 6. T * + ^=°- 
 
 0 A c c z 
 
 b 2 
 
 Ans. x— — ± 0 . 
 ac 
 
 „ 48 _ 165 _ 
 
 2 + 3 2 + 10 
 
 A 27 
 
 Ans. 2=—, 2=0. 
 5 
 
 0 2 + 12 2 26 
 
 2 ' 2 + 12 5 
 
 Ans. 2=3, 2= —15. 
 
 ax 3x 2 , 1 +b x 2 x 
 b 4 b 4 a 
 
 . 1 a 
 
 Ans. 2=-, 2= — - . 
 a' b 
 
 22—3 32—5 5 
 
 U ' 32-5 + 22-3 2 " 
 
 i y 
 
 Ans. 2=^ > 2=1. 
 
 85. Trinomial Equations. 
 
 A trinomial equation is one which contains but two different pow- 
 ers of the unknown quantity. Such equations, when simplified, 
 have the form, 
 
 x m + 2+2" = q. 
 
 Complete equations of the second degree, such as we have just 
 been considering, are a particular case of this general form, m being 
 2 and n unity. 
 
 The method just explained, of solving a complete equation of the 
 second degree, is equally applicable for the solution of any trinomial 
 equation, when m—2n; that is, when the equation has the form, 
 
 x" n + 2+2" = q. 
 
 We can make the first member a perfect square in the same man- 
 
EQUATIONS. 89 
 
 ner as in the last article, and we should have, after extracting the 
 square root of both members and transposing, 
 
 x n = — p±Vq+p z . 
 
 Now extracting the nth root of both members, we have, 
 x = y — p ± V q + p 2 ■ 
 
 To solve such an equation, then, we adopt the same method as in 
 the case of an equation of the second degree, except that we write 
 the unknown quantity with an exponent equal to half of its highest 
 exponent in the reduced equation, equal to the roots, and after that 
 extract the nth root of both numbers of these new equations. 
 
 For example : 
 
 a: 4 — 4:X 2 — 12, 
 
 whence, x 2 = 2 ± V12 + 4, 
 
 and thus, x— ± y2±'\/l6 ; ; 
 
 or, x— ± ^/2±l. 
 
 It will be observed that there are four roots in this equation. 
 Written separately, they are, 
 
 X— + a/W, x— — t/fT, x— + a/— 2, and x= — V — 2. 
 
 It will be observed, further, that two of them are imaginary. 
 
 It may be well to remark here, that every such equation has as 
 many roots as there are units in the number which indicates its de- 
 gree. Sometimes some of the roots are equal to each other and some 
 are imaginary. 
 
 Examples. 
 
 1. Solve, x 2 * — 2.t" = 8. Ans. x—'sfi, x—'\/—2. 
 
 Remark. — Where only two roots are given in tlie answers, the student may 
 find the others. 
 
 2. x 4 +x 2 = 20. Ans. x=±2, x=±V—5. 
 
 3. x G +4x 3 = 96. Ans. x=2, x= ■y / —l2. 
 
 4. (x + y) 2 — (x+y) = (j. Ans. x-\-y— 3, x + y=— 2. 
 
 Remark. — Here x+y is to he regarded as a single quantity. 
 
90 
 
 ELEMENTS OF ALGEBRA. 
 
 5. ( Z-1) 2 -(Z-1) = 6 . 
 
 6. x—Vx—3. 
 
 Ans. x=4, x= — l. 
 Ans. Vx=3, ^/x=—2. 
 Ans. 's/x—Z, / \fx=—2. 
 
 7. 's/ x — \J x—G . 
 
 86. Simultaneous Equations. 
 
 Simultaneous Equations are those in -which any quantity in one 
 of the equations is the same in quantity and quality as that quan- 
 tity in any other of the equations; thus, if 
 
 are simultaneous equations, and x in the one has a particular value, 
 as 5, it must have the same value in the other; and again, if it stand 
 for a particular kind of quantity, as pounds, in the one, it must be 
 pounds in the other. So, also, with y, or any other unknown quan- 
 tities which may enter such equations. 
 
 If the known quantities are represented by letters, the same thing 
 is true of them as well ; thus, if 
 
 are simultaneous, then a must have the same numerical value and 
 represent the same kind of quantity in both equations. 
 
 It is plain that only simultaneous equations can be combined, as, 
 for example, added member to member, or multiplied member by 
 member, for if the same symbols should represent different things in 
 the several equations, the result of such combination, though a true 
 equality, would mean nothing. 
 
 Equations are commonly combined for the purpose of getting rid 
 of one or more of the unknown quantities which enter them ; or, as 
 it is said, in order to eliminate such unknown quantities. 
 
 There are three methods, commonly in use, of combining equa- 
 tions for the purpose of eliminating unknown quantities. They are, 
 
 In general, the first thing to be done with equations preparatory 
 to combining them by any one of these methods, is to subject them 
 to the first three transformations. The same unknown quantity 
 
 5x + 3y= 10, and 
 3x — 7y=lo, 
 
 ax + by—c 
 dx—ay—b 
 
 1st. By Addition. 
 
 2d. By Substitution. 
 3d. By Comparison. 
 
EQUATIONS. 
 
 91 
 
 will then enter any one of the equations but once, if the equations 
 are of the first degree. If of a higher degree, any particular power 
 of tlie same unknown quantity will enter but once. 
 
 87. Elimination by Addition. 
 
 Let us now consider these three methods of elimination in the 
 order given. 
 
 Take two simultaneous equations of the first degree, one numer- 
 ical and the other literal, in order to make the case more general, 
 and suppose them already brought to the proper form ; thus, 
 
 ox—oy—lQ 
 ax+ by—c. 
 
 How, we may multiply the first equation through, by a, and the 
 second by 5, without disturbing the equality in either case. We 
 shall thus have, 
 
 5ax—3ay=10a. 
 
 5ax + 5by —5 c. 
 
 We may further change the signs of all the terms of either of the 
 equations, and still preserve the equality. Changing the signs of 
 the second, we have,* 
 
 5ax— 3ay=10a 
 
 5ax + 5by — oc. 
 
 How, adding the equations member by member, we have, 
 —3ay—5by=—5c + 10a. 
 
 and from this, 
 
 _5c — 10a 
 y 3 a + ob' 
 
 In like manner, we may find the value of x ; or, we may substi- 
 tute this value of y in either one of the equations, and find it in 
 that way; thus, writing this value of y in the first of the given equa- 
 tions, we have. 
 
 * It will be found best not to destroy the original signs, but simply to write the new ones 
 below, and a little to the right of the old ones, as here shown. 
 
92 
 
 ELEMENTS OF ALGEBKA. 
 
 Solving, 
 
 _106 + 3 c 
 X 3« + 56‘ 
 
 It will be observed that the object in this method of elimination 
 is to make the terms containing the particular unknown quantity 
 which we want to be rid of, cancel when we come to add the equa- 
 tions. These terms must be made entirely the same and must have 
 different signs, in order to do this. We multiply or divide the two 
 equations by any quantities which will cause the terms in question 
 to become alike, and if they have not already different signs, we 
 change all the signs of one of the equations to make them differ. 
 
 We may say, then, practically, 
 
 To eliminate a quantity by addition, 
 
 1. Subject the equations, if necessary, to the first three transform- 
 ations ; 
 
 2. Multiply or divide either or both equations by whatever will 
 rnalce the terms to be cancelled the same ; 
 
 3. Make the signs of these terms differ, if they do not already, by 
 changing all the signs in one of the equations, and then add member 
 to member. 
 
 The resulting equation will be free from the quantity in question. 
 
 Examples. 
 
 .. c , 4x _ x+2 y 
 
 2 4 
 
 -v „ 24 — 6.r 
 
 Simplifying, we have, 
 
 ”tx- 2y=20 
 4Sx + 24y=2L 
 
 Dividing the second equation by 12, 
 
 qx-2y=20 
 4x + 2y—2. 
 
 Adding and then solving, 
 
 x=2. 
 
 This value of x in any of the above equations gives, 
 
 y=~ 3 - 
 
EQUATIONS. 
 
 93 
 
 2. Solve 
 
 3. Solve 
 
 i3x-\-3y—21 
 \iy —2x—8. 
 
 ! ax=by 
 x—c—y. 
 
 Ans. x—2, y— 3. 
 
 , be ac 
 
 Ans. x — — — r , y — - 
 
 (5s-8i=7y— 44 
 4. Solve \ J 
 
 (2 x=y+%. 
 
 5. Solve ■< 
 
 ! ax + by=c 
 fx+gy-h. 
 
 a + b’ J n + b' 
 Ans. x=4£, y= 8f. 
 
 . cq—bh ali—cf 
 
 Ans. x — — — — , y = -f-,. 
 
 ag-bf J ag-bf 
 
 6. Solved 
 
 1 . 3 ^ 72 ,- 341 = If + ?|r 
 
 2z+f=l. 
 
 a 10 18 W1 35 
 
 Ans. *=-18g^,y=-71^. 
 
 7 Solye |(*+5)(y+7)=(*+i)(y-9)+ii2 
 
 ( 2* + 10=3y + l. x=3, y=5. 
 
 (sx+o y = < j = a al 
 
 I ^ b z —a z 
 
 8. Solve < 
 
 a 2 be 
 
 b z x — - + (« + b + c)ay=a z x(2a + b)ab. 
 
 . ab ab 
 
 Ans. x— , yz 
 
 b—a ’ J a + b" 
 
 88. Elimination by Substitution. 
 
 Resuming the equations in the last article, 
 
 6z-3y=10, 
 
 cix+by—c, 
 
 let us find the value of one of the unknown quantities, as x, in terms 
 of the other in, say, the first equation ; that is, let us regard all the 
 quantities in the first equation as known, except x, and then find the 
 value of x, under this hypothesis. We shall have, 
 
 104-3?/ 
 x— 
 
 5 
 
94 
 
 ELEMENTS OF ALGEBRA. 
 
 Now, we may place this value ol' x for that quantity in the second 
 equation ; thus, 
 
 o (10+%) +Jy=a 
 
 We have now an equation with but one unknown quantity in it. 
 
 Solving this, we have, 
 
 ^ __5c— 10a 
 ^ 3a + hb‘ 
 
 Substituting this result for y in either of the equations, we have, 
 
 ,_ 10£ + 3c 
 da + hb' 
 
 This is the method of elimination by Substitution. We may thus 
 say that, 
 
 To eliminate a quantity by Substitution, find the value of such quan- 
 tity in terms of the remaining quantities from one of the equations, 
 and substitute this value in the other. 
 
 The resulting equation will be free from the quantity in question. 
 
 Examples. 
 
 1. Solve < 
 
 *_i-2 _L 
 2 "3 
 
 2z=^ + 6. 
 
 O 
 
 j' x—y — 1 
 
 2. Solve ■{ y 
 
 x—~—2x—ll. 
 
 I 6 
 
 3. Solve 
 
 4. Solve 
 
 x+y—a 
 
 x—y—b. 
 
 2x 3 y __ 9 
 
 F + T _ 20 
 
 3x 2 y _ 61 
 
 T + y _ i20 < 
 
 Ans. x—4, y= 3. 
 
 Ans. x=8, y— 9. 
 
 a + b 
 
 T 
 
 Ans. x=—r~ , y- 
 
 a—b 
 
 . 1 1 
 
 Ans. x—-^, y—-. 
 
 /V O 
 
 5. Solve" 
 
 x _ v 
 
 - + 5=fr-® 
 
 a 2 
 
 V , 
 
 x +f~i —ay. 
 
 Ans. x— - - - , y- 
 
 Let the examples in the previous article be solved by this method. 
 
EQUATIONS. 
 
 95 
 
 89. Elimination by Comparison. 
 
 Take, again, the equations 
 
 5x—Zy=lQ 
 ax + by—c. 
 
 Find the value of x in both equations in terms of y ; thus, 
 
 10 + 3?/ 
 x = — - — -. 
 o 
 
 -by 
 
 Tliese must be equal to each other, and thus we have, 
 10 + 3?/ _ c— by 
 
 Solving, we have, as before, 
 
 _5c—10a 
 y 3a + 56’ 
 
 Finding x in the same ivay or by substitution, we have, 
 
 _106 + 3c 
 X 3a + 56’ 
 
 This is called the method of elimination by Comparison. Prac- 
 tically, 
 
 To eliminate a quantity by Comparison, find tlic value of the quan- 
 tity in both equations in terms of the remaining quantities, and 
 place these values equal to each other. 
 
 The resulting equation will be free from the quantity in question. 
 
 Examples. 
 
 1. Solve 
 
 16z + 30?/= 18 
 12a;— 2+?/= —2. 
 
 ;=99— 7 y 
 
 7®=51-|. 
 
 Ans. z=|-, y=\. 
 
 2. Solve < 
 
 Ans. x=H, y= 14. 
 
96 
 
 ELEMENTS OF ALGEBRA. 
 
 3. Solve 
 
 ax + by=c 
 dx+fy—g. 
 
 4. Solve 
 
 j" abx — | 
 \^aby—l—x. 
 
 Ans. x 
 
 fc— bg _cd—ag 
 fa—bd y bd—af 
 
 Ans. 
 
 U 
 
 1 _ 2 ab 
 
 2aW+l' V ~ 2a 2 b*- 1' 
 
 5. Solve -j * + ?/= 1 8.73 
 
 ( 0.56a; + 13.42?/= 763.4. Ans. x— — 39.81 
 
 y— 58.54. 
 
 Let the examples in the last two articles he used for practice by 
 this method. 
 
 90. Elimination in General. 
 
 Let us now extend the principle of elimination to the solution of 
 equations containing any number of unknown quantities. 
 
 Let us have three equations which have been already subjected to 
 the first three transformations, and containing three unknown quan- 
 tities; thus, 
 
 2^ + 3 y— z=l - - - - (1). 
 
 "5x— 2y + oz=2 - - - - (2). 
 
 x+ y— 2=3 - - - - (3). 
 
 Manifestly we may eliminate one of the unknown quantities from 
 any two of these equations, by either of the three methods already 
 given. Let us, say, combine the first two, and get rid of 2 . Multi- 
 plying the first by 3, and adding, we have 
 
 §x + 1y=o - - - - (4). 
 
 In the same way, combining the first and last, eliminating z, we 
 have 
 
 x+2y=— 2 - - - - (5). 
 
 We now have two equations (4) and (5), with hut two unknown 
 quantities. These we may now combine by any one of the three 
 methods, and shall have 
 
 24 , 23 
 
 x= - and y-- Tv 
 
EQUATIONS. 
 
 97 
 
 Substituting these in either of the three original equations, we 
 have 
 
 2 = 
 
 32 
 
 11 ' 
 
 We have thus solved the group of equations. The same course 
 may be pursued with any number of equations, provided that 
 there are as many equations as there are unknown quantities. 
 
 Thus we may say, that 
 
 To solve a group of simultaneous equations, there being as many 
 equations as there are unknown quantities, 
 
 Combine the equations two and two, being careful not toxmite the 
 same two more than once, and always eliminating the same quantity, 
 until there are as many resulting equations as there are unknown 
 quantities remaining. 
 
 Combine these resulting equations in the same way. We shall at 
 last have an equation containing a single unknown quantity, which 
 may be solved. Complete the solution by substituting the value of the 
 quantity so found in one of the resulting equations , with two unknown 
 quantities in it, and so on. 
 
 Examples. 
 
 r x+y — 10 
 
 1. Solve •< x+ 2=19 
 i y + 2=23. 
 
 At 
 
 y=± 1 -^ 
 
 A 
 
 41 2 
 
 2. Solve { x _~2 4 
 
 5- 
 
 ( x+y+z = 30 
 
 3. Solve -j 8.3 + 4?/ + 22=50 
 ( 27.3 + 9y + 3z= 64. 
 
 Ans. x—3, y— 7, 2 = 10 . 
 
 Ans. 3=18, y= 32, 2 = 10 . 
 Ans. 3 =f, y=— 7, z=36£. 
 
 4. Solve < 
 
 x+y=a—z 
 
 by—cx 
 
 dz—fx. 
 
 Ans. x 
 
 abd 
 
 bd + cd + bf’ 
 
 acd 
 
 V ~ bd+cd+bf’ 
 
 _ a ^f 
 
 bd + cd +bf 
 
 G 
 
98 
 
 ELEMENTS OP ALGEBRA. 
 
 (1,1 
 - + -=« 
 
 2 
 
 A?is. x — 
 
 
 * y 
 
 a + b-c ’ 
 
 5. Solve < 
 
 -+\=b 
 
 2 
 
 x 2 
 
 ^ a—b + c’ 
 
 
 1 1 
 — h - ~C. 
 
 2 
 
 z — 
 
 
 Ly z 
 
 b + c—d 
 
 91. Equations of Condition. 
 
 
 If we should have more unknown quantities than we have inde- 
 pendent equations, w r e shall have more than one unknown quantity 
 in the last resulting equation. 
 
 If we should have more equations than there are unknown quan- 
 tities, we may eliminate all the unknown quantities, and thus have 
 a resulting equation between known quantities; thus, let us have, 
 
 y=ax+b. 
 y—cx + d. 
 
 y =/%+!/■ 
 
 Placing the second members of the first two equal to each other, 
 we have, 
 
 ,7 ,7 i 
 
 ax + b—cx + d :.x— . 
 
 a—c 
 
 Then combining the first equation with the third, eliminating y, we 
 have, 
 
 ax + b=fx+y .*. 
 
 9~b 
 A «-/' 
 
 Placing these two values of x equal to each other, we have 
 ^ — y> an equation in which there is no unknown quantity. 
 
 Now, if this last result is not a true equality, all the equations 
 from which it was derived cannot be true. They could not all be 
 satisfied at the same time for the same set of values for the unknown 
 quantities. Such a resulting equation between constants is called an 
 Equation of Condition; since it is the condition upon which the equa- 
 tion from which it is derived can be true. 
 
 When there are more equations than there are unknown quanti- 
 ties, there must thus be a certain interdependence between the con- 
 stants which enter them. The equations from which such equation 
 of condition is derived, cannot, therefore, be independent equations. 
 
EQUATIONS. 
 
 99 
 
 Examples. 
 
 Find the equation of condition in the following groups of equa- 
 tions : 
 
 r ax=b 
 l cx=d 
 
 2 . ^ 
 
 ax— by 
 cy=dx 
 
 l fx=g • 
 
 3. 
 
 j y—ax + b 
 ( y=a'x + b. 
 
 Ans. 
 
 t_d 
 a~ c 
 
 Ans. 
 
 cJl_ag 
 ¥ ¥ ' 
 
 Ans. a=a'. 
 
 92. Simultaneous Equations of a Higher Degree. 
 
 Equations of a higher degree than the first can be combined and 
 a single equation found containing a single unknown quantity. The 
 method of elimination is altogether the same as that already ex- 
 plained ; but the resulting equation is generally of too high a degree 
 to be managed by methods falling within the province of this work : 
 
 For example, let us have, 
 
 ax 2 + by = c. 
 
 2y 2 -3x=5. 
 
 Solving the first equation with respect to y, we have, 
 
 c—cix 2 
 
 This substituted in the second, gives, 
 
 0 /h— «z 2 \ 2 
 \— r -) -3z=o, 
 
 which is an equation of the fourth degree ; and, in general, when 
 both equations are of the second degree, the resulting equation will 
 be of the fourth. 
 
 93. Simultaneous Equations of the First and Second Degrees. 
 
 There are, however, two classes of simultaneous equations, beyond 
 the first degree, which admit of ready solution. They are, 
 
 1. Equations containing twkunhnown quantities, when one is of the 
 second degree and the other of the first. 
 
 2. When the equations are both of the second degree and homogeneous 
 with respect to the unknown quantities. 
 
100 
 
 ELEMENTS OP ALGEBRA. 
 
 I. Let us take the first case. Assume, 
 
 Gx 2 — 2xy=36 
 3x + y= 12. 
 
 The value of y, in the second, in terms of x, is, 
 
 y=12— 3 x. 
 
 This substituted in the first gives, 
 
 6a; 2 — 2a;(12— 3x)=36 ; 
 whence, x 2 —2x=3. 
 
 x—3 and —1. 
 
 These values in the second equation give, 
 y — 3 and 15. 
 
 A like course may be pursued in any such case, so that we may 
 always solve two simultaneous equations containing two unknown 
 quantities, when one of them is of the first and the other of the 
 second degree, by simply eliminating one of the unknown quan- 
 tities. 
 
 94. Homogeneous Equations of the Second Degree. 
 
 II. When the equations are homogeneous Avith respect to the un- 
 known quantities. 
 
 Take the equations, 
 
 x 2 —2xy=l—3xy - - (1), 
 
 y 2 + 4a; 2 — 2 = 5a; 2 - - (2). 
 
 The solution is accomplished by using an auxiliary quantity. 
 Applying the second and third transformations and making y=2 }X ’ 
 Ave have, 
 
 x 2 +px 2 = l. .-. ** = — - - (3), 
 
 p 2 x 2 — x 2 =2. .*. x 2 — t - - (4). 
 
 p~ — l 
 
 Placing these tivo values of x 2 equal to each other, 
 
 1 _ 2 
 l+p—p 2 —l' 
 p 2 — 2p—3 
 
 p-i±V3+i 
 p>=3 and —1. 
 
 Whence. 
 
EQUATIONS. 
 
 101 
 
 Using the first value of p in either (3) or (4), we have, 
 
 x= ±-i. 
 
 The first values of ^ and x in y=px, give, 
 
 y=b 
 
 The same course may be pursued with any two equations of this 
 class. 
 
 Examples. 
 
 1. Solve x 2 + y 2 = 13 
 
 x + y— 5. 
 
 CO 
 
 II 
 
 CO 
 
 II 
 
 2. Solve x 2 +y s =41 
 
 
 x-y=-l. 
 
 Ans. x=4, y= 5. 
 
 3. Solve x 2 +y 2 = 13 
 
 
 %y-y 2 =2- 
 
 Ans. x=3, y = 2. 
 
 4. Solve x 2 + 2xy+y 2 =2o $ 
 
 
 0 , x 1 
 
 3x—4y=- + j. 
 J 4 4 
 
 Ans. x= 3, y—2. 
 
 5. Solve 3x 2 -2y 2 =57 
 
 
 3x + y 
 2 
 
 Ans. x=5, y= 3. 
 
 6. Solve 3xy—2x 2 = 10 
 
 
 1 1 T’S ,;/2 
 
 2xy + 2y 2 - 7 ■' +25. 
 
 Ans. x=2, y= 3, 
 
 7. Solve x+y — 13 
 
 
 Vx + \/y=5. 
 
 Ans. x=9, y= 4. 
 
 8. Solve ‘\fxy + x = 15 
 
 
 V'x— \/9 = l. 
 
 Ans. x=9, y= 4. 
 
 95. Solution of Simultaneous Equations of any degree. 
 
 The principles already established will enable us to solve many 
 equations of the second and higher degrees by the exercise of a little 
 ingenuity. No specific rules would prove of much service; so that 
 the general management of such cases may be best exemplified by a 
 few examples. 
 
 1. Solve x 3 —y 3 = 19 - - (1), 
 
 x -y = l - - (2). 
 
102 
 
 ELEMENTS OF ALGEBRA. 
 
 We may divide the first equation by the second, and thus have, 
 x~+xy + y~ — 19 - - (3). 
 
 Now the value of x from the second, a;=l +y, in equation (3), 
 gives, 
 
 y^+y-6. 
 
 Whence, y=—\+zy/ 6+£ 
 
 y=2, y= — 3. 
 
 These values of y in (2) give, 
 
 x— — 1, x——2. 
 
 2. Solve (x + y) 2 + (x + y) = 30 - - (1), 
 
 x 2 -y 2 =o - - (2). 
 
 From (1) we have, 
 
 x+y— — 1-± V30 + 1, 
 x+y——\± i i~, x + y=5, x + y=—6. 
 
 Dividing (2) by these values, we have, 
 
 x—ry—1, x—y= — f. 
 
 From these values of x + y and x—y, we have, 
 x=3, x——\. 
 
 y= 2, y=V—V-- 
 
 3. Solve x 3 —y 3 =11T - - (1)> 
 
 (x—y) 2 — {x—y) = (j - - (2). 
 
 From (2) we have 
 
 x—y— 3, x—y— —2. 
 
 Dividing (1) by this first value of x — y we have, 
 x 2 +xy + \y 2 — 39 - - (3). 
 
 Substituting the value of x from x—y—o, in (3) and solving, we 
 have, 
 
 */=2, y=-5, 
 
 £c=5, x— —2. 
 
 4. Solve x 3 + y 3 = 152 - - (1), 
 
 x +y =8 - - - (2). 
 
 Dividing (1) by (2), we have, 
 
 x 2 —xy + y 2 = 19. 
 
EQUATIONS. 
 
 103 
 
 Combining this with. (2), we have, 
 
 x=5, x=3. 
 
 Whence, y =3, y= 5. 
 
 5. Solve x 2 +y 2 = 74: - - (1), 
 
 xy =35 - - (2). 
 
 Multiplying (2) by 2, w r e have, 
 
 2xy=70 - - (3). 
 
 Adding this to (1), we have, 
 
 x 2 +2xy+y 2 = 144. 
 
 Whence, x+y= 12 - - (4). 
 
 This with (3) will complete the solution ; or, subtracting (3) from 
 (1), we have, 
 
 x 2 — 2xy + y 2 =4:. 
 
 Whence, x—y=2. 
 
 This with (4) gives, 
 
 x=7, y= 5. 
 
 6. Solve x— y + \/x— y=3 
 x+y= 8. 
 
 Follow the same general course as in the second example. 
 
 Ans. x=6, y= 2. 
 
 7. Solve x 2 +y z = 9 - - - (1), 
 x 2 y + xy 2 = 3 - - (2). 
 
 Multiplying (2) by (3) and adding to (1), we have, 
 x 3 + 3x 2 y + 3xy 2 +y 3 = 27. 
 
 Taking the cube root of this, 
 
 x + y=3 - - (3). 
 
 We may write (2) thus, 
 
 xy{x + y) = 3. 
 
 This with (3) gives, 
 
 3 xy=3 - - (4). 
 
 From (3) and (4) we have 
 
 x=l, y= 2. 
 
 8. Solve Sx 3 +128y 3 =3520 
 2a: + 4y=16. 
 
 Ans. x=2, y= 3. 
 
104 
 
 ELEMENTS OF ALGEBRA, 
 
 9. Solve x + y — 10 
 f \fxy= 4 . 
 
 Ans. z=8, y— 2. 
 
 10 . Solve -*-1=3 
 
 y 2 
 
 Ans. 2 = 6 , y— 3. 
 
 11. Solve x 2 +2xy + _?y 2 =64 . 
 
 x 2 —y 2 = — 16. 
 
 12. Solve re 2 — 3x + xy=li + 3y 
 
 x 2 + xy + 52=70 — 5 y. 
 
 13. Solve 3xy-\-Ax— ty 2 +10=2?/ 
 
 1 5x 2 — 62 + iy + 3 = lOxy. 
 
 14. Solve x+^/xy + ?/=14 
 
 a : 2 +xy+y 2 = 84. 
 
 2=3, ?/= 5. 
 
 Ans. 2 = 5 , y= 2. 
 
 Ans. x=l, y—2. 
 
 Ans. 2 = 2 , y— 8 . 
 
 96. Radical Equations. 
 
 We shall now give some further examples of the manner of solving 
 equations containing radical quantities. No invariable method can 
 he pointed out. 
 
 If there is but one radical in the equation, we may make it stand 
 alone in one member, and then by raising both members to a power 
 equal to the index of the radical, it will disappear. If there are two 
 radicals, by placing them in different members and raising to two 
 successive powers, they may be made to disappear; but generally the 
 resulting equation is of too high a degree to be easily managed. 
 
 When there are fractional quantities entering the equation, it may 
 often be greatly simplified by suppressing common factors. 
 
 1. Solve 5 — Vz + i=x. 
 Transposing, 
 
 \/x + i=x—5. 
 
 Squarin 
 
 2 + l= 2 2 — 10a: + 25. 
 
 Solving, 
 
 2 = S, 2 = 3. 
 
 2. Solve ~ — l = '\/2x—i. 
 
 V 2—1 
 
EQUATIONS. 
 
 105 
 
 Dividing the numerator of the fraction by the denominator, 
 V^ + l — 1 — V^x—4: 
 
 Vx—V'^x—i. 
 
 Squaring, 
 
 x=2x — 4 
 
 X—4:. 
 
 3. Solve a + Vx—b—Vx + a 2 . 
 
 Squaring, a 2 +2aVx—b + x—b^=x+a 2 . 
 
 Simplifying, 2aVx—b=b. 
 
 Squaring again, 4a 2 (a;— b) = b 2 . 
 
 b 2 +4 a 2 b 
 
 Whence, 
 
 4. Solve 
 Squaring, 
 Simplifying, 
 
 * ‘ 4a 2 
 
 a Vx + a_Va—x 
 x b ~ b 
 
 a 
 
 2 V ax + a 2 x + a_a—x 
 
 x bVx + b 2 ~ b 2 ' 
 
 a 
 
 2 Vax + a 2 
 
 bVx 
 
 = 0 . 
 
 Transposing and squaring, 
 
 4ax + 4a s 
 b 2 x 
 
 Clearing of fractions, a 2 b 2 =4 ax 2 +4 a 2 x; 
 
 whence, 
 
 „ ah 2 
 
 x 2 +cix=——. 
 
 5. Solve 
 
 a , 
 
 ■T=-j± 
 
 ab 2 a 2 
 
 ~T + T> 
 
 —a±V ab 2 +ct 2 
 *= 1 * 
 
 x—a -Vx + Vn r- 
 + — V x. 
 
 V x— V a 
 
 x—a 
 
 Getting rid of common factors, 
 
 Vx + V a + 
 
 Vx—V i 
 
 : —\/x. 
 
106 
 
 ELEMENTS OF ALGEBRA. 
 
 Clearing of fractions, 
 
 x—a + l=x— '/ax. 
 Simplifying, and transposing, 
 
 '/ax—a — l. 
 
 Squaring and dividing £>y a, x= ^ — 
 
 6. Solve '/x— / x— 5 = 1. 
 
 7. Solve a—b=Vx—2ab. 
 
 1 Vx + 1 
 
 8. Solve 
 
 's/x—l a 
 9. Solve VYx+l — 2 — //+ll. 
 
 Vx + o 
 
 /x + l. 
 
 10. Solve 
 
 x—9 
 
 -—/x — o. 
 
 Ans. a; =9. 
 Ans. x=a 2 + i 2 . 
 
 2a + 1 
 
 Ans. x- 
 
 a - f 1 
 Ans. x=o. 
 
 Ans. x =4. 
 
 11. Solve 
 
 12. Solve 1 + Vl + x=2. 
 
 13. Solve |'/ a + /x—c. 
 
 14. Solve x 2 + (x— 9)^=1 — - — . 
 
 (x—9) 2 
 
 15. Solve |/l6 — Vx 2 + 45=^. 
 
 16. Solve -+ ( aS 
 
 X X 0 
 
 Ans. x—a 3 c—i. 
 Ans. a: =8. 
 Ans. x—(c m —a)”. 
 Ans. x=12. 
 
 Ans. x=2. 
 Ans. x—{2 ai—i 2 )~. 
 
 97. Inequalities. 
 
 We shall now give a few principles witli regard to Inequalities. 
 
 Two inequalities are said to subsist in the same sense when the 
 greater quantity is in the same member in both; thus, 5>2 and 
 7>3 are in the same sense. When the greater quantity is in the 
 first member of one, and in the second of the other, they subsist m 
 a contrary sense; thus, 5>2 and 3 <7, are in a contrary sense. 
 
EQUATIONS. 
 
 107 
 
 1. We may add the same quantity to loth members of an inequal- 
 ity , or subtract the same quantity from both members, without chang- 
 ing the sense. 
 
 For example, add 5 to both members, and then subtract it from 
 both members of the following : 
 
 7>3. 
 
 We have, 
 
 12>8, and 2>— 2. 
 
 If we subtract 10, we hare, 
 
 — 3 > — 7 . 
 
 This last result is true, considered in an algebraic sense. 
 
 This principle enables us to transpose terms from one member of 
 an inequality to the other, by changing the signs, as in an equation. 
 
 2. If both members of an inequality be multiplied or divided by the 
 same positive quantity, the sense will not be changed ; but if the mul- 
 tiplier be negative, the sense will be changed. 
 
 For example, multiply both members of 7> 5 by 5. We have, 
 
 35>25. 
 
 Multiplying by —5, we have, 
 
 -35 <-25. 
 
 If we multiply by —1, we change the sense and the signs at the 
 same time ; hence, we may change the signs of an inequality if we 
 at the same time reverse the sense. 
 
 3. When both members of an inequality are positive, we may raise 
 to any power ; or ive may extract any root, provided we use only the 
 positive roots. 
 
 For example, raise both members of 7>5 to the second power. 
 We have, 
 
 19 >25. 
 
 Extracting the square root of this last inequality, we have, 
 
 7>5; 
 
 hut we could not take the negative roots and say 
 
 — 7> —5. 
 
 These principles will enable us to transform an inequality so as 
 to make any quantity stand alone in one member, greater or less than 
 a resulting quantity ; thus, let us have, 
 
 3z— 9>21. 
 
108 
 
 ELEMENTS OF ALGEBRA. 
 
 Transposing and dividing, we have, 
 
 £> 10 . 
 
 Examples. 
 
 1. 2£ + 5>12. 
 
 2. ^-l<£ + 4. 
 
 Ans. £<10. 
 
 Ans. £>f. 
 
 3. f— |<l0-6£. 
 o 2 
 
 Ans. £<-rx • 
 o7 
 
 100 
 
 . U-C/ U/U 
 
 Ans. £> — = — — . 
 d+1 
 
 ac—cib 
 
 a 
 
 a 
 
 $8. Problems. 
 
 A. problem is a question proposed for solution. 
 
 The solution consists, 
 
 1. In translating the given conditions into algebraic language, 
 and thus deriving one or more equations, involving the given and 
 the required elements ; and 
 
 2. In the solution of the resulting equation or equations. 
 
 Represent the required elements by x, y, etc., and then use these 
 
 quantities as though their values were known. 
 
 Be careful in forming an equation to put like quantities equal to 
 each other; that is, men = men, pounds=pounds, etc. Be careful, 
 also, that the quantities are expressed in a common unit. 
 
 What two numbers are those whose sum is a, and whose 
 difference is b ? 
 
 Here, although two numbers are required, it is not necessary to use 
 more than one unknown quantity ; thus, 
 
 Let x— the greater. 
 Then a—x= the less. 
 
 From the condition, we shall have 
 
 x—(a—x) = b. 
 2x—a + b. 
 
 X— — - — , the greater ; 
 
 A 
 
 a + b 
 
EQUATIONS. 
 
 109 
 
 We may also use two unknown quantities in tlie solution of this 
 problem ; thus, 
 
 Let x = the greater, and 
 
 y — the less. 
 
 Then x+y=a, 
 
 and x—y—b', 
 
 , a + b a b 
 
 whence x=— 
 
 /v A A 
 
 a—b a b 
 
 y ~ir~2~2' 
 
 These results may be translated into English thus : 
 
 The greater of two quantities is equal to the half sum, phis the 
 half difference. 
 
 The lesser of two quantities is equal to the half sum, minus the 
 half difference. 
 
 2. What number is that froniAvhich if 5 be subtracted, two-thirds 
 of the remainder will be 40 ? 
 
 Let cr=the number. 
 
 Then, x— 5=the remainder. 
 
 From the conditions, f(a— 5) =40. 
 
 x — 65. 
 
 3. A horse said to a mule: If I give you one of my sacks we 
 shall have an equal number; if I take one of yours, I shall have 
 double the number you have left. How many had each? 
 
 Let x— the number the horse had. 
 y— the number the mule had. 
 
 Then, x—l=y + l, 
 
 and x + l = 2(y — l), 
 
 whence x=7, y— 5. 
 
 4. Divide 64 into two parts which shall be to each other as 3 to 5. 
 
 Let 3x= one part, 
 
 Then ox— the other. 
 
 But 3. , r + 5x=64. 
 
 Whence, x=8. 
 
 3x=24 one part, 
 
 5.r=40 the other. 
 
no 
 
 ELEMENTS OF ALGEBRA. 
 
 5. Divide a into two parts which shall be to each other as b to c. 
 
 Ans. 
 
 ab ac 
 b + c’ l + c' 
 
 6. A man left £2,400 to be divided between two sons and a servant. 
 
 The sons’ parts were to be to each other as 3 to 2, and the servant 
 was to have half as much as the son who received the smaller sum. 
 How much had each ? Ans. $1,200, $800, $400. 
 
 7. A person upon being asked his age, replied: that J of his age 
 
 multiplied by -fa of his age would give a product equal to his age. 
 IIow old was he ? Ans. 16 years. 
 
 8. A and B have the same income; A contracts an annual debt of 
 
 % of his income ; B lives upon £ of his ; at the end of 10 years, B 
 lends A money enough to pay off his debts, and has 1607 left. What 
 was their income ? Ans. 2S0I. 
 
 9. A man, fifteen years after his marriage, was asked the age of 
 himself and of his wife at their marriage. He replied that he was 
 then twice as old as his wife, but that now he was only once and a 
 half as old. What Avere their ages ? Ans. 30 and 15. 
 
 10. A person passed £ of his age in childhood, T h in youth, i and 
 5 years besides in matrimony, at the end of which time he had a son, 
 Avho died 4 years before his father, having reached half his father's 
 age. What Avas the father’s age ? Ans. S4. 
 
 11. A privateer running at the rate of 10 miles an hour discovers 
 a ship-of-Avar 18 miles off pursuing at the rate of 8 miles an hour. 
 IIow many miles Avill the privateer make before the ship overhauls 
 her ? 
 
 Let x— the distance the privateer will make before she is overhauled. 
 
 Then, ®+18= the ship’s corresponding distance. 
 
 X 
 
 — will be the number of hours required by the privateer to reach the point of 
 
 union. will be the time required by the ship. 
 
 Since these times must be equal, we have, 
 x ® + 18 
 10 
 
 8 
 
 x=— 90. 
 
 , whence. 
 
 It will be observed that our result is negative. IV e must, then, reckon back- 
 ward 90 miles to find the place at which the vessels would have teen together. 
 
EQUATIONS. 
 
 Ill 
 
 This results from the- fact that positive and negative quantities do not differ 
 essentially, but merely in their positions with respect to the origin of reckoning. 
 It is thus that the algebra takes no note of the word “ pursuing,” except in its 
 essential sense of traveling on the same line with the object with which this 
 relation obtains. So in general, the algebra always interprets words which have 
 relative meanings in their broadest sense. It thus makes no distinction between 
 “will be” and “has been,” “up” and “ down,” “ to the right ” and “ to the 
 left,” etc. 
 
 A negative result shows that there has been some error made in the use of 
 words in the enunciation. It is to be interpreted in a directly contrary sense 
 from a positive result. 
 
 12. Find two numbers such that tlie first added to three times the 
 
 second will give 11 ; and three times the first less twice the second 
 will give zero. Ans. 2 and 3. 
 
 13. Find three numbers such that three times the first increased 
 
 by twice the second will give 16 ; and twice the third less the first 
 will give 6 ; and five times the third divided by twice the second will 
 give 10. Ans. 2, 5, and 4. 
 
 14. There are three numerals expressed by single figures 'which 
 
 added together give 8; written in a certain order they give five times 
 the number expressed by the second and third written together ; if 
 the first and third be written together, they express f- of the number 
 formed by writing the first and second together. What are the num- 
 bers ? Ans. 1, 2 and 5. 
 
 15. What three numbers are those, which if three times the first 
 be added to twice the second and this sum divided by the third, will 
 give 4; and if three times the second be added to five times the third 
 and the sum be divided by the second, will give 7 ; and if four times 
 the sum of all the numbers be divided by 2, will give 11 ? 
 
 Ans. 2, 5, and 4. 
 
 16. A man upon being asked his own and his wife’s ages, replied 
 that his age divided by his wife’s would give one-fifteenth of her age; 
 but that if 30 had been first subtracted from his own age, the result 
 would have been one-tliirtietli of his wife’s age. How old was each ? 
 
 Ans. 40 and 30. 
 
 17. The length of a rectangular field exceeds its breadth by 10 
 
 chains, and it has 2000 square chains in it. What is its length and 
 breadth ? Ans. 40 and 50. 
 
 18. What two numbers are those which if the first be taken from 
 
112 
 
 ELEMENTS OF ALGEBRA. 
 
 the second and the difference multiplied by the first, the result Mill 
 he four times the square of the first ; and if this difference be squared, 
 the result will be 64 ? Ans. 2 and 10. 
 
 19. A merchant has two sorts of tea, one worth 75 cents a pound 
 
 and the other one dollar. He wants to make a mixture of 25 pounds 
 which he can sell at 90 cents. How many pounds of each must he 
 take? Ans. 10 of the 1st and 15 of the 2d. 
 
 20. A hare is 50 leaps before a greyhound, and makes 4 leaps to 
 the greyhound’s three; but two of the hound’s leaps are equal to 
 three of the hare’s. How many leaps must the greyhound make to 
 catch the hare ? 
 
 In sucli a case as this, we must first fix upon a unit of distance, which here 
 may be either one leap of the dog or one of the hare. Let us take the length 
 of the hare’s leap as the unit ; then, the length of the dog’s leap will be £ times 
 that of the hare’s. 
 
 Now, if x be the number of leaps the hound must make, the hare will in the 
 same time make leaps. This number of leaps made by the hare, added to 
 the number of leaps it is in advance, will be the distance, in our assumed unit, 
 from where the dog is to the place where the hare is caught. But the dog will 
 pass over a distance equal to \x ; that is, the number of leaps multiplied by the 
 length of one of them ; hence, 
 
 ix+ 50=%x. 
 
 a:=300. Ans. 
 
 21. Two trains are traveling towards each other at the rate of 20 
 
 and 33 miles per hour, respectively ; they are 265 miles apart. How 
 long before they will meet ? Ans. 5 hours. 
 
 22. If the trains had been traveling at the rate of a and b miles 
 
 per hour and had been c miles apart, how long would they take to 
 come together ? . c 
 
 a + b 
 
 From this formula, tell at once what the time would have been 
 had the rates been 2 and 3, and the distance 10. Also, if the rates 
 had been \ and ^ , the distance Also, if the rates had been 5 and 
 — 7, the distance 6. 
 
 23. Two travelers set out at the same time to meet each other, 
 being 154 miles apart. If one travels at the rate of 3 miles in 2 
 hours, and the other at the rate of 5 miles in 4 hours, how far shall 
 each travel before they meet, and after what time ? 
 
 Ans. Distances, 84 and 70. Time, 56 hours. 
 
SYMBOLS 0 AND 00 . — DISCUSSIONS. 113 
 
 Let letters be used for tlie numerals, and the results he applied to particular 
 cases, as above. So with any of the preceding or following problems. 
 
 24. A can do a piece of work in 5 days, .and B can do the same 
 in 7; working together, how long will they be at it ? 
 
 Ans. 2 ih days. 
 
 25. The half of A’s fortune less B's is equal to 83,000; B’s for- 
 tune less one-fifth A’s gives zero. What sum had each ? 
 
 Ans. A, 810,000, B, 82,000. 
 
 26. What two numbers are those which cubed and added together 
 
 give 35, and which if the first be squared and multiplied by the sec- 
 ond, and this result be added to the product of the second squared by 
 the first, will give 30 ? Ans. 2 and 3. 
 
 SECTION VIII. 
 
 SYMBOLS O AND oo.— DISCUSSIONS. 
 
 99. The entire absence of value is represented by the symbol 0, 
 called zero. A man who is altogether destitute of money, has 0 
 dollars. This is the ordinary meaning of zero ; but in algebra it 
 has a somewhat more extended signification. 
 
 A man may not only be destitute of money, but he may have less 
 than none at all, in an algebraic sense, as we have seen in Art. 26; 
 he may be in debt. In such a case, zero is the point of division be- 
 tween his assets and liabilities. 
 
 Again, if we agree to reckon distances upon a right line or scale 
 from a particular point upon it, such point, having no distance from 
 the origin of distances, is the zero point. We have a familiar ex- 
 ample of this in the thermometer ; the position of the zero point 
 upon the scale being a mere matter of agreement. 
 
 Thus, zero in algebra is but the point of separation between all 
 positive quantities on the one hand, and all negative quantities on 
 the other. If a positive quantity be continually diminished, it will 
 approach zero as a limit ; but as we cannot conceive of a quantity so 
 small that there cannot be a smaller, we could never find the smallest 
 possible quantity. But that which continually approaches a limit 
 may be said to have the limit itself for ’its extreme value; and so 
 n 
 
114 
 
 ELEMENTS OF ALGEBRA. 
 
 0 is often said to be a quantity smaller than any assignable quan- 
 tity. 
 
 The symbol co , called infinity, represents a quantity than which 
 a greater is impossible. It is the limit of all augmentation. We 
 may approach it to any possible degree of approximation, but cau 
 never reach it. 
 
 100. Combinations of 0 and®. 
 
 Any product which has 0 in it as a factor, must itself be zero ; thus, 
 
 a x 0 = 0. 
 
 For if zero be taken any number of times, it will still be zero ; or 
 if any quantity be taken zero times, we shall have zero as well. 
 
 Zero divided by any finite quantity is equal to zero ; thus, 
 
 a 
 
 For, since the product of the divisor and quotient must produce 
 the dividend, we must have 0 for the quotient to give with a, the 
 dividend 0 ; or, more briefly, zero times the fractional unit, gives 
 zero. 
 
 A finite quantity divided by zero, gives infinity ; thus, 
 
 a 
 
 -- co. 
 
 For, as we diminish the divisor, we increase the quotient. When 
 we have made the divisor the least possible, the quotient must be 
 the greatest possible, that is, infinite. 
 
 Zero divided by zero, gives any quantity whatever, or, as it is 
 said, is Indeterminate j thus, 
 
 0 
 
 Q -a. 
 
 For, any quantity, a, the quotient, multiplied by the divisor 0, 
 gives 0 the dividend. 
 
 Any quantity divided by infinity gives zero ; thus, 
 
 — = 0 . 
 oo 
 
 For, as the divisor is increased, the quotient is diminished. 
 When the divisor is the greatest possible, the quotient must be the 
 least possible, or zero. 
 
 Infinity divided by a finite quantity gives infinity ; thus, 
 
 oo 
 
 • = 00 . 
 
 a 
 
SYMBOLS 0 AND CO .—DISCUSSIONS. 
 
 115 
 
 For, when the dividend is the greatest possible, the divisor being 
 constant, the quotient must also be the greatest possible, or infinite. 
 Thus we may say that, 
 
 1. Zero multiplied by any quantity is zero; 
 
 2. Zero divided by a finite quantity is zero; 
 
 3. A finite quantity divided by zero is infinite; 
 
 4. Zero divided by zero is any quantity ; 
 
 5. A' finite quantity divided by infinity is zero; 
 
 6. Infinity divided by a finite quantity is infinite. 
 
 101. "Vanishing: Fractions. 
 
 1. Sometimes an algebraic fraction reduces to the form under a 
 certain hypothesis made upon the quantities which enter it, when 
 0 is not the true value of the fraction ; thus, 
 
 (a 3 —x 3 ) 2 
 
 when a — b, becomes, 
 
 This is not the true value of y ; for, before making the hypothesis, 
 resolve the numerator and denominator into their factors ; thus, 
 _(a—x)(a 2 + ax + x 2 ) 
 
 ^ {a + x) (a — x) ’ 
 
 and cancel the common factor, x—a\ thus, 
 
 a 2 + ax+x 2 
 
 y— 
 
 J a + x 
 
 Now make a=b, and we have, 
 
 3 a 2 3 a , , ,, 
 
 y ~~. — =— , the true value ot y. 
 'Za Z 
 
 2. Again, 
 
 _ f a 2 —x 2 \ __ 0 
 
 y ~\{a-x)*) IB= a*~ 0* 
 
 But, 
 
 3. Again, 
 
 _(a + x)(a— x) _(a + x\ _2 a 
 y—{a—x){a—x)~~\a—x) x=a , 0 
 
 * This notation shows that a is to be made equal to b in the expression ; read, when x is 
 equal to a. 
 
116 
 
 ELEMENTS OE ALGEBRA. 
 
 But, 
 
 y- 
 
 (a—x)(a—x) 
 
 ( a—x_ \ _ 0 _ 
 
 \a z +ax + x 2 ') x=a ~ 3a 2 ~ 
 
 (a—x)(a 2 +ax + x 2 ) Xa^+ax + x^j 
 
 Such expressions are called Vanishing Fractions. They appear 
 to be equal to for a particular hypothesis, but really are not so. 
 They reduce to this form from the presence of a concealed common 
 factor which becomes zero under the particular hypothesis. When 
 such factor is stricken out, the true value results under the hypoth- 
 esis. As we have seen, the true value may be either 7 ,-,or 
 
 J b’ O’ a 
 
 102. The Problem of the Couriers. 
 
 The discussion of an expression consists in making every possible 
 supposition upon the arbitrary quantities which enter it, and inter- 
 preting the results. 
 
 The following problem gives rise to some interesting results. 
 
 Two couriers traveling on the same straight line, one at the rate 
 of m miles an hour, and the other at n miles an hour, are separated 
 by a distance of a miles at 12 o’clock m. When will they be 
 together ? 
 
 A B 
 
 m n 
 
 Let one of the couriers be at A and the other at B. The distance 
 from A to B will be a. 
 
 Now, let the couriers be moving to the right, and let us reckon 
 all distances from A as the origin of distances, — those to the right 
 of A being positive, and those to the left negative. 
 
 Let the courier at A be traveling at m miles an hour, and the one 
 at B at the rate of n miles an hour. Using two unknown quanti- 
 ties, let t be the number of hours, counting from 12 M. until the 
 union takes place, and x the number of miles to be traveled by the 
 courier at B to reach the place of meeting. Then we shall have, 
 
 nt—x - - - (1), 
 
 and mt—xA-a - - (2). 
 
 Combining these equations and eliminating x, we have, 
 
 mt—nt—a. 
 
 a 
 
 W hence, t — . 
 
 ’ m—n 
 
SYMBOLS 0 AMD GO . — DISCUSSIONS. 
 
 117 
 
 This value of t in (1) gives, 
 
 na 
 
 X ~ m—n 
 
 for the distance traveled by the foremost courier. Now, let us take 
 the root, t=——^ f aud make every possible hypothesis upon a, m, 
 and n, and ascertain what t will denote in each case. 
 
 I. First let &l>e a positive quantity, and m>n. 
 
 In this case the denominator of the fraction which expresses the 
 value of t, will be positive, and since a is also positive, t must be 
 essentially positive. 
 
 We shall have to add t hours, therefore, to 12 M. in order to find 
 the time at which the couriers will be together. 
 
 This is what common sense would tell us, since we simply have 
 the case of one courier pursuing the other at a more rapid pace, and 
 so must be constantly gaining on him and at last must overtake 
 him. 
 
 II. Let a he positive, and m<n. 
 
 In this case the denominator of the fraction is negative and the 
 numerator positive ; t is, therefore, negative. 
 
 We must, hence, subtract t hours from the origin of time, 12 
 o’clock; that is, reckon backwards to find the time at which the 
 couriers were together .* 
 
 This we can readily understand, also; for since the courier in ad- 
 vance is, under the hypothesis, traveling more rapidly than the one 
 in the rear, a moment before 12 o’clock, there was less distance be- 
 tween them, and less the moment before that, and so there must 
 have been a time when they were together. 
 
 III. Let a he positive, and m=n. 
 
 In this case the denominator of the fraction will be 0 : hence 
 i is qo . 
 
 This is plainly as it should be, since with a certain distance be- 
 tween them, and traveling at the same rate, they cannot be together 
 in any finite time. The result go is here equivalent to never. 
 
 IV. Let a he negative, and m>n. 
 
 In this case, the denominator of the fraction is positive, but the 
 numerator negative, t is, therefore, negative. We must reckon 
 backwards to find the time of conjunction. 
 
 * See remarks uuder problem. 11, page 110. 
 
118 
 
 ELEMENTS OF ALGEBRA. 
 
 a being negative, places the courier who is traveling at the rate 
 of n miles an hour, m rear of the other: and since he is moving 
 more slowly, they must have been already together in the past. 
 
 V. Let a be negative, and nr<n. 
 
 t will be positive, and the meeting has yet to take place. 
 
 YI. Let a be negative, and m=n. 
 
 t is infinite, and may be either positive or negative. The couriers 
 never have been, and never will be together; or we may say they 
 were together, and that they will be together an infinite time either 
 in the past or future. 
 
 VII. Let a=0, and m>n or <n. 
 
 t is zero; and the couriers were together at 12 o’clock. They 
 manifestly never could have been together before, and can never be 
 again. 
 
 VIII. Let a=0, and m — n. 
 
 t becomes 
 
 That is, any time may be added to or subtracted from the epoch, 
 12 o’clock. 
 
 Since the couriers are together at 12 o’clock, and are moving at 
 the same rate, they have always been, and will always be together. 
 
 IX. Let a=0, m=0, and n=0. 
 
 Under this hypothesis, we have again t=%. Here the couriers are 
 together and are not moving at all. Of course they always have 
 been and always will be together. 
 
 X. "We may now fix the value of any three of the four quantities, 
 t, a, in, or n, at pleasure, and thus determine the remaining one. 
 Let the instructor make such hypotheses and require the student to 
 determine and interpret the results. 
 
 We have not yet entirely exhausted the possible hypotheses upon 
 the quantities which enter the expression under consideration. Ye 
 may make in and n negative in succession; in which case the couri- 
 ers would be traveling in opposite directions; or we may make them 
 both negative at the same time, in which case they would be travel- 
 ing to the left. Let the instructor give such exercises. 
 
 O 
 
 103. The Problem of the Lights. 
 
 The discussion of the Problem of the Lights, as it is called, also 
 gives rise to many interesting and instructive results. 
 
 The problem is this : 
 
SYMBOLS 0 AND 00 . — DISCUSSIONS. 
 
 119 
 
 Given, two lights, placed anywhere upon a straight line, to deter- 
 mine the point or points which will be equally illuminated by each. 
 
 We must first know, obviously, the law which governs the illumi- 
 nating power of light; and, upon this point, physics teaches us, 
 that. 
 
 The intensity of n light at any distance is equal to its intensity at 
 the distance 1, divided by the square of the given distance j that is, 
 if a light has an intensity, say 10, at the distance of one yard from it, 
 
 at two yards its illuminating power will be ^ ; at three yards, ^ ; 
 
 at x yards, 
 
 10 
 
 Then let the lights in question be placed upon the line A B, one 
 at A, and the other at B. 
 
 A c P B 
 
 Let the distance between them be c, and let the intensities of the 
 lights at the unit of distance be, respectively, a and l. Suppose the 
 point P to be one point equally illuminated, and let its distance from 
 A, the origin of distances, be x\ its distance from B will be c—x. 
 Let distances to the right of A be positive ; those to the left will be 
 negative. 
 
 Now, the illuminating power of the light A, for the point P, will 
 
 be ~ ; and that of the light B for the same point, will be^— — 2 . 
 
 Since the quantity of light from each of the lights for this point is 
 to be the same, we must have 
 
 a _ b 
 x 2 (e — x) 2 
 
 Solving this equation, we have 
 
 cV a , 
 
 x— — — —> and 
 
 Va + yb 
 
 cVa 
 
 gr “ 
 
 Va — Vb 
 
 Since we find two roots of the equation, there must be two points 
 of equal illumination, so long, at least, as the roots are of different 
 values. 
 
120 
 
 ELEMENTS OF ALGEBRA. 
 
 Now, to discuss these two expressions, 
 
 I. Let c be positive, and a >b. 
 
 The first root will manifestly be positive ; the point corresponding 
 to this value of x will, therefore, be found somewhere to the right of 
 
 a / Cl 
 
 A. Its distance from A will be c times the fraction - — — — — But 
 
 Vn + V# 
 
 since a is greater than b in this case, the denominator is less than 
 twice the numerator, and, hence, this fraction is greater than c 
 times the fraction must, therefore, be greater than \c. x is thus 
 
 Q 
 
 greater than -, or the point P must be farther from A than it is from 
 
 /y 
 
 B. 
 
 This is what common sense would teach; for, the stronger light 
 being at A, the point equally illuminated must be nearer the lesser 
 light. 
 
 The second root under this hypothesis is also positive, and since 
 
 in the fraction, — - — — , the denominator is less than the numerator, 
 
 's/a— Vb 
 
 the fraction will be greater than unity, and consequently c times it 
 will be greater than c. x is thus greater than c ; that is, this second 
 point lies beyond B, the feebler light. 
 
 This must be so, since, as we move to the right of the lesser light, 
 the difference between the quantity of light from the two sources 
 becomes less and less. It will, thus, at last be zero ; that is, we 
 shall reach a point which will receive the same quantity of light 
 from each. 
 
 II. Let c be positive and a<b. 
 
 Under these hypotheses, the first root will still be positive, but it 
 will be less in value than c; that is, the corresponding point will lie 
 nearer to A, now the feebler light. 
 
 The second value of x will be negative ; that is, the second point 
 will lie to the left of A. 
 
 In this case we have but changed the places of the lights, and of 
 course the circumstances are just reversed from those in the pre- 
 vious case. 
 
 III. Let c be positive and a— b. 
 
 In this case the first root becomes, x— — 5 that is, the first 
 
 2 Va 2 
 
SYMBOLS 0 AND CO . — DISCUSSIONS. 121 
 
 point is midway between the lights. Since the lights are equal, 
 under the hypothesis, this is manifestly true. 
 
 The second root becomes, 
 
 c / \/~a 
 
 X ~~0 = °°* 
 
 The farther we recede from the lights, the less will be the differ- 
 ence in the quantity of light at any point ; but this difference will 
 not become zero, or, in other words, the quantity of light will not be 
 entirely equal until we have reached an infinite distance, rvhich, of 
 course, can never be. 
 
 IV. Let c he negative and a > o?’ < b, or a= b. 
 
 The effect of making c negative, is to place the light whose inten- 
 sity is b on the left of the other, the light having the intensity a 
 being still the origin of distances. The separate discussion of the 
 several relations of a and h to each other in this Case would give the 
 reverse of the results already considered. 
 
 V. Let c = 0 and a> or <b. 
 
 Both roots in this case become 0, and thus the origin is equally 
 illuminated. 
 
 [The supposition that c is 0, places the two lights together at the 
 origin. The lights, however, are of different powers. It is thus laid 
 down in several late works on the subject, that this point cannot be 
 equally illuminated by the two lights, and consequently that the 
 analysis fails for this case. Several attempts have been made to ex- 
 plain the anomaly ; but in reality it should seem that , there is no 
 failure and no real difficulty. 
 
 Let it be remembered that our equation is deduced under a postu- 
 late borrowed from physics with regard to the effect of light at dif- 
 ferent distances. The algebra accepts that law as absolutely true, 
 and its results must be interpreted strictly under that hypothesis. 
 
 Let us then approach, say, the light whose intensity is a at the unit’s 
 distance. When we have reached the distance -V, its intensity is, by 
 
 the law, at the distance T -^g-, it is 10,000«; at 0 it is go. 
 
 The light whose intensity is 5, or has any other intensity at the unit’s 
 distance, is also oo at the point zero. Since, then, under our assumed 
 law, both lights are infinite at the point 0, that point is equally il- 
 luminated, and the roots are truly zero. The results of the analysis 
 
122 
 
 ELEMENTS OF ALGEBRA. 
 
 may be farther vindicated by using mathematical lines, and thus re- 
 moving the question entirely beyond the laws of physical science. 
 See note at the end of the book.] 
 
 VI. Let c— 0 and a=b. 
 
 The first root becomes 0, and the second -gq that is, One of the 
 points is at the origin, and the other anywhere we please. 
 
 This is easily understood. 
 
 VII. Let a and b be negative in succession, or together. 
 
 In this case both roots are imaginary. 
 
 This is as it ought to be, since under our assumed law of physics 
 the absence of light, or total darkness, cannot be reached until the 
 distance from the source becomes infinite. There is thus an infinite 
 separation between absolute light on the one hand, and no light at 
 all on the other. Less than no light is, therefore, impossible. 
 
 VIII. Let a=0, b = 0, c=0. 
 
 Both roots become -g, and thus all points are equally illuminated, 
 and this must be true since no point would have any light at all. 
 
 104. General Properties of Equations of the Second Degree. 
 
 Equations of the second degree containing but one unknown 
 quantity possess some important properties which we shall now pro- 
 ceed to investigate. 
 
 Besummg the general equation, 
 
 X s +2 px=q (1), 
 
 let us complete the square and transpose all the terms into the first 
 member. We shall have, 
 
 x z + 2g>x+p~ — (q+p-)—0. 
 or, (x+jp) 2 — (q+p 2 )= 0. 
 
 Regarding this as the difference of two squares, we may write, 
 (x+ 2 J — Vq+]J 2 ){x+p+Vq+P’) = 0 (2). 
 
 This equation can be true only upon the supposition that one of 
 the factors composing the first member is equal to zero. 
 
 We may, then, have either, 
 
 x+p—\/q+p 2 = 0, or, 
 
 x+jj + Vq+2) 2 = 0. 
 
SYMBOLS 0 AND CO . — DISCUSSIONS. 123 
 
 From these we get, 
 
 x— —p + VqTp* : the first root, and 
 
 x^—p—'f q+p 2 : the second root. 
 
 As there are no other possible values of x which will satisfy this 
 equation, we may say that, 
 
 1st. Every equation of the second degree has tioo roots and only two. 
 By inspection of equation (2) we see that, 
 
 2d. The first member of every equat ion of the second degree, when 
 the second member is 0, is composed of two factors, having the un- 
 Tcnown quantity for the first term in each, and the respective roots 
 with their signs changed for the other terms. 
 
 If the roots are given, the equation can be constructed at once ; 
 thus, for example, let the roots of a certain equation be x —a and 
 x=—b. The equation will be, 
 
 [x— a)(x + b) =0, or 
 x 2 + bx—ax—ab—0. 
 
 Examples. 
 
 The roots being 4 and — 5 ; what is the equation ? 
 
 Ans. x 2 — x— 20 = 0. 
 
 The roots being ab and — c: a and a 2 : — 7 and 8: —6 and —5: 
 
 ^ and—-.: —land +1: V — 1 and — V — 1: what are the equa- 
 b d 
 
 tions in these several cases ? 
 
 105. The Sum of the Roots. 
 
 Let us now add together the two roots, 
 x' — —p + V q +p 3 
 x"=—p—Vq+p 2 
 
 using x' and x!' to distinguish them. We get, 
 x' -\-x" ——2 p. 
 
 Hence we may say, that, 
 
 3d. The sum of the two roots is equal to the co-efficient of the first 
 power of the unlcnoivn quantity with its sign changed. 
 
124 
 
 ELEMENTS OF ALGEBEA. 
 
 106. The Product of the Hoots. 
 
 Now, multiplying the roots together, 
 x' =—p + */q+p* 
 
 x"~ —p— V q +p 2 ; we have, 
 x'x " = —q. 
 
 Hence, we may say that, 
 
 Uh. The p>roduct of the two roots is equal to the second member 
 with its sign changed . 
 
 107. The Greatest Numerical Value of q when Negative. 
 
 Since the product of the two roots is always equal to q (the second 
 member), with a contrary sign, if q is negative, the roots must have 
 like signs; and when added together, the algebraic sum will be their 
 numerical sum. Now, this sum is, as we have seen, equal to 2p. 
 
 We thus have the. sum of two quantities given: and now let us 
 find how to divide this sum into two parts, so that their product 
 shall be the greatest possible. 
 
 Let d be the difference between the two parts, 2p being the sum. 
 
 Then, the greater will be (Prob. 1, Art. 98), 
 
 and the less, 
 
 d 
 
 2 ’ 
 
 Their product must be equal to q, and we shall have, 
 
 Now, as d is diminished, q will increase, until when <7=0, q will be 
 equal to p 2 , and will then be the greatest possible: that is to say, 
 when the roots are equal, their product will be the greatest possible. 
 q, therefore, can never be greater than p z . This, however, is under 
 the supposition that q is negative. 
 
 When q is positive, since the roots multiplied together must then 
 give — q, they will have contrary signs, and when added, will give 
 their numerical difference, instead of their numerical sum. There is, 
 then, no limit to the values of q in such a case. 
 
 Hence, we may say, that, 
 
SYMBOLS 0 AND 00 . — DISCUSSIONS. 
 
 125 
 
 5 th. When the second member is negative, it can never be numeri- 
 cally greater than the square of one half the co-efficient of the first 
 power of the unknown quantity. 
 
 108. Discussion of the Four Forms. 
 
 Resuming the Four Forms already given (Art. 82), viz. : 
 
 x 2 -\-2px~ q - - - - (1). 
 
 x 2 —2px— q - - - - (2). 
 
 x 2 +2 px——q - - - - (3). 
 x 2 —2 px=—q - - - - (1). 
 
 Writing the root, respectively, we have, 
 
 x=—p±Vq +p' 2 - - - (1). 
 x=+p± Vq+p 2 - - - (2). 
 x——p±V—q+p 3 - - (3). 
 x—+p±V—q+p' i - - (4). 
 
 Since V" q +p 2 is greater than p, the radical parts of the roots in 
 the first and second forms are greater, numerically, than their entire 
 parts. The radical parts will therefore govern the signs in these 
 two forms ; so that in the first and second forms the signs of the 
 roots will be unlike, and the negative root will be numerically the 
 greater in the first, while the positive root will be the greater in the 
 second. 
 
 In the third and fourth forms, the roots will be imaginary when 
 q is numerically greater than p 2 . This should be the case, since we 
 have proved that it is impossible for q, when negative, to be greater 
 thany> 2 . When q has such a value, it shows that the equation from 
 which the roots came is impossible. 
 
 When q is less than p 2 , the radical parts in the last two forms will 
 be less than p, the entire parts ; so that when the roots are real in 
 the third and fourth forms, they are both negative in the third, and 
 both positive in the fourth. 
 
 Ifg — p% in the third and fourth forms, the radical parts of the 
 roots reduce to zero. The roots will then be equal in either form, 
 both being —p in the third, and +p in the fourth form. 
 
 When_p=0, the roots in the first two forms reduce to ±Vq, and 
 in the last two ± V —q. Under this supposition, the roots are thus 
 equal with contrary signs in the first and Second forms, and are 
 
126 
 
 ELEMENTS OF ALGEBRA. 
 
 always imaginary in the third and fourth. This ought evidently to 
 be the case, since by making p=0 in the forms themselves, they re- 
 duce to two sets of incomplete equations; thus, 
 
 x 2 — q 
 x 2 — — q. 
 
 Making q— 0, the roots of the first and third forms become, 
 
 x — —p ±p. 
 
 Those of the second and fourth, 
 
 x=+p±p. 
 
 Or, x— 0, and x— —2 'p 
 
 x=+2 p, and x= 0. 
 
 Thus, under this hypothesis, one of the roots in each form be- 
 comes zero. 
 
 The reason of this readily appears, for if q be made 0 in the forms 
 themselves, we have, for the first and third, 
 
 x 2 + 2px=0 ; - - (a) 
 
 and for the second and fourth, 
 
 x 2 — 2px=0. - - (b) 
 
 These may be written, 
 
 x(x + 2p)=Q 
 x(x—2p)=0. 
 
 We may satisfy these, by making either one of the factors zero ; 
 thus, 
 
 x—0 or (x + 2p) — 0 
 x—0 or (x—2p)=0. 
 
 Whence the roots are, 
 
 ,r=0 and X——2 p 
 x=0 and x— +2p. 
 
 We might at once divide out x from equations (a) and (b), and 
 thus reduce them to equations of the first degree; and, generally, 
 whenever ivc can divide one equation through by the unknown quan- 
 tity, one of its roots is zero. 
 
 If p = 0 and <7=0, the roots are 0. 
 
 This supposition reduces the equations themselves to 
 
 x 2 =0 .•. x=0 and x=0. 
 
 Much the same discussion may be had from the forms themselves, 
 
ARITHMETICAL PROGRESSION. 127 
 
 by the use of the 3d and 4th general properties established in Arti- 
 cles 105 and 106. 
 
 For example, in the first form the roots must hare unlike signs, 
 because their product must give — q; they are unequal in numerical 
 value, and the negative is the greater, because their algebraic sum 
 must give —2 p. 
 
 Let the student carry on the discussion. 
 
 o»o- 
 
 SECTION IX. 
 
 ARITHMETICAL PROGRESSION.— RATIO AND PROPORTION.- 
 GEOMETRICAL PROGRESSION. 
 
 109. A series is a succession of terms, any one of which may be 
 derived from the preceding term or terms according to a uniform law, 
 called the law of the series. 
 
 110. Arithmetical Progression. 
 
 An Arithmetical Progression is a series in which any term may be 
 derived from the one preceding it by adding a constant quantity, 
 called the common difference ; thus, 
 
 1, 3, 5, 7, 9 - - etc., 
 
 is such a progression, the common difference being 2. When the 
 common difference is positive, as in this case, we have an increasing 
 progression. When the common difference is negative, we shall have 
 a decreasing progression ; thus, 
 
 9, 7, 5, 3, 1, -1, -3 - - etc., 
 is a decreasing progression, in which —2 is the common difference. 
 
 111. Formula for the Last Term. 
 
 Any term with which we choose to begin the series is called the 
 first term; that with which we end it, is’calledthe last term. These 
 two are the extremes. 
 
 Let 
 
 a, b, c, e, f - - etc., 
 
 be an arithmetical progression, in which cl is the common difference. 
 Then we must have, 
 
 b=ci + cl, c=b + d=a + 2d, 
 e—c + cl—a + od; etc., 
 
128 
 
 ELE SCENTS OE ALGEBRA. 
 
 and it is evident that any term may be found by adding to the first 
 term as many times the common difference as there are preceding 
 terms. 
 
 Then if l represent the last term,, and n the number of terms, we 
 must have, 
 
 l=a+(n— l)d. 
 
 If d be negative, the formula will be, 
 
 l—a—(n—l)d. 
 
 Hence, we may say that, 
 
 To find any term of an arithmetical progression, multiply the com- 
 mon difference by the number of preceding terms, and add this prod- 
 uct to the first term, if the progression is increasing, or subtract it 
 if it is decreasing. 
 
 112. The Sum of Equi-distant Means. 
 
 Let us have an increasing progression of a definite number of 
 terms. If t represent the term which has m terms before it, and d 
 the common difference, we shall have, 
 
 t—a + md - - - - (1). 
 
 How, if we reverse the order of terms, we shall have a decreasing 
 progression, with —cl for the common difference. If /' represent the 
 term which now has m terms before it, we shall have 
 
 t'—l—md - - - - (2). 
 
 Adding (1) and (2), we have 
 
 t -\-t — a "h l. 
 
 Hence, we conclude, that 
 
 In any arithmetical progression, the sum of the two terms which 
 are at equal distances from the extremes, is equal to the sum of the 
 extremes themselves. 
 
 113. Formula for the Sum. 
 
 If s represent the sum of n terms of a progression, we shall have, 
 s=-a + b + c+ - - - - j+lc+l. 
 
 Reversing the order, we shall have, 
 s=l-\-lc + 
 
 c- + b + a. 
 
ARITHMETICAL PROGRESSION. 
 
 129 
 
 Adding these equations, member to member, we have, 
 
 2s— (a + 7) + (7+ 7c) + ... - (& + 5) + (7 + ft). 
 
 The terms taken two and two, as shown, give equal sums, from the 
 principle just established; and there will be as many of these partial 
 sums as there are terms in the progression ; so that we shall have, 
 
 2 s= ( a + l)n; whence 
 (a + l)n 
 
 Hence, we may say, that, 
 
 The sum of a definite number of the terms of an arithmetical pro- 
 gression, is equal to half the sum of the extremes, multiplied by the 
 number of terms. 
 
 The two formulas, 
 
 l=a + {n—T)d - - - - (1), 
 s={a+T)n .... (2), 
 
 are sufficient to solve all ordinary questions touching an arithmetical 
 progression. There are altogether five arbitrary quantities, 
 
 a, l, d, n, s, 
 
 entering these formulas. We may assume any three of them at pleas- 
 ure, and, regarding the other two as unknown, we may combine the 
 equations, and thus deduce their separate values. 
 
 For example, let d, n, and s be given to find a and 7. 
 
 Substitute the value of l from (1) in (2), we shall have, 
 
 _[a + a + {n— 1 ) d~\ n 
 
 s- - . 
 
 From this, 
 
 _ 2s— (n— l)dn 
 a ~~ 2 n 
 
 This value of a in (1), gives, 
 
 7 _ 2s + (n— 1) dn 
 ~~ ~2n ' 
 
 Let the student be required to find the formulas for determining 
 any two of the elements when the other three are given by the 
 instructor. 
 
 H 
 
130 
 
 ELEMENTS OF ALGEBRA. 
 
 Examples. 
 
 1. In the progression 1, 2, 3, etc., of 14 terms, what is the last 
 
 term, and what the sum of the terms? Ans. Z=14, 5 = 105 . 
 
 2. In 2, 5, 8 of 17 terms, find Z and s. 
 
 Ans. 1— 50, 5 = 442. 
 
 3. In 7, 7J, 74, - - - - of 16 terms, find Z and s. 
 
 Ans. Z= 10|, s=142. 
 
 4. In f, |, of 20 terms, find Z and s. 
 
 Ans. Z= — 1-1, 5 = — 13 J. 
 
 5. In 0, 1, i, of 11 terms, find Z and s. 
 
 Ans. 1—5, s=27|. 
 
 6. In —10, -12,-14, of 6 terms, find Z and s. 
 
 Ans. Z= — 20, s=— 90. 
 
 7. Given a = 2, n = 5, l— 22, to find cl and s. 
 
 Ans. cl= 5, s=60. 
 
 8. Given cl= 2, « = 12, 5 = 96, to find a and Z. 
 
 Ans. « = — 3, Z=19. 
 
 9. One hundred stones being placed on the ground in a straight 
 
 line, two yards apart, how far will a person travel who shall bring 
 them one by one to a basket, placed at two yards from the first 
 stone? Ans. 20,200 yds. 
 
 10. A railway train moves two yards the first second, four yards 
 the second second, six yards the third. In how long a time will the 
 train be traveling at the rate of a mile a minute? 
 
 Ans. 14.33 sec. 
 
 114. Ratio. 
 
 Ratio is the relative magnitude of two quantities of the same 
 kind. The measure of this relationship, or, as it is commonly said, 
 the ratio itself, may be always found by dividing one of the quan- 
 
 b ct 
 
 titles by the other ; thus, if a and Z> are the quantities, then — or - , 
 
 expresses the number of times the one contains the other, and is 
 their ratio, or the measure of their relative magnitudes. 
 
 There is some difference in usage as to which quantity shall be 
 made the divisor ; thus, the ratio of 3 : 6 is about as often written, 
 3 6 
 
 - as - . The question depends upon which of the two is regarded 
 u o 
 
 as the standard ; the ratio in this case being 4, if 6 is taken as the 
 
RATIO AND PROPORTION. 
 
 131 
 
 measure ; or 2, if 3 is so taken. It is perhaps better to make the 
 quantity mentioned first, called the antecedent , the divisor, and the 
 second quantity, called the consequent, the dividend. At any rate, 
 we shall adopt this method. 
 
 115. Proportion. 
 
 When two ratios are equal to each other ; as, 
 
 b _d 
 a c ’ 
 
 ( 1 ) 
 
 the four quantities are said to be in proportion. They are often 
 written, 
 
 a : b : : c : d. 
 
 This and equation (1) express entirely the same truth, and the 
 one may at any time be used for the other. 
 
 We may say, then, that, 
 
 A proportion is an equality of two ratios. 
 
 The ratios are called couplets ; thus, a : b is the first couplet; 
 c : cl is the second couplet. Of the four quantities in a proportion, 
 the last one is called a fourth proportional to the other three. If the 
 second term is used also as the third, such term is called a mean 
 proportioned. In this case the last term is called a third propor- 
 tional. The first and last terms are called extremes ; the second and 
 third are called means. 
 
 116 . Let us have the proportion, 
 
 a : b :: c : d\ 
 or, writing it as an equation, 
 
 b_d m 
 a c’ 
 
 whence, be— ad. - - (1) 
 
 Hence, we may say that, 
 
 The product of the means is equal to the product of the extremes. 
 The converse of this is equally true ; for dividing each member of 
 (1) by ac, we have, 
 
 b d , y 
 
 - = - ; or a : b : : c : d. 
 a c 
 
132 
 
 ELEMENTS OE ALGEBRA. 
 
 Whence, 
 
 If the product of two quantities is equal to the product of two other 
 quantities, two of them may he made the extremes, and two the means 
 of a proportion. 
 
 117 . Assume again the proportion, 
 
 a : b :: c : d; or 
 
 b 
 
 a 
 
 ( 1 ) 
 
 We may multiply both terms of each fraction by any quantity, as 
 m ; thus. 
 
 mb md 7 
 
 ■ — = — ; /. ma : mb :: me : md. 
 ma m c 
 
 ( 2 ) 
 
 We may also divide both terms of each fraction by any quantity, 
 as m ; thus, 
 
 b_ d 
 
 m _ m . . a ' b " c ' d . , 
 
 a c ’ ” m ' m m ’ m ' ' 
 
 m m 
 
 We may extract any root, as the ?«th, of both members of (l),and 
 shall have, 
 
 = Va ■ Vb : : V c : tyd. - - ( 4 ) 
 
 V a yc 
 
 Or, again, we may raise both members of (1) to the ?nth power, 
 thus, 
 
 b m d m 7 7 
 
 — = /. a m : b :: c m : d m . (o) 
 
 a c 
 
 Hence, from (2), (3), (4), and (5), it follows that, 
 
 1. We may multiply all the terms of a proportion by the same quan- 
 tity. 
 
 2. We may divide all the terms of a proportion by the same quan- 
 tity. 
 
 JRemark . — It is plaia tliat the multiplier or divisor may he different for each 
 couplet. 
 
 3. We may extract the same root of every term of a proportion. 
 
 4. We may raise every term to the same power. 
 
RATIO AXD PROPOETION. 
 
 133 
 
 118. Dividing unity by each member of the equation 
 
 b cl . 
 
 - = - ; we have, 
 a c 
 
 b : a : : cl : c. 
 
 Whence, 
 
 Consequents may be made antecedents and antecedents consequents. 
 
 The quantities are then said to be in proportion by inversion. 
 
 Multiplying both members of 
 
 b cl . . c . 
 
 by we have, 
 
 a 
 c ' 
 
 d 
 
 V 
 
 a : c :: b : cl. 
 
 Whence, 
 
 The antecedents may be made one couplet , and the consequents 
 another. The quantities are then said to be in proportion by 
 alternation. 
 
 119 . Adding unity to both members of 
 
 b _d 
 a c’ 
 
 and then subtracting unity from both, we have, 
 
 b d b , cl , 
 
 -+ 1 =- + 1 ; 1 = — 1 . 
 
 a c a c 
 
 b+a d+c b—a d—c 
 
 Whence, 
 
 From which we may write, 
 
 a : b + a :: c : d + c, a : b—c :: c : d—c. 
 
 Hence, we may say that, 
 
 The first antecedent may be added to its consequent, provided the 
 second antecedent is added to its consequent. The quantities are then 
 said to be in proportion by composition. In the same way the conse- 
 quents may be subtracted. The quantities are then said to be in 
 proportion by division. 
 
 Let us have two proportions with a couplet the same in each ; thus, 
 b_cl b _g 
 a~ c’ a - /’ 
 
134 
 
 ELEMENTS OF ALGEBRA. 
 
 Then, 
 
 That is, 
 
 d _g 
 
 c~f 
 
 c : d :: f: g. 
 
 Jf the first couplets are the same in two proportions, the other two 
 couplets form a proportion. 
 
 „ l) d mb nd , T . . 
 
 120. From — =- we may write, — = — . Now, making m = 
 
 a c ma nc 
 
 0) V . . . 
 
 1 ± , and « = 1 ±-, in this equation we shall have, after a slight 
 
 q s 
 
 transformation, 
 
 b± 1 --b d±--d 
 q _ s 
 
 p r 
 
 a±--a c±--c 
 
 q s 
 
 Whence, b±-- a : b±!--b :: c±-- d : d±--d. 
 
 q q s s 
 
 That is, 
 
 We may increase or decrease antecedent and consequent by like 
 parts of each. 
 
 121. Let us have two proportions, 
 
 a : b :: c : d ) b d q n 
 f : g ::vn: n ) c f m 
 
 Multiplying the equations, member by member, 
 
 by _ d n 
 af ~ cm’ 
 
 Whence, af : bq : : cm : dn. 
 
 That is, 
 
 We may multiply proportions togeth er, term by term. 
 
 122. From a : b : : c : d, 
 we have bc—ad\ adding ab to both members, 
 
 bc + ab=ad + ab. 
 Whence, b(a + c)—a{b + d). 
 
GEOMETRICAL PROGRESSION. 
 
 135 
 
 And, from a previous principle, 
 
 a : b : : a + c : b+d. 
 
 Hence, 
 
 Antecedent is to its consequent , as the sum of the antecedents is to 
 the sum of the consequents. 
 
 123. Where several proportions are written together, thus: 
 
 a : b :: c : cl :: f : g , etc., 
 
 it is called a continued proportion. 
 
 It is sometimes written thus, 
 
 a : c : / : : b : d : g. 
 
 The principles already explained may be readily extended to con- 
 tinued proportions. 
 
 124. The mean proportional between two quantities may be readily 
 found : thus, let it be required to find the mean between a and b. 
 
 We have, a : x : : x : b, 
 
 x z =ab, 
 x = fab. 
 
 That is. 
 
 The mean of any tiuo quantities may be found by multiplying the 
 quantities together and extracting the square root of the product. 
 
 125. When one of two variables is expressed in terms of the recip- 
 rocal of the other, as, 
 
 1 m 
 
 x— - ; or x— — , 
 
 V V 
 
 the quantities are said to be reciprocally proportional. It is manifest 
 that one will increase as the other diminishes, and -they are thus said 
 to vary inversely. 
 
 From this expression, we have, 
 
 xy— 1, or xy—m. 
 
 Hence, the product of two such quantities is always constant. 
 
 126. Geometrical Progression. 
 
 A Geometrical Progression is a series, any term of which may be 
 
136 
 
 ELEMENTS OF ALGEBRA. 
 
 derived from the one preceding, by multiplying it by a constant 
 quantity called the ratio. It is a continued proportion. 
 
 The progression will be increasing when the ratio is greater than 
 unity: thus, 
 
 2, 4, 8, 16, 32, etc., 
 
 is an increasing progression, whose ratio is 2. 
 
 The progression will be decreasing when the ratio is less than 
 unity; thus, 
 
 32, 16, 8, 4, 2, 1, i, etc., 
 is a decreasing progression, in which | is the ratio. 
 
 127. Formula for East Term. 
 
 Let us assume the geometrical progression, 
 a : b : c : d : e : f : etc., 
 in which r is the ratio. 
 
 From the definition we shall have, 
 
 b=ar, c=br=ar 2 , d—cr=ar 3 , etc. 
 
 Whence we see that the exponent of r, being unity in the expres- 
 sion for b, goes on increasing by unity for c, d, etc. ; so that for the 
 term which has n terms before it, calling it 1, we shall have, 
 
 l—ar n ~\ 
 
 Hence, we may say that, 
 
 Any term of a geometrical progression may be found by multiply- 
 ing the first term by the ratio raised to a power whose exponent is 
 equal to the number of preceding terms. 
 
 128. Formula for the Sum. 
 
 To find a formula for computing the sum of any number of terms 
 of such a series, let us take a progression of a definite number of 
 terms, 
 
 a i b : c : d : - - - - : j : k : l, 
 the ratio being r. 
 
 Replacing each term after the first by its value in terms of the 
 first term and the ratio, and representing the sum of n terms by s, 
 we shall have, 
 
 s=a+ar+ar z + - - - ar n ~* + ar n ~\ 
 
 How, multiplying both members of this by r, 
 
 sr—ar + ar 2 +ar 3 + - - - ar n ~ x + ar n . 
 
GEOMETRICAL PROGRESSION. 
 
 137 
 
 Subtracting the first of these equations from the second, we shall 
 have, 
 
 whence, 
 
 sr—s—air—a : 
 
 ar —a . . , , _ 
 
 s= — ; or, since ar n is equal to Ir, 
 
 s=- 
 
 r — 1 
 Ir—a 
 
 r—1 ’ 
 
 an expression for the sum of any given number of terms. 
 We have thus the two formulas, 
 
 l—ar n ~ x and 
 Ir—a 
 
 r—1 
 
 , in which there are five arbi- 
 
 trary quantities. If we know all but one which enter either of them, 
 we can substitute the given values in such formula, and at once de- 
 duce the remaining one; thus, if a— 2, r= 2, and n=5, we shall 
 have from the first, 
 
 7=2 x 2 4 = 32. 
 
 If s=15, a— 1, and r= 2, from the second formula, 
 
 15 = ^ — ; whence, 1=8. 
 
 Z — 1 
 
 129. To find the Formula for any Element. 
 
 When any three of the five quantities, a, l, r, n, and s, which enter 
 these two formulas, are given, we may combine the formulas, and 
 eliminate one of the remaining quantities, and thus find the fifth 
 quantity. When a combination of the formulas is required, make 
 the combination and deduce a general expression for the desired 
 quantity, before substituting for the given quantities. 
 
 For example, let s = 62, r= 2 and n= 5. Combiningthe formulas, 
 eliminating 1, we have, 
 
 whence, substituting the given values, 
 
 2° —I 
 
 Substituting a= 2, r= 2 and n — 5 in the first formula, we have, 
 l- 32. 
 
 When n is to be found, the resulting formulas will require the use 
 of logarithms, which are yet to be explained. 
 
138 
 
 ELEMENTS OF ALGEBRA. 
 
 Examples. 
 
 1. Given a= 1, r=2, n — 7, to find l and s. Ans. 1= G4, s=127. 
 
 2. Given a — 4, r— 3, n— 10, to find l and s. 
 
 Ans. /= 78732, s=11809G. 
 
 3. Given a= 9, r=J, « = 7, to find l and s. 
 
 Ans. I— 258£, s = 591 T \ 
 
 4. Given « = 6£, r=\, n— 8, to find l and s. 
 
 Ans. J=106f, s = 3074. 
 
 5. Given a— 8, r=\, n — 15, to find l and s. 
 
 Ans. ^-g-oVg, s=15 + . 
 
 6. Given a—f, r— 1| « = 11, to find / and s. 
 
 Ans. ?=xrtfx, 
 
 130. To insert Geometrical Means. 
 
 Let it now be required to insert a given number of geometrical 
 means between any two numbers, as a and b. These two quantities, 
 together with the means, will form a geometrical progression of two 
 more terms than the number of the means. If there are to be in 
 means there will be in + 2 terms. Now, from the formula 
 
 This is the value of the ratio when there are n terms, and a and l 
 are the extremes. But in the case in hand, a and b are the extremes 
 andm + 2the number of terms. Substituting these in the above 
 expression for r, we shall have, 
 
 This being the new ratio, we may now find the means successively 
 from the first term a ; thus, 
 
 For example, let it be required to insert 3 geometrical meaus be- 
 tween 2 and 32. 
 
 In this case 
 
 m — 3, a = 2, and b= 32. 
 
GEOMETRICAL PROGRESSION. 
 
 139 
 
 These in (1) give, r=V^=='V / 16=2. 
 
 Whence the means will he 4, 8, and 16, and we shall have the pro- 
 gression, 
 
 2 : 4 : 8 : 16 : 32. 
 
 Examples. 
 
 1. Insert 5 geometrical means between 3 and 192. 
 
 Ans. 3 : 6 : 12 : 24 : 48 : 96 : 192. 
 
 2. Insert 5 geometrical means between 5 and 1215. 
 
 Ans. 5 : 15 : 45 : 135 : 405 : 1215. 
 
 131. The Sum of an Infinite Progression. 
 
 Let us now have a decreasing progression of an infinite number of 
 terms. The ratio will be less than unity. The formula, 
 
 ar n —a ... 
 
 s=— — — , may be written, 
 
 ar n a 
 S= r 
 
 r — 1 r — 1 
 
 But, since r is less than 1, r 2 will be less than r, r 3 still less, and 
 so the results will go on diminishing as the power is increased; and 
 since n is infinite, r n is zero; whence it follows, that the first term 
 of the second member of the above expression is zero, and we have 
 
 a 
 
 But, since r<l, we may write 
 
 a 
 
 S= l-? 
 
 That is to say, 
 
 The sum of the terms of a decreasing geometrical 'progression of an 
 infinite number of terms, is equal to the first term, divided by unity 
 minus the ratio. 
 
 What is the sum of the following? 
 
 1. 1 : £ : i - - - - to infinity. Ans. 2. 
 
 2. 1 : J : -$• - - - - to infinity. Ans. f. 
 
 3. 4:2:1 - - - - to infinity. Ans. 8. 
 
140 
 
 ELEMENTS OF ALGEBRA. 
 
 4 - i : is '• tIt - - - to infinity. Ans. \. 
 
 5- 1 : - : \ - - - to infinity. Ans. — — . 
 
 a a 2 J a — l 
 
 If the ratio of a geometrical progression is negative, the terms 
 will be alternately positive and negative. In substituting in the 
 formulas, be careful to give the ratio its proper sign. 
 
 G. Given, a=—2, r— — 3n—5, to find l and s. 
 
 Ans. I— — 162, s = — 121. 
 
 SECTION X. 
 
 LOGARITHMS. 
 
 132. If the number 10 be raised to the second power, we shall have 
 100 for the result; thus, 
 
 10 2 = 100. 
 
 The exponent 2 is called the logarithm of 100. Let us write 
 several exact powers of 10, thus, 
 
 (10)~ 2 , (10) -1 , 10°, 10 1 , 10 2 , 10 3 , etc. 
 
 Or, 
 
 aV )2 1, 10, 100, 1000, etc. 
 
 The exponents, 
 
 —2, —1, 0, 1, 2, 3, etc., 
 in the upper row, are the respective logarithms of 
 .01, .1, 1, 10, 100, 1000, etc., 
 
 in the lower row. 
 
 If we should take any number which is not an exact power of 10, 
 the exponent would not be a whole number, but would be made up 
 of an entire part and a fraction ; thus, the number 25 is greater than 
 the first power of 10, and less than the second pow T er ; whence 10 
 must be raised to a power greater than 1, and less than 2, to produce 
 25. Assuming that the fraction to be added to 1 to give the proper 
 exponent is .397940, we should have, 
 
 (10)’- 3,7940 =25. 
 
 Here 1.397940 is the logarithm of 25. 
 
L0GABITH1T3. 
 
 141 
 
 It is obvious that any number except unity may be used instead 
 of 10, as the fixed number, whose several powers are to be taken. 
 
 We may, then, say that. 
 
 The logarithm of a number is the exponent of the paver to which it 
 is necessary to raise a fixed number, called the base, in order to pro- 
 duce the given number. 
 
 The base of the Common System, as it is generally called, is 10. 
 
 The base of the Napier an System, so named from Baron Napier, 
 to whom the invention of logarithms is due, is 2.71828182.... 
 
 The entire part of a logarithm is called the characteristic ; the deci- 
 mal part is called the mantissa. 
 
 133. Characteristic. 
 
 The characteristic of the logarithm of a number may always be 
 written at once from the number itself. The law for writing the 
 characteristic may be discovered from the following : 
 
 ( 10) 4 = 1,0000 
 ( 10) 3 = 1,000 
 
 ( 10) 2 = 100 
 
 ( 10) 1 = 10 
 
 ( 10 )° = 1 
 
 (10)- 1 =.l 
 ( 10)- 2 =.01 
 ( 10)- 3 =.001 
 ( 10)- 4 =.0001 
 
 log 
 
 10,000 = 
 
 4. 
 
 log 
 
 1,000 = 
 
 n 
 
 O. 
 
 log 
 
 100 = 
 
 2. 
 
 log 
 
 10 = 
 
 1. 
 
 log 
 
 1 = 
 
 0, 
 
 log 
 
 .1 =- 
 
 -1. 
 
 log 
 
 .01 fj 
 
 -2. 
 
 log 
 
 .001 =- 
 
 -3. 
 
 log 
 
 .0001 =- 
 
 -4. 
 
 It thus appears that the characteristic of all numbers greater than 
 unity is positive, and that the characteristic of all numbers less than 
 unity is negative ; and that, 
 
 When the number is greater than unity, the characteristic is one 
 less than the number of digits composing the number. If the num- 
 ber is partly decimal, only the entire part is to be counted. 
 
 When the number is less than unity, and is expressed as a deci- 
 mal, the characteristic is negative and one greater than the number 
 of 0’s immediately following the decimal point. 
 
 The mantissa, or decimal part, must be found in a table computed 
 for the purpose. 
 
 134. The following is the beginning of such a table, showing the 
 logarithms from 1 to 100 : 
 
142 
 
 ELEMENTS OP ALGEBRA. 
 
 TABLE. — Logarithms from 1 to 100. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 1 
 
 0 000000 
 
 26 
 
 1 414973 
 
 51 
 
 1 707570 
 
 76 
 
 1 880814 
 
 2 
 
 0 301030 
 
 27 
 
 1 431364 
 
 52 
 
 1 716003 
 
 77 
 
 1 886491 
 
 3 
 
 0 477121 
 
 28 
 
 1 447158 
 
 53 
 
 1 724276 
 
 78 
 
 1 892095 
 
 4 
 
 0 602060 
 
 29 
 
 1 462398 
 
 54 
 
 1 732394 
 
 79 
 
 1 897627 
 
 5 
 
 0 698970 
 
 30 
 
 1 477121 
 
 55 
 
 1 740363 
 
 80 
 
 1 903090 
 
 6 
 
 0 778151 
 
 31 
 
 1 491362 
 
 56 
 
 1 748188 
 
 81 
 
 1 908485 
 
 7 
 
 0 845098 
 
 32 
 
 1 505150 
 
 57 
 
 1 755875 
 
 82 
 
 1 913814 
 
 8 
 
 0 903090 
 
 33 
 
 1 518514 
 
 58 
 
 1 763428 
 
 83 
 
 1 910078 
 
 9 
 
 0 954243 
 
 34 
 
 1 531479 
 
 59 
 
 1 770852 
 
 84 
 
 1 924279 
 
 10 
 
 1 000000 
 
 35 
 
 1 544068 
 
 60 
 
 1 778151 
 
 85 
 
 1 929419 
 
 11 
 
 1 041393 
 
 36 
 
 1 556303 
 
 61 
 
 1 785330 
 
 86 
 
 1 934498 
 
 12 
 
 1 079181 
 
 37 
 
 1 568202 
 
 62 
 
 1 792392 
 
 87 
 
 1 939519 
 
 13 
 
 1 113943 
 
 38 
 
 1 579784 
 
 63 
 
 1 799341 
 
 88 
 
 1 944483 
 
 14 
 
 1 146128 
 
 39 
 
 1 591065 
 
 64 
 
 1 806180 
 
 89 
 
 1 949390 
 
 15 
 
 1 176091 
 
 40 
 
 1 602060 
 
 65 
 
 1 812913 
 
 90 
 
 1 954243 
 
 16 
 
 1 204120 
 
 41 
 
 1 612784 
 
 66 
 
 1 819544 
 
 91 
 
 1 959041 
 
 17 
 
 1 230449 
 
 42 
 
 1 623249 
 
 67 
 
 1 826075 
 
 92 
 
 1 963788 
 
 18 
 
 1 255273 
 
 43 
 
 1 633468 
 
 68 
 
 1 832509 
 
 93 
 
 1 968483 
 
 19 
 
 1 278754 
 
 44 
 
 1 643453 
 
 69 
 
 1 838849 
 
 94 
 
 1 973128 
 
 20 
 
 1 301030* 
 
 45 
 
 1 653213 
 
 70 
 
 1 845098 
 
 95 
 
 1 977724 
 
 21 
 
 1 322219 
 
 46 
 
 1 662758 
 
 71 
 
 1 851258 
 
 96 
 
 1 982271 
 
 22 
 
 1 342423 
 
 47 
 
 1 672098 
 
 72 
 
 1 857333 
 
 97 
 
 1 986772 
 
 23 
 
 1 361728 
 
 48 
 
 1 681241 
 
 73 
 
 1 863323 
 
 98 
 
 1 991226 
 
 24 
 
 1 380211 
 
 49 
 
 1 690196 
 
 74 
 
 1 869232 
 
 99 
 
 1 995635 
 
 25 
 
 1 397940 
 
 50 
 
 1 698970 
 
 75 
 
 1 875061 
 
 100 
 
 2 000000 
 
 135. General Principles of Logarithms. 
 
 The principles already established, Art. 50, with regard to expo- 
 nents in general, govern the use of logarithms, they being but a 
 class of exponents. 
 
 Let us repeat the principles referred to, however, in this connec- 
 tion, since they form the basis of all logarithmic computations. 
 
 Let m and n be any two numbers whose logarithms are x and y. 
 Then, 
 
 (10) I = H1 - - (1) 
 
 (10 /=n - - (2). 
 
 Multiplying member by member, we have, 
 
 (10) I+!, = mn, 
 
 whence, x+y= log mn, 
 
 That is, 
 
 1. The suih of the logarithms of two numbers is the logarithm of 
 their product. 
 
LOGARITHMS. 
 
 143 
 
 Again, from (1) and (2) we have, 
 
 ( 10 )" __m m 
 (10)* ~ n’ ° r ’ 
 
 VI 
 
 (10)*"*=— ; whence, 
 n 
 
 * 
 
 , m 
 
 x ~y= lo g-« 
 
 That is, 
 
 2. The difference of the logarithms of two numbers is the logarithm 
 of their quotient. 
 
 We have again, from (1), 
 
 ((10)') ? rr»iP 
 
 m px =m p 
 px= log Vl p . 
 
 That is, 
 
 3. The product of the logarithm of a number by p is the logarithm 
 of the p power of the number. 
 
 If we extract the r root of both members of (1), we have, 
 
 (10) r = ffm :. 
 
 rQ r 
 
 —A: log V Ol. 
 r * 
 
 That is, 
 
 4. The quotient of the logarithm of a number by r, is the logarithm 
 of the r root of the number. 
 
 These four truths are called the general principles of logarithms. 
 
 Apply the foregoing principles in the transformation of the follow- 
 ing expressions into corresponding logarithm equations: 
 
 Ans. log a;=log ft + log b — log c — log d. 
 
 A ns. log x—m log « + « log b + p log c. 
 
 Ans. log x=m log a — n log b—p log c—q log d. 
 
 1. x— 
 
 ab 
 
 cd’ 
 
 2. x—a n b n c p . 
 a m b ~ n 
 
 o. x= 
 
 c p d q ' 
 
 4. x—a n b i. 
 
 Ans. log x— — log a— - log b. 
 
144 
 
 ELEMENTS OF ALGEBRA. 
 
 5. x = y // a m b~ n ci 
 
 a^y cT 
 
 6. x— 
 
 Ans. log x= — log a — log b + — log c. 
 ° n ° nq 
 
 m 
 
 7. x=s 
 
 A?is. log cr=log a -\ — log c — log b—\ log d. 
 
 bVd n 
 
 {a + b) n cA 
 
 ( c + d ) V d 3 
 
 Ans. log x—n log {a + b)+m log c — log {c + d) — f- log d. 
 
 8. x— 
 
 d/a + b 
 
 9. x— \/a 2 —x 2 . 
 10. x={a m ) n . 
 
 Ans. log x— — log {a + l). 
 
 Ans. log a=^-log {a + x)+ ^ log {ci—x). 
 
 Ans. log x—nin log a. 
 
 136. Solution of Equations. 
 
 By the aid of logarithms we are enabled to solve what would other- 
 wise be difficult equations. 
 
 Solve the following : 
 
 ( x log 5= log 25. 
 
 2. 7 V * = 17. 
 
 Whence, Vo" log 7 = log 17, 
 log 17 
 
 which gives, V x= tyY m 
 
 3. a x = b. 
 
 4. 
 
 log 25 
 Ans. x— W — — • 
 log o 
 
 Ans. 
 
 log b 
 log a 
 
 Ans. 
 
 ( log g V 
 
 \log {a + b)J 
 
 x 
 
 5. a m — b. 
 
 Ans. 
 
 x — rn 
 
 137. Logarithms of Decimals. 
 
 When a number is composed of the same figures, no matter how 
 its value may be made to change by moving the decimal point to 
 
LOGARITHMS. 
 
 145 
 
 the right or left, the mantissa will always remain the same ; the 
 only change being in the characteristic : thus, if the logarithm of 
 2975376 is 6.354189, we shall have as follows : 
 
 log 2975.376 =3.354189. 
 log 29.75376 =1.354189. 
 log 2.975376 =0.354189. 
 log .02975376 = 2.354189. 
 
 The mantissa, it will be observed, remains the same. 
 
 To prove that this is true, let n be any number whatever, and p 
 any other whole number, positive or negative. 
 
 Then, n x (lO) 71 , will represent the product of n by any exact 
 power of 10. We may write, 
 
 log (»x (10) 7> )=log ?j + log 10 r =log n+p. 
 
 Now, in the expression, log n+p, since p is entire, it will be added 
 to, or if negative subtracted from, the characteristic; so that the 
 decimal part will remain intact. But to multiply a number by any 
 entire power of 10, positive or negative, is but to remove the decimal 
 point to the right or left ; and hence it follows that the mantissa of 
 the logarithm of a number expressed by any combination of figures 
 will remain the same, whether any part of it is decimal or not. 
 
 This principle enables us to disregard the decimal point entirely in 
 finding the mantissa of the logarithms of numbers. 
 
 138. General Properties. 
 
 When the base of a system of logarithms is greater than unity, 
 the logarithm of oo is +co, and the logarithm of 0 is —go . 
 
 For assuming the exponential equation, 
 
 10 x =n, 
 
 n being any number and x its logarithm, we see that as n increases 
 x must increase, and when n is infinite x must be infinite also. 
 
 As n is made less and less, x must be diminished, until when n is 
 1, x is 0. If, now, n is still diminished, x becomes negative, and 
 when n is 0, x is again co, and becoming so from the negative side, 
 is still regarded as negative. Thus, the logarithms of positive num- 
 bers alone require for their logarithms all numbers from — go to 
 + co. There cannot, then, be any logarithms of negative quantities. 
 
 If the base is less than unity, the same thing will hold; except 
 IC 
 
146 
 
 ELEMENTS OF ALGEBRA. 
 
 that the logarithms of numbers greater than unity will then be nega- 
 tive, and those of numbers less than unity will be positive. 
 
 It will be seen that the logarithm of unity in any system is zero. 
 
 139. Moduius. 
 
 Let us assume what is called the Logarithmic Series, leaving its 
 deduction for a more advanced course. It is, 
 
 ?/2 yS y4: 
 
 l °g( 1 +y)=m(y-'.r + j-^r+ - - - - etc.), 
 
 in which we have the logarithm of a number, (1 + y), developed into 
 a series, in terms of a number, y, less by unity than the number 
 itself. 
 
 It will be observed that the second member is composed of two 
 factors, m and the series within the brackets. The factor m depends 
 for its value entirely upon the base of the system, and, for the same 
 system, is constant. It is called the modulus of the system. The 
 modulus, then, of any system of logarithms is the constant factor, 
 which is common to all logarithms of that system. 
 
 The modulus of the N apieran system is 1 ; the modulus of the 
 common system is 0.43429448. 
 
 140. Indeterminate and Identical Equations. 
 
 An Indeterminate quantity is one which has no fixed value, but 
 admits of an infinite number of values in succession. Indeterminate 
 and arbitrary quantities do not differ in nature ; but it is usual to 
 call such quantities, when represented by the first letters of the 
 alphabet, arbitrary ; and when represented by the final letters, inde- 
 terminate. 
 
 An Indeterminate Equation is one which contains at least two 
 quantities which can only be assumed in relation to each other ; 
 thus, 
 
 bx — 4y=9, 
 
 is such an equation ; x and y admitting of any number of sets of 
 values, but neither of them admitting of any specific value indepen- 
 dently of the other. Such quantities are called indeterminate, or, 
 perhaps more correctly, variables. 
 
 An Identical Equation is one which is true for all possible values 
 of the indeterminate, or arbitrary quantities, which enter it. For 
 example, 
 
LOGARITHMS. 
 
 147 
 
 ax + by—ax + by, 
 
 (a+x)~ = a 2 +2 ax + x 2 , 
 
 5x -t-7a; 2 +4=5x + 7.'C 2 +4, 
 are manifestly sucli equations. 
 
 An identical equation differs materially from the ordinary or de- 
 terminate equation, which admits of but a specific number of values; 
 and also from the indeterminate equation, which is satisfied for any 
 number of sets of relative values of the variables which enter it. In 
 an identical equation, the co-efficients of the indeterminate quanti- 
 ties (x, y, z, etc.), are no longer arbitrary, but have resultant values, 
 and so are themselves determinate. 
 
 141. Indeterminate Co-efficients. 
 
 Any identical equation containing but one indeterminate quantity 
 can be reduced to the form of 
 
 a + bx + cx 2 + dx 3 + etc.=0 - - (1), 
 
 an equation in which a, b, c, cl, etc., are called Indeterminate Co-effi- 
 cients, since they are co-efficients of the indeterminate quantity. 
 
 Now, since this equation is true whatever value x may have, it will 
 be true when x—0. Making this hypothesis in the equation, we 
 shall have, 
 
 o=0. 
 
 This value of a in (1) gives, 
 
 bx + cx 2 + dx 3 +etc.=0. 
 
 Factoring, 
 
 x(b + cx + dx 2 +etc.)=0. 
 
 This may be satisfied for x=0, or, 
 
 b + cx + dx 2 +etc. = 0 - - (2) 
 
 Now, x, in this equation, must admit of the same value as in (1); 
 hence, this must also be an identical equation, and is true for x—0. 
 This value of x in (2) gives, 
 
 b—0 ; 
 
 and in the same way we may prove, 
 
 c= 0, 
 d— 0; 
 
 and so for all the co-efficients of x. 
 
 Hence, we may say that, 
 
148 
 
 ELEMENTS OF ALGEBEA. 
 
 In cm identical equation of one indeterminate quantity, the co-effi- 
 cients of that quantity are severally equal to zero. 
 
 143. Principle of Indeterminate Co-efficients. 
 
 Let us now have the identical equation, 
 
 a + bx + ex* -\- Qtc.=za' + b' x + c x* +etc. 
 
 Transposing and factoring, 
 
 a—a' + (b—b')x + (c—c')x z +etc.=0. 
 
 Now, from the principle just deduced, 
 
 a— a'=0 .•. a— a! 
 b-b '= 0 b-V 
 
 c—c '= 0 c—c' 
 
 etc. etc. 
 
 Whence, we may say that. 
 
 In an identical equation of one indeterminate quantity, the co-effi- 
 cients of the like quivers of that quantity in the two members are sev- 
 erally equal to each other. 
 
 This principle may be readily extended to identical equations con- 
 taining any number of indeterminate quantities. It is called the 
 Principle of Indeterminate Co-efficients. 
 
 143. Development of Expressions. 
 
 Let us now apply the principle of Indeterminate Co-efficients to the 
 development of algebraic expressions into series. 
 
 Take the fraction, — r, and placing it equal to a development 
 
 1 — X — X“ 
 
 of the proposed form : we have, 
 
 "I -4- 9 p 
 
 — — a + bx+cx z +dx* + etc. - (1). 
 
 l—x—x- 
 
 Since this development must be true for all values of x, this 
 
 must be an identical equation. 
 
 Clearing it of fractions, using the vertical bar for convenience, 
 we have, 
 
 1 + 2 x=a + b 
 
 x + c 
 
 x ~ + d 
 
 —a 
 
 -b 
 
 — c 
 
 
 —a 
 
 -b 
 
 x 3 + etc. 
 + etc. 
 + etc. 
 
 Whence, by the principle of Indeterminate Co-efficients, we shall 
 have. 
 
LOGARITHMS. 
 
 149 
 
 l— a :. «=1 
 
 2=b—a .\ b— 3 
 
 0 —c—b—a :. c= 4 
 0 =d—c—b :. d—5 
 etc. etc. 
 
 These values, substituted for a, b, c, d, etc., in (1) give, 
 l + 2x 
 
 1—x—x 2 
 the desired development. 
 
 = 1 + 3x + 4x 2 + 5a; 3 + etc., 
 
 Examples. 
 
 1 | 
 
 1. Develop - — — into a series. Ans. l + 5.T + 15a; 2 + 45 .t 3 +etc. 
 
 2 ft 
 
 2. Develop — — : — - into a series. 
 
 1 + x + x 2 
 
 Ans. l—2x + x 2 +x 3 — 2x i +eto. 
 
 3. Develop into a series. 
 
 a + x 
 
 x x 
 
 Ans. 1— 1 + — — — r + etc. 
 
 a a* 
 
 2 2 ) 
 
 4. Develop - — — — into a series. 
 
 1 — 2x— 3x 2 
 
 Ans. l + .-r + Sx 3 +13a; 3 +etc. 
 
 5. Develop - 
 
 2x + x 2 
 In this case we have, 
 
 into a series. 
 
 "Whence, 
 
 =a + bx + cx 2 +etc. 
 
 2x + x 2 
 
 x 3 +etc 
 + etc. 
 
 1 = 0 
 0=2 a 
 etc. 
 
 1 = 2 a 
 
 x + 2 It 
 
 x 2 +2 c 
 
 
 + • a 
 
 + b 
 
 Here we encounter an absurdity in the result, 1=0. 
 
 This shows that the development cannot be made as proposed. 
 
 Then, let us factor the expression thus, - x - : now let us de- 
 
 CC /v “p Oj 
 
 velop 
 
 2 + x 
 
 . We have, 
 
 2 + x 
 
 -~a + bx + cx + etc. 
 
150 
 
 ELEMENTS OF ALGEBRA. 
 
 Proceeding as usual, we have, 
 
 And thus we have, 
 
 l=2a + 2b\x + 2c\x + etc. 
 
 a\ b\ +etc. 
 1 = 2 a. a — 
 
 0=2 b + ci. l — —\. 
 
 0=2c +b. c— 
 
 etc., etc. 
 
 1 _1 
 
 2 ~i~ x 2 
 
 1 1 „ 
 
 ~± x + % x +etc. 
 
 Now, multiplying both members by — , we have, 
 
 2 x + x 2 2x 4 8 
 
 5i.-r+s a; + etc 5 
 
 which is the true development of the original expression. 
 
 1 . 
 
 Here the first term — is the same as 1 — - ; so that the absurditv 
 2x 2 ' 
 
 above arose from not starting our development with a low enough 
 power of x. 
 
 So, in general, when such an absurdity develops itself, it will be 
 found due to a like cause. 
 
 143. Partial Fractions. 
 
 The Principle of Indeterminate Co-efficients affords an easy method 
 of resolving a fraction into its partial fractions. 
 
 To do this, place the given fraction equal to the sum of as many 
 partial fractions as the denominator of the fraction has factors, the 
 several numerators of such assumed partial fractions being p, q, r, 
 etc. — quantities to be determined — and the denominators being the 
 several separate factors of the denominator. 
 
 For example, let it be required to find the partial of — — 
 
 We have, 
 
 - — =— — | — — , an identical equation. 
 a 2 —x z a + x a—x 
 
 Clearing of fractions, 
 Whence, 
 
 2 a = ap —px + aq + qx. 
 
 2 a= ap + aq; 
 
 0 =—p + q. 
 
 Combining these two equations and eliminating q, we have, 
 
 P-1- 
 
This value of p gives, 
 Therefore, 
 
 LOGARITHMS. 
 
 151 
 
 2 = 1 - 
 
 2a _ 1 1 
 
 a 2 — x 2 a + x a — x" 
 
 Examples. 
 
 3^ 3 2 2 
 
 1. Eesolve — — - into partial fractions. Ans. „4 — — -. 
 
 x — 3 x+6 
 
 2. Resolve 
 
 a; 3 -9 
 
 x 2 + 4x — 11 
 (x— 1 ) (x + 2) (x—3) 
 
 into partial fractions. 
 
 , 1 1 1 
 
 Ans. — H 
 
 x — 1 x + 2 X — a 
 
 1 ' ^iCC | 'X/ ^ 
 
 3. Resolve — .1 ■ into partial fractions. 
 
 ( 1 — a ;) 3 
 
 When there are equal factors in the denominator, as in this case, 
 use all of the several powers ; thus, 
 
 1 — 2x + x 2 
 
 V 
 
 
 * - +: ' 
 
 ( 1 — a ;) 3 ( 1 — a :) 3 ( 1 — a ;) 3 1 — a ;’ 
 
 i 
 
 Ans. 
 
 1 1 
 
 + : 
 
 1— a; 1 — a: 1— a; 
 
 4. Resolve 
 
 a + x 
 
 5. Resolve - 
 
 (a + x 
 -16 
 
 — into partial fractions. Ans. - 
 •) 3 1 « + ■ 
 
 x a + x 
 
 x 3 + 2x — 15 
 
 into partial fractions. 
 
 , 2 2 
 
 Ans. . 
 
 a; + o x—o 
 
NOTE. 
 
 PROBLEM OF THE LIGHTS. 
 
 Tiie point maintained in the discussion of the problem of the lights, that c= 
 0, when a is greater or less than b, is a legitimate hypothesis, and that the an- 
 alytical result is to be interpreted in a strict and natural way, is so entirely at 
 issue with all the late writers upon algebra, that a more detailed examination 
 of the question seems to be demanded than could well be given it in the body of 
 the text. 
 
 Let us, then, consider, first, the arguments upon which we are asked to aban- 
 don the analysis in favor of what is supposed to be common-sense. 
 
 It will be remembered that the roots of the equation derived from the con- 
 
 ditions of the problem are, x=- 
 
 Va 
 
 . Now, when c is made equal to zero 
 
 Va + Vb 
 
 in these roots, rtand& remaining unequal, we must have x—0, for the value of 
 either root. The interpretation of this result, according to the general law, 
 gives us the origin of distances as the single point of equal illumination. We 
 then have the case of two lights, placed at the same point, differing, however, 
 much in intensity, and are required, by the analysis, to conclude that such point 
 is equally illuminated by the greater and the lesser light. This conclusion, we 
 are told, is not true. 
 
 The first distinguished author, whose reasoning upon the point we shall ex- 
 amine, says, in his University Algebra, “It is obvious that when two lights, 
 of unequal intensities, occupy the same place, there is no point in space equally 
 illuminated by them ; not even the point in which they are both situated.” 
 
 To show how the result, x=0, in this case fails, this author carries us back to 
 
 the original equation, — — p> 1 which, we agree with him, “truly repre- 
 
 sents the conditions of the problem.” 
 a _b 
 x 4 x * 
 
 a>b or a < b. For, by substituting any value for x, ve shall always obtain two 
 unequal fractions. If *=0, the two members are two unequal infinities.” And 
 so he concludes, that, under this hypothesis, “ the problem fails altogether, and 
 is impossible.” But can two infinite distances be unequal ? It is to be presumed 
 that no one will dispute that an infinite length is a distance than which nothing 
 can be conceived to be greater. But to say that two distances are unequal in 
 length, is incontestably to limit the less, and so to reduce it at once to the finite ; 
 hence, one of this author’s infinities must be finite — a contradiction in terms. 
 The argument failing, the analytical deduction, £=0, still demands our con- 
 fidence. 
 
 But, if possible, a more surprising turn, to avoid the difficulty, is made by 
 another eminent authority. It is the more marked, too, in this case, from the 
 fact that there has been a deliberate departure in the last edition of this author's 
 
 He says, “ If we put e=0, the result is 
 an equation which cannot be satisfied by any value of x whatever, while 
 
PROBLEM OF THE LIGHTS. 
 
 153 
 
 University Algebra, as well as in liis more elaborate work — a work which has 
 long held its place as a standard authority. In the previous editions the result 
 of the analysis upon this point was accepted as true, but without any attempt 
 at an explanation of the difficulty. We are now, however, told that 
 
 “ The hypothesis of e — 0 places the two lights at the same point, in which 
 case they form one and the same light, whose intensity is equal to the sum of 
 their intensities taken separately. The conditions of the problem involve the 
 necessity of two lights, and the equation of the problem is found under this 
 hypothesis. This equation ought not, therefore, to respond to the case of a 
 single light. For, the interpretation of the result obtained from one equation, 
 can only give the cases which fall under the hypothesis. The hypothesis of 
 two lights, and the hypothesis of a single light, are not connected by any law 
 which affords a common equation of condition. Hence, the results obtained on 
 the supposition of c=0, do not belong to the problem.” 
 
 Now this reasoning has a very plausible look ; but it should seem that it will 
 not bear investigation. In the first place, the two lights do not become one in 
 the writer’s sense, any more than two circles of unequal radii become one when 
 they have a common centre. Again, if one of the lights be removed, the other 
 will remain, and since there is nothing requiring the lights to act simulta- 
 neously, the real question in issue is to find the point or points which would be 
 equally illuminated by either light, separately, and, if you please, at times, how- 
 ever remote from each other. 
 
 Again, there is a most unmistakable connection between the two lights — the 
 very arbitrary quantity c in question. Suppose it should be required to find the 
 locus of a point moving so that its distance from two fixed points shall be equal 
 to a given line. Here are two points, but shall we be told that the distance be- 
 tween them may not be made zero, and so give us the circle — a particular case 
 of the ellipse ? or that, having the equation of a line passing through two points, 
 the two points may not be made one without destroying the equation ? 
 
 Again, this author tells us, in the works in question, that the discussion of a 
 problem consists in making every possible supposition upon the arbitrary 
 quantities which enter it, and interpreting the results. How does this agree 
 with his assertion in this case, that the “results obtained by making c=0 do not 
 belong to the problem” ? Are we to understand that he considers it impossible to 
 make this hypothesis in an equation containing c ? Indeed the doctrine here 
 put forth would make short work with much of the higher mathematics, and 
 the whole doctrine of limits. 
 
 Nor would the difficulty in point be at all obviated, if one were to admit the 
 whole of this writer’s reasoning. The equation must still respond, so long as c 
 is not absolutely zero. Well, now suppose the lights to be, the one very great 
 and the other very small, and that c is an infinitesimal. We shall have the 
 lesser light occupying the second or third consecutive point from the greater. 
 We must have one point of equal illumination between them ; but will common 
 sense with regard to lights, which seems to frighten our author into deserting 
 the analysis, be in any manner better saved ? Will not the feebler one be hope- 
 lessly outshone by the more splendid luminary ? 
 
 But a third, and the latest writer upon this point, pronounces the results of the 
 analysis under the hypothesis in question, to be “ evidently absurd.” Tie says, 
 “ In discussing this problem, some have committed the error of considering that, 
 
154 
 
 ELEMENTS OF ALGEBRA. 
 
 since for c=0 and a and 6 unequal, x=c - 
 
 
 — =0, therefore, there is a point 
 
 Ya± Vb 
 
 of equal illumination at the point where the lights are situated ! This is 
 evidently absurd, since the hypothesis is that the lights are of unequal inten- 
 sity. The error consists in not perceiving that the hypothesis, c=0, excludes 
 the hypothesis, a and b unequal. That the hypotheses, a> or <b, are excluded by 
 the hypothesis, e=0, and that there is a point of equal illumination, is self- 
 evident.” 
 
 It is to be presumed that the learned author uses the word “ self-evident” in 
 an unusual sense, since the question he disposes of so summarily in the end, 
 had already cost him quite an argument, to say nothing of having been a stand- 
 ing puzzle for many years. This conclusion is further sustained by the fact 
 that he afterwards says, “ Perhaps the student may think that these conditions 
 are no more inconsistent than those in I. 3 above, viz., c finite and a=b, and a 
 point of equal illumination,” etc. 
 
 He goes on further to prove his “ self-evident ” proposition as follows : 
 
 a 
 
 Also, that a > or < b is inconsistent with c=0 and — = 
 
 (that is. 
 
 x* (c—x) 2 
 
 that there is a point of equal illumination), appears from the fact that c=0 
 
 renders the latter ~ , whence a—b.” 
 
 x 2 x 2 
 
 This writer seems to be unconscious of the fact that he has here stricken out 
 zero as a factor from both members of the equation ; thus, ax 2 =bx 2 , a-0=6 0 .'. 
 a=b ; by which process it is easy enough to prove 2=4 ; a result which follows 
 at once, if we assume a= 2, and 6=4, in the above equation. It should seem 
 that the anxiety to escape a point which is thought to militate against common 
 sense, has led to some hasty writing. 
 
 And now let us look at the question upon its own merits. It should seem 
 that the chief difficulty arises from a most unwarrantable mixing up of physical 
 phenomena and analytical exactness. Analysis, per se, knows nothing what- 
 ever of the physical world ; nor does it in the slightest degree regard any lan- 
 guage but its own. It takes our conditions and renders its verdict, leaving us 
 to look out for ourselves and our physical applications. 
 
 The confusion in the minds of those authors who desert the analysis in this 
 question, and betake themselves to what they mistake for common sense, arises 
 from not sufficiently regarding the assumed law of physics under which the 
 equation in this problem is established. These are purely theoretical lights, 
 and the law is absolute, that their intensity shall vary inversely as the square of 
 the distance. When we pass within the distance unity, their intensities increase 
 
 with wondrous rapidity, until at zero — the limit of approach — we must have — , 
 
 for their respective intensities : that is to say, the intensity of either 
 
 light at this point is absolutely infinite ; and thus, unless we may indeed have 
 “ two unequal infinities,” when the lights occupy the same point, that point 
 must be equally illuminated, however much the lights may differ in intensities 
 at the distance unity. 
 
 This may not, probably does not, accord with physical facts, which deal with 
 
PROBLEM OP THE LIGHTS. 
 
 155 
 
 tallow candles, or at best calcium lights ; but let us not charge the failure upon 
 the analysis, without first examining our assumed law of optics. That law 
 should undoubtedly read, “ the intensity of a light at any [sensible] distance is 
 equal to its intensity at the distance unity, divided by the square of that dis- 
 tance.” We should then be saved from the hypothesis in question, and from 
 the physical anomaly. 
 
 But in further support of the analysis, and to show how wondrously it re- 
 sponds to every hypothesis, let us rid the question entirely of physical phenom- 
 ena, and present it as an abstract question. 
 
 Let y represent the general ordinate of a curve which shall be equal to a 
 when the abscissa is made equal to unity, and which shall vary inversely as the 
 square of the abscissa. Then, 
 
 will be the equation of such curve. This equation may be readily constructed, 
 and we shall have a curve with two branches, BC and de as shown in Fig. 1. 
 
 By making x and y zero in succession in equation (1), it appears at once that the 
 axes are asymptotes to either branch, and thus that y is infinite when * is zero, 
 and zero when x is infinite. 
 
 Let us now have auother such curve whose equation is, 
 
 b 
 
 (c — X ) 2 
 
 (2). 
 
 This curve gives y equal to b, when c—x is equal to unity ; and the ordinate 
 whose abscissa is equal to c, is an asymptote to the curve. 
 
 Now, let it be required to find the common points of these two curves ; that 
 is, the points whose ordinates (they corresponding to the intensities in the prob- 
 lem) are equal. From (1) and (2), we must have, 
 
 a _ b 
 x 2 (c—x) 2 ’ 
 
 the old equation, but with no question of light in it. The values of x may now 
 be found as before, and the points constructed. The curves would, under the 
 first hypothesis in the problem, take the general form shown in Fig. 2. 
 
156 ELEMENTS OP ALGEBRA. 
 
 These curves will show to the eye the results of all the hypotheses which 
 may he made in the problem. We shall make only the one in dispute, namely, 
 c=0, and a> or <b. The hypothesis c=0 makes the asymptotical ordinate co- 
 incide with the axis of Y, and the curves will assume the position shown in 
 Fig. 3. 
 
 It will be seen that both curves approach the axis of Y, and that neither can 
 ever reach it — the ordinate at the point zero being infinite — that is, resuming 
 the question of lights, the intensities of both lights are infinite at the origin, 
 and that point is equally illuminated. It will hardly be said that one of these 
 curves reaches the axis before the other. 
 
 THE END. 
 

 
 
 
 
Date Due 
 
 L. B. Cat. No. 1 137 
 
512.03 S559E 
 
 StlQUT) 
 
 5518 
 
 -Ele ments of Alggh-ra 
 
 512.02 S559E 
 
 5518