IN MEMORIAM 
 FLOR1AN CAJOR1 
 
Digitized by the Internet Archive 
 
 in 2008 with funding from 
 
 Microsoft Corporation 
 
 http://www.archive.org/details/firstlessonsinalOObailrich 
 
FIRST LESSONS 
 
 jf, (^\(nri 
 
 ALGEBRA, 
 
 BEING AN ' 
 
 EASY INTRODUCTION TO THAT SCIENCE; 
 
 DESIGNED FOR THE USE OF 
 
 ACADEMIES AND COMMON SCHOOLS. 
 
 BY EBEJTEZER BAILEY, 
 
 Principal of the Young Ladies' High School, Boston 3 Author of 
 
 " Young Ladies' Class Book," &c. 
 
 KmjjrobelJ Stmotspe IBoTtfon. 
 
 BOSTON: 
 CARTER, HENDEE, AND CO. 
 
 1834. 
 
 
4/i 
 
 153 
 
 At a meeting of the Sckool Committee of the Otfy 
 of* Boston-March II, 1834,, ' 
 
 ♦ Voted, That "BaileyJs Algebra" be used in the 
 Writing Schools in which Algebra is allowed to be taught. 
 Attest, 
 *!f :'\ \ •' •/• ;. k: F. McCLEARY, Secretary. 
 
 Entered according to Act of Congress, in the year 1834, 
 
 By Ebenezer Bailey, 
 
 In the Clerk's Office of the District Court of Massachusetts. 
 
 CAJORi 
 
 STEREOTYPED AT THE 
 BOSTON TYPE AND STEREOTYPE FOUNDRY. 
 
 BOSTON: 
 
 Printed by Lewis and Pennlraai) ; 
 
 Bromfiel 1 Street. 
 
PREFACE 
 
 This treatise is especially intended for the use of 
 beginners. I have long wished that Algebra might 
 be introduced into common schools, as a standard 
 branch of education ; and there seems to be no good 
 reason why the study of this most interesting and 
 useful science should be confined to the higher semi- 
 naries of learning. The upper classes, at least, in 
 common schools, might be profitably instructed in its 
 elements, without neglecting any of those branches 
 to which they usually attend. 
 
 This work pretends to no original investigations, no 
 new discoveries. My labor has been the very humble 
 one of selecting such materials as belong to the ele- 
 ments of Algebra, and of arranging them in such a man- 
 ner as may render the introduction to the science easy. 
 If there be any peculiarity in this work, it is its sim- 
 plicity. I have endeavored to make it as plain and 
 
4 PREFACE. 
 
 intelligible as possible. There is little danger that the 
 student will find the beginning of any art or science 
 too easy ; and, in Algebra, he is required to learn a 
 peculiar language, to determine new principles, and 
 to accustom himself to an abstract mode of reason- 
 ing, with which he has been little acquainted. Let 
 the explanations, therefore, be as full and diffuse as 
 they may, he will still find difficulties enough to exer- 
 cise his mind. I have aimed to prepare a work, which 
 any boy of twelve years, who is thoroughly acquaint- 
 ed with the fundamental rules of Arithmetic, can un- 
 derstand, even without the aid of a teacher. 
 
 The following are the leading principles which I 
 have observed, in preparing this treatise : — 
 
 To introduce only such parts of the science, as 
 properly belong to an elementary work ; 
 
 To adhere strictly to a methodical arrangement, that 
 can be easily understood and remembered ; 
 
 Never to anticipate principles, so as to make a clear 
 understanding of the subject under consideration, de- 
 pend upon some explanation which is to follow ; 
 
 To introduce every new principle distinctly by it- 
 self, that the learner may encounter but one difficulty 
 at a time ; 
 
 To deduce the rules, generally, from practical ex- 
 ercises, and to state them distinctly and in form ; 
 
PREFACE. 5 
 
 To give a great variety of questions for practice 
 under each rule ; 
 
 To solve or fully explain all questions which in- 
 volve a new principle, or the new application of a 
 principle already explained ; 
 
 To show the reason of every step, without perplex- 
 ing the learner with abstruse demonstrations; 
 
 To illustrate the nature of algebraic calculations, 
 and their correctness, by a frequent reference to 
 numbers ; > 
 
 And, finally, to advance from simple to difficult 
 problems in such a manner as may fully exercise the 
 powers of the learner without discouraging him. 
 
 As this little book professes to be merely an intro- 
 duction to more full and scientifical treatises upon 
 Algebra, it was not my original, design to extend it 
 beyond Equations of the First Degree. The subse- 
 quent Chapters, on Evolution and Equations of the 
 Second Degree, have been added with a particular 
 reference to schools for young ladies. It is presumed 
 that the work, in its present form, contains as much 
 of Algebra as this class of learners will, in general, 
 
 find time to study. 
 
 E. Bailey. 
 Boston, July, 1833. 
 1* 
 
ADVERTISEMENT 
 
 TO THE STEREOTYPE EDITION 
 
 The favor with which this treatise has been received 
 by the public, — as manifested by the sale of the first edi- 
 tion, consisting of two thousand copies, in a few months, 
 and by commendatory notices from teachers and others in 
 almost every section of the country, — has induced the pub- 
 lishers to stereotype the work, and thus put it into a per- 
 manent form. 
 
 In preparing this edition, the author has made such addi- 
 tions and alterations as experience has suggested ; and he 
 trusts that he has rendered The First Lessons in Alge- 
 bra still more worthy of public favor. The arrangement 
 of several parts of the work has been changed ; some of 
 the Chapters have been nearly rewritten, especially those 
 on Powers and Evolution ; the errors of the first edition 
 have been carefully corrected; and many questions for 
 practice have been added. It has, also, been thought ad- 
 visable to omit the answers to the questions. These are 
 given in a Key, published separately, together with solu- 
 tions of all the difficult problems. 
 
 Boston, January, 1834. 
 
TABLE OF CONTENTS. 
 
 CHAP. I. 
 
 INTRODUCTION. 
 
 Sect. I. Letters and Coefficients 13 
 
 II. Algebraic Signs 17 
 
 III. Simple, Compound, Similar, Positive and Nega- 
 tive Quantities 20 
 
 CHAP. II. 
 
 ADDITION. 
 
 Sect. I. Simple Quantities that are Similar 24 
 
 v All the Quantities being Positive 24 
 
 All the Quantities being Negative 25 
 
 Some of the Quantities being Positive and others 
 
 Negative 27 
 
 II. Simple Quantities that are Dissimilar 30 
 
 III. Compound Quantities 33 
 
 General Rule for Addition .' 35 
 
 CHAP. III. 
 
 SUBTRACTION. 
 
 Sect. I. To subtract a Simple Quantity , 37 
 
 The Quantities being Similar 37 
 
CONTENTS. 
 
 The Quantities being Dissimilar 38 
 
 II. To subtract a Compound Quantity 39 
 
 All the Terms being Positive 39 
 
 Some of the Terms being Negative 40 
 
 III. General Rule for Subtraction 42 
 
 CHAP. IV. 
 MULTIPLICATION. 
 
 Sect. I. Simple Quantities • 45 
 
 To multiply a Literal Quantity by a Number ... 46 
 
 To multiply one Literal Quantity by another ... 47 
 
 Rule for multiplying Simple Quantities 49 
 
 II. Compound Quantities 50 
 
 To multiply a Compound by a Simple Quantity . 50 
 
 To multiply two Compound Quantities 51 
 
 III. Signs in Multiplication 52 
 
 The Product of two Positive Quantities 53 
 
 The Product of two Quantities, one being Posi- 
 tive, the other Negative 54 
 
 The Product of two Negative Quantities 55 
 
 IV. General Rule for Multiplication 56 
 
 CHAP. V. 
 
 DIVISION. 
 
 Sect. I. Simple Quantities 58 
 
 II. Signs in Division 61 
 
 The Quotient of two Positive Quantities 61 
 
 The Quotient of two Quantities, one being Posi- 
 tive, the other Negative 62 
 
 The Quotient of two Negative Quantities 63 
 
 General Rule for the Signs . . . i 63 
 
 III. Compound Quantities 64 
 
CONTENTS. 9 
 
 To divide a Compound by a Simple Quantity ... 64 
 
 To divide one Compound Quantity by another. . 65 
 
 IV. General Rule for Division 68 
 
 CHAP. VI. 
 
 FRACTIONS. 
 
 Sect. I. Introduction and Definitions 70 
 
 To change an Integral Quantity into a Fraction 72 
 To change an Integral Quantity into a Fraction, 
 
 having a given Denominator 73 
 
 II. Reduction 74 
 
 A. To reduce an Improper Fraction to a Mixed 
 Quantity 74 
 
 B. To reduce a Mixed Quantity to an Improper 
 Fraction 75 
 
 C. To reduce Fractions to a Common Denom- 
 inator 76 
 
 D. To reduce Fractions to their Least Terms . . 77 
 
 III. The Signs of Fractions 78 
 
 IV. Addition 80 
 
 v_ V. Subtraction 83 
 
 VI. Multiplication 85 
 
 To multiply a Fraction by an Integral Quantity 85 
 
 To multiply one Fraction by another 86 
 
 VII. Division 87 
 
 To divide a Fraction by an Integral Quantity ... 88 
 To divide one Fraction by another 89 
 
 CHAP. VII. 
 
 POWERS. 
 
 Sect. I. Involution of Simple Quantities 90 
 
 Signs of Powers 92 
 
10 CONTENTS. 
 
 II. Involution of Compound Quantities 93 
 
 To involve a Fraction 95 
 
 III. The Binomial Theorem 96 
 
 Application of the Binomial Theorem to Polyno- 
 mials 99 
 
 IV. Addition 101 
 
 V. Subtraction 103 
 
 VI. Multiplication 104 
 
 VII. Division 106 
 
 Negative Exponents 107 
 
 To divide Compound Quantities of different 
 
 Powers 108 
 
 CHAP. VIII. 
 
 EQUATIONS OF THE FIRST DEGREE. 
 
 Sect. I. Introduction 109 
 
 II. To remove Coefficients 113 
 
 III. To remove Denominators 120 
 
 IV. Transposition 126 
 
 V. Questions for Practice 134 
 
 VI. Two Unknown Quantities 147 
 
 Elimination by Comparison 147 
 
 Elimination by Substitution 151 
 
 Elimination by Addition and Subtraction 155 
 
 VII. Three or more Unknown Quantities 161 
 
 VIII. Questions for Practice , 168 
 
 CHAP. IX. 
 
 EXERCISES IN GENERALIZATION. 
 
 To add, subtract and multiply the Sum and Difference of 
 
 two Numbers 172 
 
 i 
 
CONTENTS. 11 
 
 Two Numbers being given, to find the Difference of their 
 
 Squares 174 
 
 Formulas of Second Powers 175 
 
 To find two Numbers, their Sum and Difference being 
 
 given 176 
 
 Formula for Barter 178 
 
 To divide a given Quantity into unequal Parts 179 
 
 To prepare Questions whose Answers shall be free from 
 
 Fractions 181 
 
 Formula for Single Fellowship 183 
 
 Formula for Double Fellowship 185 
 
 Miscellaneous Formulas t 187 
 
 Formulas for Interest and Discount 191 
 
 CHAP. X. 
 
 EVOLUTION. 
 
 Sect. I. Introduction 196 
 
 II. Roots of Simple Quantities 200 
 
 To find the Root of a Fraction 201 
 
 Signs of Roots 202 
 
 v III. To extract the Square Root of a Compound 
 
 Quantity . . 203 
 
 IV. To extract the Square Root of a Number 209 
 
 V. Approximate Roots 212 
 
 Of Integral Numbers 212 
 
 Of Fractions 213 
 
 Of Surds 215 
 
 CHAP. XL 
 
 EQUATIONS OF THE SECOND DEGREE. 
 
 Sect. I. Pure Equations 216 
 
 Rule for solving Pure Equations 218 
 
 
12 CONTENTS. 
 
 II. Affected Equations 223 
 
 Common Method of Completing the Square . 226 
 
 Rule for solving Affected Equations 229 
 
 Another Way to Complete the Square 230 
 
 III. Questions producing Affected Equations 234 
 
 To find the Second Power of a Surd 238 
 
 CHAP. XII. 
 
 MISCELLANEOUS QUESTIONS 241 
 
 
 ■ 
 
4 
 
 FIRST LESSONS IN ALGEBRA 
 
 CHAPTER I. 
 INTRODUCTION. 
 
 SECTION I. 
 
 Letters and Coefficients. 
 
 1. A boy bought a peach and a melon for 12 cents, 
 and the melon cost three times as much as the peach. 
 What was the price of each ? 
 
 Let the letter x represent the number of cents the 
 boy gave for the peach ; then, as he gave one x for 
 the peach, whatever the value of x may be. and as 
 the melon cost three times as much, he must have 
 given three x*s for that ; and, of course, he gave one 
 x and three a?'s, that is, four x's for both. But, by 
 the question, he gave 12 cents for both ; therefore, 
 the four x's must be equal in value to the 12 cents. 
 But, if four x's are equal to 12 cents, one x must be 
 equal to one fourth part of 12 cents, or 3 cents, 
 which was the price of the peach ; and if one x be 
 2 
 
14 FIRST LESSONS IN ALGEBRA. 
 
 equal to 3 cents, three x's must be equal to three 
 times 3 cents, or 9 cents, the price of the melon. 
 
 It will be observed, that, in this operation, the 
 ansiver of the question, the thing unknown, is assumed 
 and represented by the letter x, which is therefore 
 jcalled the unkn.own\ quantity. Any other letter, mark 
 'or* character,*' may be used with equal propriety, 
 *j^#jjleU*alwa)*s/that ;its value be indefinite. This 
 wilf he' evicferi£ "if the word share or part be substi- 
 tuted for the letter x, in the above operation. It is 
 usual, however, to represent unknown quantities by 
 the last letters of the alphabet, as x, y, z. 
 
 It is sometimes Necessary to express quantities, 
 whose values either are, or are supposed to be, deter- 
 mined, by letters. These are called known quantities, 
 and are usually represented by the first letters of the 
 alphabet, as a, b, c. 
 
 2. John is four times as old as James, and the 
 sum of their ages is 20 years. What is the age of each ? 
 
 Let x represent the age of James ; then, as John 
 is four times as old, four x's will rer*esent his age ; 
 and their joint ages must be one x and four x's, that 
 is, five x's. But the sum of their ages is 20 years, by 
 the question ; therefore, five x's must be equal to 20 
 years, and one x to one fifth part of 20, namely, 4 
 years, which is the age of James ; and, if one a: be 4 
 years, four x's must be four times 4, or 16 years, the 
 age of John. 
 
 Instead of writing one x, three x's, four x's, five x's, 
 (fee, as in these examples, we use the expressions 
 a?, 3 x, 4 a?, 5 x, &c. The numbers placed before the 
 
INTRODUCTION. 15 
 
 letters, as 3, 4, 5, are called their coefficients. When 
 no number is placed before a letter, as x, its coeffi- 
 cient is always understood to be 1. 
 
 In the last example, the algebraic expression for 
 John's age was 4 x, and the value of x was found to 
 be four years. To find the age of John in years, this 
 value of x was multiplied by 4, its coefficient. Any 
 quantity is always supposed to be multiplied by its co- 
 efficient. Thus, if the value of a: be 6, 3 # will be 3 
 times 6, or 18 ; and if the value of x be 10, then 3 x 
 will be 3 times 10, or 30 ; and 7 x will be 70, 9 x will 
 be 90, and so on. 
 
 It is often convenient, in algebraic calculations, to 
 use a letter for a coefficient, instead of a number, as 
 m x, where m is regarded'as the coefficient of x ; thus, 
 if m be 3, and x be" 5, m x will be 3 times 5, or 15. 
 
 3. A leaves Boston, and walks three miles an hour, 
 and B leaves Newburyport, at the same time, and 
 walks 5 miles an hour. In how many hours will they 
 ^meet, the places being 32 miles apart? 
 
 In this question, the thing required is, in how many 
 hours A and B will meet ; that is, how many hours 
 they will travel. Let it be assumed that they will 
 meet in * hours. Then if A walk 3 miles in 1 hour, 
 in x hours he will walk x times 3 miles, that is, x 3 
 or 3 x miles ; and if B walk 5 miles in 1 hour, in x 
 .hours he will walk x 5 or 5 x miles; and they will 
 both walk 3 x and 5 x, or 8 x miles, which is the whole 
 distance. But the distance given in the question is 
 32 miles ; therefore, 8 x miles must be the same as 32 
 miles ; or, to use a general expression, 8 x is equal in 
 
16 FIRST LESSONS IN ALGEBRA. 
 
 value to 32. And if 8 a? be equal to 32, x must be one 
 eighth part of 32, or 4. They will meet in 4 hours. 
 The expressions x 3 and 3 x, used in this operation, 
 mean the same thing; for any two or more quanti- 
 ties are supposed to be multiplied together, when they 
 are not separated; and, of course, it is of no conse- 
 quence which is placed first. Thus, if the value of 
 x be 4, as in the question, 3 a? is 3 times 4, or 12; 
 and xS is 4 times 3, or 12. But it is more con- 
 venient to place the number before the letter, which is 
 always done. 
 
 4. A farmer sold a calf, a sheep and a cow, for 
 36 dollars.; for the sheep he received twice as much 
 as for the calf, and for the cow three times as much 
 as for both the calf and the sheep. What was the 
 price of each ? 
 
 Let x represent the price of the calf; then 2 a? will 
 be the price of the sheep ; x and 2 a?, or 3 x, will be 
 the price of the two, and three times 3 x, or 9 x, will 
 be the price of the cow. The three animals were, 
 therefore, sold for x, and 2 x, and 9 x, that is, for 12 a?. 
 But, according to the question, they were sold for 36 
 dollars; 12 a: must, therefore, be equal to 36 dollars, 
 and the value of x must be one twelfth part of 36, 
 namely, 3 dollars, which is the price of the calf: if the 
 value of # be 3 dollars, 2 a; is twice 3, or 6 dollars, 
 the price of the sheep ; and 9 a: is 9 times 3, or 27 
 dollars, the price of the cow. 
 
 5. A gentleman gave a purse, containing a certain 
 sum of money, to his three children, to be divided 
 among them in such a manner, that Mary should have 
 
INTRODUCTION. 17 
 
 twice as much as Ellen, and John as much as both 
 his sisters. What was the share of each ? 
 
 As the sum contained in the purse is not named, we 
 will call it a. 
 
 Let x denote Ellen's share ; then Mary's share is 
 twice as much, or 2 a?; and John's, a? and 2 x, that is, 
 3 x ; and the sum of their shares is x and 2 x and 3 x, 
 or 6 x, which must be equal to a, the sum to be divid- 
 ed, whatever the value of a may be. And if 6 a? is 
 equal to a, x is equal to one sixth part of a, which is 
 Ellen's share. 
 
 If the purse contained 18 dollars, Ellen's share was 
 
 3, Mary's 6, and John's 9 dollars. 
 
 If the purse contained 24 dollars, Ellen's share was 
 
 4, Mary's 8, and John's 12 dollars. 
 
 In this manner the share of each may be determined, 
 whatever be the sum indicated by a. 
 
 This section will serve to give the learner a general 
 notion of the nature and use of Algebra, and the 
 manner in which it is applied to the performing of 
 questions. 
 
 SECTION II. 
 
 Algebraic Signs. 
 
 Besides letters, certain other signs are used in Alge- 
 bra, some of which are also used in Arithmetic, though 
 less frequently. It is by means of these and other 
 arbitrary signs, that calculations in Algebra are per- 
 
 2* 
 
18 FIRST LESSONS IN ALGEBRA. 
 
 formed. They enable us to express a train of rea- 
 soning in a short and exact manner, so that the whole 
 may be presented readily to the eye and the mind. 
 
 -|- (Plus) signifies addition ; as, 4 + 5 is 9. This 
 may be read, 4 plus 5 is 9, or 4 and 5 is 9, or in any 
 other way which will indicate that 4 and 5 are to be 
 added together. 
 
 Thus, too, the expression x + y signifies that the 
 quantities x and y are to be added, whatever their val- 
 ues may be. Suppose that the values of a? and y are 6 
 and 4 ; then x + y will be equal to 6 + 4, or 10. 
 
 — (Minus,) placed before a quantity, signifies that 
 it is to be subtracted. Thus, 9 — 6 is 3 ; which may 
 be read, 9 minus 6 is 3, or 9 less 6 is 3, or Gfroro, 9 is 3, 
 or in any other manner which will show that 6 is to 
 be subtracted from 9. 
 
 So, too, in the algebraic expression x — y, the 
 value of y is to be subtracted from the value of x. 
 Suppose x, for instance, to be 12, and y to be 7 ; 
 then x — y will be the same as 12 — 7, which is 5. 
 
 The signs + and — affect only those numbers and 
 quantities which immediately follow them. Of course, 
 the expressions, 
 
 9 + 6 + 5 — 7 
 
 9-1-6 — 7 + 5 
 
 9 — 7 + 6 + 5 
 
 are all of the same value, each being equal to 13 ; for 
 
 the sign + is understood before 9, the first number. 
 
 c± Two horizontal lines signify equality; that is, 
 that the quantities between which they are placed, 
 are equal to each other; as, 3 + 4 = 9 — 2. If we 
 
INTRODUCTION.- 19 
 
 have y -\- z = x, we know that the value of x is 
 equal to the values of y and z added together. Thus, 
 if y be 6, and z be 4, the value of x must be 10 ; for 
 6 + 4 e= 10. 
 
 X This sign signifies multiplication ; as, 3 X 4 = 
 12; that is, 3 multiplied by 4 is equal to 12. This 
 character is often omitted when multiplication is 
 implied ; as in the expression x y, which is the prod- 
 uct of x multiplied by y. Thus, if x — 5, and y = 3, 
 x y =. 15 ; for 3 X 5 = 15. But it is never omitted 
 between two numbers which are to be multiplied. 
 
 -7- This sign expresses that the quantity which precedes 
 it is to be divided by that which follows it. Thus, 
 12-f 4cr 3; that is, 12 divided by 4 is equal to 3. 
 
 But division is more frequently expressed in the 
 form of a fraction ; thus, -\ 2 - = 3, which may be read 
 in the same manner. 
 
 So, too, x -f- y, or — , expresses the quotient of x 
 divided by y. Thus, if x = 10, and y = 2, % - = 5 ; 
 for f = 5. 
 
 A vinculum is used to connect two or more quan- 
 tities together. Thus, 4 X a -\- b implies that the sum 
 of a and b is to be multiplied by 4. Suppose the 
 value of a to be 5, and of b to be 6 ; then, 
 
 4 X 5 + 6 == 44 ; 
 for 5 + 6 = 11, and 4x11= 44. 
 
 Again, a -f- b X c -f- d signifies that the sum of a -j- b 
 is to be multiplied by the sum of c -\- d. Let a = 2, 
 b = 3, c = 4, and d = 5 ; then, 
 
 2 + 3 X 4 + 5 = 5x9, or 45. 
 
20 FIRST LESSON3 IN ALGEBRA. 
 
 A parenthesis ( ) is often used instead of a vincu- 
 lum, to indicate that several quantities are to be taken 
 together. Thus, 3 (x -f- y) expresses that the sum of 
 x andy is to be taken three times. If the value oix 
 be 6, and of y be 4, 
 
 3 (a? + y) = 3 (6 + 4) = 3 X 10, or 30. 
 
 The several quantities under a vinculum, or in- 
 cluded in a parenthesis, may be taken collectively, 
 and regarded as a simple quantity, of which the 
 number prefixed is the coefficient. 
 
 SECTION ill. 
 
 Simple, Compound, Similar, Positive and Negative 
 Quantities. . 
 
 A Simple quantity consists of a single term, that is, 
 of one letter or number, or of several letters joined to- 
 gether ivithout the sign -J- or — ; as, x, 3 y, a b c, and 
 x y, each of which is a simple quantity. 
 
 A Compound quantity consists of two or more sim- 
 ple quantities joined together by the sign -{-or — ; as 
 x -\- y, a — b -{- 3 c, a b — 7 -f z — 8 a?, each of 
 which is a compound quantity. 
 
 A compound quantity, which consists of two terms 
 only, as x -f y, or a — b, is called a binomial. The 
 latter expression, a — b, is also called a residual quan- 
 tity, because it expresses the residue or remainder, 
 after one of the terms has been taken from the other. 
 
 Similar quantities are such as differ only in their 
 coefficients or signs. Thus, 3 a and 5 a are similar 
 
INTRODUCTION. 21 
 
 quantities ; so are 3 x y and 7 x y ; as also 2 a b and 
 — 8 a b ; and the compound quantities 3 a b -f- 4 a?, 
 and 5 « & — 9 a?. 
 
 All the quantities used in an algebraic calculation, 
 are considered, in relation to each other, either as 
 positive or negative. 
 
 A Positive quantity has the sign -j- prefixed to it ; 
 it is, in general, something to be added. When a 
 positive quantity stands alone, as x, or is the first term 
 of a compound quantity, as a -j- b, the sign is com- 
 monly omitted ; but the sign -f- ls always understood 
 in such cases. Thus, x is the same as -f- %, and a -}- b 
 the same as -f- a -f- b. 
 
 A Negative quantity is one to be subtracted, and al- 
 ways has the sign — prefixed to it. Thus, in the ex- 
 pression a — b, — b is a negative quantity, because 
 its value is to be subtracted from a. 
 
 As the subject of positive and negative quantities is 
 very apt to perplex beginners, a few examples will be 
 given, by way of illustration. 
 
 1. William has 12 apples, and gives 5 of them to 
 Samuel. How many has he left ? 
 
 In this question, 12, the number of apples which 
 William had in the first place, is a positive quantity ; 
 and 5, which must be subtracted to obtain the an- 
 swer, is a negative quantity. 12 — 5. 
 
 2. William has 12 apples, and Samuel gives him 5 
 more. How many has he then ? 
 
 Here, as 5 must be added to 12 to obtain the an- 
 swer, it is a positive quantity. 12 -f- 5. 
 
 3. A man bought a watch for 25 dollars, and sold 
 
 
22 FIRST LESSONS IN ALGEBRA. 
 
 it again for 30 dollars. How much did he gain by 
 the bargain ? 
 
 To find his gain, we must subtract what he gave 
 for the watch from the sum for which he sold it : 30 
 is, therefore, a positive, and 25 a negative quantity. 
 30 — 25. 
 
 4. A man sold a watch for 30 dollars, by which 
 bargain he gained 5 dollars. What did the watch 
 cost him ? 
 
 Here, the gain must be subtracted from the price 
 of the watch: 5 is, therefore, a negative quantity. 
 30 — 5. 
 
 5. A merchant went into trade with a certain sum, 
 say a dollars ; and, at the end of the year, he found 
 himself worth b dollars. How much did he gain dur- 
 ing the year ? 
 
 We must subtract what he had at the beginning of 
 the year, a dollars, from what he had at the end of it, 
 b dollars, to ascertain his gain : a is, therefore, a nega- 
 tive quantity, and the state of his affairs may be ex- 
 pressed thus, b — a. 
 
 In this question, if we suppose the merchant to 
 have lost instead of gained by his business, it is evi- 
 dent that the value of b will be less than that of a, f 
 and we shall be required to subtract a greater number 
 from a less, which is impossible. But it is perfectly 
 easy to represent such a subtraction, as, for instance, 
 18 — 32 ; and hence it frequently happens in Alge- 
 bra, that a negative quantity stands alone, as — x, 
 when tMJ is no quantity from which it is to be ac- 
 tuallyj| 
 
INTRODUCTION. 23 
 
 6. A man has in his possession 200 dollars, and 
 owes debts to the amount of 500 doliars. How 
 much is he worth ? , 
 
 Here, the money the man has is a positive quanti- 
 ty, and the amount of his debts, which is to be sub- 
 tracted, is a negative quantity ; therefore, the expres- 
 sion 200 — 500 will represent the state of his prop- 
 erty. Now, if he pay oft' his debts, as far as his 200 
 dollars will go, there will still be $ — 300 left ; that 
 is, he will be 300 dollars worse than nothing. 
 
 In the last question, the amount of the debts might 
 be regarded as the positive quantity $ and then the 
 opposite quantity, the money on hand, would be 
 negative, and 500 — 200 would represent the amount 
 of debts which the man could not pay. 
 
 It is evident, therefore, that positive and negative 
 are merely relative terms, which are, in general, op- 
 posed to each other. In any calculation, whatever 
 quantity is assumed as positive, all other quantities of 
 a similar nature, or which tend to increase it, are also 
 positive ; and whatever quantities are opposed to it, 
 in any way, or which serve to diminish it, are negative. 
 
24 FIRST LESSONS IN ALGEBRA. 
 
 CHAPTER II. 
 ADDITION. 
 
 SECTION I. 
 Simple Quantities that are Similar. 
 
 1. A man gave to one poor person a dollars, to a 
 second 3 a dollars, and to a third 4 a dollars. How 
 much did he give them all ? Ans. 8 a dollars. 
 
 In this question, we have three simple quantities : 
 they are all similar, and they are all positive ; for it 
 must be remembered that, when no sign is used, the 
 sign -f- is always understood. The first quantity is a ; 
 the second, 3 a, which may be written a -f- a -f- a ; 
 and the third, 4 a, or a -f- a -f- a -\- a. Now, by 
 counting, we find there are eight a's or 8 a. But the 
 sum of the coefficients, I -f- 3 -f 4, is also 8. Then, 
 to perform questions of this kind, Add together all the 
 coefficients, and place the sum before the common letters. 
 Suppose the value of a, in this question, to be 5 ; 
 then we shall have 
 
 a = 1 x 5 = 5 
 
 3«r=3x5=15 
 
 4a = 4 X 5 = 20 
 
 ~8a = 8x5 =T40 
 
ADDITION. 25 
 
 By assigning any other value to the letter 0, the stu- 
 dent will obtain a similar result. 
 
 2.) 4 5 3.) x y 4.) babe 5.)7y« 
 
 35 2 x y 6 a b c y z 
 
 5 b Ivy a b c 6y« 
 
 6 b 3a?y 2 ab c 2ys 
 
 18 5 13 xy 
 
 6. Add together a x, and 3aa;, and 5 x a. 
 
 It is of no consequence in what order the letters are 
 given ; for 5 a? a is evidently of the same value as 5 a x. 
 Let x = 6, and o=4, for instance ; then b a os 
 will be 5 X 4 X 6, or 120; and b x a will -be 
 5x6x4, which is also equal to 120. It is usual, 
 however, to arrange the letters according to their or- 
 der in the alphabet. 
 
 7. Add together 3 a b c, and 2 a be, and 4 a b c. 
 
 8. What is the sum of 5 a r x, and 3 a r x, and 
 a r x ; and 1 a r x, and 17 a r x 1 
 
 9. A merchant is indebted 4 a dollars to A, 5 a 
 dollars to B, 6 a dollars to C, and 8 a dollars to D. 
 How much does he owe them all ? ,- 
 
 10. What is the amount of a b c, and babe, and 
 7 a b c, and 12 a b cl 
 
 The student will find it a useful exercise to prove 
 his answers, by assigning definite values to the letters 
 given in the questions, in the manner exhibited above. 
 
 11. Add together — x, and — 3 a?, and — 5 x. 
 
 Ans. — 9 x. 
 This example differs from the foregoing in only one 
 particular ; that is, the quantities are all negative : 
 3 
 
26 FIRST LESSONS IN ALGEBRA. 
 
 the sign — must, therefore, be prefixed to i.'ie sum ; for 
 the whole must evidently be of the same character as 
 the parts of which it is composed. 
 
 Let us suppose that x, in this example, stands for 
 100 dollars ; then — x = — 100, — 3 x = — 300, 
 
 — 5« = — 500, and the answer — 9 x = — 900 
 dollars. But it may be asked, How can 100, or 300, 
 or 500, or 900 dollars be subtracted from nothing ? 
 Such a subtraction may be represented, although it 
 cannot be performed. 
 
 A merchant, for instance, wishes to ascertain the 
 profits of his business. His gains are positive quan- 
 tities ; and his losses, because they must be subtracted 
 from his gains before his clear profits can be known, 
 must be negative. Now, it is evident, if he has lost 
 100 dollars by one speculation, 300 dollars by an- 
 other, and 500 dollars by bad debts, that these sums 
 should be written — 100, — 300, and — 500; and 
 that they may be added together, as if they were 
 positive quantities, their amount being written — 900, 
 to show that it is to be subtracted from some other 
 quantity ; that is, from the amount or sum total of 
 his profits. 
 
 12. Add together — 6 xy, — x y, and — 16 x y. 
 
 13. What is the amount of — 1 a b c, — 12 a b c, 
 
 — ab c, — 6a b c, and — 24 a b c ? 
 
 14.)— Ixyz 15.)— 12abx \6.)—labcd 
 
 — Sxyz — Aabx — 12 abed 
 
 — 4xyz — 6abx — 16 abed 
 — ISx yz — abx — 5 abed 
 
ADDITION. 27 
 
 17. What is the sum of — 5 a b a:, and — 3 a b a?, 
 and — 7 a b x, and — 2 a b xl 
 
 18. What is the sum of — 12 a b x y, — 7 ab x y, 
 
 — 17 a b x y, — a b x y, and — 2 a b x y ? 
 
 19. Add together — 3 a, and — la, and — 12 a, 
 and — 16 a. 
 
 20. What is the sum of — x y, and — S x y, and 
 
 — 1 x y, and — 12 x yl 
 
 21. Add together 3 x, and — 2 x. Ans. x. 
 This example contains both a positive and a negative 
 
 quantity. If the Dr. column of a leger amount to 
 3 x dollars, and the Cr. column to 2 a? dollars, the con- 
 dition of the account may be expressed ' 3 x — 2 a?; 
 and the sum due is evidently x dollars. * 
 
 Suppose the value of x to be 10 dollars ; then 3 x 
 == 30, and — 2 x =z — 20, and the account will 
 be Dr. 30 dollars, Cr. 20 dollars, due 10 dollars. The 
 20 dollars credited will cancel the same amount of 
 the debt ; that is, — 2 x will cancel -|- 2 x, leaving 
 -\- x due. 
 
 22. Add together 2 x, and — 3 a?. Ans. . — x. 
 This example is like the last, only the negative is 
 
 the larger quantity, and the answer must have the sign 
 
 — prefixed. For it is evident, if the Dr. side of an 
 account be 2 x, or 20 dollars, and the Cr. side be 3 x, 
 or 30 dollars, the debt has all been paid, and x, or 10 
 dollars more, which is to be paid back. 
 
 From these examples we derive the following Rule 
 for adding together two similar quantities, when their 
 signs are not alike : Subtract the less coefficient from 
 
28 FIRST LESSONS IN ALGEBRA. 
 
 the greater. The answer must have the same sign as 
 the greater quantity. 
 
 Hence it appears, that what is called Addition in 
 Algebra, is sometimes performed by Subtraction. This 
 is an apparent contradiction, which is very apt to per- 
 plex beginners. The whole difficulty will vanish, how- 
 ever, if they remember the object of Addition ; which 
 is, to express the value of two or more quantities, of 
 any kind, in the most simple manner possible. 
 
 23. What is the sum of 5 a? y, and — 3 x yl 
 
 Ans. 2 x y. 
 24.) + 4 a b 25.) + 7 a b x 26.) + 16 a y 
 
 — 3 a b — 5 a b x — 16 a y 
 
 27. Add together 19 a b x m, nnd ■ — 27 a b m x. 
 
 28. What is the sum of — 3 a m x, and 5 a m a? ? 
 
 29. What is the sum of 5 x, and — 3 a?, and 4 x, 
 and — 2 a:? Ans. 4 a?. 
 
 Let us suppose the positive quantities in this ques- 
 tion, namely, 5 x and 4 x, to represent sums of money 
 received, and the negative quantities, — 3 x and — 2 a?, 
 payments made. Then* 9 a? dollars are received, and 
 5 x dollars paid away; and, of course, 4 x dollars 
 remain on hand. 
 
 If the value of a? be 100 dollars, then 900 dollars 
 are received, 500 dollars paid away, and a balance 
 of 400 dollars remains. 
 
 30. Add together 4 x, and — 8 a?, and — 7 a?, 
 and 6 a?. Ans. — 5 a?. 
 
 This example is like the last ; but here, the sum of 
 the negative quantities is greater than that of the posi- 
 
ADDITION. 29 
 
 live, and the difference is marked with the sign — . 
 The sum of 4 a? and 6 a? is 10 a?; — 8 x and — 7 a: 
 is — 15 a?; the difference, or balance between the two 
 sums, is — 5 a?. 
 
 If a trader gain 40 dollars in one week, lose 80 dol- 
 lars the next, lose 70 dollars the third, and gain 60 
 dollars the fourth, in the four weeks he gains 100 and 
 loses 150 dollars; that is, he loses 50 dollars more 
 than he gains, and his amount of loss and gain stands 
 
 — 50. In this case, we suppose the value of a? to be 
 10 dollars. 
 
 31. Add together 5 ab c, — 7 a be, 3 a be, — a be, 
 and 2 a b c. Ans. 2 a b c. 
 
 The sum of the positive quantities is 10 a be; and 
 that of the negative quantities is — 8 a be; and 
 10 ab c — 8abc = 2abc. 
 
 When several similar quantities, some of which are 
 -f- and others — , are to be added, it will generally be 
 found the most convenient way to proceed as in the 
 last three examples, namely : Reduce them to two 
 terms, by bringing all the positive quantities into one 
 sum, and all the negative quantities into another, and, 
 then balance these two terms in the usual manner. 
 
 32. Add together 8 x, and — 3 x, and 4 x, and 
 
 — 6 x. 
 
 33.) 
 
 — lab 
 
 34.) — 2 b c 
 
 35.) abed 
 
 3 ab 
 
 hbc 
 
 — 6 a b c d 
 
 6 a b 
 
 — Abe 
 
 — 8 a b c d 
 
 — ab 
 
 6bc 
 
 4 a b c d 
 
30 FIRST LESSONS IN ALGEBRA. 
 
 36.) x y 37.) 73 b c x 38.) — mxy 
 
 3 x y — 60 b c x 16 m x y 
 
 1 x y 2b c x — 20 m x y 
 
 — 10 x y — 15 6 ex 2 mxy 
 
 39. Add together 4 ax, and 5 ax, and — 3 ax, and 
 lax, and — 6 a x, and — % ax, and 9 a x, and 
 
 — 19 ax. 
 
 40. Add together 14 a b x, and — 6 a b x, and 
 
 — 7 a b x, and 3 a b x, and 7 abx, and — 5 a b x, 
 and 3 a b x, and — 2 a b x, and — 1 a b x, and 
 3 a 6 a?, and 1 a b x, and — a 6 a?, and — 8 a b x. 
 
 SECTION II. 
 
 Simple Quantities that are Dissimilar. 
 
 1. What is the amount of 4 a, and 3 x, and 5 y ? 
 
 Ans. 4a-f-3a?-|-5y. 
 Ii this question, the quantities are dissimilar, that 
 is, different from each other ; and, of course, they 
 cannot be actually added. All that can be done is, 
 to connect them together by their proper signs. 
 
 2. What is the sum of 3 b, and 7 c, and — 5 j? 1 
 
 Ans. 3 b -\- 7 c — 5 a;. 
 
 3. Add together 6 x, and — by, and 7 x y, and 
 — 4 z. 
 
 4. Add together lab, and — 4 y, and 12 b x, and 
 
ADDITION. 31 
 
 5.) 6abx 6.) I9ax 7.) — 4y 8.) ar 
 
 — ba — Samn z — Sz 
 
 — x y xyz axy — ab 
 27 — am — mn — 61 
 
 9. What is the sum of 4 a b, and — 5 a x, and 
 
 — 3 a b, and 6 a x, and ab c, and — ab, and — 16 ? 
 
 Ans. a a? -}- a6c — 16. 
 
 In this question, we have -j- 4 a b ; also — 3 a b, 
 
 and — ab, which, together, make — 4 ab. As these 
 
 two quantities, -f- 4 a b and — 4 a b, cancel each other, 
 
 they do not appear in the answer. We also have 
 
 — 5 a x and -f- 6 « a?, which balance each other in 
 part, the difference between them being -}- a x, which 
 is the first term of the answer. The remaining terms, 
 -\- a b c and — 16, being dissimilar, are annexed. 
 
 In all cases, the addition is performed, if we connect 
 together the several given quantities by their respective 
 signs. Thus, A ab — h ax — Sab -}- 6 a x -\- a b c 
 
 — ab — 16, is a true answer to the last question, 
 and expresses the value of the several quantities as 
 perfectly as the answer given above. What remains 
 is a species of reduction, by which we unite together 
 as many of the terms as we can, for the sake of di- 
 minishing their number, and rendering the answer 
 more simple. 
 
 It is evident that two equal similar quantities whose 
 signs are unlike, as -}- 5 x and — 5 a?, cancel each 
 other. 
 
 10. What is the sum of 16 a b, and — Sax, and 
 
32 FIRST LESSONS IN ALGEBRA. 
 
 ab c, and — 12 a b, and Sax, and — 75, and 40, 
 
 and — 4 a b1 Ans. a b c — 35. 
 
 a.) 
 
 - 4 a b c 
 
 12 
 
 •) 
 
 12 ax 
 
 13.) 
 
 — 5 a? 
 
 6 a x y 
 
 
 
 — ay 
 
 
 — 2z 
 
 1 ab c 
 
 
 
 Sax 
 
 
 Sx 
 
 - xy 
 
 
 
 — 14 a x 
 
 
 ab c 
 
 14. Add together babe, and 12 a?, and — 14 a x, 
 and — 4a b c, and — 16 x, and 3 y, and — 5 a x, 
 and 76, and — ab c, and 5 x, and 83, and 16 a x. 
 
 15. What is the sum of 15 a # y, and a b, and 
 — -2 b m n, and — 12 a x y, and — 24, and 3 bran, 
 and 17, and — ab, and 3 a x y, and y z ? 
 
 16. What is the sum of 14 a x, and 3 b, and 2 #, 
 and 17 y, and 37, and — 4 b, and — 15 a x, and 
 
 — 12 y, and a a?, and 37, and — by, and — 74, 
 and — xl 
 
 17. Add together 6 a b, 5 x y z, 14, — Sab, 21, 
 
 — 3 x y z, c d, 4 a b, — 35, and — 2 x y z. 
 
 18. What is the amount of 8 x y, — 5 b c d, 18, 
 — - 3 x y, — 9, b c d, — 12, and 4 b c d 1 
 
 19.) — 3 h m 
 
 20.) —Ixyz 
 
 21.) 
 
 754 
 
 54 
 
 — 3 m n 
 
 
 abx 
 
 + 9 h m 
 
 Qyz 
 
 
 — 649 
 
 — 60 
 
 2 m n 
 
 24.) 
 
 m 
 
 22.) 8 x y z 
 
 23.) — abx 
 
 9yz 
 
 — *y 
 
 — 7 23 
 
 
 — lyz 
 
 — 3 x y z 
 
 5 a b x 
 
 
 y* 
 
ADDITION. 33 
 
 SECTION III 
 
 Compound Quantities. 
 
 1. Add together 3 a + 2 b and 5 x + 3 y. 
 
 Ans. 3a + 26 + 5# + 3y. 
 In this example, compound quantities are given ; but 
 they are added in the same manner as simple quanti- 
 ties ; that is, all the terms of which they are composed 
 are connected together with their proper signs. If we 
 were required to add together 4 -f- 5 and 3 — {— T, we 
 might either say, 4 + 5 + 3 + 7 = 19-; or 4 + 5 
 = 9, and 3 + 7=10, and then add these two sums, 
 9 + 10 = 19. The result, both ways, is the same ; 
 but, in Algebra, the latter mode is not practicable, 
 unless the quantities are similar. 
 
 2. Add together 5 b + 2 c, and 4 d — 3 y. 
 
 Ans. 56 + 2c + 4d — 3 y. 
 
 This example is like the last, excepting that one of 
 the terms is affected with the sign — , which must be 
 retained in the answer; for we are not requirecfcto 
 add the whole value of 4 d to 5 b + 2 c, but only the 
 difference between 4 d and 3 y ; and when we add 
 4 d, as in the answer above, we add 3 y toof^much, 
 which must be subtracted. 
 
 This may be rendered more intelligible, perhaps, by 
 numbers. Add together 3+4 and 8 — 6. First, 
 3+4 = 7, and 8 — 6 = 2; and 7 + 2 = 9, which 
 is the amount of the numbers given. Again, 3+4 
 + 8 = 15, which is too much by 6, which must be 
 subtracted : thus, 3 + 4 + 8 — 6 = 9, as before. 
 
34 FIRST LESSONS IN ALGEBRA. 
 
 3. Add together 5 a + 2 c, and 3 a? — Ay, and 
 
 2 6 — 2?. Ans. ba + 2c + 3o? — 4y + 2 6 — «. 
 
 4. Add together 5 a + 2 6 + d, and 3 x — 2 6, 
 and Ad — 3* — 6 a. Ans. h d — a. 
 
 By adding all the terms in these several quantities, 
 we obtain 
 
 5a_^26 + d+3# — 26 + 4 d—Sx — 6 a. 
 By cancelling + 26 and — 26, and + 3 x and — 3 x t 
 5a_j_d + 4d_ 6«. 
 By adding the d and 4 d, 
 5 a + 5 c? — 6 a. 
 By balancing the + 5 a and — 6 a, 
 5 d — a, as above. 
 The answer should always be reduced, in this man- 
 ner, to the least number of terms possible. 
 
 5. What is the sum of 5 a x + 3 6 c, and lax 
 -—4 6 c, and — 3 a x + 17 y? 
 
 6. Add together 3 (a — 6), and 2 (a — 6). 
 
 Ans. 5 (a — 6). 
 
 f n this question, 3 (a — 6) =i 3 a — 3 6; and 
 2 (a — b) =2 a — 2 6; and the answer, 5 (a — 6) = 
 5 a — 5 6. If a — 4, and 6 = 2, (3 a — 3 6) = 
 (12 J>6) = 6 ; and (2 a — 2 6) = (8 — 4) = 4 ; 
 and 6 + 4 = 10: but (5 a— 5 6) = (20—10) =10 
 also. WAen tfAe compound quantities, included in pa- 
 rentheses, are alike, they are added like simple quanti- 
 ties that are similar. The numbers or letters before 
 the parentheses, are regarded as coefficients, and 
 added as such ; but the quantities included undergo 
 no change. 
 
ADDITION. 35 
 
 7. What is the sum of 6 (5 a — a? y), and 
 
 3 (5 a — x y) ? 
 
 8. Add 4 (« 6 — a? + 3 ?/), 2 (a 6 — a? + 3 y), 
 and 5 (a 6 — « -j- 3 y), together. 
 
 9. Required the sum of 3 a (m n — 6 + y z), 
 2 a (m n — 6 -|- y z), and 4 a (m n — 6 -\- y z). 
 
 10. Add together 2 (6 c — x y), 3 (6 c — a? y), 
 
 4 (6 c — x y), 5 (6 c — a? ?/) and 2 (6 c — a? y.) 
 From the several examples which have been given 
 
 in the course of this Chapter, and the explanations 
 with which they have been accompanied, may be de- 
 rived the following general Rule for performing all 
 questions in Addition. Connect together all the terms 
 of the given quantities, by their proper signs, and unite 
 such as are similar. 
 
 1 1 . Add together 3 a b + 2 c, and 5 ax — c + 16, 
 and 14 — Sax, and 5 a 6 + 4 c — y, 
 
 12. Add together 9a? — 4 y + 6 z — mn + S — 
 26 + 7 + 46 — 3a? + 7y — 4m?i — x— 86 + 
 7_52 + 2y_L.6a; + z + 96— 18 + 7m/i — 
 10a? — 4 — 36 — 5 y — 2 z — m n. 
 
 This example may be conveniently arranged for 
 adding, in the following manner : — 
 
 9a? — 4y + 6z— m n + 8 — 26 
 
 — 3 a? + 7 y — 4 m n + 7+46 
 
 — a? + 2y — 5* + 7 — 86 
 6o? + z + 1 mn— 18 + 96 
 
 — 10 a? — 5y — 2: — mn — 4 — 36 
 
36 FIRST LESSONS IN ALGEBRA. 
 
 13. Add together 5b c + 2 ax — 3 y — 12 -f 
 6a c x — 8 a x -f 2 y — 6 6 c -f 18 — 9 a c a? -f 
 U+9bc — 8ax — 1 y + 11 + 1 be— 12 y — 
 18 -j- 7 ac x — Qy -\- a x. 
 
 14.) 8x — ly+ 6m + 4 — 2x 
 
 — 4y_9_ 2x — 6z-{-9m 
 
 7m-9ic + 12—8 y -f 4 z 
 
 10 a? — 4m-j-3y — 6 -{- 4 a; 
 
 15.) 8 ab c — 5 a (b — x) — 5 -f- x y z 
 
 3 a (b — x) — 4abc-\-4xyz — 10 
 %abc+ 9— 15xyz-\- 24 
 
 7 a? yz — 2abc + 4a(b — x) — 18 
 
 16.) 5 a— 17 — 7a + Sxy 
 
 A{x — y) — 13 — 10 x y -f- 21 
 
 8 a -{- 8 (a? — y) — 2 x y -{- a (x — y) 
 
 8 xy+9— 3a— 10 (* — y) 
 
 17.) 16 a 6 — 3*2 + 9 — 5m 
 
 9 a? z — 3 -}- ?tt — 9 ab 
 4—3 wi — 18 a 5 — Sx z 
 Gm-\- 12 a b + 2xz-~ 11 
 
 18.) 9a#y— 15-}-a(a — &) — 96c 
 6 c — 8 a (a— 6) + 36— 5 a x y 
 4a (a — 6) — 55c-f aa;y — 13 
 2 — 4 a a? y -|- 8 6 c -|- 5 a (a — b) 
 
SUBTRACTION. 37 
 
 CHAPTER III. 
 SUBTRACTION. 
 
 SECTION I. 
 To subtract a Simple Quantity. 
 
 1. If a man receive 5 x dollars, and pay away 3 x 
 dollars, how much has he left ? Ans. 2x dollars. 
 
 In this example, where the quantities are similar, we 
 obtain the answer by subtracting the coefficients. The 
 work is written thus, 5 x — 3 a?, and then the quanti- 
 ties are balanced ; that is, the terms are reduced, ac- 
 cording to the directions given in Addition. Let it be 
 observed, that the sign of the quantity to be subtracted, 
 is changed from -\- to — . But the sign of the quan- 
 tity from which the subtraction is made, is not changed. 
 
 Suppose the value of # to be 5 ; then 5 x = 25, 
 
 and 3 x — 15 ; and 25 — 15 pz 10, or 2 x, as above. 
 
 2.) From 3 x 
 
 3.) 5 b 4.) 16 y 5.) 9 * 
 
 Take x. 
 
 2 b. 4 y. 2x. 
 
 2x 
 
 
 6.) From 16 a b c 
 
 7.) 4 x y z 8.) 14 m y 
 
 Take 2 a b c. 
 
 x y z. 3 m y. 
 
 9. Subtract 5 ax from lax. 
 10. Take 9 x y z from 10 x y z 
 4 
 
38 TIRST LESSONS IN ALGEBRA, 
 
 11. Take 6 a b x froir 6 a b x. 
 
 12. From 19 h m p take 7 h mp. 
 
 13. Subtract b from a. Ans. a — b. 
 The quantities a and 6 being dissimilar, we are 
 
 obliged to represent the subtraction, as in the answer 
 given. 
 
 That the above expression, a — b, contains the true 
 answer, may be proved by referring to numbers. Let 
 a = 12, and b = 8 ; then a — b, that is, 12 — 8 = 4, 
 which is the difference of the given quantities. Here, 
 too, the sign of the number to be subtracted is changed 
 from -J- to — . 
 
 ■14. From 5 x take 2 
 
 * 
 
 Ans. 5 x 
 
 -2y. 
 
 15.) From 5 a 16 
 Take b. 
 
 .)! x 
 
 17.) 2 m 
 3 w. 
 
 18.) 7 a* 
 2y. 
 
 5 a — 6. 
 
 
 
 
 19.) From 14 a b c 
 Take 2 x y. 
 
 20.) 
 
 12 ax z 
 29 A m. 
 
 21.) 19« 
 
 22. Subtract 5 x y from 4 z. 
 
 23. Take 13 5 c from ®. 
 
 24. From 1 mx take ^?. 
 
 25. Subtract m n from 6 c. 
 
 26. From x take 13 6 c. 
 
 27. Take 1 mx from #. 
 
 28. Subtract b c from m n. 
 
 29. From a?y2 take x y. 
 
 30. Take 16 from 32 a be. 
 
 31. From 8 a b c x take 4 a b c. 
 
SUBTRACTION. 39 
 
 SECTION II. 
 To subtract a Compound Quantity. 
 
 1. A man, who has 4 x dollars in his pocket, pays 
 one debt of 3 x dollars, and another debt of y dollars. 
 How many has he left ? Ans. x — y dollars. 
 
 It is here required to take the whole value of the 
 compound quantity, 3 x -f- y, the sums paid away, 
 from 4 x. Now, if only 3 a? be taken, it is evident 
 that not enough is subtracted by the value of y, what- 
 ever that may be. The work may, therefore, be 
 expressed thus, 4 x — 3 x — y ; which, reduced, 
 gives x — y. 
 
 To illustrate this by figures, let x 3= 5, and y = 3 ; 
 then 4 x = 20, and 3 x -f- y = 15 -f- 3, A or 18 : now, 
 20 — 18 = 2, and 20 — 15 — 3 is also 2. In this 
 last expression, we may either subtract 15 from 20, 
 and then subtract 3 from the remainder ; or we may 
 add — 15 and — 3 together, and subtract their sum 
 from 20. The result is the same both ways. Here, 
 the signs of both the quantities to be subtracted are 
 changed from -j- to — . 
 
 2.) Subtract 3 a -\- b from 5 x. Ans. 5 a? — 3 a — 6. 
 
 3.) From 3 a 4.) 5 a b c 5.) 4 a? 
 
 Take 2 a -f- 6. a b c -f- y. 6-j-z. 
 
 a — b. 
 
 6.) From abed 7.) 47 8.) abx 
 
 Take 2 m + 3 y. x -f- b d. 3 a x + 12. 
 
40 FIRST LESSONS IN ALGEBRA. 
 
 9. From 6 a take 3 a -\- 6. 
 
 10. Subtract 2 x + y from 4 «. 
 
 11. Take x -{- 16 from a?. 
 
 12. Take 23 + a? from 40. 
 
 13. Subtract 4 y -\- x from 6 c. 
 
 14. From y take y -\- x. 
 
 15. Take 4 + a? from 5. 
 
 16. From 5 a? y take a? y -|- 5. 
 
 17. Subtract 36 + 2 y from 48. 
 
 18. Take x -\- y from 2 a?. 
 
 19. From 5 a? take 3 a? — y. Ans. 2 a? -f- y. 
 In this example, the value of the compound quantity 
 
 3 x — y is to be taken from 5 x. The whole value of 
 3 x is not to be subtracted, but the difference between 
 that value and the value of y. If, therefore, we sub- 
 tract the whole of 3 x, we subtract too much by the 
 value of y> which must afterwards be added, to give 
 the true answer. The work may be expressed thus, 
 5 a? — 3 x -f- y l which, reduced, is 2 x -j- y. 
 
 Perhaps this will be better understood, if illustrated 
 by figures. Let x == 6, and y == 4 ; then 5i= 30, 
 and 3a: — y = 18 — 4; that is, we are required to 
 take 18—4 from 30. Now, 18 — . 4 = 14, and 
 30 — 14 = 16, which is the true answer. But if we 
 take the whole of 18 from 30, we take too much by 
 4, as we are required to subtract only the excess of 18 
 over 4 ; we must, therefore, add 4 to the remainder, 
 to obtain the true answer ; thus, 30 — 18 -\- 4 = 16. 
 We may either add 4 to 30, and subtract 18 from the 
 sum ; or we may subtract 18 from 30, and add 4 to 
 
SUBTRACTION. 41 
 
 the remainder. Here, too, both the signs of the quan- 
 tities to be subtracted are changed, the -f- to — ■ and 
 the — to -\-. 
 
 20.) From 14 a b 21.) 17 a? y 
 
 Take 12 ab — xy — z. 12 x y — abm. 
 
 2 ab -\- x y ~j- z. 
 
 22.) From 5 ab 23.) 27 mx 
 
 Take h ab — 7 -fa;. 19 ma? — y — 3 a b. 
 
 24. Subtract b c ■— • 8 from 16. 
 
 25. Take x y z — a from b c x. 
 
 26. From a b take y — x. 
 
 27. From 48 take x— 16. 
 
 28. Take x z — y from b x. 
 
 29. Subtract x y z — 9 a? from a*b c. 
 
 30. From y take y — 4. 
 
 31. Subtract a b -+- 7 from a x. 
 
 32. Take x y — 9 a from 2 a? y. 
 
 33. From Sad take 2 « d + 9. 
 
 34. Subtract 5 m — ■ 5 a from 4 m. 
 
 35. From 18 take x — 18. 
 
 36. Take a -{- b from a. 
 
 37. Subtract 12 + 4 a from 27. 
 
 38. Subtract a -j- 12 from 19. 
 
 39. From 5 (a + 6) take 2 (a -f- 5) —a?. 
 
 40. From a? (a? — y) take 2 a? (a? — y) — - a? y. 
 
 41. Take o a? -|- 6 from 3 a a?. 
 
 4* 
 
42 FIRST LESSONS IN ALGEBRA. 
 
 SECTION III. 
 
 General Rule for Subtraction. 
 
 1. From x subtract — x. Ans. 2 x. 
 According to the principles already explained, — x 
 
 becomes -f- x, when it is subtracted from any quan- 
 tity ; we have, therefore, .r -j- x = 2 x; that is, sub- 
 tracting a negative quantity is the same thing as adding 
 a positive quantity of the sam,e value. If A is in debt 
 1000 dollars, we should subtract that sum in forming 
 an estimate of his property ; but if B cancels that 
 debt for him, that is, subtracts that — quantity, he 
 evidently increases or adds to the amount of his prop- 
 erty as much as if he had actually given 1000 dollars 
 into his hand. 
 
 2. From a -f- b subtract x — y. 
 
 Ans. a -f- b — x -\- y. 
 
 It is here required to subtract the difference of two 
 quantities, x and y, from the sum of two other quanti- 
 ties, a and b. Suppose a = 8, b = 6, x = 11, and 
 y ss 2 : we then have 8 + 6, from which we are to 
 subtract 11 — 2; that is, 14 — 9, or 5, which is the 
 answer. But 8-J-6 — 11 -f- 2 is also equal to 5, 
 which corresponds with the answer as expressed above. 
 The signs of the quantities subtracted are changed 
 as before ; but, in all cases, the signs of the other quan- 
 tities, from lohich the subtraction is made, remain un- 
 changed. 
 
 From the several questions proposed in this chapter, 
 
 
SUBTRACTION. 43 
 
 and the reasoning which follows them, we derive the 
 following general Rule for Subtraction in Algebra : 
 Change all the signs in the quantity to be subtracted, 
 each -f- to — and each — ■ to -f- ; and unite the terms 
 that are similar, as in Addition. 
 
 The subtraction is, in fact, performed, when the signs 
 of the terms to be subtracted are changed. The object 
 of the remaining part of the operation, is, to reduce 
 the number of terms, by uniting or cancelling such 
 as are similar, that the answer may be presented in 
 its simplest form. 
 
 3. What is the value of a? — (a -}- b — c) ? 
 
 Ans. x — a — b -\- c. 
 The expression used in this question implies, that 
 the whole quantity included in the parenthetical 
 marks ( ), namely, a -f- b —r* c, is to be subtracted 
 from x; of course, all the signs must be changed. 
 
 4. What is the value of a b — ( — c x -f- d — 16) ? 
 
 Ans. ab -\- c x — d -\- 16. 
 
 5.) From 6x + 3y 6.)\lxy z + U — ab 
 
 Take 5 x ~ 4 y 13 a b + 24 + I8xy z. 
 
 It is recommended to the student, in performing 
 these examples, actually to change the signs ; at least, 
 until he becomes perfectly familiar with the operation. 
 The last two examples, thus prepared, will stand 
 
 5.) 6a? + 3y 6.) llxyz + 14— ab 
 
 — 5#-f 4y — 13 a 6 — 24 — 18 xyz. 
 
 x + ly — 14 ab — 10 — xyz. 
 
44 FIRST LESSONS IN ALGEBRA. 
 
 7.) From 17 xy~ 102— 12 a bm 8.)lxm-\-14y 
 Take 47 a bm — xy-{-3ar. 14 y-j- 6 m. 
 
 p.) From abx-\-3yz 10.) Zlxyz — 34 -f z 
 
 Take 3yz-\-abx. 21 -f- z — < 14 xyz. 
 
 11.) From an-j-op 12.) 19«5a? — 2amn — 1 bed 
 Take x yz — 60. 3abc-\- 5b c d — amn. 
 
 13.) Fromx-\-y 14.) ryz— 15 + 7 ab + 25* 
 
 Take a? — y. 24 -f- 2 r y 2 — 8a?. 
 
 15.) From 24x -\-3y z — 12abc — 6mnp -\-5 
 Take 16 a be — 6mnp -f- 23 x -j- y z -\- 12 — xy. 
 
 16.) From a? + y + ft 17.) 3 a 6 a? — 14 + 6a 
 
 Take a? — y — z. 3 ab x — 14 — 6 a. 
 
 18.) From 3 x -f- 5 y — « 19.) 8 « m — x y 
 
 Take 2 x -\- 5 y -\- z. 3am — z. 
 
 20.) From 6 m p — 8 a z -\- 12 — 14 m x y 
 Take — 6m])-|-8fl2 — 12 + 14 m a? y. 
 
 21 .) From 3 ax -\- 5 ay — 2 ab take 2 a x -f- 5 o y 
 — 3ab. 
 
 22.) Take 18 + 2 « m from 5 a m — 17. 
 
MULTIPLICATION. 45 
 
 CHAPTER IV. 
 MULTIPLICATION 
 
 SECTION 1. 
 Simple Quantities. 
 
 1. What will 5 oranges come to, at a cents apiece ? 
 
 Ans. 5 a cents. 
 It is evident that 5 oranges must cost five times as 
 much as one orange ; therefore, if one orange cost a 
 cents, whatever the value of a may be, 5 oranges will 
 cost 5 a cents. Let a = 2 cents ; then 5 a = 5 X 2 
 or 10 cents. 
 
 Multiplication is merely a short way of perform- 
 ing Addition, when the quantities to be added happen 
 to be equal. When these quantities are unequal, their 
 united value or amount can be found only by adding 
 them all together. And it is evident that, when they 
 are equal, their amount can be found in the same way. 
 Thus, 
 
 2 + 2 = 4, or 2X2= 4, 
 
 2 + 2 + 2 = 6, or 2X3= 6, 
 
 2 + 2 + 2 + 2 = 8, or 2X4= 8, 
 
 2 + 2 + 2 + 2 + 2 = 10, or 2X5 = 10; 
 where, in the first column, the same results are ob- 
 
46 FIRST LESSONS IN ALGEBRA. 
 
 tained by addition, as are obtained, in the second, by 
 multiplication. 
 
 So, too, in literal quantities ; if one orange cost a 
 cents, two oranges will cost a -\- a = 2 a; three, 
 a -\- a -\- ar=3«; four, a -\-a-\-a-±-a = 4 a ; 
 and five, fl-fa-ffl-fa-f« = 5fl cents, as in the 
 question. 
 
 To multiply a literal quantity by a number, there- 
 fore, all that is required is, to make the number the 
 coefficient of the quantity ; for the coefficient shows 
 how many times the value of the quantity is to be 
 taken. 
 
 In Algebra, as in Arithmetic, the number to be mul- 
 tiplied is called the multiplicand; the number by 
 which we multiply, the multiplier ; and the result of 
 the operation, the product. The multiplier and mul- 
 tiplicand, when spoken of together, are called factors. 
 
 2. If a yard of cloth is worth a dollars, what is the 
 value of 6 yards ? Ans. $6 a. 
 
 It is recommended to the student to prove his an- 
 swers to the questions, by substituting such numbers 
 for the letters given, as he pleases. 
 
 3. If there be x apples in a bushel, how many are 
 there in 9 bushels ? Ans. 9 x apples. 
 
 4. If the interest of a given sum of money be y 
 dollars per annum, what will the interest be for 8 
 years ? 
 
 5. How many bushels of corn are there in a field 
 of 12 acres, which produces a bushels to the acre ? 
 
 6. How much will 10 yards of cloth come to, at c 
 dollars per yard ? 
 
MULTIPLICATION. 47 
 
 7. A man gave away a dollars every day ; how 
 much did he give away in a week ? 
 
 8. What are a man's expenses for a year, who ex- 
 pends a dollars a day ? 
 
 9. If there be x yards of cloth in one piece of linen, 
 how many yards are there in 15 pieces? 
 
 10. How many rods of wall can a man build in 
 three weeks, if he build a rods a day ? 
 
 11. How many pines of glass are required for 14 
 windows, each window having c panes ? 
 
 12. What is the value of a yards of calico, at c cents 
 a yard ? Ans. a c cents. 
 
 It is evident that the price, c, must be multiplied by 
 the quantity, a ; which may be expressed thus, a X c, 
 or thus, without the sign, a c. Let a = 7, and c=:\2; 
 then a X c, or a c = 7 X 12, or 84. 
 
 If the price had been c d cents, we should have 
 multiplied that quantity by a, and the answer would 
 have been a c d. 
 
 Or, if the quantity had been a b yards, and the 
 price c d cents, we should still have multiplied the one 
 by the other, and the answer would have been abed. 
 
 Hence, to multiply one simple literal quantity by 
 another, we write all the letters in both quantities to- 
 gether. 
 
 13. If x men can do a piece of work in y days, how 
 lo \g will it take one man to do it ? Ans. x y days. 
 
 14. Multiply a x by y. Ans. a x y. 
 
 15. Multiply b c by m n. 
 
 16. Multiply a b x by y z. 
 
 17. Multiply a m n by x y z. 
 
48 FIRST LESSONS IN ALGEBRA. 
 
 18. Multiply a b c by h m n. 
 
 19. Multiply x by a b c. 
 
 20. Multiply m x z by a b y. 
 
 21. If a horse travel b miles in one hour, how far 
 will he travel in 2 a hours ? Ans. 2 a b miles. 
 
 Let b == 6, and « = 5 ; then 2 « = 2 X 5, or 10 ; 
 and 2 a 6 = 2 X 5 X 6, or 60. Or we may say, 
 in a hours he will travel a b miles ; in 2 a hours he 
 will travel twice as far, that is, 2 a b miles. 
 
 22. What will y loads of hay come to, at 4 a dol- 
 lars a load ? 
 
 23. Multiply 4 a b by x y. 
 
 24. Multiply p, x by 7 6 c. 
 
 25. Multiply y by 9 a b m. 
 
 26. Multiply 7 a b n by & y z. 
 
 27. Multiply « r x by 12 6 c y. 
 
 28. Multiply 9 a m w by #. 
 
 29. Multiply 4 y by a b m. 
 
 30. Multiply 17 a m x by b c y z. 
 
 31. What will 5 barrels of flour cost, at 2 a dollars 
 per barrel ? Ans. $ 10 a. 
 
 If one barrel cost 2 a dollars, 5 barrels will cost 5 
 times 2 «, that is, 10 a dollars. Let a = 3 ; then 
 2 « will be twice 3, or 6 dollars a barrel ; and 5 bar- 
 rels will cost 5 X 6 = 30 dollars. But 10 a = 10 X 3, 
 or 30 dollars also. 
 
 32. A cistern has 4 cocks, each of which will dis- 
 charge 5 x gallons in an hour. How much will flow 
 from them all in 2 hours? Ans. 40 x gallons. 
 
 33. Multiply 8 a x by 4. 
 
 34. Multiply 12 a bx by 6. 
 
MULTIPLICATION. 49 
 
 35. Multiply 9 a b c by 2. 
 
 36. Multiply 15 a by 3. 
 
 37. Multiply 8 x y z by 14. 
 
 38. Multiply 12 b c m y by 18. 
 
 39. Multiply 73 b m y by 42. 
 
 40. Multiply 19 a b x z by 12. 
 
 41. If a man's income be 2 6 dollars per day, what 
 will it be in 6 a days ? Ans. 12 a b dollars. 
 
 Here the income of one day must be multiplied by 
 the number of days, and the answer may stand 6 a 2 b. 
 Let a — 2, and b = 4 ; then we shall have 2 6 = 8, 
 and 6 a = 12 ; and 12x8= 96. But 6 a 2 b, that 
 is, 6 X 2 X 2 X 4, is also 96. This expression is 
 made more simple, however, if we multiply the two 
 numbers, 2 and 6, together, and use their product 
 12 a b, as in the answer given above ; for 12 a b, that 
 is, 12 X 2 X 4 = 96. 
 
 And let it be observed, that, although it matters not 
 in what order any two letters are written, two numbers 
 must always be separated, either by a letter, or by the 
 sign X j for, if a = 2 and b = 4, 
 
 6a26 is 6x2x2x4= 96. 
 26 a b is 26 X 2 X 4 = 208. 
 62 a 6 is 62 X 2 X 4 = 496. 
 
 From these examples we derive the following Rule 
 for multiplying simple quantities : Multiply the num> 
 bers or coefficients, and annex all the letters in tm 
 flveral quantities to their product. 
 
 42. Multiply 5 a b by 6 x. 
 
 43. Multiply 8 a x by 3 m y. 
 
 44. Multiply 19#y.by 4 b m 
 
50 FIRST LESSONS IN ALGEBRA. 
 
 45. Multiply 24 x y z by b x. 
 
 46. Multiply 12 m n by 52. 
 
 47. Multiply 19 x y by 13 a b z. 
 
 48. Multiply x y by 14 a m. 
 
 49. Multiply 71 a m by 22 b h y. 
 
 50. Multiply 19 y by 12 a x. 
 
 SECTION II. 
 
 Compound Quantities. 
 
 1. A man gave to his son a dollars, and to his 
 daughter b dollars, every week. How much did he 
 give them both in c weeks? Ans. a c -f- b c dollars. 
 
 To his son and daughter he gave a -f- b dollars in 
 one week. Here we have a compound quantity, a -f- b, 
 to be multiplied by a simple quantity, c, the number 
 of weeks. To the son he gave a c dollars, to the 
 daughter b c dollars, in c weeks ; that is, a c -f- b c 
 to the two. 
 
 Let a = 4, b =3, and c = 5 ; then a c = 20, 
 b c = 15, and a c + b c = 20 -f- 15 = 35. Again, 
 a -f b = 4 -f 3 = 7 ; and 7 X 5 = 35, as before. 
 
 Hence, to multiply a compound quantity by a sim- 
 ple one, we must multiply every term of the former by 
 the latter. 
 
 2.) Multiply 4a + 56 3.) a b + 3 a x 
 
 By 3 c 5 a c 
 
 12 a c -\- 15 b c. 5aabc-\~ 16 a a ex. 
 
MULTIPLICATION. 
 
 51 
 
 4. Multiply 6 + 6c + 5#y + 13 by a b. 
 
 5. Multiply 6a& + 8y+19&by4a#. 
 
 6. Multiply 12 b + 8 * + 15 by 7 a a? y. 
 
 7. Multiply 11 «&-f» + yby9flffi«. 
 
 8. Multiply cy-\-m-{-4~\-z by abx. 
 
 9. Multiply 6dm-|- x + 5 y + 1 by la m. 
 
 10. Multiply 5 a x z -{- 1 + 3# + ybya&#. 
 
 11. Multiply lnx-\-5y + 8a by 4bcm. 
 
 12. Multiply 9xy + l-\-6azbya. 
 
 13. Multiply cd + 6 + c + a + #bya 6. 
 
 14. Multiply lab-\-9abxy-\-2 by cd. 
 
 15. What is the product of a -f- b multiplied by 
 x -\- yl Ans. ax-\-bx-\-ay-\-by. 
 
 In this question, both the multiplicand and the mul- 
 tiplier are compound quantities. Now, if we multiply 
 a -\- bby x only, the product is a x + b x ; but as x 
 is smaller than the whole multiplier by the value of 
 y, we must multiply by this quantity also, and add 
 the product, a y -\- b y, to the former. 
 
 To illustrate this by numbers, let a = 2, b = 3, 
 x ss 4, and y = 5. Then we have 2 + 3 to be mul- 
 tiplied by 4 -f- 5. Now, 2 + 3 = 5, and 4 + 5 = 9, 
 and 5 X 9 = 45. Or the work may be expressed 
 thus : 
 
 2+3 
 4+5 
 
 8 + 12 
 
 10+ 15 
 
 8 + 12 + 10 + 15 = 45. 
 
52 FIRST LESSONS IN ALGEBRA. 
 
 16. What is the product of 2 a-f- 3 b multiplied by 
 3 a -f- 2 b? 
 
 2a + 3 b 
 3a + 2b 
 
 6 a a -f- 9 « b 
 I Aab + 6bb 
 
 Ans. 6aa-\- 13 a b -\~ 6 b b. 
 
 The terms 9 a b and 4 a b being similar, they are 
 united as in Addition. When both the multiplicand 
 and multiplier are compound quantities, each term of 
 the* former must be multiplied by each term of the 
 latter; and the several products must then be added 
 together. 
 
 17. Multiply 9 ay -{- b d by b d -\- a y. 
 
 18. Multiply 3abc-{-8ay -{• x by a b c -\- x. 
 
 19. Multiply b + c -f 8 by b -j- c -f 10. 
 
 20. Multiply a-f-y-f-jz-^bya-f-c. 
 
 21. Multiply 3 a b -\- c -\- x y by x y -\- c -{- a b. 
 
 22. Multiply 4«6 + 26 + 5aby3a + 7 5. 
 
 23. Multiply a b -f 4 by 5 -f a 6. 
 
 SECTION III. 
 ♦S%ws in Multiplication. 
 
 As all the examples hitherto proposed in this Chap- 
 ter, consist of positive quantities, nothing has been 
 said on the subject of the signs ; for it is evident, that 
 
MULTIPLICATION. 53 
 
 the product of two positive quantities must also be 
 positive. When you multiply -\- a by -\- b, you 
 merely take -f- a as many times as there are units in 
 b, without making any alteration in the nature of the 
 quantity expressed by it. 
 
 1. A gentleman's income being a dollars, and his 
 expenses b dollars per week, how much will he save 
 in c weeks ? Ans. a c — be dollars. 
 
 Here we have a — b to be multiplied by c. His 
 income is a c dollars, and his expenses b c dollars, for 
 the given time ; then a c — be expresses the excess 
 of the former over the latter. 
 
 Let a = 12, b = 10, and c = 5 ; then we shall 
 have 
 
 a — b 12 — 10 = 2 
 
 c 5 5 
 
 ac—bc. 60 — 50 = 10. 
 
 Here, — multiplied by -}-, gives — in the product. 
 
 2. Multiply 3 x — b x by a d. 
 
 3. Multiply 8 a b — I by a x. 
 
 4. Multiply 9 a x — 17 x by 3 a c. 
 
 5. Multiply a — bx — x y by c m n. 
 
 6. Multiply a b -}- x — 1 -\^ aby x y z. 
 
 7. Multiply 3 x y — b x by 5 c d. 
 
 8. Multiply a b c — 1 -\- 6 x by 9 m n. 
 
 9. Multiply ax — 4 — x by a x -j- x. 
 
 10. Multiply b c x — 16 by a c -{- m z. 
 
 11. What is the product of a -f- b multiplied by 
 x — y 1 Ans. a x -\- b x — ay — by. 
 
 We first multiply a -j- b by x, which gives ax -f- bx; 
 5* 
 
54 FIRST LESSONS IN ALGEBRA. 
 
 but this product is evidently too large, as we are not 
 required to multiply by the whole value of x, but only 
 by its excess over the value of y. We must, there- 
 fore, multiply by y, and subtract the product, a y -\- 
 b y, from the former ; but when subtracted, it becomes 
 — ay — by. 
 
 Let a = 7, b = 5, x = 10, and y = 4 ; then we 
 shall have 
 
 « + & 7+5 rr 12 
 
 a? — y 10-^ 4 = 6 
 
 a x + 6 a? 70 + 50 72 
 
 — ay — by —28 — 20 
 
 a;r +&tf_« y _5 y . 70 + 50 — 28 — 20=72. 
 
 In this operation, + multiplied by — , gives — in 
 the product. 
 
 12. Multiply 3a + bybyd— f. 
 
 13. Multiply afc + c + a? by 2c — d. 
 
 14. Multiply 5 a + 2 6 by 6 a — 3 6. 
 
 15. Multiply x -{- y -\- zby x — >y — z. 
 
 16. Multiply b x +5 by 3a? — 3. 
 
 17. Multiply ab x -\- c m by 1 — by. 
 
 18. Multiply r x + a + 16 by — 8 x. 
 
 19. Multiply a + 5 + a? by a — a? — 5. 
 
 20. Multiply Am + 16+yby96 — 8 6. 
 
 21. Multiply a — b by x — y. 
 
 Ans. fl« — b x — a y + b y. 
 
 We first multiply a — b by x, and the product is 
 
 a x — b x. But we are not required to multiply 
 
 a — b by the whole value of a?, but only by its excess 
 
 over the value of y ; we must, therefore, multiply by 
 
MULTIPLICATION. 55 
 
 y, and subtract this product from the former. The 
 product of a — ■ b multiplied by y, is a y — b y, and 
 when subtracted, it becomes — a y -\- b y. 
 
 Let a = 15, b =. 5, x = 12, and y = 8 ; then we 
 have 
 
 a — b 15 — 5 = 10 
 
 x — u 12 — 8 = 4 
 
 a x — b x 180 — 60 40 
 
 — ay + by — 120 -j- 40 
 
 ax ^hx — ay + by. 180 — 60 — 120+40 = 40. 
 
 Hence it appears, that — multiplied by — gives -\- 
 in the product. This is a principle which is very apt 
 to perplex beginners. To use a common expression, 
 they cannot understand how the multiplying of less 
 than nothing by less than nothing, can give a real or 
 positive quantity. Thus stated, the subject is, indeed, 
 quite inexplicable ; but the whole difficulty vanishes 
 when it is remembered that there are, in fact, two 
 operations carried on at the same time, namely, Mul- 
 tiplication and Subtraction. We first multiply by 
 the negative quantity, as if it were positive ; and then, 
 by changing the signs of all the terms, subtract the 
 product from the quantity already obtained. 
 
 22. Multiply b -f- a c — b cby f — < c. 
 
 23. Multiply a — ab — 4 by 6 — 5. 
 
 24. Multiply — 6 a by — 6 a. 
 
 25. Multiply 12 a b — 8 by a — 12. 
 
 26. Multiply — 8 by — 6. 
 
 27. Multiply 8a& — 3#by2a& — 5 x. 
 
56 FIRST LESSONS IN ALGEBRA. 
 
 SECTION IV. 
 General Rule for Multiplication. 
 
 In the preceding sections of this Chapter, have 
 been developed all the rules to be observed in the 
 multiplication of algebraic quantities. To facilitate 
 practice, they will now be repeated together. 
 
 1. Multiplication. Multiply each term of the 
 multiplicand by each term of the multiplier. 
 
 2. Signs. When both terms have the same sign, the 
 product has the sign -f- ; but when they have different 
 signs, the product has the sign — . 
 
 3. Coefficients. Multiply the coefficients of both 
 terms together, and use their product. 
 
 4. Letters. Write the letters of both terms in or- 
 der, one after the other. 
 
 5. Reduction. Add together the several products 
 by their proper signs, and unite such as are similar into 
 one term. 
 
 1. Multiply x-]-2xy-\-ybyx — y. 
 
 Ans. xx -f- 2 x x y — 2 x y y — y y. 
 
 2. Multiply x-\-xy-\-ybyx — x y -f- y. 
 
 3. Multiply 3a?— 2xy + 5byx-]-2xy — 3. 
 
 4. Multiply 2 a — 3 ax-\-4xby5a — 6 ax — 2x. 
 
 5. Multiply a? -f 3 6 — 6 by 4 a: — 86 — 8. 
 
 6. What is the product of 7 Z — 2 m — 9 multi- 
 plied by 3 Z— 11 ml 
 
 7. Multiply a + b+6 by a — b — 6. 
 
 8. Multiply a b — bc-\-cd by ab-j-bc — c d. 
 
MULTIPLICATION. 57 
 
 9. What is the product of 5 -f- b — c multiplied 
 by a — b -f 9 ? 
 
 10. What is the product of 7 a — 3 a? -f- 5 multi- 
 plied by 4 a -J- rf? 
 
 11. Multiply 66 — 6c-j-ccby6 + c. 
 
 12. Required the product of 3 m m x — m x -\- x 
 multiplied by 2 m — y. 
 
 13. What is the product o£xx-\-xy-\-Sz mul- 
 tiplied by x -j- 1 ? 
 
 14. Multiply a -f- 6 — c by 1 — #. 
 
 15. What is the product of a a — 4 a x -f- ## 
 multiplied by a -j- a?? 
 
 16. Required the product of a a — 2 a y -{- y y 
 multiplied by a — y. 
 
 17. Multiply a -f- c — g-by4aa-f3cg\ 
 
 18. What is the product of w — n -\- z multiplied 
 by 5 m to — 2 y z? 
 
 19. Required the product of 16 am -f- 3 a c multi- 
 plied by — a y -f- 4 m. 
 
 20. Multiply a a — bb -j- 8 by # — y y y. 
 
 21. What is the product oi x x -f- y y y — 7 multi- 
 plied hy x — y ? 
 
 22. Required the product ofa-\-b-\-c-\-d mul- 
 tiplied by a — b — c — d. 
 
 23. Multiply 13 x y — 12 x by y — 1. 
 
 24. Multiply 3 a + 5 — 26 by 50 + 26 — 5. 
 
 — f-f ". 
 
 i 1 1 
 
58 FIRST LESSONS IN ALGEBRA. 
 
 CHAPTER V 
 DIVISION. 
 
 SECTION I. 
 
 Simple Quantities. 
 
 1. If you divide 15 cents equally among 3 boys, 
 how many cents will each boy have? Ans. 5 cents. 
 
 When any given quantity is to be separated into a 
 certain number of equal parts, the value of one of 
 those parts is determined by Division. Thus, if we 
 were to count off 1 5 cents into three equal piles, we 
 should find that each of those piles would contain 5 
 cents; that is, 3 is contained 5 times in 15. 
 
 The,quantity to be divided is called the Dividend; 
 the quantity, denoting the number of equal parts into 
 which the dividend is to be divided, is called the Di- 
 visor ; and the value of one of those parts is called 
 the Quotient. Thus, in the above example, 
 
 15 is the Dividend, 
 3 is the Divisor, 
 5 is the Quotient. 
 
 As Division is the reverse of Multiplication, the 
 divisor and quotient being multiplied together, will 
 
DIVISION. 59 
 
 reproduce the dividend. Thus, in the question above, 
 3x5 = 15. Indeed, the dividend may be regarded 
 as a product, of which one of the factors, the divisor, 
 is known ; and the whole object of Division is, to find 
 the other factor, namely, the quotient. 
 
 2. 6 a -r- a = 6 ; for 6 X a =z 6 a. 
 
 3. ab -^ b =. a; for a X h = a b. 
 
 4. abc-^-a = bc; for b c X a = a b c. 
 
 5. abc-±-b = ac; for a c X b = a b c. 
 
 6. abc-±-c — ab; for a b X c = a 6 c. 
 
 7. 16 a -r- 4 = 4 a ; for 4 a X 4 = 16 a. 
 
 8. 20 a b -r- 5 a = 4 6 ; for 4 b X 5 a = 20 a b. 
 
 9. 32a6c-r-8a6 = 4c; for 4 c x8ab = 32 abc. 
 
 In each of these examples, our object has been, to 
 find some quantity, which, being multiplied by the di- 
 visor, would produce the dividend. The mode in 
 which this may be done, is evident, namely : Divide 
 the coefficient of the dividend by that of the divisor ; 
 and omit in the quotient all those letters which are com- 
 mon to the given terms. 
 
 10. Divide 18 a b m x by 9 a b x. 
 
 11. Divide 27 a b c m y by 3 b c y. 
 
 12. Divide 9hmnphy3hm. 
 
 13. Divide a b x by a b x. 
 
 14. Divide 39 a z by 13 *, 
 
 15. Divide 108 ab chmn x by \2 a c m x. 
 
 16. Divide 24 a x y z by 12 x y z. 
 
 17. Divide 36 a x y by 9 a x y. 
 
 18. Divide 72 a b c x y by 8 a b x. 
 
 19. Divide 18 h m n x z by 6h m x z. 
 
60 FIRST LESSONS IN ALGEBRA. 
 
 20. Divide 56 a x y z by 7 x y z. 
 
 21. Divide a by b. Ans. — . 
 
 Here, the given quantities being dissimilar, the di- 
 vision cannot be performed, but only represented ; we 
 therefore write the divisor under the dividend, in the 
 form of a fraction. 
 
 22. Divide x by y. 
 
 23. Divide a b by c x. 
 
 24. Divide 4 a b c by 5 x y. 
 
 25. Divide 17 a b x by 21 c d y. 
 
 26. Divide a b by a x. Ans. — . 
 
 This may be written a — x , and then reduced, as in 
 
 the answer, the two a's being cancelled. Let a = 4, 
 5 = 5, and x = 6 ; then a b = 4 X 5, and also 
 
 a* = 4x6;and-:=— =-= 7 ,as above. 
 
 27. Divide a b c by a b x. 
 
 28. Divide a b m x by a b x y. 
 
 29. Divide a b m by a b m. 
 
 30. Divide abmx by abmxy. 
 
 31. Divide a b x y by a b x. 
 
 32. Divide b c m x y by a b c x z. 
 
 33. Divide 14 « 6 by 7 c. 
 
 We may first express the division thus, -=^- , and 
 reduce the result by dividing both terms by 7. 
 
 34. Divide 16 a b by 4 x. 
 
 35. Divide 20 a b by 4 a x. 
 
 36. Divide 15 x y by 5 a x y. 
 
 37. Divide 25 a b c m by 10 a b x. 
 
DIVISION. 61 
 
 38. Divide 18 abfm by 6 a bfm n x. 
 
 39. Divide 21 a x by 7 a x y. 
 
 40. Divide 5 g h by 3 x y. 
 
 41. Divide 14 a b x by 2 a b x. 
 
 42. Divide abfghbyafhx. 
 
 43. Divide 49abcxyby7abcdmx. 
 
 44. Divide abmxby4abmy. 
 
 45. Divide 12 (a — b -f 9) by 8 (a — 6 + 9). 
 
 46. What is the quotient of a b (19 + x — y) di- 
 vided by b (19 -f x — y) I 
 
 47. Divide 12 a b (x — y -\- z) by 3 a (x — y -f- z ). 
 
 48. Divide 4 (a b — 10 -f- x) by 7 (a 6 — 10 -f- x). 
 
 49. Divide 3 (a y — 12 b) by a (a y — 12 6). 
 
 SECTION II. 
 jS£gn« t» Division. 
 
 The process of dividing one algebraic quantity by 
 another, consists of two parts : the first is, to ascertain 
 the proper expression of the quotient, in letters and 
 figures ; the other is, to determine the character of 
 that expression, either as positive or negative. In the 
 last section, the mode of dividing one simple quantity 
 by another, was considered alone, without any regard 
 to the signs. The mode of determining these, will 
 form the subject of the present section. 
 
 1. Divide a b by b. Ans. a. 
 
 In this example, both of the given quantities are 
 positive. And the divisor, b, being -{-, the quotient, 
 6 
 
62 FIRST LESSONS IN ALGEBRA. 
 
 a, must be -f- also; for, multiplied together, they 
 must give the product -f- a b, that is, the dividend. 
 But if we suppose the quotient to be — a, we shall 
 have — a X -f- b, which will give — a b. Hence, 
 -f- divided by -\- gives -f- in the quotient. 
 
 2. Divide a b by — b. Ans. — a. 
 Here, the divisor, b, being — , the quotient, a, must 
 
 be — also ; for -f- a X — b will give — a b, and not 
 -f- a b, which is the dividend. Hence, -f- divided by 
 — — , gives — in the quotient. 
 
 3. Divide 16 a b x by — Sax. Ans. — 2 b. 
 
 4. Divide 36 a h m n y by — 12 a m n. 
 
 5. Divide 72 a b x y by — 9 a xy. 
 
 6. Divide « 6 x by — ai«. 
 
 7. Divide 12 a m n by — 4 a m n. 
 
 8. Divide 16 a b r z by — 8 a b r. 
 
 9. Divide 28a6mwpby — lamp. 
 
 10. Divide 48 a h y by — 16 a y. 
 
 11. Divide — a bby b. Ans. — a. 
 
 In this example, as the dividend, a b, is a — quan- 
 tity, and the divisor, b, is -f-, the quotient must be — ; 
 for — a X -f b gives — a b, the dividend. Hence, 
 — divided by -\-, gives — in the quotient. 
 
 12. Divide — 16 a x by 4 a x. Ans. — 4. 
 
 13. Divide — 21 abmn by 1 am n. 
 
 14. Divide — 35 x y z by 7 x z. 
 
 15. Divide — 32 a b c m y by 16 a b m y. 
 
 16. Divide — a b by a b. 
 
 17. Divide — 28 by 7. 
 
 18. Divide — 49 a x z by 7 a z. 
 
 19. Divide — 96 a b c x by 12 c x. 
 
DIVISION. 63 
 
 20. Divide — 28 a m n p by 7 a m n. 
 
 21. Divide — a b by — b. Ans. a. 
 Here, as the divisor is — , the quotient must be -f- ; 
 
 for their product must produce the dividend, namely, 
 
 — a b, which is the product of -f- a X — b. But if 
 we suppose the quotient to be — , then we shall have 
 
 — a X — b = -f- a b, which is evidently wrong. 
 Hence, — divided by — gives -f- in the quotient. 
 
 22. Divide — 18 a b m x by — 6 a b m. Ans. 3 x. 
 
 23. Divide — 25 h m n x by — 5 h m. 
 
 24. Divide — 32 a b c y by — 8 a c. 
 
 25. Divide — 45 h m r x by — 9 m x. 
 
 26. Divide — 12 b c by — S c. 
 
 27. Divide — b c x by — 6 c. 
 
 28. Divide — 27 y z by — 9 y. 
 
 29. Divide — 64 m n by — 16 m. 
 
 30. Divide — 9 x y zby — Z x y z. 
 
 From the preceding examples and remarks, we de- 
 rive the following general Rule for the signs in Di- 
 vision : 
 
 When the signs of the divisor and dividend are 
 alike, the sign of the quotient is -\- ; when they are 
 not alike, it is — . 
 
 31. Divide 18 a b x y by — 9 a x. Ans. — 2 b y. 
 
 32. Divide — 64 a m x by 8 a m. 
 
 33. Divide 81 a b c m n by 9 a b c. 
 
 34. Divide — 63 h m n p by — 1 h m. 
 
 35. Divide 84 a y by \2 a y. 
 
 36. Divide — 21 a b x y z by — 9 a b x. 
 
 37. Divide 42 h m x by — 7 a?. 
 
64 FIRST LESSONS IN ALGEBRA. 
 
 38. Divide a b by c d. 
 
 39. Divide a by — b. 
 
 40. Divide — 9 a x by A x. 
 
 41. Divide 9 x y by — 4 y. 
 
 42. Divide — 12 b c d by — 3 b c x. 
 
 43. Divide 18 x z by — 6 x y. 
 
 44. Divide 7mnby — 3n. 
 
 45. Divide — 2 a? 2 by 3 z. 
 
 46. Divide — 7 a b by — 3 a. 
 
 47. Divide 7 « 6 by 3 a. 
 
 48. Divide — 8o6cby3a6c. 
 
 SECTION III. 
 Compound Quantities. 
 
 1. Divide ab -\- a cby a. Ans. 6 -j- c. 
 We have here a compound quantity, a b -f- « c, to 
 
 be divided by a simple quantity, a. We first divide 
 b by a, and the quotient is b ; we then divide a c by 
 a, and the quotient is c : that is, we divide each term 
 of the compound quantity separately. 
 
 2. Divide 12 a c + 9 b c by 3 c. Ans. 4 a + 3 b. 
 
 3. Divide 18 a x -f 15 — 21 a b by 3 a. 
 
 4. Divide 3abc-{-12abx — 9a6by3a6. 
 
 5. Divide 10 a x — 15 a: by 5 a?. 
 
 6. Divide a r x -\- a h x — a by a x. 
 
 7. Divide ax — a y -\- a z by a. 
 
 8. Divide 6 a b + 12 a c by 3 a. 
 
DIVISION, 65 
 
 9. Divide 35 d n -f 14 d x by 7 d. 
 
 1 0. Divide I2aby + 6 ab x — 18 b m-\- 24 bby 6 b. 
 
 11. Divide 16 a — 8 -f 12 y — 20 a da?+mby4. 
 
 12. Divide 15 a b — 25 b x -f 10 a m by 10 a. 
 
 13. Divide 28 a 6 # — 14 6 # — 49 b x y by 7 b x. 
 
 14. Divide 16 a b — 12 a c by 4 a. 
 
 15. Divide 12 x y — 18 y by 6 p. 
 
 16. Divide 24 a b — 24 a by 8 a. 
 
 17. Divide 21 a b — 14 a? by — 7 a. 
 
 18. Divide a + b — 8 by a. 
 
 19. Divide 15 m n -f- 35 # by 5 a?. 
 
 20. Divide 17 — 12 a + a 6 by 3 ff. 
 
 21. Divide a -\- bby x -\- y. Ans. -. 
 
 In this example, both the divisor and dividend are 
 compound quantities ; but, as all their terms are dis- 
 similar, it is evident that the division can only be rep- 
 resented, and not actually performed. 
 
 22. Divide a c -\- b c by a -\- b. 
 
 a-{-b)tic-\-bc(c 
 a c -f- b c 
 # # 
 
 In this example, we divide the first term of the 
 dividend, a c, by the first term of the divisor, a, and 
 obtain c for the quotient. We then multiply the 
 whole divisor, a -\- b, by c, to ascertain whether it be 
 the whole quotient, or only a part of it ; and the pro- 
 duct is a c -j- b c, that is, the dividend. 
 
 23. Divide b x -\- c x by b -\- c. 
 
 24. Divide a a -\- a bby a -{- b. 
 
 6* 
 
66 FIRST LESSONS IN ALGEBRA. 
 
 25. Divide 3 a « — 2 a b by S a — 2 6. 
 
 26. Divide 12 aa + 6ab by 4 a + 2 6. 
 
 27. Divide 18 flafl-)-6a ai — 12 a a # by 6 a 
 4-26 — 4a?. 
 
 26. Divide 14 a6 + 21 ac + 49 a# by 26+3c 
 + 1 oc. 
 
 29. Divide 66-j-36c + 2ccby6 + c. 
 
 b + c) 66+36c-f2cc (6 + 2 c. 
 66 + 6c 
 
 26c -f 2 cc 
 2 6c + 2cc 
 
 In this example, as in the preceding, we divide the 
 first term of the dividend by the first term of the di- 
 visor. The quotient of 6 6 divided by 6, is 6. "We 
 then multiply the whole divisor, 6 -{- c, by 6, and ob- 
 tain the product 6 6 -f- 6 c, which is not equal to the 
 dividend. We are, therefore, certain that 6 is not 
 the whole quotient. This product is then subtracted 
 from the dividend, and the remainder is 2 o c -f- 2 c c, 
 which must also be divided. We begin as before, 
 and divide the first term of this remainder by the first 
 term of the divisor ; that is, — - =2c, which is the 
 
 b 
 
 second term of the quotient. We next multiply the 
 divisor, 6 -\- c, by 2 c, and the product is equal to the 
 remainder of the dividend. The whole quotient is 
 b -f- 2 c, which, being multiplied by the divisor, will 
 reproduce the dividend. 
 
DIVISION. 67 
 
 30. Divide aa + Zah + bb by a + b. 
 
 a + b) aa + 2ab + bb ( a + K 
 a a -\- a b 
 
 ab + bb 
 
 ab + bb 
 
 # # 
 
 31. Divide a a — b bby a -\- b. 
 
 a -\- 6 ) a a — b b (a — b. 
 a a -f- ab 
 
 — a b — bb 
 
 — ab — bb 
 
 This example is like the former, excepting that it 
 contains negative quantities. We obtain the second 
 dividend, — ab — b b, by subtracting a a -f- a b from 
 a a — b b; and the sign of the second term of the 
 quotient must be — , because the signs of the divisor, 
 -f- «, and of the dividend, — a b, are not alike. It 
 is obvious that the first term of the quotient is too 
 large ; for (a -f- b) a = a a -f- a b, which is evidently 
 a larger quantity than the dividend, a a — b b, what- 
 ever may be the values of a and b. 
 
 32. Divide 8 x x x — y y y by 2 x — y. 
 
 2x — y ) 8 x x x — y y y (4:XX-\-2xy-\-yy. 
 8 x x x — Ax x y 
 
 Ax x y — y y y 
 
 Ax x y — 2 x y y 
 
 Zxyy — yyy 
 
 2xy y — yyy 
 
68 FIRST LESSONS IN ALGEBRA. 
 
 SECTION IV. 
 General Rule for Division. 
 
 From the foregoing examples and observations, we 
 derive the following general Rule for Division in Al- 
 gebra, when both the divisor and dividend are com- 
 pound quantities : 
 
 Divide the first term of the dividend by the first term 
 of the divisor, for the first term of the quotient. 
 
 Multiply' the whole divisor by this term, and subtract 
 the product from the dividend. 
 
 Divide the first term of the remainder by the first 
 term of the divisor, for the second term of the quotient. 
 
 Multiply the whole divisor by this second term, and 
 subtract the "product from the remainder. 
 
 Continue this series of operations as long as the na- 
 ture of the question may require. 
 
 1. Divide a a -f- a b -f- a c -j- 5 a -f- 5 6 -f- 5 c 
 by a -f- b -f- c. 
 
 2. Divide a a — 6 a ~\- ab — a c -}- 6 c — be by 
 a — c. 
 
 3. Divide bbb -\- c c c by b -\- c. 
 
 4. Divide 6 m m m x — 3 mmxy—~2mmx 
 -\ r mxy-\-2mx — x y by 2 m — y. 
 
 5. Divide xxx-\-xx-\-xxy-\-xy-\-Sxz-\- 
 3 z by x + 1. 
 
 6. Divide a -f- b — c — ax — bx-\-cxbya-\- 
 b — c. 
 
 7. Divide a a a — 3a ax — Saxx-^-xxxby 
 a -f- x. 
 
DIVISION. 69 
 
 8. Divide a a a — 3 a a y -\- 3 a y y — y y y by 
 a — y. 
 
 9. Divide 4 a a a -\- 3 a cg-\-4aac-\-3ccg 
 — 4 a ag — 3 eg gby a + c — g. 
 
 10. Divide 5 m m n — 2 my z — 5 mnn -{-2ny z 
 -\- 5 m n z — 2 y z z by m — n -\- z. 
 
 11. Divide x x x — 3 x x y -{- A x y y — 4 y y y 
 by## — 2 x y -\- y y. 
 
 xyy — 3y y y 
 
 xx—2zy+yy)xzx—3xzy + 4zyy—4yyy(x — y+— I ^—^ 
 zzz — 2zzy-\-zyy 
 
 —zzty +3xyy—4yyy 
 —xxy +2xyy— yyy 
 
 zyy — 3yyy 
 
 As there is a remainder in this example, it is an- 
 nexed to the quotient in the form of a fraction. 
 
 12. Divide ax-{-bx-\-ay-\-by-{-z by a -\-b. 
 
 13. Divide c d — c h -\- d e — e h -\-xby d — h. 
 
 14. Divide x x x y — xxyy-\-5yyy by x — y. 
 
 15. Divide 3 a b b b -\- 2 a a b be — abb -f- bbbc 
 -{-b b x — Saab — 2 a a a c -\- a a — a b c — ax, 
 by 3ab-\-2aac — a -\- b c -{- x. 
 
 16. Divide 18 a a b -f- 3 a a. c c — 3 ax x -\- 
 3 a b b c — 6 ab b — a be c -f- b x x — b b b c -\- 
 ax-\-c, by Qab-\-acc — x x -\-b b c. 
 
 17. Divide — 16 a a m y — 3aacy-\-Q4amm 
 -\- 12 a c mby 16 a m -\- 3 a c. 
 
 18. Divide 12 a a b — 8 a — 144 a b + 96 by 
 a— 12. 
 
70 FIRST LESSONS IN ALGEBRA. 
 
 CHAPTER VI. 
 FRACTIONS. 
 
 SECTION I. 
 
 Introduction and Definitions. 
 
 1. Let it be required to distribute a dollars equally 
 among b poor persons. What will be the share of 
 
 each ? Ans. $ - . 
 
 The number of dollars must be divided by the 
 number of persons ; but, as the division cannot be 
 actually performed, all that can be done is, to repre- 
 sent it as above. So, too, if we suppose a = 3, and 
 b = 5, we must indicate the answer in a similar man- 
 ner ; thus, f . 
 
 These expressions, -, 7, and others like them, are 
 
 called Fractions, from a Latin word, which signifies 
 broken, because the value of a fraction is always ex- 
 pressed in parts of a whole one. If, for instance, we 
 cut an apple into 5 equal parts, and give 3 of those 
 parts to a boy, he will have f of the apple, which ex- 
 pression is read three fifths. Other fractions are read 
 in a similar manner ; as, ±, one half; f , two thirds ; f , 
 three fourths ; f , five sixths ; f-, seven eighths, &c. 
 
FRACTIONS. 71 
 
 The number below the line, which shows into how 
 many parts the unit or quantity is divided, is called the 
 Denominator. 
 
 The number above the line, which shows how many 
 of the parts are taken, is called the Numerator. 
 
 In the fractions j,^, - , - , ££ -, and - , the nu- 
 merators are a, m, 6, 19, a -|- b, and 3 ; and the de- 
 nominators are b, n, 7, 4, x — y, and z. 
 
 A Proper fraction is one whose value is less than a 
 unit ; that is, whose numerator is less than its de- 
 
 2 3 7 a x~-y „ 
 
 nom.nator;as,-, -, -, — , — ,..&c. 
 
 An Improper fraction is either equal to or greater 
 
 -, .~j 2 14 49 a ah ab~\-b s T . 
 
 than a unit; as, - , -, -, -, — , — -— , &c. It 
 
 is evident that the division here represented can be 
 performed, either wholly or in part. 
 
 Expressions consisting of a whole number and a 
 
 fraction, are called Mixed numbers ; as, 2 — , and b -f- - ; 
 
 and 2 and b are called integers, or integral quantities. 
 It often happens, both in Arithmetic and Algebra, 
 that there is a remainder after division, which should 
 be written above the divisor, and annexed to the quo- 
 tient in the form of a fraction. Hence the origin of 
 mixed numbers. [See Chap. V. Sec. III. and IV.] 
 
 Since fractions always imply division, any quotient 
 may be expressed in the form of a fraction, the divi- 
 dend being the numerator, and the divisor the denom- 
 inator. 
 
72 FIRST LESSONS IN ALGEBRA. 
 
 Express the answers of the following examples in 
 the form of fractions: 
 
 2. Divide 476 by 19. Ans. ™ . 
 
 3. Divide 6 a b c by d. 
 
 4. Divide 8 a -f- b — xby c — d. 
 
 5. Divide 17 x y — ■ m -f- 7 by x -f 10. 
 
 6. Divide a — m -f 7 — 5 y by 1.0 6 — x + 14. 
 
 As the value of any quantity is not altered when 
 it is divided by a unit, we can convert an integer into 
 a fraction by making 1 the denominator. ( 
 
 Convert the following quantities into fractions : 
 
 7.)8 = {. 8.)« = f. 9.)a + b= a -±± 
 
 10.) 24 11.) a — b. 1%) a + b—mn. 
 
 The numerator of a fraction being a dividend, and 
 the denominator a divisor, it follows, that when these 
 are equal to each other, as f , the value of the frac- 
 tion is 1 ; for 2 divided by 2 gives- 1. Therefore 
 
 12345a , ab „ , \ . 
 
 ~ > V > T y T > T > H > and T h > are a11 ec l ual to each 
 
 1 7 2'3'4 7 5'« 7 . ab 
 
 other, the value of each fraction being unity, or 1. 
 Hence, it is evident, that if both the numerator and de- 
 nominator be either multiplied or divided by the same 
 number or letter, the value of the fraction is not changed. 
 On this principle are founded the rules for bringing 
 fractions to a common denominator, and for reducing 
 them to their least terms. 
 
 13. Express 6 in the form of a fraction, having 4 
 
 for its denominator. Ans. —. 
 
 4 
 
 It is evident that the numerator must be 4 times 
 
FRACTIONS. 73 
 
 6 ; for, when divided by 4, the denominator, the quo- 
 tient must be 6. 
 
 14. Given the quantity x to be expressed in the 
 form of a fraction, having y for a denominator. 
 
 Ans. — . 
 y 
 
 We may first convert x into this fraction, — , and 
 
 then multiply both terms by y, according to the prin- 
 ciples already given. In other words, to change a 
 whole quantity to a fraction, having a given denomi- 
 nator, we multiply the whole quantity by the given de- 
 nominator ; and the product is the numerator of the 
 fraction required. 
 
 15. Change 8 to a fraction, having 5 for its de- 
 nominator. 
 
 16. Change a -f- b to a fraction, having c for its de- 
 nominator. 
 
 17. Express x -f- y in the form of a fraction, having 
 5 z for its denominator. 
 
 18. Change a — 5 to a fraction, having 2 -f- b for 
 its denominator. 
 
 19. Change 5 a — b c to a fraction, whose denom- 
 inator shall be 4 a:. 
 
 20. Express x -J- b — 5 in the form of a fraction, 
 having 3 a — ■ z for the denominator. 
 
 21. Change 5a — 2 -{- y to a fraction, having 6 
 for its denominator. 
 
 22. Change 28 — x -f- 4 y to a fraction, whose de- 
 nominator shall be 2 a. 
 
 23. Express 2 a b -f- x — 5 in the form of a frac- 
 tion, having 5 a b for its denominator. 
 
 fl 
 
1A FIRST LESSONS IN ALGEBRA- 
 
 SECTION II. 
 
 Reduction. 
 
 It is often convenient, and sometimes necessary, to 
 change the form in which a fraction or a mixed quan- 
 tity is expressed. For instance, to prepare different 
 fractions for addition or subtraction, we must always 
 express them in other fractions, which shall have a 
 common denominator. The process, by which the 
 form of a fraction is changed, its value remaining the 
 same, is called Reduction. 
 
 A. To reduce an Improper Fraction to a Mixed 
 Quantity. 
 
 1. What is the value of f ? Ans. 1. 
 
 2. What is the value of *£1 Ans. 3|. 
 As 5 fifths are equal to 1, 19 fifths must be equal 
 
 to 3, and 4 fifths more. 
 
 3. What is the value of — 1 Ans. b. 
 
 a 
 
 4. What is the value of ? Ans. b -f- — . 
 
 To reduce an improper fraction to a whole or 
 mixed quantity, Divide the numerator by the denom- 
 inator, and annex the remainder , if any, to the quo- 
 tient, in the form of a fraction. 
 
 Reduce the following fractions to whole or mixed 
 quantities : 
 
 «.)=. 6.)|. 7.)f. a)f. 
 
FRACTIONS. 75 
 
 9.)t- W-)*TT' HO °4^- 12.) '4^ 
 
 7 a ' b ' a-\-b ' 6 x 
 
 13. What is the value of * ab ~- bb l 
 
 a 
 
 14. What is the value of aaa ~ — ? 
 
 a — b 
 
 15. Reduce - — - to a mixed quantity. 
 
 a — b 
 
 B. To reduce a Mixed Quantity to an Improper 
 Fraction. 
 
 16. Reduce 7§ to an improper fraction. 
 
 Ans. ffr* 
 In 1 there are 3 thirds, and in 7 there are 7 times 
 3, or 21 thirds ; and 2 thirds added to 21 make 23 
 thirds. 
 
 17. Reduce b -\- — to an improper fraction. 
 
 Ans. — — . 
 
 a 
 
 The integral quantity, 5, reduced to a fraction, be- 
 comes — ; [See Sec. I.] and the whole quantity is 
 
 a — -}- — , or , which expresses the same value. 
 
 Hence, to reduce a mixed number to an improper 
 fraction, multiply the integer by the denominator, and 
 add the numerator to the product. The sum will be 
 the numerator of the fraction required. 
 
 Reduce the following mixed numbers to improper 
 fractions : 
 
 18.) 9 1. 19.)« + f. 20.)xy + W + j- 
 
76 FIRST LESSONS IN ALGEBRA. 
 
 24.) a — 12 -f b -f -^- . 25.) m — x-\-b 
 
 b ' ' ' Z» + 14 
 
 26. Reduce a -{- 6 -| — — to an improper fraction. 
 
 27. Reduce a -f- ~4ta to an * m P r0 P er fraction. 
 
 C. To reduce Fractions to a Common Denominator 
 
 3 2 
 
 28. Reduce — and — to a common denominator. 
 
 4 5 
 
 3 15 , 1 8 
 
 AN S . T = -,and-^- 
 
 We are here required to find two other fractions 
 having the same denominator, which shall be equiva 
 lent to f- and f . If both terms of the first fraction, 
 f , be multiplied by the denominator of the second, it 
 becomes jft ; and if both terms of the second frac- 
 tion, f , be multiplied by the denominator of the first, 
 it becomes ^. 
 
 12 3 
 
 29. Reduce — , — and — to a common denomma- 
 
 2 7 3 4 
 
 A 1 12 2 16 , 3 18 
 
 tor. Ans. - = -, - = -, and - = -. 
 
 We multiply both terms of each fraction by the de- 
 nominators of the other fractions. 
 
 30. Reduce — and — to a common denominator. 
 
 A a ay , x bx 
 
 Ans. — = — , and — = -—-<« 
 
 b by' V by 
 
 Hence, to reduce several fractions, having different 
 denominators, to other fractions of equal value with 
 a common denominator, we have the following Rule* 
 
 For the numerators : Multiply each numerator by 
 all the denominators except its own. 
 
FRACTIONS. 77 
 
 For the common denominator : Multiply all the de- 
 nominators together. 
 
 12 7 
 
 31. Reduce—, — and — to a common denominator. 
 
 32. Reduce—, — and — to a common denominator. 
 
 ha j 
 
 33. Reduce — and — to a common denominator. 
 
 a 5b 
 
 34. Reduce — , - and - to a common denominator. 
 
 b ' c a 
 
 35. Reduce — , — and y to a common denominator. 
 
 36. Reduce — , -j , -^ and °— — to a common de- 
 nominator. » 
 
 37. Reduce - and ^ to a common denominator. 
 
 38. Reduce x. — and 8 to a common denominator. 
 
 D. To reduce Fractions to their Least Terms. 
 
 As small numbers are more convenient to work 
 with than large ones, a fraction should always be kept 
 in its least terms. For this purpose, the following 
 general Rule will be found useful : Divide both the 
 numerator and denominator by any quantity which 
 will divide all the terms of both without a remainder. 
 Experience will suggest various expeditious ways of 
 applying this principle. 
 
 Reduce the following fractions to their lowest terms : 
 
 on \ 127 X An. \ ab b /M \ Sax l 
 
 7* 
 
78 FIRST LESSONS IN ALGEBRA. 
 
 40 \ ax + a v _ x +v 40 \ 11? 
 
 ^ - v 3arw . ,,_, \ 2aa<& — 4aa.z 
 
 44.) 2 — 45.) 
 
 ' 9ax + 27*?/ ' 6 aaa-|-10aai/ 
 
 ~ v 24aa — 12 ~ v 8 a x y — 4 ay^-l6aay^ 
 
 '' 16a?/ + 20* ' 12o*y — 20ay + 24a*v« 
 
 48. Reduce a , y ~"*^ to its least terms. 
 
 a 9T c ?T flJl: 
 
 49. Reduce g? ",/,.; "/ to its least terms. 
 
 20 az — 10 am-\~5ab 
 
 SECTION III 
 
 The Signs of Fractions. 
 
 The learner will remember, that the numerator is s 
 dividend, that the denominator is a divisor, and that 
 the value of the fraction is to be found in the quotient. 
 The signs prefixed to the several terms of a fraction, 
 affect those terms only ; but when a sign is prefixed 
 to the fraction itself, it affects its whole value, that is, 
 the value of all the terms taken collectively. With- 
 out regarding the signs, we know that the value of 
 
 the fraction a — is b. Now, by the common rules of 
 
 division, we have the following cases : 
 
 — ab , 
 
 — a 
 
FRACTIONS. 79 
 
 . Whence we infer, that the signs of both the nume- 
 rator and the denominator may be changed without al- 
 tering the value of the fraction ; and that the value is 
 altered when the sign of either is changed, and not 
 that of the other. 
 
 an 
 
 1. Let it be required to reduce lb ^T to 
 
 c . 14a£ — c-\-d 
 
 improper traction. Ans. 2a . 
 
 In this example, the value of the fraction -y^- is to 
 
 be subtracted from the integral part, 7 b. For the 
 sign — , prefixed to the fraction, does not apply to the 
 first term of the numerator, nor to the denominator, 
 but to the value of the whole fraction. If we reduce 
 the integral part to the form of a fraction, we shall 
 
 have -j^ Ya~ ' ^et us now re J ect tne denomina- 
 tor, 2 a, and subtract the numerator, c — d, from 
 14 a b, as if they were integral quantities ; and we 
 shall obtain 14 a b — c -\- d, by the rules for subtrac- 
 tion. It only remains to restore the common denom- 
 inator, an^ we have the answer given above. 
 
 c-k d 
 
 2. Reduce lb j^- to an improper fraction. 
 
 • I4>ab — c — d 
 
 ANS - 2a • 
 
 Here the quantity c -f- d is subtracted from 14 a b, 
 and both signs are changed, as before. 
 
 c d 
 
 3. Let 7 b -{- yt ^ e reduced to an improper frac- 
 
 A 14 a b -f- c — d 
 
 tion. Ans. ^ • 
 
80 FIRST LESSONS IN ALGEBRA. 
 
 4. Let 7 6+ lu~ ^ e re duced to an improper frac- 
 
 . Uab + c + d 
 
 tion. Ans. g^ . 
 
 In the last two examples, as the value of the frac- 
 tion c ~^- in one, and of -j^- in the other, is to be 
 added to lb, none of the signs are changed. 
 
 5. Reduce 3 a — to an improper fraction. 
 
 6. Reduce x y ^- to an improper fraction. 
 
 7. Reduce a b + 3m to an improper fraction. 
 
 8. Reduce 8 b x + m ^ a v to an improper fraction. 
 
 6 x — 12 v 
 
 9. Reduce 9 m ^— - to an improper fraction. 
 
 10. Reduce x + ^- to an improper fraction. 
 
 SECTION IV. 
 Addition. 
 
 1. What is the sum of — and — ? Ans. — . 
 
 As these fractions have a common denominator, we 
 evidently obtain the true answer by adding together 
 their numerators, which are similar quantities ; and 
 3 a + 2 a =. 5 a. For, if a =. 4, and c = 2, then 
 
 2a 8 . ,3a 12 - , . ' **'■ ' 
 
 — = -, or 4 ; and — = -, or 6 ; and 4 + 6 = 10 : 
 but -^ z= — , or 10, also. 
 
FRACTIONS. 81 
 
 2. Add together -£- , -£- , ~ and £■ 
 
 o txri . • -i r a H"** 3a-4-2£ , 4a — 6£ -, 
 
 3. What is the sum of - J — , — — — and ? 
 
 x — y J x — y x — y 
 
 a a 1 1 x xi x — y-\r z %y — z — % x i 2x — y-\-z 
 
 4. Add together —^- , — ^— - and -^- . 
 
 5. What is the sum of •?. and - ? Ans. ^— — • 
 
 c c c 
 
 Although these fractions have a common denom- 
 inator, still, as the numerators are dissimilar, they can 
 only be added by means of the sign -{-. Let fl = 4, 
 
 b = 6, and c = 2; then (| + |) = (£+{) = 
 2 + 3, or 5 ; also — - = -5- = — , or 5. 
 
 6. 
 
 What is the sum of 
 
 a 
 
 b c , 2d n 
 *> * and lT ? 
 
 
 7. 
 
 Add together 
 
 2a 
 
 X'—y y 
 
 3b 3a 
 x — y ' x — y 
 
 ***£}• 
 
 
 8. 
 
 Add together 
 
 3a — I 
 
 x+1- 
 
 i 
 
 c + 2b 
 x+l — z 
 
 , a — c — 
 &nd * + 7- 
 
 -b 
 
 •z 
 
 9. 
 
 What is the 
 
 sum 
 
 of 
 
 x + y 
 
 e 3A+18 + t 
 
 » x+y 
 
 and 
 
 «_4,£ — 10 — 3c n 
 
 
 
 
 
 
 10. What is the sum of y and -j ? Ans. ^ • 
 
 These fractions having different denominators, they 
 can only be added, as they are proposed, by means 
 of the sign -f- 5 we, therefore, reduce them to a com- 
 mon denominator, and proceed as before. 
 
 From these examples and observations, we derive 
 the following Rule for adding algebraic fractions : 
 Reduce the given fractions to a common denominator; 
 
82 FIRST LESSONS IN ALGEBRA. 
 
 add their numerators together; and place the sum, for 
 a new numerator, over the common denominator. 
 
 11. Add together ^, ^ and -^. 
 
 5 « X b X cd—5abcd,-> 
 
 
 
 be 
 
 
 
 3 c # 3 X be X cd = 3bccd,y the 
 be~ —'J | 
 
 numerators. 
 
 4>b X bcX b = 4<bbbc,J 
 
 
 cd X be X b =b b c c d r the 
 
 denominator. 
 
 _,, „ ha 5 abed 5 ad 
 
 Therefore, bc = bbccd = bcd ; 
 
 
 3c 3bccd 3cd 
 
 
 bc b bc cd bed/ 
 
 
 4Z» 4>bbbc bbb 
 
 
 cd ~~~bbccd bed' 
 
 
 We obtain the values of the fractions given in the 
 last column, by dividing all the terms by b c. The 
 fractions are thus reduced to their least common de- 
 
 . 5ad + 3cd + 4,bb 
 
 nominator. Ans. j^j 1 . 
 
 12. What is the sum of -g| , -^ and ^ ? 
 
 13. Add together — , —and — . 
 
 14. Add —^— and ^ together. 
 
 15. Add« + ^to5« + 3 ^. 
 
 16. What is the sum of 6 x, -~- $ 5, and ? 
 
 i>* a 11 ^ 1x4-1 3.r + 5 . _, , . 
 
 17. Add — , — |— , — - — and 2 x -f- y together. 
 

 FRACTIONS. 83 
 
 18. 
 
 U j 3a — 2 4a + 7 , , a — b 
 
 Add a: — ,3 a? — , and b -\ jp 
 
 getl 
 
 ler. 
 
 19. 
 
 Add together « 6, 15, —^ and — . 
 
 20. 
 
 Add together j- , -r , 4 a? and — . 
 
 21. 
 
 13 5 a: 
 
 Add together — a?, and — a;. Ans. — . 
 
 It is evident that — a? expresses the same value as 
 — , and that — a? is equivalent to — ; and — -f- — 
 
 5x u 1 
 
 = T ,or 1-*. 
 
 12 3 5 
 
 22. Add together — a?, and — x, and — a?, and — x. 
 
 12 1 
 
 23. What is the sum of — «, — a, and - a ? 
 
 4 3 Z 
 
 2 5 3 1 
 
 24. Add together — ab,— ab, — ab, — a b, and 
 
 25. Add together — (« — a?), — (a — a?), and 
 { (a — x). 
 
 SECTION V. 
 
 Subtraction. 
 
 Subtraction is performed like addition ; but the 
 signs of the quantities to be subtracted must all be 
 changed. 
 
84 FIRST LESSONS IN ALGEBRA. 
 
 1. Subtract Sg from 5 -^. Ans. ^ 
 
 2. Subtract — from — . 
 
 y v 
 
 3. Subtract — from — . 
 
 b a 
 
 4. Subtract -^ from -^ . 
 
 5. Subtract x + — p from 3 2/ + ■*-- . 
 
 J: ~ , 3a4-5 r 7a 
 
 6. Subtract — J- from j* 
 
 7. Subtract -~T from -^py . 
 
 8. Subtract^ from ^. 
 
 9. Subtract ^ from a ■ — ^ . 
 
 10. Subtract 6a — ^y- from 9 a — ^- . 
 
 1 1 . Subtract 3 y from x — -7* . 
 
 12. Subtract 4 y from 5 y — -^ — . 
 
 13. Subtract^ from S^±a. 
 
 4 « 
 
 14. From — r— . take 
 
 a -f- M 6 * 
 
 15. Take ^from^. 
 
 16. From b a x take — . 
 
 17. From a -f- b take -j- . 
 
 18. Take a — ^ from x — 
 
 27 x 
 
 19. Subtract a — b from ^-r^ 
 
> FRACTIONS. 85 
 
 section yi. 
 
 Multiplication. 
 
 1. If a man gives to each of 3 beggars £ of a dol- 
 lar, what does he give to the whole ? 
 
 Ans. £ of a dollar. 
 For i + I + i, or i x 3 = f . If we substitute 
 
 the algebraic fraction — for £, and let c represent the 
 number of beggars, the process will be the same : 
 thus,— x c = — -. That is, we multiply the nume- 
 rator by the integral quantity. 
 
 2. Multiply 2^ by 4 x. 
 
 Ans --*f. 
 
 3. Multiply < ~ r by 4 x y. 
 
 
 a— —2 
 
 4. Multiply ^-t^ by a; — 2. 
 
 
 5. Multiply — ^ — by 5 x y. 
 
 
 6. Multiply £* t by — hay. 
 
 
 7. Multiply £ b y &• 
 
 Ans. f . 
 
 In this example, instead of multiplying the nume- 
 rator, which would give ^j, we divide the denomina- 
 tor. In general, to multiply a fraction by an integral 
 quantity, we divide the denominator, when we can do 
 so without a remainder ; but when there would be a re- 
 mainder, we multiply the numerator. 
 8 
 
86 FIRST LESSONS IN ALGEBRA. 
 
 When a fraction is multiplied by a quantity equal to 
 its denominator, the numerator is the product ; for, in 
 that case, it is both divided and multiplied by the 
 same quantity. Thus, JX 2= 1; }X4=B; 
 j x b — a ; £~ X « + 3 = x — z, &c. 
 
 8. Multiply f by 4." Ans. § = l£. 
 
 9. Multiply ^ by 2 6. Ans. J§ = f . 
 
 10. Multiply 2T ^^- by 7. 
 
 11. Multiply j-^i^ by 5 6. 
 
 12. Multiply -£^h b y — * 
 
 13. Multiply j by y. Ans. £. 
 If we multiply - by y, that is, by c, we obtain the 
 
 product — . But, as we are required to multiply by 
 
 the quotient of c divided by d, and not by the whole 
 value of c, this product is evidently too large, and 
 must be divided by d. To multiply a fraction by a 
 fraction, therefore, we multiply the numerators for a 
 new numerator, and the denominators for a new denom- 
 inator. 
 
 The value of a Compound Fraction, which is the 
 fraction of a fraction, as % of f , is found by this rule. 
 Thus, £ of £ is f ; and^ of £ is •&. 
 
 I 14. Multiply H by *-f. 
 
 15. Multiply"-^ by ^. 
 

 
 
 FRACTIONS 
 
 16. 
 
 Multiply 
 
 a — b 
 V 
 
 h y b^i- 
 
 17. 
 
 Multiply 
 
 a + b 
 3+~* 
 
 b y S5i« 
 
 18. 
 
 Multiply 
 
 ah — 
 4 
 
 3 , 4c 
 
 b y „-.*• 
 
 p 
 
 87 
 
 19. Multiply ^, by *=5. 
 
 20 . Multiply 3 -^=± by & 
 
 % SECTION VII. 
 
 Division. 
 
 1 . A man distributed f of a dollar equally among 
 3 beggars. What did he give to each ? 
 
 Ans. £ of a dollar. 
 
 One third part of three apples, three dollars, three 
 
 units, or three fourths, is evidently one apple, one 
 
 3 — 3 1 
 
 dollar, one unit, or one fourth ; that is, -r ■" = — . 
 So, too, if ^j- dollars be equally distributed among 
 
 c persons, each one will have — - ' == - dollars. Here, 
 
 to obtain the quotient, we divide the numerator by 
 the integer. 
 
 2. Divide -~ by 3 a. Ans. — . 
 
 3. Divide ^ by a + y. 
 
 4. Divide r~ by 4 y. 
 
88 FIRST LESSONS IN ALGEBRA. 
 
 5 ' Divide ^Ty h Y 7 « 
 
 6. Divide 16 a a ~*x h Y 4 * 
 
 7. A man divided ^ a dollar equally between 2 
 persons. What was the share of each ? 
 
 Ans. £ of a dollar. 
 Each person has £ of the sum given ; and ^ of ^ 
 is £. Here, the denominator is multiplied by the in- 
 teger. In general, to divide a fraction by an integral 
 quantity, we divide tfie numerator, when we can do so 
 without a remainder ; and when we cannot, we multiply 
 the denominator. 
 
 8. Divide ~ by b. Ans. y. 
 
 9. Divide ££3j by 3 x. 
 
 10. Divide 32x a ~\ bx by 8 *.. 
 
 11. Divide ^by3 6. 
 
 12. Divide —^ by 3 y. 
 
 13 . Divide 8 -^* by 4. 
 
 9 3 
 
 14. Divide — by — . Ans. 3. 
 
 As these fractions have the same denominator, we 
 divide the numerator of the dividend by the numera- 
 tor of the divisor. It is clear that T 9 ^ contains ^ 
 3 times. 
 
 15. Divide j by j. Ans. ^ . 
 
FRACTIONS. 89 
 
 We first reduce the fractions to a common denom- 
 inator, and we find that 4" = rr , and -r = rr i and 
 
 b b d ' d Ad ' 
 
 we then divide the numerator of the one by the nu- 
 merator of the other, which gives us r-. But if we 
 
 multiply the numerator of the dividend by the de- 
 nominator of the divisor, and the denominator of 
 the dividend by the numerator of the divisor, we 
 
 shall obtain the same result: thus, — X — = rr. 
 
 ' i) c ft 
 
 Hence, to divide one fraction by another, we invert 
 the divisor, and proceed as in multiplication. 
 
 Ans. -rr • 
 
 ox 
 
 16. 
 
 Divide 
 
 2a , 
 
 Tx b y 
 
 b 
 3» * 
 
 
 17. 
 
 Divide 
 
 3a — b 
 c 
 
 by- 
 
 b 
 
 18. 
 in 
 
 Divide 
 
 B(a — b) 
 
 xy 
 
 4>ab 
 
 by 
 
 z{a + b) 
 xy 
 
 yz 
 
 3.*,. "Mn,v x (a + 8) "J 8(8m-«)' 
 
 20. Divide &±i by §=f . 
 
 •* — y J y + 4 
 
 21. Divide ~~ by a z. 
 
 x -\- y J 
 
 22. Divide x y by "-ffEhjT"' 
 
 23. Divide =^ by ^. 
 
 24. Divide ^- by ^-. 
 
 25. Divide - by -. 
 
 26. Divide ^ by ^. 
 
 8* 
 
90 FIRST LESSONS IN ALGEBRA, 
 
 CHAPTER VII. 
 POWERS. 
 
 SECTION I. 
 Involution of Simple Quantities. 
 
 1. What is the product of a a multiplied by a? 
 
 Ans. a 3 . 
 
 The answer to this question, obtained by the com- 
 mon process of multiplication, is a a a. Instead of 
 repeating the letter, we indicate the number of times 
 it occurs in the product, by the small figure placed 
 above it on the right hand. This figure is called its 
 Exponent. It shows the Power of the letter ; that 
 is, how many tithes it is used as a factor in the multipli- 
 cation. When a letter has no number annexed, the 
 exponent is always a unit, or 1, and the letter is said 
 to be the first power or Root ; thus, 
 
 a is the root, or first power ; 
 a X a = a 2 is the second power ; 
 a X a X a =z a 3 is the third power ; 
 aX«X«X«==a 4 is the fourth power, &c. 
 
 The second power is sometimes called the square ; 
 the third power, the cube ; and the fourth power, the 
 
POWERS. 91 
 
 biquadrate. If we suppose the value of a to be 3, 
 a =. 3, the first power; 
 a 2 — 3 2 , or 3 X 3 = 9, the second power; 
 a 3 = 3 3 , or 3 X 3 X 3 5= 27, the third power ; 
 a 4 = 3 4 , or 3 X 3 X 3 X 3 = 81, the fourth power, &c. 
 
 The coefficient and the exponent of a quantity, be- 
 ing very different things, must not be confounded to- 
 gether. Thus, the values of 5 a? and x 5 are far from 
 being equal. Let the value of x be 6 ; 
 then 5# = 5x6=:30; 
 but x 5 = 6 X 6 X 6 X 6 X 6 = 7776. 
 It is evident that we raise a single letter to any pro- 
 posed power by giving it the exponent of that power. 
 
 2. What is the sixth power of x ? the fourth power 
 of d ? the fifth power of c ? the eighth power of a 1 
 the seventh power of x ? 
 
 3. What is the product of a b multiplied by a b; 
 that is, what is the second power of a b ? 
 
 a b X a b = a a b 6, or a 2 6 2 . 
 Here, the quantity to be raised consists of two fac- 
 tors, a and b ; and the required power is expressed by 
 the same factors, with the exponent of that power writ- 
 ten above each. 
 
 4. What is the fourth power of x y 1 
 
 5. Raise a b c to the sixth power. 
 
 6. What is the ninth power of a b x? 
 
 7. Involve m n y to the seventh power. 
 
 8. What is the square of 4 a bl Ans. 16 a 2 6 2 . 
 It will be remembered, that the square, or second 
 
 power of any quantity, is the product of that quantity 
 
92 FIRST LESSONS IN ALGEBRA. 
 
 multiplied by itself; consequently, the second power 
 of 4 a b is 4 a b X 4 a b = 16 a a b b, or 16 a 2 6 9 . 
 
 Hence, coefficients must be raised to any required, 
 power, by actual multiplication. 
 
 9. What is the third power of 5 a x 1 
 
 10. What is the fourth power of 7 a b c 1 
 
 1 1 . Raise 6 x y z to the fifth power. 
 
 12. What is the fourth power of — 2 a b c? 
 
 — ■ 2 a b c, first power. 
 ~ 2 ab c 
 
 -f 4 a 2 6 2 c 2 , second power. 
 . — 2 a b c 
 
 — ■ 8 a 3 b 3 c 3 , third power. 
 
 — 2 a b c 
 
 + 16 a 4 6 4 c 4 , fourth power. 
 Hence.it appears, that when the root or first power 
 is a negative quantity, the odd powers are negative, 
 and the even powers are positive. 
 
 13. What is the second power of a 3 ? Ans. a 6 . 
 Here, the quantity to be involved is already a 
 
 power, and the exponent is multiplied by the exponent 
 of the power proposed; thus, o 3X2 == a 6 ; for a 3 =3 
 a a a, and a a a X a a a = a 6 . 
 
 14. What is the third power of a P ? Ans. a 3 b 9 . 
 
 15. What is the second power of a 3 b x 2 ? 
 
 16. Raise a 2 b 3 x to the third power. 
 
 17. What is the fourth power of m 4 x 2 y b ? 
 
 18. Involve 6 a 2 c 3 x A to the second power. 
 
 19. What is the third power of 3 x 2 y 3 z 4 ? 
 
 20. Required the fourth power of 4 a 2 x 3 y 2 z. 
 
POWERS. 
 
 93 
 
 SECTION II. 
 
 Involution of Compound Quantities. 
 
 1. It is required to find the fifth power of the bi- 
 nomial quantity a -\- b. 
 
 a -\- b, first power. 
 a +b 
 
 « 2 -j- a b 
 
 a 2 -j- 2 a b -f- 6 2 , second power. 
 a +6 
 
 a 3 + 2o 2 H a 6 2 
 
 a 2 6 -j- 2 a J 2 _|_ J3 
 
 a 3 + 3 a 2 6 -f 3 a 5 2 + 6 3 , third power. 
 a +b 
 
 a 4 _|_ 3 fl 3 5 _j_ 3 a 2 52 _|_ a £3 
 
 a 3 5 + 3 a 2 6 2 + 3 a b 3 + 6 4 
 
 «4 _f_ 4 a 3 5 + 6 a 2 6 2 + 4 a b 3 + 6 4 , fourth power. 
 a +6 
 
 a 5 _j_ 4 a 4 J _j_ g a 3 Q 2 _|_ 4 a 2 53 _|_ ft 64 
 
 «4 J _|_ 4 ft 3 J2 _f_ 6 a 2 6 3 + 4 a ¥ -f- 6 5 
 a 5 + 5a 4 6 + 10 a 3 6 2 + 10 a 2 5 3 + 5 a b 4 + b 5 , Ans. 
 
 The powers of compound quantities are obtained by 
 actual multiplication. These powers, however, are 
 often expressed by means of an exponent. Thus, 
 (a -f- b -f- c) 3 indicates the third power of a \-\- b -f- c 
 
94 FIRST LESSONS IN ALGEBRA. 
 
 2. It is required to find the fifth power of the re- 
 sidual quantity a — • 6. 
 
 a — b, first power. 
 a — b 
 
 a? 
 
 — ab 
 
 — ab + b* 
 
 secon 
 
 
 a 2 
 a 
 
 — 2 a b + 6 2 , 
 
 — b 
 
 d power. 
 
 fl3 
 
 — 2 a 2 6 + 
 
 — a 2 6 + 2 
 
 a 6 2 
 
 -6 3 
 
 a3 
 a 
 
 — 3 a 2 6 + 3 
 
 a6 2 - 
 
 - 6 3 , third power. 
 
 4 
 
 — 3 a 3 6 + 3 
 
 — ; ; a 3 6 + 3 
 
 a 2 6 2 - 
 a 2 6 2 - 
 
 - a6 3 
 
 - 3 a 63 + J4 
 
 a 4 
 a 
 
 — 6 
 
 a 2 6 2 - 
 
 - 4 a 6 3 + 6 4 , fourth p 
 
 a* 
 
 _4a 4 6 + 6 
 
 — a 4 6 + 4 
 
 a 3 6 2 - 
 « 3 6 2 - 
 
 - 4 a 2 6 3 + a 6 4 
 -6a 2 6 3 + 4a6 4 — 6 5 
 
 a 5 _ 5 a 4 j +10 a 3 6 2 — 10 a 2 6 3 + 5 a 6 4 — 6 5 , Ans. 
 
 3. What is the second power of 3 a — 6 + 2 c ? 
 
 3 a — 6 + 2 c 
 3 a — 6 + 2 c 
 
 9 a s — 3 a 6 + 6 a c 
 
 — 3 a 6 + 6 2 — 26c 
 
 + 6 a c — 2 6 c + 4 c 2 
 
 9 a 2_ 6 a 6 + 12 a c + 6 2 — 4 6 c + 4 c 2 . 
 
POWERS. 95 
 
 4. What is the third power of a -f- 1 ? 
 
 5. Raise x -\- y -f- ar to the second power. 
 
 6. Involve a + ^ + c + ^ to tne second power. 
 
 7. What is the third power of 2 a — x -j- c 2 ? 
 
 8. Involve 3 « -f~ 2 b to the third power. 
 
 9. What is the second power of a -\- b — c 1 
 
 10. What is the third power of a — x -f- 8 ? 
 
 11. What is the second power of ~? Ans. ^. 
 
 For - X -t = 7T, or ' — , according to the princi- 
 ples already explained. Therefore, we involve a frac- 
 tion, by raising both the numerator and denominator to 
 the required power. 
 
 12. What is the second power of -s-J Ans. a —— . 
 
 13. What is the second power of sr—j- ? 
 
 14. What is the second power of 3d2 ? 
 
 15. What is the second power of ^- ? 
 
 16. What is the third power of ^-? 
 
 3 x^ v 
 
 17. Raise -y^- to the third power. 
 
 2 a — 3 
 
 18. Required the second power of b2 . c . 
 
 19. Required the fourth power of ^r - - 
 
 20. Required the second power of — - - — . 
 
 21. What is the second power of ^-jrl 
 
96 FIRST LESSONS IN ALGEBRA. 
 
 SECTION III. 
 
 The Binomial Theorem. 
 
 The first and second examples of the last section 
 exhibit the method of involving binomial and residual 
 quantities, by actual multiplication. It is very appa- 
 rent that, when a high power is required, the process 
 must become tedious. From these operations, how- 
 ever, may be derived rules for raising such quantities 
 to any proposed power, without the intervention of 
 the lower powers. This method ©f involving quanti- 
 ties was discovered by Sir Isaac Newton, and is called 
 the Binomial Theorem. 
 
 When we would express any required power of a 
 given binomial or residual quantity, four things re- 
 quire attention ; namely, the number of terms in that 
 power, the signs, the exponents, and the coefficients. 
 
 The Terms. It will be observed, that the number 
 of terms in every power, is greater by one than the ex- 
 ponent of that power. Thus, the second power con- 
 sists of three terms ; the third power, of four terms ; 
 the fourth power, of five terms, &c. 
 
 The Signs. The corresponding powers of these 
 two quantities, a -f- b and a — • b, differ only in their 
 signs. All the terms of the binomial quantity are posi- 
 tive ; whereas the even terms of the residual quantity 
 are negative, and the odd terms positive. 
 
 The Exponents. The first term of every power 
 consists of the first letter of the given binomial, a, 
 
POWERS. 97 
 
 raised to that power ; and the other exponents of that 
 letter diminish by one towards the right. The first let- 
 ter of the binomial, a, is called the leading quantity ; 
 and its exponents for the seventh power, are 
 7,6,5,4,3,2,1. 
 
 The last term of every power is the last letter of the 
 given binomial, 6, raised to that power ; and the other 
 exponents of that letter diminish by unity towards the 
 left. The last letter of the binomial, b, is called the 
 following quantity ; and its exponents for the seventh 
 power, are 
 
 1, 2, 3, 4, 5, 6, 7. 
 
 The sum of the exponents of the two letters, is 
 always equal to the exponent of the power in which 
 they occur. 
 
 To apply these principles, let it be required to de- 
 termine the number of terms, the signs and expo- 
 nents of a -f- o and a — b, in the seventh power. 
 The number of terms will be eight, that is, one more 
 than the power proposed ; and their signs and expo- 
 nents, without the coefficients, will stand thus : 
 
 (a+by=za 1 +aSb+a 5 bz+a A b*+a*¥+ans-\-a¥+b'' 
 (a—b) 7 = a 7 — a 6 6+a 5 5 2 — a 4 b 3 +a 3 b 4 ~a? 6 5 + " ° 6 — ol 
 
 The Coefficients. It will be seen, by an inspec- 
 tion of the examples above referred to, that the coeffi- 
 cient of the first term is always 1, and that the coeffi- 
 cient of the second term is the exponent of the power. 
 The remaining coefficients may be found by the fol- 
 lowing Rule : If the coefficient of any term be multi- 
 plied by the exponent of the leading quantity in that 
 9 
 
98 FIRST LESSONS IN ALGEBRA. 
 
 term, and the product be divided by the number which 
 denotes its place from the left, it will give the coeffi- 
 cient of the next term. 
 
 The coefficients of a -\- b, raised to the seventh 
 power, are 
 
 _ 7X6 21 X 5 35X4 35x3 21 X 2 7Xl 
 ' V 2 ' 3 ' 4 ' 5 ' 6 ' 7 ' 
 
 1, 7, 21 , 35 , 35 , 21 , 7 , 1 . 
 
 If we prefix these coefficients to the terms of a -f- b 
 and a — b, already obtained, we shall have the sev- 
 enth powers of those quantities complete : 
 
 (a -j- by = a 1 + 7 a 6 b + 21 a 5 6 2 + 35 a* 6 3 + 
 
 35 a 3 ft 4 + 21 a 2 6 5 + 7 a ¥ + 6 7 . 
 (a — by = a 1 — 7 a 6 5 -f- 21 a 5 6 2 — 35 a* ¥ -f 
 
 35 fl 3 frt _ QJ a 2 £5 _J_ 7 a &6 _ £7. 
 
 It will be observed, that the coefficients are equal 
 in the first and last terms, also in the second and last 
 but one, the third and last but two, and so on. It 
 will be sufficient, therefore, in practice, to find the 
 coefficients of half the terms, if their number be even, 
 or of one more than half, if it be odd, and apply them 
 to the rest. 
 
 3. What is the fourth power of x -{- y 1 
 
 Ans. x 4 -f- 4 x 3 y -j- 6 x% y% -{- 4 x y 3 -f- y 4 . 
 
 4. Raise x — y to the sixth power. 
 
 5. What is the fifth power of a -\- cl 
 
 6. What is the eighth power of m — n ? 
 
 7. What is the fourth power of m -f- n? 
 
 8. Required the seventh power of x -f- y. 
 
 9. What is the ninth power of a -\- b? 
 10. Raise c -\- d to the tenth power. 
 
POWERS. 99 
 
 11. Required the thirteenth power of a — b. 
 
 12. Required the sixth power of x -f- z, 
 
 13. What is the seventh power of c — d 1 
 
 14. What is the third power of a + 3 b c? 
 
 This quantity, in its present form, can be involved 
 only by actual multiplication ; for the rules given 
 above require each term of the binomial to consist 
 of a single letter, whose coefficient must be 1. But 
 if the value of the second term, 3 b c, be represented 
 by some letter, a binomial will be formed, which can 
 be involved in the usual manner. 
 
 Suppose m, for instance, to be equivalent to 3 be; 
 then, instead of a -\- 3 b c r we have the binomial 
 a -f- »*, the third power of which is 
 
 a 3 + 3 a 2 m -f- 3 a ?« 2 + m 3 . 
 
 In the place of m, put down its value, and the re- 
 quired power will be obtained. We supposed m — 
 3 b c; therefore, m 2 =s 9 5 2 c 2 , and m 3 = 27 b 3 c 3 . 
 
 The first term is a 3 . 
 
 The second term is 3 a 2 m, or 3 a 2 X 3 b c = 9 a 2 b c. 
 The third term is 3 a m 2 , or 3 a X 9 6 2 c 2 = 27 a W- c 2 . 
 The fourth term is m? == 27 b 3 c 3 . 
 
 Ans. a 3 + 9 a 2 6 c + 27 a ¥ c 2 + 27 b 3 c 3 . 
 
 15. What is the fourth power of 2 a b — x? 
 Suppose m = 2 a b ; and involve m — x. The 
 
 fourth power of this binomial is 
 
 m 4 — 4 m 3 x -\- 6 m 2 a? 2 — 4 »i x 3 -f a; 4 . 
 
 In the place of m, use its value ; observing that 
 m = 2 a b, m? = 4 a 2 5 2 , w 3 = 8 a 3 5 3 , and w 4 = 
 16 a 4 6 4 . 
 
 Ans. 16 a 4 b 4 — 32 a 3 J 3 a;+24 a 2 6 2 a^ 2 — 3 a b x 3 +x 4 . 
 
100 FIRST LESSONS IN ALGEBRA. 
 
 16. What is the second power of 4 a b — 5 c 2 ? 
 Let m = 4 a b, and n = 5c 2 ; and, instead of the 
 
 given binomial, involve m -f- n. 
 The second power of m -f- n is 
 
 m 2 -f- 2 m » -f- 7i 2 . 
 Instead of m and w, put their respective values into 
 this power. Observe, m = 4 a b, and »i 2 = 16 a 2 6 2 ; 
 also n = 5 c 2 , and w 2 = 25 c 4 . 
 
 Ans. 16 a 2 6 2 + 40 a 6 c 2 -f 25 c*. 
 
 17. What is the third power of x -f- y -f- z ? 
 
 Let m = y -\- z, and involve x -}- m, which is equiv- 
 alent to a? -f- y + *• 
 
 The third power of x -j- m is 
 
 To obtain the required power, we must restore the 
 value of m. 
 
 m = y -f- z. 
 
 m 2 = y 2 + 2 y z + * 2 . 
 
 ra 3 -j/ 3 + 3y 2 2-f-3i/2 2 -j-^ 
 
 Ans. x 3 -\- 3 x 2 y -\- 3 x 2 z -\-3 x y 2 -\- 6 x y z -f- 
 
 3«2 2 + ^ 3 -j-3l/ 2 2 + 3^ 2 + 2 3 . 
 
 18. What is the second power ofa-{-b-\-c-\-d? 
 Let m = a -f- &, and w = c -|- d; and then involve 
 
 )n -|- n. 
 
 (« -f- n) 2 = w 2 -|- 2 « 7i -f- ?i 2 . 
 
 In this quantity, substitute the values of m and /t 
 
 respectively, and the required power will be complete. 
 
 m 2 = « 2 + 2 « 6 + J 2 . 
 n 2 ■= 6 - 2 _L.2cd + d 2 . 
 Ans. « 2 -f-2«^4-6 2 4-2 ac + 2b c + 2 a d ■+■ 
 2bd + c* + 2cd + d*. . 
 
POWERS. 101 
 
 In this way, any quantity whatever can be raised 
 to any required power, by means of the binomial 
 theorem. The given quantity must first be reduced to 
 a binomial, consisting of two single letters ; and after 
 it has been involved, the respective values of these let- 
 ters must be restored. 
 
 19. What is the second power of a — 3 b? 
 
 20. Required the third power of 5 a 2 -f- b. 
 
 21. What is the second power of 3 a 2 -f- 5 6 3 ? 
 
 22. What is the second power of a -f- b 2 — 7 ? 
 
 23. Required the third power of 2 # — y -J- z 2 . 
 
 24. What is the fifth power of a 2 — c — 2d? 
 
 25. What is the fifth power of 7 x 2 y 5 — 10 x?> z 2 ? 
 
 SECTION IV. 
 Addition. 
 
 1. What is the sum of a 2 and b 2 ? Ans. a 2 -f b 2 . 
 
 As the quantities a and b are dissimilar, any pow- 
 ers of these quantities must also be dissimilar ; and 
 they can only be added by means of the sign +• 
 
 2. What is the sum of a 2 and a 3 ? Ans. a 2 -\- oh 
 It is evident that different powers of the same let- 
 ter are dissimilar quantities, and must be added as 
 above. If we suppose the value of a to be 4, a 2 ~ 
 4 X 4, or 16; and a 3 =4 X 4 X 4 = 64 ; and 
 a 2 + a 3 — 16 + 64, or So! But 2 a 2 = 16 X 2, or 
 32; and 2 a 3 = 64 X 2, or 128; therefore, neither 
 2 a 2 nor 2 a 3 is the true sum of a 2 and a 3 . 
 
 9* 
 
102 FIRST LESSONS IN ALGEBRA. 
 
 3. What is the sum of 2 a 2 and 3 « 2 ? Ans. 5 a 2 . 
 For 2 a 2 = 2 X a 2 , and 3 a 2 = 3 X a 2 ; and 
 
 (2 + 3) X a 2 = 5 a 2 . Let the value of a be 4 ; then 
 2 a 2 = 2 X 4 X 4, or 32 ; and 3 a 2 =3 X 4 X 4, 
 or 48 ; and 32 + 48 = 80 ; but 5 a 2 == 5 X 4 X 4, 
 or 80 also. 
 
 Observe that the exponents are not changed, when 
 the powers of quantities are added together. 
 
 From all which we infer, that the same letters under 
 the same exponents are similar quantities, and must be 
 added by the common rules ; but different letters, or the 
 same letters under different exponents, are dissimilar 
 quantities, and must be added by means of the sign -{-. 
 
 4. What is the sum of a 3 and ft 3 ? Ans. a 3 -f- ^ 3 » 
 
 5. What is. the sum of a 3 and a 4 ? 
 
 % 6. Add together 2 a 2 , 3 a 2 , 4 a 2 , and 5 a 2 . 
 ?. What U the sum of 2 a 3 , a: 2 , 3 a 3 , and 5 a? 2 ? 
 8. What is the sum of a x*, b a? 2 , c x 2 , and d a? 2 ? 
 
 9.) 5a 3 ft + 3a 2 ftc — 7aft 3 10.) a 3 a? 2 -f-fty 2 
 
 — 8a 3 ft — 8a 2 ftc + 7aft 3 erf a? 2 +dy 2 
 
 9a 3 ft + 5a 2 ftc — aft 3 — 2« 3 ^- rfy 2 
 
 11. What is the sum of a (a? 2 — !/ 3 ) 2 , 3 a (# 2 — y 3 ) 2 , 
 and 5 a (a; 2 — y 3 ) 2 ? 
 
 12. What is the sum of 8 (a 2 — ft 3 ) 2 and4(a 2 — ft 3 ) 2 ? 
 
 13. Add together 5 a 2 ft c 3 , 3 a 2 ft c 3 , a 2 ft c 3 , and 
 
 2 a 2 ft 3 c. 
 
 14. Required the sum of 3 a 3 + ft c 2 , 5 a 3 -f- 2 ft c 2 , 
 a 3 + 5 ft c 2 , and 6 a 3 + 2 ft c 2 . 
 
 15. What is the sum of \ (a 2 — ft xf, f (a 2 — ft a?) 3 , 
 
 3 ( a a __ & ff)3 ? an d £ (a 2 — ft a:) 3 ? 
 
POWERS. 103 
 
 16. Add together a 2 b 3 -f 7 + # 2 , 3 a 2 63 -f 9, 
 * 2 -f 14, 3 a 2 6 3 — 19 — 5 a; 2 , 12 — 2 a 2 6 3 , and 
 4 a 2 6 3 — 8 + 2 a: 2 . 
 
 17. Required the sum of a 3 (6 2 — 12 a 3 -f y 4 ) 3 , 
 3 a 3 (J8 _ 12 a 3 + 2/4 )3 > 2 a 3 (# — 12 a 3 + y 4 ) 3 , 
 and 8 o 3 (6 2 — 12 a 3 -f y 4 ) 3 . 
 
 18. What is the sum of a; 2 y 3 + 14 — 13 6 2 , and 
 12 + 3 a? y 3 — 8 6 2 , and 15 6 2 — 20 — a: 2 y 3 ? 
 
 SECTION V. 
 
 I. What is the difference between a 3 and 6 2 ? 
 
 Ans. a 3 — 6 2 . 
 
 The questions in this section are solved like those 
 in the last. The signs of the quantities to be sub- 
 tracted must, however, be changed. 
 
 Observe, that the exponents remain unaltered in 
 subtraction, as in addition. 
 
 2. Subtract 2 a 3 from 5 a 3 . Ans. 3 a 3 . 
 
 3. From 5 (a 2 — b 3 ) take 3 (a 2 — b 3 ). 
 
 4. From 6 a 2 (a 3 — 6 2 + 3) 2 take 5 a 2 (a3_62_|_ 3 ) 2# 
 
 5. Subtract 3 a 5 b 3 from 7 a 5 b 3 . 
 
 6. Subtract 2 a 2 from 3 a 3 . 
 
 7. Subtract 3 a 3 from 5 a 3 . 
 
 8. Subtract 5 a 2 b from 2 « 3 a?. 
 
 9. Subtract 7« 2 i from 9 a 2 y. 
 
 10. Subtract 5 a 2 cc -f 3 d 2 x from 8 a 2 # — 4 d 2 *. 
 
 II. From 8 a 2 a? y — a? 3 take 5 a? x y -{- 2 x 3 . 
 
104 FIRST LESSONS IN ALGEBRA. 
 
 12. From 9 A 3 — l2+20a 2 ta 3 13. ) llaW+Zxf 
 Take 3 a^x 3 -\-16— 1 2« 2 ^ 3 . 1 2«% 3 — ixy^. 
 
 14. From 5(a 2 — 6 3 ) 2 -f ^ I5.)2x(a— &) 3 +(a— &) 3 
 Take 3(o 2 — & 3 ) 2 +2# 2 y. a: (a— 6) 3 +(«— i) 3 . 
 
 16. From 3 a 3 (&2 _ 8)3 take a 3 (6 2 — 8) 3 . 
 
 17. Subtract 2 a: 2 (6 2 — c 3 ) from 5 * 2 (6 2 — c 3 ). 
 
 18. Take 3 a 2 6 — 3 a 2 i/ 3 + 8 from 8«H|^ 
 
 2/3—17. - 
 
 '19. From 5 a: 3 y (a 6 2 — 17) 3 take3a: 3 y (a& 2 — 17) 3 . 
 20. Subtract 6 a 6 2 c 3 — 6 2 c 3 + 7 a from 8 a 6 2 
 c 3 + 2 6 2 c 3 — 3 a 2 . 
 
 SECTION VI. 
 Multiplication. 
 
 1. What is the product of 4 a 2 multiplied by 5 a 3 ? 
 
 Ans. 20 a 5 . 
 
 This product may be expressed thus, 20 « 3 a 2 ; that 
 is, 20 a a a a a, or 20 a 5 . But as the sum of the ex- 
 ponents is 5, the exponent of the product may be 
 obtained, at once, by adding together the exponents 
 of the factors. Thus, a 2 X a 3 = a? + 3 , or a 5 ; also 
 x* X # 5 = oc 4 + 5 , or x 9 ; and a 2 6 3 X « 3 & 4 = a 2 + 3 
 6 3 + 4 , or a 5 6 7 . 
 
 Hence, when the same letter, under any power 
 whatever, occurs in both the multiplier and multipli- 
 
105 
 
 cand, add together its exponents, and the sum will be 
 the exponent of the product. 
 
 2. Multiply 3 a 2 b by 6 a b 3 . Ans. 18 a 3 6 4 . 
 
 3. Multiply 8 a 3 & x by 2 a 2 62 jp. 
 
 4. Multiply 3 a 5 6 2 x y by 5 a b 3 x 3 y 5 . 
 
 5. Multiply 9 a 3 c 2 m by 9 a 3 c^ ro. 
 
 6. Multiply x 3 + 5 « by # 2 . 
 
 7. Multiply 3 a 2 — 5 a a; 3 by 2 a 3 a? 2 . 
 
 8. What is the product of a y — 6 -{- 2/ 3 multi- 
 plied by a 3 1/ 2 ? 
 
 9. Required the product of a 3 — b 2 multiplied by 
 a 3 + 6 2 . 
 
 10. Multiply a 2 be 3 — 12 by 2 a 2 6 3 . 
 
 11. Multiply x 3 -\- x 2 y -\~ x y 2 -\- y 3 by x — y. 
 
 12. Required the product of a 3 -f- a — 6 multi- 
 plied by 2 a 2 + a -f 1. _^ 
 
 13. Multiply a 2 -f a 4 + a 6 by a 2 — 1. 
 
 14. Multiply 3 (a* — y 3 ) 2 by a (a: 2 — y 3 ) 2 . 
 
 Ans. 3 a (x* — y 3 ) 4 . 
 
 15. Required the product of a 2 (# 3 — 12 -}- a: 2 ) 3 
 multiplied by a b (x 3 — 12 + a? 2 ) 3 . 
 
 16. Multiply 3 (a 2 + 6 3 ) 2 by 5 (a 2 -f- b 3 f. 
 
 17. Multiply a (a: 3 -- y 2 + 4) 3 by 6 (a; 3 — y* -f-4) 3 . 
 
 18. Multiply 3 a 2 (> 3 — y 3 ) 2 by 4 a 2 (a? — ?/3)2. 
 
 19. Multiply 4 a 2 — 16 a x -f 3 # 2 by 5 a 3 — 2 a 2 a\ 
 
 20. Multiply a 4 — 2 a 3 6 + 4 a 2 6 2 — 8 a b 3 + 
 16 & 4 by a + 2 5. 
 
 21. Multiply 2 -f- 6 a 2 — 6 3 by z 2 — • a. 
 
 a-3 — y* u rf + y'* 
 a 2 + A 
 
 22 . Multiply^ by ^ 8 
 
106 FIRST LESSONS IN ALGEBRA. 
 
 SECTION VII. 
 Division. 
 
 1. Divide a 5 by a 3 Ans. a 2 . 
 The exponent of the quotient added to that of the 
 
 divisor, must be equal to the exponent of the divi- 
 dend ; for these two quantities are multiplied together 
 by the adding of their exponents ; and the product 
 of the divisor and quotient must always give the 
 dividend. 
 
 In the above example, the dividend a 5 = a a a a a, 
 and the divisor a 3 = a a a; and the division may be 
 
 expressed thus, a aaa . Now, if we cancel a a a in 
 
 the numerator and denominator, there will be left in 
 the dividend a a, that is, cP, as in the answer given 
 above. 
 
 Hence, if we are required to divide a power of any 
 letter, as a 5 , by another power of the same letter, as 
 a 3 , we must subtract the exponent of the divisor from 
 the exponent of the dividend, and the remainder will 
 be the exponent of the quotient. 
 
 2. Divide a 6 by a 3 . Ans. a 3 . 
 
 3. Divide a 3 b 4 by a b 3 . 
 
 4. Divide a 6 b 5 c 2 by a 6 ¥ c. 
 
 5. Divide 16 a 10 b 4 x* by 4 a 5 b 3 x*. 
 
 6. Divide 39 a 2 m 3 y 4 by 13 a 2 m 2 y 3 . 
 
 7. Divide a 2 by « 2 . . Ans. 1. 
 By the rule above obtained, o 2 -7- a 2 = a ; for 
 
POWERS. 107 
 
 fl 2 — 2 _ a o j7 ut <L __ i . f or if we divide any quan- 
 tity by itself, the quotient is 1. Hence we see, thct 
 a is always equal to 1, whatever may be the value of 
 a; and the same is true of any other quantity which 
 has ° for its exponent. 
 
 8. Divide a 5 by a 5 . 
 
 9. Divide 5 a 3 by 5 a 3 . 
 
 10. Divide a 2 by a 3 . Ans. a— 1 . 
 Here the exponent of the divisor is greater than 
 
 that of the dividend ; but the general rule must be 
 observed. Thus, a 2 -=- a 3 — a 2 — 3 = a — 1 . In the 
 expression a — 1 , a is said to have a negative exponent. 
 So, also, a -r- a? = a— 1 ; a -f- a 3 = a— 2 ; a -£- a 4 
 == «— 3 ; « -^- a 5 = a— 4 ; and so on. 
 
 11. Divide a 7 by a 9 . Ans. a— 2 . 
 
 12. Divide a 2 6 4 a? 3 by ar> 6 6 * 7 . 
 
 13. Divide a by a. Ans. a— i. 
 The division of a by a, that is, by a 1 , may be thus 
 
 expressed; a° -^ a 1 = a — 1 =s a— 1 . But it will be 
 
 remembered, that a = 1 ; therefore, — is the same 
 
 as — ; and, consequently, a— 1 = — . 
 
 Hence we learn the value of those powers which 
 have negative exponents. Thus, a° = 1 ; a—* 1 =cp -; 
 
 a 
 
 _2_1. rt _ 3 _ 1. „ — 4 _ 1. „ 5 
 
 If the value of a be 2, then a = 1 ; a -1 == £; 
 
 a _2 __ J.. fl — 3 — £. a — 4 _ y l F . a -5 — . ^1_. & c . 
 
 14. Divide (a + 6) 3 by (a + 6) 3 . Ans. 1. 
 
 15. Divide (a 2 — j* + J) 4 by (a 2 — • ^ -f- £) 2 . 
 
108 FIRST LESSONS IN ALGEBRA. 
 
 16. Divide (a 3 + 5 — y) 3 by (a 3 fl 5 — y)?. 
 
 17. Divide a b 3 + 2 a 3 b + 2 a 2 6 2 + a 4 by a 6 2 + 
 a 3 + a2 6. 
 
 Before we begin to divide compound quantities, we 
 should arrange the terms of the divisor and dividend 
 according to the powers of their letters, as this will 
 greatly facilitate the work. The highest power of a 
 letter should come first, and the lower powers should 
 succeed in order. The first term of the divisor and 
 the first term of the dividend should contain the same 
 letter. 
 
 To arrange this question, we place the letter of the 
 divisor which is of the. highest power, first, and the 
 other terms in order, thus ; a 3 -{- a? b -\- a b*. Now, 
 as the first term of the divisor is a, the first term of 
 the dividend should also contain a, and the whole 
 should be arranged thus ; a 4 -}- 2 a 3 b -f- 2 a 2 b 2 + a b 3 . 
 
 a 3 + «2 i + a 52 ) a 4 + 2 a 3 b + 2 a 9 b* + a b 3 ( a + b. 
 
 a 4 _|_ a 3 I _L. 
 
 a 2 & 2 
 
 a 3 b + 
 a 3 b + 
 
 a 2 62 + a b 3 
 a*b 2 + a b 3 
 
 18. Divide a b 3 — b 3 c by a — c. 
 
 19. Divide 3 a* + 16 a 4 b — 33 a 3 b* + 14 a 2 b 3 
 by a 2 + 7 a 6. 
 
 20. Divide a? 3 -f- 9 # 2 + 4 * — 80 by x -f- 5. 
 
 21. Divide b 8 — 16 c 8 by 6 2 — 2 c 2 . 
 
 22. Divide a 2 a? — 6 2 x + 8 a? — a 2 y 3 + J 2 y 3 — 
 8 y 3 by a? — y 3 . 
 
 23. Divide a 6 — a 4 a? — a 2 a? 3 + 2 ^ by a 4 — - x 3 . 
 
EQUATIONS OF THE FIRST DEGREE. 109 
 
 CHAPTER VIII. 
 EQUATIONS OF THE FIRST DEGREE. 
 
 SECTION I. 
 Introduction. 
 
 When two equal quantities, differently expressed, 
 are compared together by means of the sign *= between 
 them, such an expression is called an equation. Thus, 
 8 -j- 4 = 18 — 6 is an equation; for the sums are 
 equal, though expressed in different numbers. So, 
 too, if x -{- 5 and a — 7 represent equal quantities, 
 we have the equation x -j- 5 = a — 7. 
 
 It is by means of equations that most of the inves- 
 tigations of Algebra are carried on ; and the pre- 
 ceding chapters may be regarded as merely prepara- 
 tory to this part of the science. 
 
 An equation of the first degree contains only the first 
 power of the unknown quantity, as x. When some 
 higher power of the unknown quantity, as x 2 , or x 3 , 
 enters into the equation, it is said to be of the second 
 or third degree. 
 
 The terms on the left of the sign —, taken together, 
 are called the first member of the equation ; those on 
 the right, the second member. Thus, in the equation 
 10 
 
J 10 FIRST LESSONS IN ALGEBRA. 
 
 x + 6 = a ■ — 5, the first member is x + 6, and a — 5 
 is the second. 
 
 Any changes, which convenience requires, may be 
 made in the members of an equation, provided the 
 same be made in both, so that their equality is pre- 
 served ; as may be seen in the following examples: 
 
 Given the equation 8 -f- 4 = 18 — 6. 
 
 Add 10 to each member; 
 
 8 + 4 + 10 = 18 — 6 + 10. 
 
 Subtract 12 from each member ; 
 
 8 +4 + 10 — 12 = 18 — 6 + 10 — 12. 
 
 Multiply each term by 2 ; 
 
 16 + 8 + 20 — 24 = 36 — 12 + 20 — 24. 
 
 Divide every term by 4 ; 
 
 l_|_2+5 — 6 = 9- 3 + 5 — 6. 
 
 Although the members of the given equation, 8 + 4 
 s* 18 — 6, are changed in form and value by each 
 successive operation, it will be seen that their equality 
 is preserved throughout. Whence we infer that, 
 
 The same quantity may be added to both members of 
 an equation; 
 
 The same quantity may be subtracted from both mem- 
 bers of an equation; 
 
 All the terms of an equation may be multiplied by 
 the same quantity: and 
 
 All the terms of an equation may be divided by the 
 same quantity; without affecting, in either of these 
 cases, the equality of the two members. 
 
 The application and use of these and other changes 
 in the terms and members of equations, will be ex- 
 
EQUATIONS OF THE FIRST DEGREE. Ill 
 
 plained hereafter, as it is found necessary to introduce 
 them. 
 
 When a question is proposed, to be solved by Al- 
 gebra, the first step is, to express its conditions in the 
 form of an equation ; or, in common language, toput 
 the question into an equation. 
 
 The next step is, to find the value of the unknown 
 quantity from the terms with which it is associated; 
 which process is called the reducing or resolving of an 
 equation. 
 
 The rules for reducing equations are few and sim- 
 ple ; and they will be given in the subsequent sec- 
 tions. But no particular mode of putting questions 
 into an equation, can be prescribed, as the process must 
 vary with the conditions of every question. The fol- 
 lowing general directions may be of some service : 
 
 When a question is proposed, before its solution is 
 attempted, get a clear and distinct understanding of 
 its nature and design. 
 
 Let the thing required, the answer to the question, 
 be represented by some letter, as x or y. 
 
 Regard the letter used as the answer of the ques- 
 tion, and perform the same operations on it as would 
 be necessary to prove the real answer to be correct. 
 
 The result thus obtained will be one member of 
 the equation ; and the other member will be found 
 in the corresponding conditions of the question. 
 
 To illustrate these general directions by a single 
 example: Let us. suppose that a man gave 175 dol- 
 lars for his watch, chain and seal ; that the chain cost 
 twice as much as the seal ; and the watch twice as 
 much as the chain. What was the price of each ? 
 
112 FIRST LESSONS IN ALGEBRA. 
 
 This question has three required quantities, or an- 
 swers, namely, the prices of the watch, the chain, and 
 the seal, either of which may be assumed and repre- 
 sented by a letter, as x; and, either being known, the 
 others can readily be found. 
 
 First, let us suppose the watch cost x dollars ; then 
 
 the chain cost —dollars, or half as much as the watch ; 
 
 and the seal cost — dollars, or half as much as the 
 
 chain ; and, according to the answer assumed, they 
 
 all cost x -f- y -j- j dollars. But they actually cost, 
 
 by the question, 175 dollars ; we have, therefore, this 
 equation : 
 
 z+i +7= 175 - 
 
 Again, let us suppose the chain cost x dollars ; then 
 the watch cost twice as much, or 2 x dollars ; and 
 
 the seal cost half as much as the chain, or — dollars. 
 According to this supposition, the cost of the whole 
 was 2 x -f- x -J- — dollars ; and we have this equa- 
 tion : 
 
 2* + * + Y = 175. 
 
 • Finally, let us suppose the seal cost x dollars ; then 
 the chain cost 2 x dollars ; and the watch cost, twice 
 as much as the chain, or 4 # dollars. By this suppo- 
 sition, the price of the whole was 4 x -[- 2 x -f- x dol- 
 lars ; and we have the following equation : 
 
 4x -\-<2x + x = 175. 
 
 Either of these equations will give one answer to 
 
EQUATIONS OF THE FIRST DEGilEE. 113 
 
 the question, from which the other answers may be 
 obtained by common multiplication or division. The 
 last form, being the most simple, is to be preferred. 
 
 It will be observed, that the value of x is different 
 in each of these equations. In the first, it is 100 dol- 
 lars, the price of the watch ; in the second, it is 50 
 dollars, the price of the chain ; in the third, it is 25 
 dollars, the price of the seal. 
 
 To find the numerical value of the unknown quan- 
 tity, we make it stand alone, neither multiplied nor di- 
 vided by any number or quantity, as one member of the 
 equation ; and collect all the known quantities together 
 for the other member. 
 
 SECTION II. 
 To remove Coefficients. 
 
 1. A man gave 381 dollars for a horse and chaise ; 
 and the chaise cost twice as much as the horse. Re- 
 quired the price of each. 
 
 Let x represent the price of the horse. Then 2 m 
 will be the price of the chaise ; and x -\- 2 x will be 
 the price of both. We have, therefore, the following 
 equation : 
 
 x + 2x = 381. 
 Add them's, 3 x = 381. 
 
 Divide by 3, x =s= 127, the price of the horse. 
 
 And 2 x = 254, the price of the chaise. 
 
 To obtain the second equation, viz. 3 x = 381, we 
 10* 
 
114 FIRST LESSONS IN ALGEBRA. 
 
 collect all the terms containing the unknown quantity 
 into one term. 
 
 We then divide this equation by 3, to find the value 
 of x ; for if 3 a? be equal to 381, x is evidently equal 
 to ^ of 381 ; that is, we divide the terms of the equa- 
 tion by the coefficient of the unknown^quantity . 
 
 In general, to reduce an equation, when the unknown 
 quantity is multiplied, by any number, we must divide 
 all the other terms of the equation by that number. 
 
 This principle must be observed, when the unknown 
 quantity occurs in more than one term, and is multi- 
 plied by letters instead of numbers. 
 
 2. Given the equation a x -f- b x =. 38. 
 
 As the first member, a x -\- b x, is the product of 
 a -f- b multiplied by x, it may be expressed thus : 
 (a -f- b) x ; where a -f- b is the coefficient of x. 
 
 a x -f- b x = 38 
 (a + b) x = 38 
 
 38 
 X — a + b* 
 
 L'et us suppose that a — 10, and b = 9, and sub- 
 stitute these numbers for the letters ; we shall then 
 have 
 
 10 x + 9 x = 38 
 
 (10 -f 9) x = 38 
 
 38 
 * = 10 + 9 = 2 - 
 
 * 3. Two men, A and B, trade in company, and 
 
 * The questions thus marked are referred to in a subsequent 
 chapter. 
 
EQUATIONS OF THE FIRST DEGREE. 115 
 
 gain $141, of which B is to have twice as much as 
 A. What is the share of each ? 
 
 Let x =. A's slnre. 
 Then 2x = B's share. 
 And x -f 2 x = 141. 
 
 Ans. A's share, $47; B's, $94. 
 
 4. A man distributed 70 cents among four poor 
 persons ; giving the second twice, the third three . 
 times, and the fourth four times, as much as he gave 
 the first. "What did he give to each ? 
 
 * 5. Said a father to his son, " Our joint ages are 
 78 years, and I am 5 times as old as you." What 
 were their ages ? 
 
 Let x = the son's age. 
 Then 5 x = the father's age. 
 
 And 5 x -f- x = 78. 
 
 # 6. Three men, A, B and C, trade in company, 
 and gain $696, of which B is to receive 3 times as 
 much as A, and C as much as both A and B. What 
 is the share of each ? . 
 
 Let x — A's share. 
 
 Then 3 x = B's share, 
 
 and x -\- 3 x, or 4 x := C's share. 
 
 And x + S x + 4x = 696. 
 
 7. A farmer hired two men and a boy to do a cer- 
 tain piece of work ; agreeingto pay one of the men 
 5 shillings, the other 4 shillings, and the boy 3 shil- 
 lings, a day. When the work was finished, he paid 
 them $54. How many days were they employed? 
 
116 FIRST LESSONS IN ALGEBRA. 
 
 Let x s= the number of days. The first man earns 
 5 x shillings ; the other, 4 x shillings ; and the boy 
 3 x shillings ; which, together, are equal to 324, the 
 shillings in $54. 
 
 . 5^+4^ + 3^ = 324. 
 
 8. A farmer sold an equal number of oxen, cows 
 and sheep, for $632. For the oxen he received $47 
 apiece; for the cows, $25; and for the sheep, $7. 
 How many did he sell of each sort ? 
 
 9. A merchant, failing in trade, owes to A, B, C 
 and D, $3597. To B he owes twice as much as to 
 A ; to C, as much as to A and B ; and to D, as much 
 as to B and C. How much does he owe to each of 
 them ? 
 
 10. A boy bought 2 oranges, 3 pears and 4 apples, 
 for 22 cents. He gave as much for a pear as for 2 
 apples ; and twice as much for an orange, as for a 
 pear and an apple. What was the price of each ? 
 
 * 11. The age of A is double that of B ; the age 
 of B is three times that of C ; and the sum of all 
 their ages is 140. N What is the age of each? 
 
 # 12. A farmer sold a quantity of wood for 104 
 dollars ; one half of it at $6 per cord ; the other 
 Half at $7. How many cords did he sell ? 
 
 Let x = half tHe number of cords. 
 Then 6 x -\- 1 x = 104. 
 
 13. A man bought three equal lots of hay for 
 $366: for the first lot he gave $19 a ton; for the 
 second lot, $20; for the third, $22. How many 
 tons did he buy ? 
 
EQUATIONS OF THE FIRST DEGREE. 117 
 
 * 14. How long will it take three masons to lay 
 1800 bricks, if one can lay 50, another 60, and the 
 third 70 bricks an hour ? 
 
 15. A man left an estate of $60000, to be so di- 
 vided between his widow, 3 sons, 2 daughters and a 
 ward, that each daughter should receive twice as 
 much as the ward, each son as much as the ward and 
 a daughter, and the widow twice as much as each son. 
 What was the share of each ? 
 
 16. How long will it take two men to build 387 
 rods of wall, if one build 4 and the other 5 rods a day ? 
 
 Let x — the number of days. 
 
 * 17. Four brothers gained, in a year, $4755 ; of 
 which B gained three times as much as A ; C gained 
 as much as A and B ; and D gained as much as B 
 and C. What sum was gamed by each ? 
 
 18. In how many hours will a cistern, containing 
 264 gallons, be emptied by 3 cocks ; one of which 
 discharges 2 gallons in 15 minutes ; the second, 5 
 gallons in 30 minutes, and the third, 3 gallons in 45 
 minutes ? 
 
 * 19. One man leaves New York for Boston, and 
 travels 9 miles an hour ; another man leaves Boston for 
 New York, and travels 7 miles an hour. In how many 
 hours will they meet, the cities being 224 miles apart ? 
 
 20. Four boys, A, B, C and D, upon counting their 
 money, found they all had $30; of which sum A's 
 share was three times greater than B's ; C's share was 
 equal to B's, and one third of A's ; and D's share 
 was equal to A's, arid half of C's. What was the 
 share of each ? 
 
118 FIRST LESSONS IN ALGEBRA. 
 
 Let x r= B's share. 
 Then 3i = A's share, 
 
 x -f- •—, or 2 x = C's share, 
 
 2- x 
 
 and Sx-\--~,ov4xz= D's share. 
 Therefore, x + 3x-{-2x + 4x = 30. 
 
 21. A asked B how much money he had ; who re- 
 plied, that if he had seven times as much, he could 
 lend four times what he then had, and have $69 left. 
 How much had he? 
 
 Let x =: B's money. 
 Then 7 x — 4 x = 69, 
 or 3 x = 69. 
 
 22. A rather is seven, and a mother five, times as 
 old as their son ; and the difference of their ages is 
 16 years. How old is the son ? 
 
 23. A man directed, in his will, that his property 
 should be so divided, that his son should have three 
 times as much as his daughter, and his widow twice 
 as much as both her children ; by which division she 
 received $8235 more than the son. What was the 
 share of each? 
 
 24. A farmer bought a horse, a chaise and a house, 
 for $800. Now, he paid 4 times as much for the 
 chaise as for the horse, and 5 times as much for the 
 house as for the chaise. What was the price of each? 
 
 * 25. Two men, A and B, travel the same way; 
 A at the rate of 45 miles, and B 30 miles, a day. In 
 how many days will they be 300 miles apart ? 
 
 * 26. If they were ' to travel in different direc- 
 
EQUATIONS OF THE FIRST DEGREE. 119 
 
 tions, in how many days would they be 300 miles 
 apart ? 
 
 27. A man has six children, whose united ages are 
 42 years ; and the common difference of their ages 
 is equal to the age of the youngest child. What are 
 their several ages ? 
 
 *28. A gentleman bought 3 kinds of wine, of each 
 an equal quantity. For the first kind he paid 7s. , for 
 the second 8s., and for the third 10s., a gallon ; and 
 the price of the whole was $50. How many gallons 
 did he buy ? 
 
 *29. A farmer employed two men to build 105 
 rods of wall; one of whom could build 1 rods, and 
 the other 3 rods a day. How many dayg did they 
 work ? 
 
 30. A laborer, who spent every week as much as he 
 earned in 2 days, saved 32 dollars in 4 weeks. What 
 were his daily wages ? 
 
 31. A farmer sold a quantity of cider for 84 dol- 
 lars ; one half of it at $3 per barrel, the other half 
 at $4. How many barrels did he sell? 
 
 32. A man left an estate of $21546 ; one third of 
 .which he bequeathed to his widow, and directed the 
 remainder to be so divided between his 2 sons and 2 
 daughters, that each son might receive as much as 
 both the daughters. What was the portion of each ? 
 
 33. Three travellers found a purse, containing 
 $54 ; of which B secured three times as much as A, 
 and C secured half as much as both of the others. 
 What was the share of each ? 
 
120 FIRST LESSONS IN ALGEBRA. 
 
 SECTION III. 
 To remove Denominators. 
 
 1. A man received $18 towards a debt, which 
 was only § of the sum due. What was the whole 
 debt? 
 
 Let x denote the amount of the debt. 
 
 x 1 2 # 2 
 
 Then ■— must be — , and — must be — of it. 
 
 Therefore, 2 -f = 18. 
 
 It has already been shown, [See Sec. I.] that, if 
 all the terms of an equation be either multiplied or 
 divided by the same quantity, the equality of the 
 members will not be affected ; therefore, to remove 
 the denominator of the first term, 
 
 Multiply by 3 ; 2 x = 54. 
 
 Divide by 2; x = 27. Ans. $27. 
 
 We first multiply the whole equation by 3, the de- 
 nominator of the fraction, and then find the value of 
 x as before. It has been shown, in a former chap- 
 ter, [See Chap. VI. Sec. VI.] that a fraction is mul- 
 tiplied by its denominator, when that denominator is 
 removed. 
 
 2. What number is that, £ and f of which are 44 ? 
 Let x indicate the number. Then — is half of it, 
 
 2 ' 
 
 3 x 
 
 and — three fifths of it. 
 
 o 
 
EQUATIONS OP THE FIRST DEGREE. 121 
 
 Therefore, ^ -f- -f- = 44. 
 
 Multiply by 2 ; x + y- = 88. 
 
 Multiply by 5; 5« + 6a: = 440. 
 Add the*'*; 11 a; = 440. 
 Divide by 11 ; x = 40. Ans. 40. 
 
 We first free the equations from fractions, by mul- 
 tiplying all the terms by the denominators, one at a 
 time, and then find the value of x as before. 
 
 3. What number is that, which, being increased by 
 |, £ , f and T \ of itself, becomes 146 ? 
 
 Let x — the number. 
 
 Then, by the question, # + f" + "f + ^ + -^ = *46. 
 
 Multiply by 2; 2 x + x . + ^ + ~ + | = 292. 
 
 Multiply by2;4^+2^-|-3a:-fy + y-=: 584. 
 
 Multiply by 7 ; 28 # -f 14 a; -{- 21 # + 8 * +2# z= 4088. 
 
 Add the x's ; 73 x = 4088. 
 
 Divide by 73 ; x = 56. 
 
 Ans. 56. 
 
 3x x 
 
 Observe that we multiply the fractions ^rand — 
 by dividing their denominators by 2 ; and we multi- 
 ply the fractions — and — , by removing their de- 
 nominators. 
 
 To free an equation from fractions, we multiply all 
 the terms by all the denominators, taking them in such 
 order as may be most convenient. 
 11 
 
122 FIRST LESSONS IN ALGEBRA. 
 
 4. Says A to B, " If to my age £ and £ of my age 
 be added, the sum will be 81 ; what is my age ? " 
 
 Let x denote A's age. 
 
 x 3 x 
 
 Then x -f- — -f- ~r = 81, by the question. 
 
 * 5. If f of a ship cost $4000, what is the whole 
 ship worth ? 
 
 6. A man, driving his geese to market, was met by 
 another, who said, " Good morrow, master, with your 
 hundred geese." Said he, " I have not a hundred ; 
 but if I had as many more, and half as many more, 
 and two geese and a half, I should have a hundred." 
 How many had he ? 
 
 x + x + | = 100 — ty. 
 
 7. A post, standing in a pond, is £ of its length 
 under water, and £ above water, it being 14 feet from 
 the top of the post to the bottom of the pond. What 
 is the whole length of the post . ? 
 
 8. What number is that, whose sixth part exceeds 
 its eighth part by 20 ? 
 
 Let x — the number. 
 Then £ — -f- = 20. 
 
 b o 
 
 * 9. A man, having spent three fifths of his estate, 
 had $ 978 left. How much had he at first ?/ / ,.^ 
 
 10. A man spent £ of his life in England, £ of it 
 in Scotland, and the remainder of it, which was 20 
 years, in the United States. What was his age ? 
 
 11. What number is that, £ of which is greater 
 than f of it by 21 ? 
 
EQUATIONS OF THE FIRST DEGREE. 123 
 
 12. A man sold 75 bushels of wheat to two per- 
 sons ; to one, £\ and to the other, f of all he had. 
 How many bushels had he ? 
 
 13. A man gave to three poor persons $6 ; to the 
 first, £, to the second, £, and to the third, £ of all the 
 money he had in his pocket. How much had he ? 
 
 14. A and B divide $320 between them, of which 
 B has three times and f- as much as A. How much 
 has each ? 
 
 15. A says to B, " Your age is twice and § of my 
 age, and the sum of our ages is 54 years." What is 
 the age of each ? 
 
 * 16. If you divide $50 between two persons, giv- 
 ing one f as much as the other, what will be the share 
 of each ? 
 
 17. A stranger in Boston spent, the first day, £ of 
 the money he brought with him ; the second day, £ ; 
 and the third day, £; when he had only $26 left. 
 How much money did he bring? 
 
 18. A man, having invested f of his property in 
 bank stock, by which he lost § of the sum invested, 
 had stock worth $723 remaining. How much prop- 
 erty had he at first ? 
 
 19. A merchant retired from business when he had 
 passed y 9 ^ of his life, during § of which period he had 
 been engaged in trade, having commenced at the age 
 of 21 . At what age did he die ? ' 
 
 20. The age of A is £ that of B, and the age of C is 
 £ that of A, and the sum of all their ages is 120 
 years. What is the age of each ? 
 
 21. A farmer wishes to mix 90 bushels of proven- 
 
124 FIRST LESSONS IN ALGEBRA. 
 
 der, consisting of barley, oats and corn, so that there 
 may be f as many bushels of oats as of corn, and ^ 
 as many bushels of barley as of oats. How many 
 bushels of each sort must he use f 
 
 22. Three boys spent 98 cents for fruit. B spent 
 | as much as A, and C spent i as much as B. What 
 did each spend ? 
 
 23. If you divide $75 between two men, in the 
 proportion of 3 to 2, what will each man receive ? 
 
 One man will receive 2 dollars as often as the other 
 receives 3 ; that is, one will receive § as much as the 
 other. 
 
 Let x — the share of one. 
 
 Then — = the share of the other. ' 
 
 o 
 
 And x -f- -£■=. 75. 
 
 o 
 
 * 24. Divide 84 into two numbers, which shall be 
 to each other as 7 to 5. 
 
 25. What two numbers are to each other as 4 to 
 9, their sum being 91 ? 
 
 * 26. Three men trade in company, and gain $780. 
 A put in $2 as often as B put in $3 and C put in 
 $5. What part of the gain must each one receive? 
 
 Let x = A's share. 
 
 Then — = B's share, 
 
 5 x 
 
 and — = C's share. 
 
 27. Divide 234 into 4 numbers, which shall be to 
 each other in the proportion of 5, 6, 7 and 8. 
 
 28. Divide 36 into three such parts, that £ of the 
 
EQUATIONS OF THE FIRST DEGREE. 125 
 
 first, £ of the second, and £ of the third, shall be equal 
 to each other. 
 
 29. Four towns are situated in the order of the four 
 letters, A, B, C, D. The distance from A to D is 
 102 miles. The distance from A to B is to the dis- 
 tance from C to D as 2 to 3 ; and £ of the distance 
 from A to B, added to £ the distance from C to D, is 
 3 times the distance from B to C. How far are the 
 towns apart ? 
 
 30. Two young men began to trade at the same 
 time, the capital of A being to that of B as 7 to 6. 
 The first year, A gained a sum equal to -£ of his capi- 
 tal, and B lost £ of his. The second year, A lost f 
 of what -he then had ; and B's <*ain was to what re- 
 mained of his original capital, as 2 to 5. At the be- 
 ginning of the third year, the two had $'2450. What 
 was the original capital of each ? 
 
 31. A gentleman, by his will, divided his property 
 equally between his son and daughter ; and the son, 
 having spent f of his portion, had $7268 left. * What 
 was the whole amount of property left ? 
 
 32. A man, being asked the age of his daughter, 
 replied, " My age is to hers as 4 to 1, and her moth- 
 er's age is to mine as 7 to 8, and the sum of all our 
 ages is 102 years." What was the daughter's age ? 
 
 33. A trader, having increased his capital by J of 
 itself, lost £ of what he then had; he afterwards 
 gained a sum equal to £ of the remainder, when he 
 was worth $3935. What was his capital? 
 
 11* 
 
126 FIRST LESSONS IN ALGEBRA. 
 
 SECTION IV. 
 
 Transposition. 
 
 1. If you divide $36 between two persons, in such 
 a manner that B shall have $12 more than A, what 
 is the share of each ? 
 
 Let x = A's share. 
 
 Then x + 12 = B's share. 
 
 a. And x -\- x -{- 12 == 36, by the question ; 
 
 b. or 2 x -{- 12 = 36, by adding the x's. 
 
 In the last equation, w% find that 2 x -f- 12 = 36 ; 
 that is, 2 # js 12 less than 36. We may free the first 
 member from known quantities, by subtracting 12 
 from each side of the equation. We shall then have 
 
 c. 2i -f 12— 12 = 36— 12. 
 
 But as the -{-12 and — 12, in the first member, can- 
 cel each other, they may be omitted ; thus, 
 
 d. 2x = 36— 12; 
 
 e. or 2 -x = 24, by subtraction. 
 
 f. * = 'y. s= 1$, A's part; 
 
 and a? + 12 = 24, B's part. 
 
 Compare equations b and d ; whence it appears, 
 that -f- 12 is transferred from the first to the second 
 member of the equation, the sign being changed from 
 -f to — . 
 
 Therefore, to remove a positive quantity, produces 
 the same result as to annex a negative quantity of the 
 same value. 
 
EQUATIONS OF THE FIRST DEGREE. 127 
 
 The removing of a term from one side of an equa- 
 tion to the other, is called Transposition. 
 
 2. A trader has bills aga'»nst A and B for $187, 
 the bill of A being $25 more than that of B. What 
 is the amount of each ? 
 
 Let x denote B's bill. 
 
 Then x -f- 25 = A's bill. 
 
 And x -\- x -\- 25 = 187, by the question. 
 
 * 3. A spent $87 more than B, and they both spent 
 $325. How much was spent by each ? 
 
 4. A gentleman gave to two beggars 67 cents, 
 giving to the second 13 cents less than to the first. 
 How many cents did each receive ? 
 
 Let x = the cents given to the first. 
 
 Then x — 13 = the cents given to the second. 
 
 a. And x -\- x — 13 ;= 67, by the question ; 
 
 b. or 2 a? — 13 = 67, by addition of x's. 
 
 As 2 # — 13 = 67, it is evident that 2 a? is greater 
 than 67 by 13. If we add 13 to each side of the 
 equation, we shall have 
 
 c. 2 x — 13 + 13 = 67 + 13. 
 
 Since — 13 and -j- 13 cancel each other, in the 
 first member, we have 
 
 d. 2x =*67 + 13; 
 
 e. or 2 x = 80, by addition. 
 
 f. x = *£• = 40, the share of the first ; 
 
 and x — 13 = 27, the share of the second. 
 
 Compare equations b and d, ' as before. It appears 
 
123 FIRST LESSONS IN ALGEBRA. 
 
 that — 13 is transferred from the first to the second 
 member, the sign — being changed to -}-. 
 
 Whence it appears, that to remove a negative quan- 
 tity, has the same effect as to add a positive quantity 
 of equal value. 
 
 5. Three men, A, B and C, trade in company, upon 
 a capita] of $3981 ; of which B furnished $337 more, 
 and C $181 less, than A. What was the share of 
 each ? 
 
 6. A man left an estate of $9931, to be divided be- 
 tween his widow, son and daughter, in such a man- 
 ner, that the son should have $522 more than the 
 daughter, and $592 less than his mother. Required 
 the portion of each. 
 
 7. Divide 42 into four such parts, that the first shall 
 be 5 more than the second, 8 less than the third, and 
 9 more than the fourth. 
 
 * 8. An express had been travelling 5 days, at the 
 rate of 60 miles a day, when another was despatched 
 after him, who travelled 75 miles a day. In how 
 many days did the latter overtake the former ? 
 
 Let x = the number of days. Then the second 
 express travels 75 x miles ; and the first express travels 
 60 x miles, after the departure of the second, and was 
 300 miles (60 X 5) in advance of him when he 
 started. Now, as they both travel the same distance, 
 we have 
 
 a. 75 x = 60 x + 300. 
 
 Here, the unknown quantity is found on both 
 sides of the equation ; but all the terms containing 
 the unknown quantity must be brought into one mem- 
 
EQUATIONS OF THE FIRST DEGREE. 129 
 
 ber, and all those containing known quantities, into the 
 other. 
 
 If any commodity to be weighed, as coffee, for in- 
 stance, were mixed with the iveights of a grocer, al- 
 though his scales might be perfectly balanced, it is 
 evident that he could not determine how many pounds 
 of coffee they contained. His only way would be, 
 to collect all the coffee into one of the scales, and 
 all the weights into the other. Now, an equation is 
 not unlike the grocer's scales : the commodity to be 
 weighed is the unknown, and the leaden weights, the 
 known quantity. 
 
 If 60 x be subtracted from each side of the equa- 
 tion, we shall obtain 
 
 b. 75 x — 60 x = 60 x -f- 300 — 60 x. 
 
 If we cancel -\- 60 x and — 60 x, in the second 
 member, the equation will be 
 
 c. 75 x — 60 x = 300; 
 or 15 x = 300, by subtraction. 
 
 x-— 20. 
 
 Ans. 20 days. 
 
 Compare equations a and c, where a term contain- 
 ing the unknown quantity, 60 x, is removed from the 
 second to the first member ; the sign -j- being chang- 
 ed to — . 
 
 9. The water had been flowing from a full cistern 6 
 hours, at the rate of 12 gallons an hour, when a pipe 
 was conducted into it, which restored 21 gallons an 
 hour. In how many hours was the cistern full again? 
 
 10. A man, being asked his age, answered, that if 
 
130 FIRST LESSONS IN ALGEBRA. 
 
 his father's age, which was 52 years, were added to 
 three times his own, the sum would be five times his 
 age. How old was he ? * £ 
 
 * 11. A father is three times as old as his son ; but 
 in 20 years he will be only twice as old. What is 
 the age of each ? ■>, J? , 
 
 12. When a boy would buy a certain number of 
 oranges at 6 cents apiece, he found they would come 
 to 12 cents more than he had ; he, therefore, bought 
 the same number at 5 cents each, and had 6 cents 
 left. How many oranges did he buy? 
 
 * 13. Divide 64 into two such parts, that five times 
 the first shall be equal to three times the second. 
 
 Let x = the first part. 
 
 Then 64 — x = the second. 
 
 And 5« = (64 — x) X 3, by the question ; 
 
 a. or 5 a; = 192— 3 a?. 
 
 It appears that 5 a? is not equal to 192 by 3 x. If 
 we add 3 a: to each side of the equation, we have 
 
 b. 5 x -f- 3 x = 192 — 3 x + 3 x. 
 
 And if we cancel the — 3 a: and -^ 3 x, in the 
 second member, the equation becomes 
 
 c. 5 x -f 3 x =z 192; 
 
 or 8 x = 192, by addition. 
 
 x js= 24, the first part ; 
 
 and 64 — x = 40, the second part. 
 
 Observe, in the equations a and c, that — 3 a:.is re- 
 moved from the second member to the first, tne sign 
 — being changed to -}-• 
 
EQUATIONS OF THE FIRST DEGREE. 131 
 
 From the preceding examples and observations we 
 derive the following principle : Any term may be trans- 
 posed from one member of an equation to the other, if 
 the sign be changed. 
 
 14. A pole is 4 feet in the ground, £ of its whole 
 length under water, and ^ above water. Required 
 its length. 
 
 15. The head of a fish weighs 8 lbs.; his tail weighs 
 as much as his head and half his body, and his body 
 weighs as much as his head and tail. What is the 
 weight of the fish ? 
 
 Let x = the weight of the body, 
 
 and — -j- 8 = the weight of the tail. 
 
 Then* =|+16. 
 
 16. A man, being asked his own and his wife's 
 age, said, that his youngest child was 4 years old ; 
 that the age of his wife was twice the age of the 
 child and £ of his own age ; and that his own age 
 was equal to the united ages of his wife and child. 
 How old were they ? 
 
 * 17. A merchant has wines at 9 shillings, and at 
 13 shillings, per gallon ; and he would make a mix- 
 ture of 100 gallons, that shall be worth 12 shillings 
 per gallon. How many gallons of each must he 
 take ? 
 
 * 18. How many gallons of wine, at 9 shillings a 
 gallon, must be mixed with 20 gallons at 13 shillings, 
 that the mixture may be worth 10 shillings a gallon? 
 
 19. A merchant, having mixed 10 gallons of wine, 
 at 8 shillings a gallon, with 25 gallons at 10 shillings, 
 
 7 
 
 ) 
 
132 FIRST LESSONS IN ALGEBRA. 
 
 wishes to add as much wine at 15 shillings as shall 
 make*the whole mixture worth 2 dollars a gallon. 
 How many gallons must he take ? 
 
 20. How many gallons of water must be mixed 
 with 35 gallons of wine, at 9 shillings, and 45 gallons 
 at 13 shillings, a gallon, that the whole mixture may 
 be worth 10 shillings a gallon? 
 
 21. A clerk spends § of his salary for board and 
 lodging, f of the remainder in clothes, and saves $150 
 per annum. What is his salary . ? 
 
 22. What is that number, £ and £ of which is 35 
 more than its sixth part ? 
 
 *«23. Two men, A and B, have each $80. A 
 spends $5 more than twice as much as B, and has 
 then half as much as B, wanting $13. How much 
 did each spend ? 
 
 24. Divide 84 into two such parts, that if £ of the 
 less be subtracted from the greater, and ^ of the great- 
 er be subtracted from the less, the remainders shall 
 be equal. 
 
 Let x s= the greater number, 
 and 84 — x = the less. 
 
 Then x jp-« is £ the less subtracted from the 
 
 greater ; and 84 — x — — is £ of the greater sub- 
 tracted from the less ; and, by the terms of the ques- 
 tion, their values are equal. 
 
 84 — x DA x 
 
 2x— -84 +x = 168 — 2a? — f . 
 
 4 
 
 Ans. 48 and 36, 
 
EQUATIONS OF THE FIRST DEGREE. 133 
 
 Observe that the sign — , before the fraction — - — , 
 
 affects the value of the whole fraction, and not any 
 particular term of it. We are not required to sub- 
 tract half of 84 from the preceding term, but only 
 half of the excess of 84 above x. Therefore, if the 
 sign — is prefixed to 84, when the denominator is re- 
 moved, the sign — before x must be changed to -f- ; 
 otherwise, too much, by the value of x, would be 
 taken from the first term of the equation. Were the 
 
 sign -f- before the fraction — —• , it would not be ne- 
 cessary to change the sign before x from — to +• 
 [See Chap. VI. Sec. ILL] 
 
 25. Divide 72 into two unequal numbers, so that, if 
 | of the less be subtracted from the greater, the re- 
 mainder may be equal to f of the number which re- 
 mains, after the excess of the greater above the less 
 is subtracted from the less. 
 
 26. Two clerks, A and B, sent ventures in different 
 vessels, A's being worth only f as much as B's. A 
 gained and B lost $23; then & of B's returns, sub- 
 tracted from A's, would leave £ of the value of A's 
 venture. How much did each send ? 
 
 27. A gentleman bought a watch and chain for 
 $160. If f of the price of the watch be subtracted 
 from 6 times the price of the chain, and T V of the 
 price of the chain be subtracted from twice the price 
 of the watch, the remainders will be equal. What 
 was the price of each ? 
 
 28. A person, being asked the time of day, replied, 
 " If to the time from noon be added its J-, $, i and -£, 
 
 12 
 
134 FIRST LESSONS IN ALGEBRA. 
 
 the sum will be equal to the time till midnight. Re- 
 quired the hour. 
 
 29. There are two numbers in the proportion of 3 
 to 4 ; but if 24 be added to each of them, the sums 
 will be in the proportion of 4 to 5. What are the 
 numbers ? 
 
 # 30. What number is that which, being added to 5, 
 and also multiplied by 5, the product shall be 4 times 
 the sum ? 
 
 31. A man, having spent $10 more than ^ of his 
 money, had $15 more than £ of it left. How much 
 had he ? 
 
 * 32. What number is that, which, being divided 
 by 16, will amount to 84, when the quotient, dividend 
 and divisor are added together ? 
 
 SECTION V. 
 
 Questions for Practice. 
 
 1. What number is that, to which if there be added 
 £, J, and i of itself, the sum will be 50 ? 
 
 2. A person, upon being asked his age, replied, that 
 | exceeded £ part of it by 5 years. What was his 
 age? 
 
 3. A trader gave three checjts, amounting to 
 the first for £, the second for £, the third for -£, of 
 the money he had in the bank. How much had he ? 
 
 4. To find two such numbers, in the proportion of 
 
EQUATIONS OF THE FIRST DEGREE. 135 
 
 2 to 1, that, if 4 be added to each, they will be in pro- 
 portion of 3 to 2. 
 
 5. In the composition of a certain quantity of gun- 
 powder, § of the whole and 10 pounds were nitre; £ 
 of the whole, wanting 4^ pounds, was sulphur; and 
 the charcoal was t of the nitre, wanting 2 pounds. 
 How many pounds were there ? 
 
 * 6. A person, being asked the time of day, an- 
 swered, that the time past from noon was equal to f 
 of the time to midnight. What was the hour t 
 
 Let x = the hour. 
 Then 12 — x = time to midnight. 
 
 . | 48 — 4a; 
 
 And x = — - — . 
 
 5 
 
 7. A privateer, running at the rate of 10 miles an 
 hour, discovers a ship 18 miles off, making way at the 
 rate of 8 miles an hour. In how many hours will the 
 ship be overtaken ? 
 
 8. Required two numbers, which are to each other as 
 
 3 to 2 ; and whose sum- equals £ part of their product. 
 
 Let x denote the larger number, and the smaller 
 
 2 v 2 x 
 
 number will be — , and x -j- — = their sum. 
 
 o o 
 
 2 x x 2 .r 2 
 
 Now, — X ~ — -r t which is the product of the 
 
 1 2 a -2 2 a: 2 x 2 
 
 two numbers ; and — part of — is — , or — . 
 
 Then — z=. x -\- —-, by the question. 
 
 a? 2 = 9 x -f- 6 x, by removing the denominators. 
 
 This is an equation of the second degree, as it con- 
 tains the second power of the unknown quantity, 
 
136 FIRST WESSONS IN ALGEBRA. 
 
 namely, £ 2 ; but if we divide all the terms by oc, we 
 
 shall obtain 
 
 x = 9 + 6, 
 
 which is an equation of the first degree. 
 
 Ans.~ 15 and 10. 
 
 9. Divide 48 into two such parts, that the one part 
 may be three times as much above 20 as the other 
 wants of 20. 
 
 10. A father leaves $1600, to be divided between 
 his widow, son and daughter, in such a manner that 
 the widow is to have $200 more than the son, and the 
 son $ 100 more than the daughter. What is the share 
 of each ? 
 
 11. A man leaves $11000, to be divided between 
 his widow, two sons, and three daughters. By his will 
 the mother is to receive twice as much as one of the 
 sons ; and each* son is to receive twice as much a& a 
 daughter. How much is each of them to receive ? 
 
 12. An estate is divided between three men in such 
 a manner, that A receives $1000 less than £, B $800 
 less than £. and C $600 less than £ of the whole. 
 What is the value of the whole estate, and what is 
 the share of each individual? 
 
 13. A father leaves his property to four sons, who 
 share it in the following manner : A has $3000 less 
 than i; B has $1000 less than $; C has £; and D 
 has $600 more than $ of the property. What is the 
 whole amount bequeathed ? and the share of each of 
 the sons ? 
 
 * 14. A father, being asked the age of his daugh- 
 ter, replied, that, if her age were multiplied by 5, the 
 
EQUATIONS OF THE FIRST DEGREE. 137 
 
 product would be as much less than 40, as her age 
 was less than 12. What was her age ? 
 
 * 15. Divide 76 into two such parts, that the quo- 
 tient of the greater, divided by the less, may be 37. 
 
 Let x = less number. 
 Then 76 — a? ''= the greater. 
 
 And 7 -5=i = 37. 
 
 X 
 
 * 16. A man has two silver cups,, having but one 
 cover for both. The first cup weighs 12 ounces ; and 
 when it is covered, it weighs twice as much as the 
 other cup ; but if the second cup be covered, it weighs 
 three times as much as the first. Required the weight 
 of the cover and of the second cup. 
 
 * 17. If a certain number be divided by 7,. the sum 
 of the dividend, divisor and quotient, will.be 71. What 
 is the number ? 
 
 18. What is that number, of which ^ is as much 
 smaller than 65, as twice the number is greater than 
 640? 
 
 19. Two persons, A and B, are 320 miles apart, 
 and travel towards each other; A at the rate of 9 
 miles an hour, and B at the rate of 7 miles. In what 
 time will they meet? How many miles does each 
 travel ? 
 
 20. Two brothers, A and B, had the same income. 
 A spent all of his, and | more ; B saved % of his. At 
 the end of 10 years, B paid A's debts, and had $160 
 left. What was their income ? 
 
 21. A farmer planted corn in 4 fields. The third 
 produced 9 bushels more than the fourth ; the second, 
 
 12* 
 
138 FIRST 'lessons in algebra. 
 
 12 bushels more than the third ; the first, 18 bushels 
 more than the second ; and the whole produced 6 bush- 
 els more than 7 times as much as the fourth. What 
 was the whole number of bushels ? 
 
 * 22. Two men have equal sums of money. One 
 having spent $39, and the other $93, the one has but 
 half as much left as the other. How much had each ? 
 
 23. What is that number, which being added to 
 24 and 36, the sums will be to each other as 7 to 9 ? 
 
 * 24. A laborer receives 3s. 6d. every day he works, 
 and forfeits 9d. every day he is idle. At the end of 
 24 days, there is due to him the sum of £3 2s. 9d. 
 How many days was he idle ? 
 
 * 25. Two persons, A and B, have each an annual 
 income of $400. A spends, every year, $40 more 
 than B ; and, at the end of 4 years, they both to- 
 gether save a sum equal to the income of either. 
 Whai do they spend annually ? 
 
 26. A gentleman leaves $315, to be divided among 
 four servants in the following manner : B is to receive 
 as much as A, and £ as much more ; C is to receive 
 as much as A and B, and ^ as much more ; D is to 
 receive as much as the other three, and £ as much 
 more. What is the share of each ? 
 
 27. What number is that, which being multiplied 
 by 4, and 30 subtracted from the product, and being 
 divided by 4, and 30 added to the quotient, the sum 
 and difference shall be equal ? 
 
 28. A person, being asked his age, replied, " If f 
 of my age be multiplied by £ of my age, the product 
 will be equal to my age." What was his age ? 
 
EQUATIONS OF THE FIRST DEGREE. 139 
 
 29. Two numbers are to each other as 2 to 3 ; but 
 if 50 be subtracted from each, one will be ^ of the 
 other. What are the numbers ? 
 
 * 30. A waterman went down a river and returned 
 again in 6 hours. Now, with the stream, he can row 9 
 miles an hour ; but, against it, he can make a head- 
 way of only 3 miles an hour. How far did he go ? 
 
 31. In a mixture of wine and cider, £ of the whole 
 and 25 gallons was wine ; and £ of the whole, wanting 
 5 gallons, was cider. Required the quantity of each 
 in the mixture. 
 
 32. A gentleman visited several poor families. At 
 the first house, he gave 3 shillings less than \ of all 
 his money ; at the second, 4 shillings less than $ of 
 what he had left ; at the third, \ of what he had left ; 
 at the fourth, 3 shillings less than £ of what he had 
 left; at the fifth, 2 shillings more than £ of what he 
 had left ; and at the sixth, all he had remaining. Now, 
 he gave the same sum to each family. How much 
 did he leave at each house ? 
 
 33. In a certain river is a post that stands 4 feet in 
 the ground ; at high water, f of the remainder is cov- 
 ered, while only £ of the whole post appears above 
 the surface of the water ; but, at low water, the length 
 above the surface is just equal to that which, at high 
 water, is immersed. Required the length of the post, 
 and the rise of the tide. 
 
 * 34. What number is that, to which if I add 13, 
 and from T \ of the sum subtract 13, the remainder 
 shall be 13? 
 
 35. Three men, A, B and C, build 318 rods of 
 
140 FIRST LESSONS IN ALGEBRA. 
 
 wall : A builds 7 rods, B 6 rods, and C 5 rods a day ; 
 B works twice as many days as A, and C works ^ as 
 many days as both A and B. How many days does 
 each work ? 
 
 36. A farmer has his sheep in four pastures : in the 
 first, | of his flock ; in the second, |; in the third, Jj 
 and in the fourth, 18 sheep. How many sheep has 
 he? 
 
 * 37. If a man fills a certain chest with corn, at 5s. 
 a bushel, he will spend all his money; but if he fills 
 it with oats, at 3s. 6d. a bushel, he will have £1 4s. 
 left. How many bushels does the chest hold ? 
 
 38. Three merchants, A, B and C, engage in a 
 speculation, by which they gain .$960. A put in $3 
 as often as B $7, and C $5. What is each man's 
 share of the gain ? 
 
 # 39. Says John to William, " I have three times 
 as many marbles as you." " Yes," says William ; 
 " but if you will give me 20, I shall have 7 times as 
 many as you." How many has each ? * 
 
 40. A woman sells eggs at 5 cents a dozen more 
 than apples ; and 8 dozen of eggs are worth as much 
 as 13f dozen of her apples. What is the price of 
 each? 
 
 '"*41. Two travellers found some five-dollar bills in 
 the road, of which A secured twice as many as B ; 
 but had B secured 5 more of the bills, he would have 
 had 3 times as much money as A. How much did 
 each find ? 
 
 42. A man's age, when he was married, was to that 
 of his wife as 6 to 5 ; and after they had been ma*- 
 
EQUATIONS OF THE FIRST DEGREE. 141 
 
 ried 8 years, her age was to his as 7 to 8. What 
 were their ages when they were married ? 
 
 43. A gentleman gave $44 more for his chaise than 
 for his horse. Now, if f of the price of the horse be 
 subtracted from the price of the chaise, the remainder 
 will be the same as if f of the excess of the price of 
 the horse above $84 be subtracted from the price of 
 the horse. What did he give for the horse ? 
 
 44. A hare is 50 leaps before a hound, and takes 4 
 leaps to the hound's 3 ; but 2 of the hound's leaps 
 are equal to 3 of the hare's. How many leaps must 
 the hound take to catch the hare ? 
 
 45. A man and his wife usually drank a cask of 
 beer in 12 days ; but when the man was from home, 
 it lasted his wife 3(kdays. How many days would it 
 last the man alone ? 
 
 Let x denote the number of days it would last the 
 man alone. Now, as the cask would last the man 
 and woman together 12 days, in one day the two 
 would drink ^ of it. Also, as it would last the woman 
 30 days, in one day she would drink ^ of it ; and 
 t l — J^ expresses the proportion of the cask which 
 the man drank in a day. But if he drank the whole 
 
 of it in x days, in one day he drank — part of it. 
 
 Therefore,^ -£ = 7- 
 
 * 46. A and B together can build a piece of wall in 
 8 days ; and, with the assistance of C, they can build 
 it in 5 days. In how many days could C build it 
 alone ? 
 
142 FIRST LESSONS IN ALGEBRA. 
 
 * 47. A cistern has two cocks, one of which will 
 empty it in 7 hours, the other in 9 hours. How long 
 will it take both to empty it? 
 
 Let x denote the number of hours required to empty 
 the cistern. 
 
 The two cocks will discharge \ -f- ^ of the water 
 in an hour. Also, if they will empty the cistern in x 
 
 hours, in one hour they will discharge — part of the 
 water. 
 
 Therefore, 7 + 7 = 7. 
 
 48. If a reservoir can be exhausted by one engine 
 in 7 hours, by another in 8 hours, and by a third in 9 
 hours, in what time will it be exhausted, if they are 
 all worked together? 
 
 *49* A reservoir can be filled by two hose compa- 
 nies in 12 hours, and by one of them alone in 20 
 hours. In what time could the ether fill it? 
 
 50. In an orchard of fruit-trees, -^ of them bear ap- 
 ples, I of them pears, £ of them peaches, 7 trees bear 
 cherries, 3 plums, and 2 quinces. How many trees 
 are there in the orchard ? 
 
 51. A boy, being asked his age, answered, that if £ 
 and I of his age and 20 years more were added to 
 his age, the sum would be three times his age. How 
 old was he ? 
 
 * 52. A father is 40 years old, and his son is 8. In 
 how many years will the father's age be 3 times the 
 son's ? 
 
 53. Two travellers, A and B, find a purse with dol- 
 lars in it. A takes out $2 and £ of what remains; 
 
EQUATIONS OF THE FIRST DEGREE. 143 
 
 and B takes out $3 and J of what remains ; when 
 they have equal shares. How much money did they 
 find? 
 
 * 54. There is a certain number consisting of two 
 digits or figures, and their sum is 6. If 18 be added 
 to the number, the sum will consist of the same digits 
 inverted. What is the number? 
 
 Since the number consists of two digits, the first 
 must be in the place of tens. Let thte be denoted by 
 x. Then, since their sum is 6, the other digit must 
 be 6 — x ; and the number is 10 x -}- 6 — x. Now, 
 if 18 be added to the number, the digits will be in- 
 verted. 
 
 Therefore, 10 x -f- 6 — x + 18 = 10 (6 — - x) -f- ft, 
 
 or 10 a? + 6 — # + 18 = 60 — 10 x + x. 
 
 10 a? — x + 10 # — x = 60 — 6 — 18; 
 
 18 x = 36. 
 
 and6-* = 4^ hedi S its -' 
 The number is 24 ; and 24 -f 18 = 42. 
 
 55. Required the number, from which if 27 be sub- 
 tracted, the digits of which it is composed will be in- 
 verted ; the sum of the digits being 9. 
 
 56. A grocer bought a quantity of oats at the rate 
 of 2 bushels for a dollar, and as many more at the 
 rate of 3 bushels for a dollar; and he sold them 5 
 bushels for 3 dollars, by which he gained $10. How 
 many bushels did he sell ? 
 
 Let x = the bushels of each sort. 
 
144 FIRST LESSONS IN ALGEBRA. 
 
 Then ~ = the cost of the first purchase, 
 
 and -| = the cost of the second purchase. 
 
 — -f- --> -f- 10 = the whole cost and profits. 
 
 As he sold 5 bushels for $3, he sold each bushel 
 for $f ; and, of course, he received for 2 x bushels, 
 
 6 x 
 
 that is, for x bushels of each sort, $ — . 
 
 Therefore, ^ = f + f + 10. 
 
 Ans. 54 t 6 t bushels. 
 
 57. Two clerks, A and B, have the same income. 
 A saves | of his ; but B, by spending $80 a year 
 more than A, at the end of 4 years finds himself $220 
 in debt. What was their income f 
 
 58. After spending £ of my money, and a of the 
 remainder, I had $96 left. How much had I at first ? 
 
 59. A traveller spent £ of his money in Boston ; } 
 of the remainder in Providence ; f of what was left 
 in New York ; f of the balance in Philadelphia, and 
 had $80 left. How much had he at first? 
 
 60. Divide 26 into three such parts, that, if the first 
 be multiplied by 2, the second by 3, and the third by 
 4, the products shall all be equal. 
 
 61. Divide 56 into two such parts, that, the larger 
 being divided by 7, and the smaller by 3, the sum of 
 their quotients may be 10. 
 
 62. A cistern has three cocks ; the first will fill it 
 in 5 hours, the second in 10 hours, and the third will 
 empty it in 8 hours. In what time will the cistern be 
 filled, if all the cocks are running together ? 
 
EQUATIONS OF THE FIRST DEGREE. 145 
 
 63. A school-boy, being asked his age, replied, that 
 £ of his age multiplied by T V of his age, would give a 
 product equal to his age. How old was he ? 
 
 64. A person has a lease for 99 years ; and, being 
 asked how much of it had expired, he replied, that § 
 of the time past was equal to f of the time to come. 
 How many years had the lease run ? 
 
 65. What number is there which may be divided 
 into either two or three equal parts, and the continued 
 product of those parts shall be equal? 
 
 66. A shepherd, driving a flock of sheep in time of 
 war, meets with a company of soldiers, who plunder 
 him of half his flock and half a sheep over ; and a 
 second, third and fourth company treat him in the 
 same manner, each taking half the flock left by the 
 last and half a sheep over, when but 8 sheep remain- 
 ed. How many sheep had he at first ? 
 
 67. A gentleman has two horses, and a chaise worth 
 $ 150. Now, if the first horse be harnessed, the horse 
 and chaise together will be worth twice as much as 
 the second horse ; but if the second horse be har- 
 nessed, they will be worth three times as much as 
 the first horse. What is the value of each horse ? 
 
 * 68. Divide 54 into two such parts, that, if the 
 greater be divided by 9, and the less by 6, the sum 
 of the quotients shall be 7. 
 
 69. A farmer sells a quantity of corn, which is to 
 the quantity left as 4 to 5. After using 15 bushels, 
 he finds he has £ as much left as he sold. How many 
 bushels had he at first ? 
 
 70. Divide 84 into two such numbers, that the quo- 
 
 13 
 
146 FIKST LESSONS IN ALGEBRA, 
 
 tient of the greater, divided by their difference, may 
 be four. 
 
 # 71. A laborer agreed to work for a gentleman a 
 year, for $72 and a suit of clothes; but at the end 
 of 7 months, he was dismissed, having received his 
 clothes and $32. What was the value of the clothes ? 
 
 72. A laborer reaps 35 acres of wheat and rye 
 For every acre of rye he receives 5 shillings ; and 
 what he receives for an acre of wheat, if it were 1 
 shilling more, would be to what he receives for an acre 
 of rye as 7 to 3. For the whole he receives £13. 
 How many acres are there of each sort ? 
 
 * 73. A man and his wife consumed a sack of meal 
 in 15 days. After living together 6 days, the woman 
 alone consumed the remainder in 30 days. How long 
 would the sack last either of them alone ? 
 
 74. A company of men, women and children con- 
 sists of 75 persons. The number of the men exceeds 
 that of the women by 5, and there are 13 more chil- 
 dren than adults. Required the number of men, 
 women and children. 
 
 75. Three adventurers, A, B and C, bought 10170 
 acres of wild land. By the terms of the contract, B 
 had 549 acres less than A, and C had 987 acres more 
 than B. How many acres had each ? 
 
 76. A man, being asked how much money he had, 
 replied, " If you multiply my money by 4, add 60 to 
 the product, divide the sum thus obtained by 3, and 
 then subtract 45 from the quotient, the remainder 
 will be the number of dollars I have." How much 
 money had he ? 
 
EQUATIONS OF THE FIRST DEGREE. 147 
 
 SECTION VI. 
 
 Two Unknown Quantities. 
 
 Elimination by Comparison. 
 
 1. A fruiterer sold to one person 6 lemons and 3 
 oranges for 42 cents ; and to another, 3 lemons and 8 
 oranges for 60 cents. W,hat was the price of each . ? 
 This question can be solved, without difficulty, by 
 means of only one unknown quantity : the solution 
 will be more simple, however, if two unknown quan- 
 tities are used. When the conditions of a question are 
 such as require two or more unknown quantities, our 
 first object is to obtain an equation involving but one 
 unknown quantity. This process is called Elimina- 
 tion. There are several methods of elimination, with 
 which the learner should become familiar, as he may 
 often find it convenient to use them all in the same 
 operation. 
 
 Let x = the price of a lemon, 
 
 and y s= the price of an orange. 
 
 a. Then 6 x -f- 3 y = 42, by the first sale. 
 
 b. 6 x = 42 — 3 y, by transposition. 
 
 c. x =i — - — , by division. 
 
 d Again, 3 x -f- 8 y = 60, by the second sale. 
 e. 3 x — 60 — 8 y, by transposition. 
 
 60—8?/ , ,. . .', 
 
 f. x = — — , by division. 
 
 Now, as things, which are equal to the same thing, 
 are equal to each other, we can form a new equation 
 
148 FIRST LESSONS IN ALGEBRA. 
 
 from the value of x, as determined in equations c 
 and f, which will contain but one unknown quantity. 
 
 42 — 3 y 60 — 8y 
 
 h. 126 — 9 y =s 360 — 48 y, by multiplication. 
 
 i. 48 y — 9 y = 360 — 126, by transposition. 
 
 j. 39 y ar 234, by uniting terms. 
 
 k. y =. 6, the price of an orange. 
 
 By substituting this value of y in equation c, we 
 have 
 
 _ 42— (3X6 ) ___ 42—18 
 X — 6 ~ — 6 ' 
 
 or x = -% 4 - = 4, the price of a lemon. 
 
 When a question involves two unknown quantities, 
 its conditions must admit of two equations, or it can- 
 not be solved. By this method of elimination, we 
 find the value of one of the unknown quantities in 
 each of the equations, and make the expressions of its 
 value, thus found, equal to each other. An equation is 
 thus obtained, involving but one unknown quantity. 
 
 * 2. A gentleman has two silver cups, and a cover 
 adapted to each, which is worth £10. If the cover 
 be put upon the first cup, its value will be twice that 
 of the second ; but if it be put upon the" second, its 
 value will be three times that of the first. What is 
 the value of each cup? 
 
 Let x = the value of the first cup, 
 and y = the value of the second cup. 
 Then 2 y = x -f- 10, by the question. 
 _ x + 10 
 
 y 2 * 
 
EQUATIONS OF THE FIRST DEGREE. 149 
 
 Again, y -f- 10 = 3 x, by the question. 
 y — 3 x — 10. 
 
 And 3 x — 10 =z — - — , by comparing values of y. 
 
 Ans. First cup, £6 ; second, £8. 
 
 *3. There are two numbers whose sum is 120; 
 and if 4 times the less be subtracted from 5 times the 
 greater, the remainder will be 150. Required the 
 numbers. 
 
 4. If the greater be added to half the less of two 
 numbers, the sum is 48 ; but if the less be added to 
 half the greater, the sum is 42. What are the num- 
 bers? 
 
 5. A vintner has two sorts of wine, which, if mixed 
 in equal parts, will be worth 15 shillings a gallon ; but 
 if 2 gallons of the first be mixed with 3 gallons of 
 the second, a gallon of the mixture will be worth only 
 14 shillings. What is each sort worth per gallon ? 
 
 6. If we add 7 to the numerator of a fraction, its 
 value becomes 2 ; if we add 7 to its denominator, it 
 becomes |. What is the fraction? 
 
 7. It is required to find two numbers, such that $ 
 of the first and £ of the second shall be 87, and £ 
 of the first and £ of the second shall be 55. 
 
 8. Says A to B, " 6 years ago, your age was double 
 mine ; and, in 4 years, my age will be £ of yours." 
 What is the age of each ? 
 
 9. There is a number consisting of two figures ; 
 and, if the number be divided by the sum of the fig- 
 ures, the quotient is 4 ; but if the figures be inverted, 
 
 13* 
 
150 FIRST LESSONS IN ALGEBRA. 
 
 and the number divided by 1 more than their sum, the 
 quotient will be 6. What is the number ? 
 
 Let x = the first figure, 
 
 and y = the second figure. 
 
 Then 10 x -f- y = the number. 
 
 As the first figure is in the place of tens, it must be 
 multiplied by 10, to express its local value. 
 
 io *+ r ky tne conditions of the question. 
 
 *+y+i = 6 J 
 
 10. A farmer sold to one man 10 bushels of corn 
 and 12 bushels of potatoes for 54 shillings ; and to 
 another, 2 bushels of corn and 4 bushels of potatoes 
 for 14 shillings. What was the price of each per 
 bushel ? 
 
 11. A gentleman gave to his two sons, A and B, 
 9800 dollars. At the end of a year, A finds that he 
 has spent -J- of his share ; but B, having spent only £ 
 of his, has just as much left as his brother. What 
 was the share of each ? 
 
 12. Says A to B, " Give me 6 dollars, and I shall 
 lia\e 4 times as much as you." " Rather give me 3 
 dollars," says B, "and I shall have just as much. as 
 you." How many dollars has each ? 
 
 *13. If I take 10 apples from A, he will still have 
 twice as many as B ; but if I give them to B, they 
 will each have the same number. How many have 
 they ? 
 
 14. If you multiply the greater of two numbers by 
 
EQUATIONS OF THE FIRST DEGREE. 151 
 
 2 and the less by 3, the sum of their products is 101. 
 And if you divide the greater by 4 and the less by 5 r 
 the sum of their quotients is 10. Required the num- 
 bers. 
 
 Elimination by Substitution. 
 
 15. A draper sold a yard of broadcloth and 3 yards 
 of velvet for 25 dollars ; and afterwards he sold 4 
 yards of broadcloth and 5 yards of velvet for 65 
 dollars. What was the price of each per yard ? 
 
 Let x — the price of the broadcloth, 
 
 and y 5a the price of the velvet. 
 
 a. Then x -f 3 y = 25, by the first sale, 
 
 b. and 4 x -f- 5 y = 65, by the second sale. 
 
 c. 4 x =r 65 — 5 y, by transposition. 
 
 65 — 5y . .... \ 
 
 d. a? = — - — , by division. 
 
 Now, if we substitute the value of x for x itself in 
 equation a, we shall obtain a new equation with but 
 one unknown quantity. 
 
 65 — 5 y + 12 y = 100 
 
 7 y — 100 — 65 = 35 
 
 y = 5, the price of the velvet. 
 
 d. a? = ~~° y = 4jp- — 10, the price of the broadcloth. 
 
 Either of the unknown quantities may be thus made 
 
152 FIRST LESSONS IN ALGEBRA. 
 
 to disappear ; but it will be found convenient to elimi- 
 nate that quantity which is the least involved. 
 
 16. A builder paid 5 men and 3 boys 42 shillings 
 for working a day ; he afterwards hired 7 men and 
 5 boys a day for 62 shillings. What were the wages 
 of each ? 
 
 Let x = the wages of a man, 
 
 and y i±= the wages of a boy. 
 
 Then 5 x -j- 3 y = 42, paid the first day. 
 
 5 x = 42 — 3 y, by transposition. 
 
 42 — 3w 
 
 Again, 7 £ -}- *> V — 62, paid the second day. 
 
 294 21m 
 
 ~ — -\- 5 y s= 62, by substitution. 
 
 294 — 21 y -f 25 y — 310 
 
 4 y = 16 
 
 ?/ = 4s. a boy's wages. 
 
 42 — 3 y 42 — 12 ^ , 
 
 x zz — - — =. — - — =z bs. a man s wages. 
 
 Let the following questions be solved in the same 
 manner, namely : Find the value of either of the un- 
 known quantities in one of the equations, as if the other 
 were known; and use the value thus found, instead of 
 the quantity itself in the other equation. This method 
 of making one of the unknown quantities disappear, 
 is called elimination by substitution. 
 
 17. Says A to B, " Give me $8, and I shall have 
 twice as much money as you will have left ; but if 
 I give you $6, my money will be equal to but \ of 
 yours." How much has each ? 
 
 18. A gentleman has two horses, and a chaise worth 
 
EQUATIONS OF THE FIRST DEGREE. 153 
 
 $250. If the first be harnessed, the horse and chaise 
 will be worth twice as much as the second horse ; but 
 if the second be harnessed, they will be worth three 
 times as much as the first horse. What is the value 
 of each horse ? 
 
 19. A merchant bought two lots of flour for 
 $576; the first lot for $5, and the second for $6, 
 per barrel. He then sold £ of the first lot and f of 
 the second for $353, by which he gained $11. How 
 many barrels were there in each lot ? 
 
 20. What fraction is that, whose numerator being 
 doubled, and denominator increased by 7, the value 
 becomes i; but the denominator being doubled, and 
 the numerator increased by 7, the value becomes 1 ? 
 
 21. A teacher, being asked the dimensions of his 
 school-room, answered, that if it were 5 feet broader 
 and 3 feet longer, the floor would contain 422 feet 
 more ; but if it were 3 feet broader and 5 feet longer, 
 the floor would contain 400 feet more. What were 
 the dimensions of the room ? 
 
 Let x — the length of the room, 
 and y — the breadth. 
 Now, to find the area, we must multiply the length 
 by the breadth. Consequently, x y = the area of 
 the floor. 
 
 Then + 5) X (y, + 3) == * y f 400, 
 and (a? + 3) X (y + 5) = a; y + 422. 
 
 22. If 6 feet were added to each side of a hall, the 
 breadth would be to the length as 6 to 7 ; but if 6 feet 
 were taken from each of the sides, they would be to 
 
154 FIRST LESSONS IN ALGEBRA. 
 
 each other as 4 to 5. Required the dimensions of 
 the hall. 
 
 23. A farmer has 86 bushels of wheat at 4s. 6d. a 
 bushel, with which he wishes to mix rye at 3s. 6d. a 
 bushel, and barley at 3s. a bushel, so as to make 136 
 bushels, that shall be worth 4s. a bushel. How much 
 rye and barley must he take ? 
 
 24. A merchant put 13 crates and 33 bales of goods 
 into a warehouse, which was all that it would hold. 
 After he had removed 5 crates and 9 bales, he found 
 that the house was two thirds full. How many crates 
 or bales would it take to fill it ? 
 
 25. If you multiply the greater of two numbers by 
 3 and the less by 4, the difference of their products 
 is 48 ; but if you divide the greater by 4 and the less 
 by 3, the sum of their quotients will be 14. What 
 are the numbers ? 
 
 26. A gentleman, having a quantity of gold and 
 silver coins, finds that 24 pieces of gold and 40 pieces 
 of silver, will pay a certain debt ; of which 5 pieces 
 of gold and 15 pieces of silver, will pay i part. How 
 many pieces of gold, and how many of silver, will pay 
 the whole debt? 
 
 - 27. In an election, the, two candidates received 
 1384 votes ; but if the successful candidate had re- 
 ceived but half his number of votes, and three times 
 as many as he received had been given for the other, 
 the whole number of votes would have been 2102. 
 How many votes were given for each ? 
 
 28. The length of a certain garden, which contains 
 128 square rods, is twice as great as its width ; and 
 
EQUATIONS OF THE FIRST DEGREE. 155 
 
 if the garden were 4 rods longer, it would contain an 
 acre. Required its length and width. 
 
 Elimination by Addition and Subtraction. 
 
 29. A man bought 3 bushels of wheat and 5 bush- 
 els of rye for 38 shillings ; he afterwards bought 6 
 bushels of wheat and 3 bushels of rye for 48 shil- 
 lings. What did he give a bushel for each ? 
 
 Let x = the price of a bushel of wheat, 
 
 and y = the price of a bushel of rye. 
 
 a. Then 3 x 4- 5 y = 38, by the first purchase, 
 
 b. and 6 a?4-3y = 48, by the second purchase. 
 
 If we multiply all the terms of equation a by 2, 
 [See Sec. I. of this Chap.] we shall have 
 
 c. 6 x -f- 10 y — 76. 
 
 Now, if we subtract equation b from equation c, 
 the remainder will be a new equation, containing only 
 one unknown quantity, whose value may be found as 
 before. Thus, 
 
 c. 6 x + 10 y — 76 
 
 b. 6 x -f- 3 y = 48 
 
 d. * 1 y = 28, by subtraction. 
 
 y =z 4, the price of the rye. 
 
 By substituting this value of y, in either of the 
 above equations, we shall obtain the value of x. 
 
156 FIRST LESSONS IN ALGEBRA. 
 
 a. 3 x + (5 X 4) = 38 
 
 3 x - 38 — 20 = 18 
 
 x =. 6, the price of the wheat. 
 
 The student will now understand why equation a 
 is multiplied by 2. Having determined to eliminate 
 x y l wish to make the coefficients of the terms con- 
 taining x, in the equations a and b, equal, that they 
 might cancel each other when subtracted. It mat- 
 ters not which of the unknown quantities is elimi- 
 nated in this way. 
 
 30. A boy bought 7 oranges and 5 lemons for 55 
 cents ; and afterwards let one of his companions have 
 4 oranges and 3 lemons for 32 cents, which was their 
 cost. What was the price of each ? 
 
 Let x = the price of an orange, 
 
 and y = the price of a lemon. 
 
 a. Then 7 x -\- 5 y = 55 3 ~) by the conditions of the 
 
 = 55,> 
 = 32, J 
 
 b. and 4 x -f- 3 y = 32, 5 question. 
 
 We will first eliminate y. As we cannot multiply 
 either equation by any number, which will make the 
 coefficients of the terms containing y alike, we must 
 multiply both equations by such numbers as will pro- 
 duce this result. 
 
 c. 21 x -\- 15 y = 165, by multiplying equation a by 3. 
 
 d. 20 a? -f- 15 y = 160, by multiplying equation b by 5. 
 
 x * = 5, by subtraction. 
 b. (4x5) +3 2/= 32, by substitution. 
 
 Sy = 32 — 20 = 12 ; and y = 4. 
 
 Ans. Oranges, 5 ; lemons, 4 cents. 
 
EQUATIONS OF THE FIRST DEGREE. 157 
 
 31. A gentleman paid for 6 pair of boots and 8 pair 
 of shoes $52 ; he afterwards paid for 3 pair of boots 
 and 7 pair of shoes $32. How much were the boots 
 and shoes a pair ? 
 
 Let x = the price of the boots, 
 and y = the price of the shoes. 
 a. Then 6 x -f- 8 y = 52, ) by the conditions of the 
 b. and 3 x -f- 7 y = 32, > question. 
 
 c. 3 x -\- 4 y = 26, by dividing equation a by 2. 
 
 e. * 3 y = 6, by subtracting equation c 
 from b. 
 y =. 2, the price of a pair of shoes. 
 c. 3 x -J- (4 X 2) = 26, by substitution. 
 
 . "^ a> = 26 -— 8 = 18, 
 and x = 6, the price of a pair of boots. 
 
 The equation a is divided by 2, to make the coeffi- 
 cient of x, the quantity to be eliminated, equal to the 
 coefficient of the same quantity in equation b. 
 
 32. If twice A's money be subtracted from 3 times 
 B's, the remainder is $38; but if twice B's money 
 be subtracted from 3 times A's, the remainder is $83. 
 How much has each ? 
 
 Let x =z A's money, 
 and y = B's money. 
 a. Then 3 y — 2 x = 38, ") by the conditions of the 
 b. and 3 x — 2 y = 83, ) question. 
 
 If we multiply equation a by 3, and equation b 
 by 2, the coefficients of the terms containing x will 
 be alike. 
 
 14 
 
158 FIRST LESSONS IN ALGEBRA. 
 
 c. 9 y — (} x =b 1 1 4, by multiplying equation a by 3. 
 
 d. — 4y -\- 6 x = 166, by multiplying equation b by 2. 
 
 5 y f "== 280, by addition, 
 
 and ?/ = 56, B's money. 
 
 b. 3 a? — ( 2 X 56) = 83, by substitution. 
 
 3 x = 83 + 112 = 195, 
 
 and a? == 65, A's money. 
 
 As the terms of the equations c and t>, containing 
 x, have unlike signs, they are cancelled by addition. 
 
 From these operations may be derived the follow- 
 ing Rule for removing one of the unknown quan- 
 tities : 
 
 Having determined which of the unknown quanti- 
 ties you will eliminate, make the coefficients of the 
 terms, containing that quantity, the same in both equa- 
 tions, either by multiplication or division. 
 
 If the signs of these terms are unlike, add both equa- 
 tions together: if they are alike, subtract the smaller 
 from the larger equation. 
 
 When the student does not readily find a proper 
 multiplier, he can multiply each equation by the co- 
 efficient of that term, in the other, which contains 
 the unknown quantity to be removed. 
 
 33. Says A to B, " § of the difference of our money 
 is equal to yours ; and, if you give me $2, I shall 
 have 5 times as much as you." How much has 
 each? 
 
 34. There is a certain number consisting of two 
 figures ; and if 2 be added to the sum of its digits, 
 the amount will be 3 times the first digit ; and if 18 
 
EQUATIONS OF THE FIRST DEGREE. 159 
 
 be added to the number, the digits will be inverted. 
 What is the number ? 
 
 35. A man has money in dwo drawers, and $25 in 
 his purse. Now, if he put his purse into the first 
 drawer, it will contain f as much as the second ; but 
 if he put his purse into the second drawer, the sum 
 in the first will be to the sum in the second as 5 to 
 13. How much is there in each drawer ? 
 
 36. Two clerks, A and B, send ventures, by which 
 A gained $20, and B lost $50, when the former had 
 twice as much as the latter; but had B gained $20, 
 and A lost $50, then B would have had 4 times as 
 much as A. What sum was sent by each ? 
 
 37. A farmer, having mixed a certain quantity of 
 barley and oats, found that, if he had mixed 6 bushels 
 more of each, he would have put into the mixture 7 
 bushels of barley for every 6 of oats ; but if he had 
 mixed 6 bushels less of each, he would have put in 6 
 bushels of barley for every 5 of oats. How many 
 bushels did he mix? 
 
 38. A person has a gold watch and a silver one, 
 and a chain for both worth $8. Now, the silver watch 
 and chain are together worth half as much as the 
 gold watch ; but when the chain is on the gold watch, 
 they are together worth 3 times as much as the silver 
 watch. What is the value of each ? 
 
 39. If a certain volume contained 12 more pages, 
 with 3 lines more upon a page, the number of lines 
 would be increased 744 ; but if it contained 8 pages 
 less, and the lines on a page were not so many by 4, 
 the whole number of lines would be diminished 680. 
 
160 FIRST LESSONS IN ALGEBRA. 
 
 How many pages are there in the book ? and how 
 many lines on a page ? 
 
 40. Two neighbors, A and B, possess 562 acres of 
 land. If A's farm were 4 times, and B's 3 times, as 
 large as each of them is, they would both together 
 have 1924 acres. How many acres has each ? 
 
 41. Two men owe more money than they can pay. 
 Says A to B, " Give me £ of your property, and I 
 shall be able to pay my debts." " If you will give 
 me i of yours," replies B, " I shall be able to pay 
 my own." The amount of A's debts is $1500, and 
 of B's, $2125. How much property has each in his 
 possession ? 
 
 42. A trader bought at auction two pipes contain- 
 ing wine. For one he gave 8s. a gallon ; for the 
 other, 105. 6d: ; and the whole came to $160. Hav- 
 ing sold 25 gallons from the first pipe, and 13 gallons 
 from the second, he mixed the remainder together, 
 and added 15f gallons of water. Afterwards, 5f gal- 
 lons of the mixture leaked out; and the remainder 
 was worth 8s. a gallon. How many gallons did each 
 pipe contain ? 
 
 43. A man had 32 gallons of wine, in two barrels. 
 Wishing to have an equal quantity in each, he poured 
 out of the first into the second as much as it already 
 contained ; again, he poured out of the second into 
 the first as much as it then contained ; and, finally, 
 he poured out, of the first into the second as much as 
 still remained in it. Each barrel then contained the 
 same quantity. How many gallons did they contain 
 originally ? 
 
EQUATIONS OF THE FIRST DEGREE. 161 
 
 SECTION VII. 
 
 Three or more Unknown Quantities. 
 
 1. Three boys, A, B and C, bought fruit at the same 
 time. A bought 4 oranges, 7 peaches and 5 pears, 
 for 51 cents ; B bought 6 oranges, 8 peaches and 10 
 pears, for 74 cents ; and C bought 9 oranges, 3 
 peaches and 2 pears, for 58 cents. What was the 
 price of each ? 
 
 Let x = the price of an orange, 
 
 y x= the price of a peach, 
 and z =s the price of a pear. 
 a. Then 4 x -f- 7 y -|- 5 z = 51, by A's purchase. 
 b. 6 x + 8 y + 10 z =p 74, by B's, 
 c. 9 x + 3 y + 2 z = 58, by C's. 
 
 These unknown quantities must be made to disap- 
 pear, one at a time. Either method of elimination, 
 explained in the last section, may be used. We must, 
 in the first place, deduce, from these three equations, 
 two others, which shall contain but two unknown quan- 
 tities each. 
 
 If we multiply equation a by 2, the coefficient of 
 z will be the same as it is in equation b. 
 
 d. 8a?+14y+10s= 102 
 b. 6 x + 8 y + 10 z = 74 
 
 e. 2 a? + 6 y * = 28, by subtraction. 
 
 f. x -f- 3 y = 14, by division. 
 14* 
 
162 FIRST LESSONS IN ALGEBRA. 
 
 Again, if we multiply c by 5, the coefficients of z 
 in b and c, will be alike. 
 
 G. 
 11. 
 
 45 
 6 
 
 * + 
 
 * + 
 
 15 y 
 
 + 
 
 + 
 
 10 2 
 10^ 
 
 — : 
 
 290 
 74 
 
 H. 
 
 39 
 
 * + 
 
 iy 
 
 
 ■K 
 
 3= 
 
 216, 
 
 216, by subtraction. 
 
 We have now two equations, namely, f and h, 
 which contain only two unknown quantities. 
 
 f. x -f 3 y = 14, 
 i. and x .= 14 — 3 y, by transposition. 
 
 If we substitute this value of x in equation h, we 
 shall have 
 
 k. 39 (14 — 3 y) + 1 y = 216. 
 
 546 — 117 y + 7 y == 216, by multiplication. 
 
 546 — 216 =i 117 y — ■ 7 y, by transposition. - 
 
 330 is 110 y, by reduction of terms. 
 
 l. y = 3, by division. 
 
 If we substitute this value of y in equation f, we 
 shall have 
 
 x + (3 X 3) ==' 14. 
 m. x = 14 — 9 = 5, by transposition. 
 
 And if we substitute the values of x and y, as de- 
 termined in l and m, in any of the preceding equations 
 containing z, we shall obtain the value of that quan- 
 tity. Take equation c, for instance. 
 
 c. 9x5 + 3x3 + 22 = 58, 
 or 45 + 9 + 2 z = 58. 
 
EQUATIONS OF THE FIRST DEGREE. 163 
 
 2 z = 58 — 45 — 9 — 4, by transposition, 
 and i'sss 2. 
 
 Ans. An orange, 5 cents ; 
 A peach, 3 cents ; 
 A pear, 2 cents. 
 
 2. A fruiterer sold to A 5 oranges, 6 peaches and 
 7 pears, for 75 cents ; to B 8 oranges, 9 peaches and 
 5 apples, for 94 cents ; to C 2 oranges, 8 pears and 
 10 apples, for 56 cents ; and to D 3 peaches, 6 pears 
 and 9 apples, for 48 cents. What was the price of 
 each ? 
 
 To solve this question, we must use four unknown 
 quantities ; but their values may be found according 
 to the principles already explained. It will be ob- 
 served, that all the unknown quantities do not enter 
 into each of the equations. 
 
 Let v = the price of an orange, 
 x = the price of a peach, 
 y =r the price of a pear, 
 
 and z =. the price of an apple. 
 
 a. Then 5 v -\- 6 x + 7 y = 75, by A's purchase, 
 
 b. 8 v + -9 x + 5 z = 94, by B's, 
 
 c. 2 v + 8 y + 10 z = 56, by C's, 
 
 d. 3 x -f 6 y + 9 z = 48, by D's. 
 
 c. 2 v -f 8 y -f 10 z = 56, 
 
 or 2 v = 56 — 8 y — 10 z, by transposition, 
 
 e. and v = 28 — 4 y — 5 z, by division. 
 
 By substituting this value of v, in equations a and 
 
 b, we have 
 
164 FIRST LESSONS IN ALGEBRA. 
 
 5 (28 — 4 y — 5 z) + 6 x + 7 y = 75, from equation a. 
 140 — 20 ?/ — 25 *-}- 6 a? + 7 y = 75, by multiplication. 
 
 f. 6x — 13y — 25^= — 65, by transposition and addition. 
 And 8(28 — 4 y — 5 z)-\-9 x-\-5 2=94, from equation b. 
 224 — 32 y — 40 2+9 # + 52= 94, by multiplication. 
 
 g. 9x — 32y — 352= — 130, by transposition & addition. 
 
 We have now three equations, namely, d, f and 
 
 g, containing but three unknown quantities, x, y and 2. 
 
 If we multiply equation d by 2, the coefficients of 
 
 ,c, 
 
 in 
 
 H. 
 F. 
 
 d and 1 
 
 6x + 
 6 x — 
 
 ', will be made 
 
 12 y + 18 2 - 
 
 13 y — 25 2 : 
 
 equal. 
 
 96 
 = — 65 
 
 1. * 25 y -\- 43 2 = 161, by subtraction. 
 
 Again, if we multiply equation d by 3, the coeffi- 
 cients of x, in d and g, will be the same. 
 
 k. 9#+ 18 y + 27 2 = 144 
 g. 9 x — 32 y — 35 2 = — 130 
 
 l. * 50 y + 62 2 = 274, by subtraction. 
 
 The two equations, 1 and l, contain but two un- 
 known quantities, which may be eliminated in the 
 usual manner. 
 
 If we multiply equation 1 by 2, we shall have 
 
 m. 50 y + 86 2 = 322 
 
 l. 50 y + 62 2 = 274 
 
 * 24 2 = 48, by subtraction, 
 
 and 2 = 2, the price of an apple. 
 
 By putting this value of 2 in equation l, we have 
 
EQUATIONS OF THE FIRST DEGREE. 165 
 
 l. 50 y -f 124 — 274, 
 
 or 50 y = 274 — 124 = 150, 
 
 and y = 3, the price of a pear. 
 
 By substituting the values of y and z in equation 
 d, we obtain 
 
 d. 3 x -f 18 -f 18 = 48, 
 
 or Sx = 48 — 36 = 12, 
 
 and x = 4, the price of a peach. 
 
 And, by putting the values of x and y in equation 
 
 a, we have 
 
 5 „ _|_ 24 -f 21 == 75, 
 
 or 5 « = 75 — 45 = 30, 
 and v = 6, the price of an orange. 
 
 When the value of one of the unknown quantities 
 is determined, it may be substituted for that quantity 
 in either of the equations. Any or all of the differ- 
 ent methods of elimination may be used ; and it mat- 
 ters not in what order the equations are compared 
 together. 
 
 3. A miller sold to one man 12 bushels of wheat, 
 10 of rye and 16 of barley, for £9 2s. ; to another, 
 7 bushels of wheat, 20 of rye and 10 of barley, for 
 £8 19s. ; and to a third, 15 bushels of wheat, 8 of 
 rye and 20 of barley, for £10 5s. What was the 
 price of each per bushel ? 
 
 4. Divide 125 into four such parts, that, if the first 
 be increased by 4, the second diminished by 4, the 
 third multiplied by 4, and the fourth divided by 4, the 
 sum, difference, product and quotient shall all be 
 equal. 
 
166 FIRST LESSONS IN ALGEBRA. 
 
 5. Find 3 such numbers, that the first with $ of the 
 other two, the second with } of the other two, and 
 the third with £ of the other two, shall each be equal 
 to 34. 
 
 6. If A and B can perform a piece of work in 8 
 days, A and C in 9 days, and B and C in 10 days, 
 in how many days can each alone perform the same 
 work ? 
 
 7. Says A to B and C, " If each of you will give 
 me 10 cents, my money will be, to what you will both 
 have left, as 4 to 5." Says B to A and C, " If each 
 of you will give me 10 cents, my money will be, to 
 what you will then have, as 5 to 4." Says C to A 
 and B, " If you will give me 10 cents each, I shall 
 have twice as much money as both of you." How 
 many cents has each ? 
 
 8. Three persons divided a sum of money between 
 them in such a manner, that the shares of A and B 
 together amounted to $900; the shares of A and C, 
 to $'800 ; and the shares of B and C, to $ 700. Re- 
 quired the sum divided, and the share of each. 
 
 9. A man, with his wife and son, talking of their 
 ages, said, that his age, added to that of his son, was 
 16 years more than that of his wife ;- the wife said., 
 that her age added to that of her son, made 8 years 
 more than that of her husband ; and that all their 
 ages together amounted to 88 years. -What was the 
 age of each ? 
 
 10. There are two such fractions, that if 3 be added 
 to the numerator of the first, its value is double that 
 of the second ; but if 3 be added to the denominator, 
 
EQUATIONS OF THE FIRST DEGREE. 167 
 
 their values a:e equal. Now, the sum of the two frac- 
 tions is 9 times as great as their difference; and if the 
 numerator of their product be increased by 10, its 
 value will be equal to that of the first fraction. Re- 
 quired the fractions. 
 
 11. A grocer has four kinds of tea, marked A B, 
 C and D. When he mixes together 7 pounds of A, 
 5 of B and 8 of C, the mixture is worth $1,21 a 
 pound. When he mixes together 3 pounds of A, 10 
 of C and 5 of D, the mixture is worth $1,50 a pound. 
 At one time he sold 8 pounds of A, 10 of B, 10 of 
 C and 7 of D, for $48 ; and, at another time, he sold 
 18 pounds of A and 15 of D, for $48. What was 
 a pound of eacii worth ? 
 
 12. There are two fractions haying the same de- 
 nominator. Now, if 1 be subtracted from the. nume- 
 rator of the smaller, its value will be ^ of the larger 
 fraction ; but if 1 be subtracted from the numeiator 
 of the larger, its value will be double that of the 
 smaller. And if the numerator of the smaller be sub- 
 tracted from that of the larger, the value of the frac- 
 tion will be £. What are the fractions . ? 
 
 13. Three travellers, A, B and C. have a bill at a tav- 
 ern, which neither of them alone is able to pay. Says 
 A to B, " If you will lend me ^ of your money, I can 
 pay the bill." B says to C, " I will pay it, if you will 
 lend me £ of your money." But C concludes to 
 borrow \ of A's money, and pay the bill himself, 
 which amounts to 6£ dollars. How much money has 
 each of the travellers ? 
 
168 FIRST LESSONS IN ALGEBRA. 
 
 SECTION VIII. 
 
 Questions for Practice. 
 
 1. A vintner sold to one man 16 dozen of sherry 
 wine and 19 dozen of port, for $382 ; and to another 
 man, 24 dozen of sherry and 17 dozen of port, for 
 $458 ; — the prices being the same to both. What 
 was the price of each kind of wine ? 
 
 2. Two sisters having bought some lace, says Mary 
 to Ann, " Give me a yard of yours, and I shall have 
 as much as you." Ann replied, " If you will give me 
 a yard of yours, I shall have twice as much as you." 
 How many yards had each ? 
 
 3. Required two such numbers, that, if £ of the 
 first be added to |- of the second, the sum shall be 66 ; 
 but if f of the first be added to £ of the second, their 
 sum shall be 60. 
 
 4. Two persons, A and B, talking of their ages, 
 says A to B, " 12 years ago I was twice as old as 
 you ; and in 12 years my age will be to yours as 3 to 
 2." What is the age of each ? 
 
 5. Three young men, A, B and C, speaking of their 
 money, A says to B and C, " If each of you will give 
 me $5, I shall have just half as much as both of you 
 will have left." B says to A and C, " If each of you 
 will give me $5, I shall have just as much as both of 
 you will have left." C says to A and B, " If each 
 of you will give me $5, I shall have twice as much 
 as both of you will have left." How much has each ? 
 
 6. If you add 2 to the numerator of a certain frac- 
 
EQUATIONS OF THE FIRST DEGREE. 169 
 
 tion, its value becomes £ ; but if you add 2 to its de- 
 nominator, the fraction will be equal to £. What is 
 the fraction ? 
 
 7. Three factory girls, A, B and C, weave 62 yards 
 of cloth in a given time. A weaves 4 times as many 
 yards as C, added to twice the yards woven by B ; 
 and twice A's part, added to 3 times B's, is equal to 
 17 times Cs. How many yards does each weave ? 
 
 8. Find three such numbers, that f of the first, f 
 of the second, and f- of the third, shall be 50 ; § of 
 the first, and twice the difference of the third and 
 second, shall be 40 ; and 10 less than ^ of the sum of 
 all the numbers shall be 30. 
 
 9. Two men, A and B, are employed to set up 220 
 rods of fence. If A work 9 days and B 8, the fence 
 will not be completed by 2 rods ; but if A work 8 
 days and B 9, they will be able to finish the fence 
 and 4 rods more. How many rods can each build in 
 a day? 
 
 10. A farmer employed three laborers, A, B and 
 C, to work at different times. At one time, A and 
 B together earned $56 in 8 weeks ; at another time, 
 A and C earned $54 in 9 weeks ; and, at another, 
 B and C earned $50 in 10 weeks. What did he pay 
 each man for a week's work ? 
 
 * 11. If you divide the greater of two numbers by 
 the less, the quotient will be 7 ; and the amount of 
 the numbers is 1008. Required the numbers. 
 
 12. A farmer, to pay a debt to a trader, agrees to 
 fill a certain chest with a mixture of corn and oats ; 
 the corn being 5 shillings a bushel, and the oats 3 
 15 
 
170 FIRST LESSONS IN ALGEBRA. 
 
 shillings. If he delivers 7 bushels of oats, and the 
 balance of the debt in corn, the chest will not be full 
 by 2 bushels ; but if he delivers 6 bushels of corn, 
 and then fills the chest with oats, 6 shillings of the 
 debt will remain unpaid. What is the amount of the 
 debt ? How many bushels does the chest hold ? How 
 many bushels of each kind of grain must he deliver ? 
 
 13. A boy bought, at one time, 5 apples, 6 pears 
 and 4 peaches, for 44 cents ; at another time, 7 pears, 
 5 peaches and 3 oranges, for 56 cents ; at another ; 8 
 apples, 12 peaches and 5 oranges, for 89 cents ; and 
 at another, 10 apples, 3 pears and 9 oranges, for 74 
 cents. What did he pay for each kind of fruit . ? 
 
 14. Three persons, A, B and C, talking of their 
 money, A says to B and C, " Give me | of your money, 
 and I shall have $85." B says to A and C, " Give 
 me i of your money, and I shall have $85." C says 
 to A and B, " Give me £ of your money, and I shall 
 have $85." What has each ? 
 
 15. A gentleman gave $4350 for a house-lot, the 
 land being valued at $2 a foot. If it had been 6 feet 
 wider, it would have cost $5394. What were the 
 length and breadth of the lot ? 
 
 16. I have a certain number of cents in each hand. 
 If I put 10 out of my left hand into my right, there 
 will be twice as many in my right as remain in my 
 left ; but if I put 10 out of my right hand into my left, 
 there will be three times as many in my left hand as 
 remain in my right. How many cents have I in each 
 hand? 
 
 . 17. Three boys, A, B and C, play with marbles. 
 
EQUATIONS OF THE FIRST DEGREE. 171 
 
 First, A loses to B and C as many as each of them 
 has. Next, B loses to A and C as many as each of 
 them now has. Lastly, C loses to A and B as many 
 as each of them now has. After all, each of them 
 has 16 marbles. How many had each at first? 
 
 18. The fore-wheel of a coach makes 5 revolutions 
 while the hind-wheel is making 4 ; but if the circum- 
 ference of each were one yard greater, their revolu- 
 tions would be to each other as 6 to 5. What is the 
 circumference of each in feet ? 
 
 19. Three sportsmen, before they separated in the 
 fields, agreed to make an equal division of whatever 
 game they might take. During the day, they bagged 
 96 birds ; and, in order to divide them equally, A 
 gave B and G as many as they took; next, B gave A 
 and C as many as they then had ; and, finally, C 
 shared in the same manner with A and B, when they 
 all had the same number. How many birds were 
 taken by each ? 
 
 20. A merchant sent ventures to sea for his three 
 children, A, B and C, upon the condition that they 
 should make an equal division of the proceeds. When 
 the ship arrived, A gave to B and C a sum equal to ^ 
 of their ventures. In like manner, B gave to A and 
 C a sum equal to i of what they then had. And C 
 found, if he gave to A and B a sum equal to ^ of what 
 they already had, that they would each have 64 dol- 
 lars. To how much did each of the ventures amount ? 
 
172 FIRST LESSONS IN ALGEBRA. 
 
 CHAPTER IX. 
 EXERCISES IN GENERALIZATION 
 
 We have already seen that Algebra enables us to 
 solve various questions with facility, and to obtain 
 their true answers in numbers. But this is not its 
 only, nor even its highest office. It also enables us to 
 deduce general truths from particular instances, and 
 thus to form rules for conducting numerical calculations. 
 This use of Algebra will be illustrated by a few mis- 
 cellaneous examples. 
 
 1. Add together the sum and the difference of 
 294 and 175. 
 
 Their sum, 294 + 175 = 469 
 
 Their difference, 294 — 175 — 119 
 
 Ans. 588. 
 
 To generalize this question, let a represent the 
 greater quantity, and b, the less. Then a -\- b will 
 be their sum, and a — b, their difference ; which we 
 are required to add together. 
 a -f b 
 a — b 
 Ans. 2 a. 
 
GENERALIZATION. 173 
 
 To express the result of this operation in language : 
 If we add together the sum and the difference of any 
 two quantities, the amount will be equal to twice the 
 greater quantity. 
 
 2. Add together the sum and the difference of 90 
 and 65, by this formula or rule. 
 
 3. Add together the sum and the difference of 136 
 and 97. 
 
 4. Add together the sum and the difference of 375 
 and 129. 
 
 5. Subtract the difference of 324 and 278 from 
 their sum. 
 
 Their sum, 324 -f- 278 = 602 
 
 Their difference, 324 — 278 = _46 
 
 Ans. 556. 
 Indicating the two quantities as before, we must 
 subtract a — b from a -\- b. 
 
 a -f- b 
 a — b 
 
 2 b. 
 
 That is, if we subtract the difference of two quanti- 
 ties from their sum, the remainder is equal to twice the 
 smaller quantity. 
 
 6. From the sum of 77 and 25 take their difference. 
 
 7. From the sum of 139 and 62 take their differ- 
 ence. 
 
 8. From the sum of 827 and 364 take their dif- 
 ference. 
 
 9. Multiply the sum of 139 and 87 by their dif- 
 ference. 
 
 15* 
 
174 FIRST LESSONS IN ALGEBRA. 
 
 Their sum, 139 -f 87 = 226 
 
 Their difference, 139 — 87 = 52 
 
 452 
 1130 
 Their product, 11752 
 
 If we represent the two quantities by the same let- 
 ters as before, we shall have a -f- b, their sum, to be 
 multiplied by a — b, their difference. 
 
 a + b 
 
 a — b 
 a 2 + a& 
 
 — a b — M 
 
 Their product, a 2 * — 6 2 . 
 
 To express this formula in words : The product of 
 the sum of two numbers, multiplied by their difference, 
 is equal to the difference of their second powers. 
 
 10. Multiply 9 + 7 by 9 — 7, according to this 
 rule. 
 
 11. Multiply 12 -f 8 by 12 — 8. 
 
 12. Multiply 24 -f- 15 by 24 — 15. 
 
 13. Given 9310 and 9298, and required the differ- 
 ence of their squares. 
 
 The answer to this question can be far more easily 
 obtained by means of the above formula, than by ac- 
 tually involving each of these numbers, and subtract- 
 ing the second power of the one from the second 
 power of the other. For, if we multiply their sum by 
 their difference, the product will be the difference of 
 their second powers. 
 
GENERALIZATION. 175 
 
 The last question is solved by means of the for- 
 mula, in the following manner : — 
 
 9310 
 9298 
 
 18608, the sum of the numbers. 
 9310 — 9298 == 12, their difference. 
 
 223296, the difference of their squares. 
 
 This principle should be remembered ; as calcula- 
 tions of this sort may be very much abridged by an 
 application of it, when the given numbers are large. 
 
 14. Given 23136 and 21947, and required the dif- 
 ference of their second powers. 
 
 15. Given 15925 and 14987, and required the dif- 
 ference of their second powers. 
 
 16. Given 987264 and 978351, and required the 
 difference of their second powers. 
 
 17. Given 999999 and 999993, and required the 
 difference of their second powers. 
 
 18. What is the second power of 28 ? Ans. 784. 
 
 Let a = 20, and 6 = 8; then a -f b = 28. 
 a + b 
 a + b 
 
 a 2 + ab 
 
 + a b -f 6 2 
 
 a 2 _L.2 a 5_|_fl2. 
 
 Hence it appears, that the second power of the sum 
 of two quantities is equal to the sum of their second 
 powers, increased by twice their product. 
 
176 FIRST LESSONS IN ALGEBRA. 
 
 19. What is the second power of a — b 1 
 
 a — b 
 
 a — b 
 a? — ab 
 
 — ab + b* 
 az—2ab + b*. 
 
 That is, the second power of the difference of two 
 quantities is equal to the sum of their second powers, 
 diminished by twice their product. 
 
 20. An artist sold to A and B two pictures for 573 
 dollars ; and B's picture cost 57 dollars more than A's. 
 What was the price of each ? 
 
 Let x = the price of A's picture. 
 
 Then a? -|- 57 = the price of B's. 
 
 And x -\- x -J- 57 == 573, by the question, 
 
 or 2 x + 57 = 573 ; 
 
 2 x = 573 — 57 = 516, by transposition, 
 
 and x = 258. 
 
 x + 57 == 315. 
 
 Ans. A's, $258; B's, $315. 
 
 In this question, we are required to divide 573 into 
 two such parts, that the greater shall exceed the less 
 by 57. In other words, haviiig the sum and the differ- 
 ence of two numbers given, we are required to find those 
 numbers. 
 
 Let the sum of the numbers be represented by a, 
 and their difference by b; and let x = the smaller 
 number. Then, as the difference of the numbers is 
 b, we have x -\- b = the larger number. 
 
GENERALIZATION. 177 
 
 Therefore, x -\- x -\- b = a, 
 
 or 2 x -\- b = a, 
 
 and 2 x = a — b ; 
 
 x b= — — , the smaller number. 
 
 To express this formula in words : If from the sum 
 of two numbers we subtract their difference, half of the 
 remainder will be equal to the smaller number. 
 
 Again, x =z — ^— , as above. 
 
 Therefore, x -\- b == -^ -f- b, by adding b to each 
 
 ■ b + 2b [member. 
 
 b . ib 
 
 Orx + b=~ + j = 
 
 x -f- b = — r~, the greater number. 
 
 That is, iy to tfAe smwi of two numbers we add their 
 difference, half of the amount ivill be equal to the 
 greater number. 
 
 21. The sum of two numbers is 175, and their dif- 
 ference 81. What are the numbers ? 
 
 Here, a = 175, and b ±= 81. 
 
 a — b 175 — 81 94 'l! . „ , 
 
 = — - — = — 2= 4i, the smaller number. 
 
 2 2 2 7 
 
 —^- = — ^- ='-—'= 128, the larger number. 
 
 2 2 2 > b 
 
 22. The salaries of two men, A and B, amount to 
 $3529 per annum ; and A receives $721 more than 
 B. What is the salary of each ? 
 
 23. A gentleman paid $387 for a horse and chaise; 
 and the chaise cost $75 more than the horse. What 
 was the price of each ? 
 
 24. A man left an estate of $9134, to be so divided 
 
178 FIRST LESSONS IN ALGEBRA. 
 
 I 
 
 between his widow and son, that the former should 
 receive $1486 more than the latter. What was the 
 share of each ? 
 
 25. A's farm contains 12 acres more than B's ; and 
 the two farms together contain 162 acres. Required 
 the number of acres in each. 
 
 26. Says A to B, " If you will give me 16 dollars, 
 I shall have as much money as you ; and we both 
 have 130 dollars." How much money has each? 
 
 27. A cistern has two cocks, which together will 
 discharge 203 gallons of water in an hour, and one 
 cock will discharge 13 gallons more than the other. 
 How much will each discharge if running alone ? 
 
 28. How much tea, at 5s. per pound, must be given 
 in barter for 75 gallons of wine at 8s. per gallon ? 
 
 Let x = the pounds of tea. 
 
 Then 5 x = 75 X 8 = 600 ; 
 
 and x =3 120. 
 
 Ans. 120 lbs. 
 
 In general terms. How much tea, at a per pound, 
 must be given in barter for b gallons of wine at c per 
 
 gallon ? 
 
 Let x == the pounds of tea. 
 Then a x = b c, 
 
 . be 
 
 and x == -. 
 
 To express this formula for Barter in words : Mul- 
 tiply the given quantity by its price, and divide the 
 product by the price of the quantity required. 
 
 29. How much oil, at 4s. a gallon, must I give in 
 barter for 58 barrels of flour at 5 dollars a barrel ? 
 
GENERALIZATION. 179 
 
 30. How much loaf-sugar, at 25 cents a pound, 
 must be given in barter for 625 yards of cotton, at 13 
 cents a yard ? 
 
 31. A man sells 17 acres of land at 30 dollars per 
 acre, and is to receive his pay in flour at 5 dollars a 
 barrel. How many barrels must he receive ? 
 
 32. How many gallons of wine, at $1,75, must I 
 give in exchange for 1295 pounds of shot, at 5 cents 
 a pound? 
 
 33. Two merchants, A and B, engage in a specu- 
 lation, and gain $750, of which B is to have 4 times 
 as much as A. What is the share of each ? 
 
 Let x = A's share, 
 and 4 x = B's share. 
 Then a? + 4 a? == 750, 
 
 or 5 x — 750, 
 
 and x = 150, A's share. 
 
 4 x = 600, B's share. 
 
 In this question, a given sum is to be divided be- 
 tween two persons in unequal proportions. It may 
 be generalized as follows : — 
 
 Represent the sum to be divided by a. 
 
 Let x == A's share, 
 
 and 4i= B's share. 
 
 Then x -\- 4 x = a, 
 
 or 5 x = a, 
 
 and x = —, A's share. 
 4 x =: — , B's share. 
 
180 FIRST LESSONS IN ALGEBRA. 
 
 Now, whatever sum may be represented by a, it is 
 evident that A is to receive -i part of it, and B f of it 
 
 If a, the sum to be divided, be $ 1000, J- == 200, 
 
 and & = 800. 
 o 
 
 If a represent $1500,y = 300, and ^- = 1200. 
 
 This question may be still farther generalized, if it 
 be announced as follows : — 
 
 Two merchants, A and B, engage in a speculation, 
 and gain a dollars, of which B is to have b times as 
 much as A. What is the share of each t 
 
 Let x == A's part of the gain. 
 
 Then b x = B's part. 
 
 And x -\- b x = a, by the question, 
 
 or (1 -j- b) x = a, 
 
 • 
 
 and x = TT~ b ) A's part. 
 
 h x — TTb x h ~ Wi f B ' s part ' 
 
 In the equation x -\- b x = a, the first member, 
 a? -j- b x, is equal to (1 -\- b) x ; therefore, (1 -}- b) 
 may be regarded as the coefficient of x. 
 
 34. By this formula, divide $1200 between two 
 men in such a manner, that one shall have 5 times as 
 much as the other. 
 
 Here, a = 1200, and b = 5. 
 
 rr»L C a 1200 1200 «™ 
 
 Therefore, — — = - — - =s -— = 200. 
 
 1 ~\- b 1 -\-o D 
 
 . j ab 6000 6000 .*w** 
 
 And — — = — — = -— - — 1000. 
 
 l-\-b 1 + 5 6 
 
 Ans. $200; and $1000. 
 
GENERALIZATION. 181 
 
 35. A father says to his son, " I am four times as 
 old as you, and the sum of our ages is 85 years." 
 Required the age of each. 
 
 36. Divide 2592 dollars between two men in such 
 a manner, that one shall have 7 times as much as the 
 other. 
 
 37. Says A to B, " Our estates together are worth 
 36465 dollars ; but my estate is worth only one fourth 
 part as much as yours." What is the value of each 
 estate ? 
 
 38. A merchant sold a quantity of blue, black and 
 mixed broadcloths, for $5000 ; the black at $10, the 
 blue at $1L, and the mixed at $7 per yard. There 
 were twice as many yards of the blue as of the black, 
 and as many yards of the mixed as of both the others. 
 How many yards of each color did he sell ? 
 
 Let x — the yards of black, 
 2 x = the yards of blue, 
 and 3 x = the yards of mixed. 
 Then 10a?-r-22a? + 21a? = 5000, by the question, 
 or 53 x — 5000. 
 And x — ^§Q- = 94£§, yards of black. 
 2x = 188ff , yards of blue. 
 3« = 283^5 yards of mixed. 
 
 All the answers to the above question contain frac- 
 tions. I now wish to ascertain what sum I must sub- 
 stitute for $5000, in order that the answers may con- 
 sist of whole numbers. For this purpose, I will solve 
 the question by using a instead of the given sum. 
 
 If I denote the yards of black by x, the yards of 
 16 
 
182 FIRST LESSONS IN ALGEBRA. , 
 
 blue by 2 x, and the yards of mixed by 3 x, as above, 
 I shall have the following equation: — 
 10 x -f 22 x + 21 x — a, 
 or 53 x = a, 
 
 and x — — . 
 
 Any number, divisible by 53 without a remainder, — 
 that is, the product of 53 multiplied by any number 
 whatever, — may be put in the place of a, and the 
 answer will be free from fractions. 
 
 Suppose, for instance, the value of the whole quan- 
 tity sold to be 2491 dollars, which is the product of 53 
 multiplied by 47. Then he would have sold 47 yards 
 of black, 94 yards of blue, and 141 yards of mixed. 
 
 39. In a certain school, £ of the pupils learn navi- 
 gation, | learn geometry, £ learn algebra, and all the 
 rest, a, learn arithmetic. How many pupils are there 
 in all? 
 
 Let x = the number of pupils. 
 
 Then* = f +f + f + «, 
 
 or 60 a? = 15 x + 12 x -f 10 x + 60 a; 
 
 60 x z=z 37 x -\- 60 a, by addition ; 
 
 23 x sa 60 «, by transposition ; 
 
 , 60 a 
 
 and x = ——. 
 
 23 
 
 Instead of a, in the question, use any number that 
 can be divided by 23 without a remainder, and the 
 answer wiU be a whole number. 
 
 Substitute, for instance, 46 in the place of a. 
 
 Ans. 120 pupils. 
 
 The last two examples are intended to exhibit the 
 
GENERALIZATION. 183 
 
 manner in which questions may be prepared, whose 
 answers shall be free from fractions. 
 
 40. Three men, A, B and C, trade in company, and 
 gain $1350. Now, if A put into the joint stock $7 
 as often as B put in $6, and B put in $6 as often as 
 C put in $5, what is each man's share of the gain? 
 
 Let x = A's share. 
 
 Then — = B's share, 
 
 and — = C's share. 
 
 Therefore, x -\- ~ -\- -^ =i 1350, by the question, 
 
 or 7 x -\-S x -f- 5 x = 9450, by multiplication. 
 
 18 x — 9450, by adding terms, 
 
 and x =: 525. 
 
 Therefore, - ? - = 450, 
 
 and ^ — 375. 
 
 Ans. A, $525; B, $450; C, $375. 
 
 To generalize this question, we will suppose A put 
 in a dollars as often as B put in b dollars ; and that B 
 put in b dollars as often as C put in c dollars ; that is, 
 we will use the letters «, b, c, instead of the numbers 
 7, 6, 5. Let the amount gained be represented byg. 
 
 Let x = A's share of the gain. 
 Then — - = B's share, 
 
 and — = C's share. 
 
 a 
 
 Then x -j — - -|- — — g, by the question. 
 
184 FIRST LESSONS IN ALGEBRA. 
 
 ax-\-bx-\-cx = a g,by multiplication, 
 
 or (a -J- & -j- c ) x — a g> 
 
 and x = — —-, — , A's share. 
 
 a-\- b-\-c 7 
 
 Then ~, or — x = , f, , B's share, 
 
 fit * « a + A + c' 
 
 a + A + c 
 
 ex c 
 
 and ~. or — x = -7 C f— — , C's share. 
 
 « ' a a -\- b + c ' 
 
 That is, To ^md eacA man's share of the gain, we 
 must, multiply the whole gain by his share of the stock, 
 and divide the product by the whole stock. 
 
 If there has been a loss, instead of a gain, each 
 ma l's share of the loss can be ascertained in a similar 
 ma nner. This is a general rule for Fellowship, when 
 all the stock has been employed the same time. 
 
 11. Divide 8736 dollars among three men in such 
 a manner, that their shares shall be to each other as 
 the numbers 3, 4 and 5, respectively. 
 
 42. Two traders, A and B, found that they had 
 gained, at the end of the year, 3792 dollars. A hav- 
 ing put $5000 and B $7000 into the joint stock, 
 what is each man's share of the gain ? 
 
 43. Three merchants, A, B and C, made up a joint 
 voyage, by which they lost 2595 dollars. A furnish- 
 ed $300 of the capital, as often as B furnished $500, 
 and C, $700. What is the loss of each ? 
 
 44. The premium on a policy of insurance is $513, 
 which is to be divided between three underwriters. A 
 took 3 shares of the risk, B 7 shares, and C 9 shares. 
 What part of the premium belongs to each ? 
 
 45. Three merchants, A, B and C, traded in com- 
 pany, and gained $8840. They severally contributed 
 
GENERALIZATION. 185 
 
 to the joint stock, in the proportion of 5, 6 and 7 ; 
 and A's money was used 4 months ; B's, 3 months, 
 and C's, 2 months. What part of the gain belongs 
 to each . ? 
 
 Let x = A's share of the stock. 
 
 Then — =z B's share, 
 and — = C's share. 
 
 5 
 
 Now, as A furnished x dollars for 4 months, he fur- 
 nished what was equal to 4 x dollars for one month. 
 
 And B's — - dollars for 3 months = — - dollars for one 
 
 5 5 
 
 month ; and-C's — dollars for 2 months = — dol- 
 lars for one month. 
 
 Therefore, 4 a; r: A's share of the gain, 
 — - = B's share, 
 
 o 
 
 and — - = C's share. 
 
 a 
 
 io r 14, i- 
 
 Then 4« + y + y= 8840, by the question. 
 
 20 x + 18 x + 14 x = 44200, 
 
 or 52 x — 44200, 
 
 and x = 850. 
 
 Then 4« = 3400, A's share of the gain, 
 
 ^ = 3060, B's share, 
 
 o 
 
 and ~ = 2380, C's share. 
 
 Let us now suppose that the sum gained was s dol- 
 lars ; that A put in a dollars ; B, b dollars ; and C, c 
 16* 
 
186 FIRST LESSONS IN ALGEBRA. 
 
 dollars ; and that A's money was in h months ; B's, 
 m months ; and C's, n months. 
 
 Let x ±= A's share 1 of the stock. 
 Then — == B's share, 
 
 a 
 
 and — = C's share. 
 
 a 
 
 Now, if we multiply each man's stock by its time, 
 as before, we shall have 
 
 h x == A's share of the gain, 
 = B's share, 
 
 a 
 
 and = C's share. 
 
 a 
 
 Then h x -\ 1 = s, by the question. 
 
 ahx-\-bmx-]-cnx = as, by multiplication, 
 
 or (a h -\- b m -\- c n) x = a s, 
 
 , as 
 
 and x = — ; — . 
 Therefore, h x = , , ; — ; , A's share, 
 
 b m x b in s T ,, . 
 
 — 7 , . — ; — ? B's share, 
 
 a ah-\- b m~\-cn 
 
 ■, cnx ens -*. , 
 
 and = — — ■ — ■ — , Is share. 
 
 a ah-\- b m-f- en 
 
 Having found the value of x, we obtain the formu- 
 las for the shares of A, B and C, by substitution. 
 For instance, A's share of the gain being h a?, we 
 
 multiply the value of a? by h, and obtain — — r-^r — ; 
 
 and so with the others. 
 
 Hence, to find the share of either of the copart- 
 
GENERALIZATION. 187 
 
 ners, Multiply the sum to be divided by the product of 
 his share of the stock and the time it uas used, for a 
 dividend. 
 
 Multiply the several shares by their respective times, 
 and divide the dividend by the sum of their products. 
 
 46. Three men, A, B and C, traded in company, 
 and gained 2160 dollars. A put in $400 for 5 
 months; B, $500 for 2 months; and C, $600 for 4 
 months. What is each man's share of the gain ? 
 
 47. Three men, A, B and C, hired a pasture in 
 common, for 82 dollars. A put in 3 horses for 4 
 months ; B, 4 horses for 2 months ; and C, 7 horses 
 for three months. How much must each man pay ? 
 
 48. Two merchants, A and B, engaged in a specu- 
 lation, by which they gained 2100 dollars. The capi- 
 tal employed was $14000, of which A furnished 
 $5000, and B, $9000; but A's money was used a 
 year, whereas B's was used only 5 months. Requir- 
 ed the gain of each. 
 
 49. A farmer would mix corn at a shillings per 
 bushel with oats at b shillings per bushel, so that a 
 bushel of the mixture may be worth c shillings. How 
 many bushels of each sort must he take, to make a 
 composition of n bushels ? 
 
 Let x = the bushels of corn. 
 Then n — x — the bushels of oats. 
 Also, a x = the value of the corn, 
 on — b x = the value of the oats, 
 and c n = the value of the mixture. 
 Then a x -\-b n — 6a? = cw, by the question. 
 
188 FIRST LESSONS IN ALGEBRA. 
 
 i 
 
 ax — b x = c n — b n, by transposition, 
 
 or (a — ty x = c n — b n, 
 
 and x e± — = — , the bushels of corn. 
 
 a — b 
 A1 en — b n an< — en ,, 
 
 Also n — x = n — > or — , the oats. 
 
 a — b a- — b 
 
 50. A grocer would mix sugar at 10c?. per pound, 
 with another sort, worth Id. per pound, so as to make 
 1 cwt. worth 8d. per pound. How many pounds of 
 each sort must he take ? 
 
 51. A dealer in oils would furnish 36 gallons at 4 
 shillings a gallon ; and he has but two kinds on hand, 
 one of which is worth a dollar, and the other, 3 shil- 
 lings, a gallon. How many gallons of each must he 
 take. 
 
 52. Two persons, A and B, set out from one place, 
 and both go the same road j but A goes a hours be^- 
 fore B, and travels n miles an hour ; B follows, and 
 travels m miles an hour. If m be greater than n, in 
 how many hours will B overtake A ? and how many 
 miles will he travel ? 
 
 Let x =s the hours B travelled. 
 
 Then x -\- a = the hours A travelled. 
 
 Also, m x = the miles B travelled, • 
 
 and n x -f- n a = the miles A travelled. 
 
 Therefore, m x = n x ,-j- n a, 
 
 or 7n x — n x = n a, 
 
 and x = , the hours B travelled. 
 
 m — n 7 
 
 «+« = \- a, or , the hours A travelled. 
 
 1 m — n ■ ' m — n 7 
 
 m x = - — - X m, or — ^-. the miles travelled. 
 
 m—n ' m — n ' 
 
GENERALIZATION. 189 
 
 53. If A sets out 10 hours before B, and travels' 3 
 miles an hour, and B travels 4 miles an hour, how 
 long and how far will they travel before they come 
 together ? 
 
 54. Suppose that A sets out 6 hours before B, and 
 travels 3 miles an hour, and that B travels 5 miles an 
 hour ; how long and how far will they travel ? 
 
 55. Divide a line c feet long into two parts, so that 
 one part may be to the other as m to n. 
 
 Let x = one of the parts. 
 
 Then — =s the other part. 
 
 Therefore, x -}- — = c, by the question, 
 
 or nx-\-mx = cn, 
 
 and x ■=. ^qr^, one of the parts. 
 
 Also — - == —7— , the other part. 
 
 n n-\- to' r 
 
 56. Let the line be 20 feet long, and be so divided 
 that one part shall be to the other as 3 to 4. 
 
 57. Divide a line 36 inches long into two parts, 
 which shall be to each other as 4 to 5. 
 
 58. Divide any number, a, into two such parts, that 
 their product shall be to the square of the greater 
 part, as m to n. 
 
 Let x = the greater part. 
 
 Then a — x = the smaller part. 
 
 Also ax — # 2 = their product, 
 
 and a? 2 = the square of the greater part. 
 
190 FIRST LESSONS IN ALGEBRA. 
 
 anx — n x* 
 
 Therefore, a: 2 = 
 
 or m x 2 = a n x — n x 2 , 
 
 and m x = a n — n x, by dividing by x. 
 
 m x -\- n x =l a n,by transposition, 
 
 and x = — — , the* greater part. 
 a — x = a — — ; — , or — -— , the less part. 
 
 m-\-n m -\- n x 
 
 59. Let the number to be divided by these formu- 
 las be 100, and the ratio given, as 2 to 3. 
 
 60. Divide 100 into two such parts, that their prod- 
 uct shall be i the square of the greater part. 
 
 61. What number is that to which if we add a, 
 subtract b from the sum, multiply the remainder by c, 
 and divide the product by d, the quotient shall be 
 equal to the number ? Ans. ac ~ b \ 
 
 d — c 
 
 62. What number is that, to which if we add 40, 
 subtract 25 from the sum, multiply the remainder by 
 6, and divide the product by 9, the quotient and the 
 number shall be equal ? 
 
 63. Required the number, to which if we add 84, 
 subtract 38 from the sum, multiply the remainder by 
 5, and divide the product by 7, the quotient shall be 
 equal to the number sought. 
 
 64. A trader, having gained $4051 by his business, 
 and lost $3186 by bad debts, found that f of what 
 he had left equalled the capital with which he com- 
 menced trade. What was his capital ? 
 
 65. A farmer paid a men and b boys m dollars for 
 
GENERALIZATION. 191 
 
 working a day ; he afterwards paid c men and d boys 
 n dollars for working a day. What was the pay of 
 each? 
 
 Let x == the wages of a man, 
 
 and y = the wages of a boy. 
 
 a. Then a x -\- b y =z m, 
 
 b. and c x -j- d y = n. 
 
 Multiply equation a by c, and equation b by a, to 
 make the coefficients of x equal ; and we have 
 
 c. acx-\-bcy=.cm 
 D.acx-\-ady=an 
 
 * bey — ady=cm — a n, by subtraction, 
 
 or (b c — ad) y =. c m — an, 
 
 , cm — an , . 
 
 and y = _ , a boy's wages. 
 
 By substituting this value of y in equation a, we 
 obtain 
 
 b n — dm , 
 
 x =. ^_ , a man s wages. 
 
 66. A farmer paid 5 men and 8 boys $9 for work- 
 ing a day ; he afterwards paid 7 men and 6 boys $10 
 for working a day. What were the wages of a man 
 and a boy ? 
 
 67. A man lent a certain sum of money ; and, at 
 the end of 4 years, he received, for principal and in- 
 terest, $775. What was the sum lent, the interest 
 being reckoned at 6 per cent. ? 
 
 Let x = the principal. 
 
 Then, since 6 per cent, implies $6 for the use of 
 $100, or T |^ of the principal, as the annual interest, 
 
192 FIRST LESSONS IN ALGEBRA. 
 
 — == the interest for 1 year, 
 
 and — p = tie interest for 4 years. 
 
 24 x 
 
 Of course, x -j- ■— = the amount. 
 
 Therefore, x -f- — j = 775, by the question. 
 
 100 a? -f- 24 a? = 77500, by multiplication, 
 
 or 124 x = 77500, 
 
 and x = 625. 
 
 Ans. $625. 
 
 68. To generalize this question, we have the 
 amount, rate and time given, to find the principal. 
 
 In solving this problem, and others of a similar 
 sort, we shall find it convenient to indicate the several 
 quantities, given and required, by their initials, in- 
 stead of the letters commonly used ; although this is 
 not essential. 
 
 Let p =z the principal, or sum lent. 
 r = the annual rate per cent. 
 t =. the time. 
 i =i the interest, 
 a =z the amount, which always — p -j - *• 
 
 In the question above, the value of p is required. 
 
 r p =z the interest for one year. 
 
 t r p =. the interest for t years. 
 
 p -f- t r p = the amount. 
 
 Therefore, p 4- t r p = a, 
 
 or (1 -j- t r) p =. a, 
 
GENERALIZATION. 193 
 
 To express this formula in words : Add 1 to the 
 product of the time multiplied by the rate, and divide 
 the amount by the sum. The quotient will be the 'prin- 
 cipal. 
 
 This is a general rule for calculating Discount. 
 When a man pays a sum of money before it has be- 
 come due, he is evidently entitled to some reduction 
 from the debt. Equity requires that he should pay 
 such a sum as would amount to the sum due, if put 
 at interest during the time for which it is advanced, 
 at any rate per cent, agreed upon by the parties. The 
 difference between such a sum, which is called the 
 present worth of the debt, and the debt itself, is the 
 discount. 
 
 69. What is the present worth of 392 dollars, due 
 in 2 years, the discount being reckoned at 6 per cent. ? 
 
 It is here required to find what sum will amount 
 to 392 dollars, in 2 years, at 6 per cent. In other 
 words, the amount, time and rate are given, to find 
 the principal. 
 
 70. A gentleman hired a sum of money at 5 per 
 cent. ; and at the end of 3 years, he paid 8234 dol- 
 lars, for principal and interest. What was the sum 
 hired ? 
 
 71. A merchant sold an invoice of goods, amount- 
 ing to 1961 dollars, on a year's credit. What dis- 
 count should he make for present payment, allowing 
 money to be worth 6 per cent. . ? 
 
 72. Required the present worth of 713 dollars, dis- 
 counted for 4 years at 6 per cent. 
 
 17 
 
194" FIRST LESSONS IN ALGEBRA. 
 
 73. What sum will amount to $667 in 3 years, at 
 5 per Cent. ? 
 
 74. Given the amount, time and principal, to find 
 the rate. 
 
 r p = the interest for 1 year. 
 
 t r p = the interest for t years. 
 
 p -f- t r p =i the amount. 
 
 Then p -\- t r p = a, 
 
 or t r p z=z a — p, 
 
 and r == - = ^. 
 t P 
 
 That is, From the amount subtract the principal; 
 and divide the remainder by the product of the princi- 
 pal multiplied by the time. The quotient will be the 
 rate. 
 
 75. A man lent $420 ; and, in 5 years, he received 
 in payment $546. At what rate per cent, was the 
 money lent ? 
 
 76. At what rate per cent, will $380 amount to 
 $513, in 7 years? 
 
 77. Given the interest, time and rate, to find the 
 principal. 
 
 r p = the interest for 1 year. 
 
 t r p = the interest for t years. 
 
 Therefore, t r p = i } 
 
 Expressed in words : Multiply the time by the rate, 
 and divide the interest by the product. The quotient 
 will be the principal. 
 
 78. A paid B $126 for the use of a certain sum of 
 
GENERALIZATION. 195 
 
 money 3 years ; the interest being reckoned at 5 per 
 cent. What was the sum lent ? 
 
 79. In the course of 4 years, a man paid interest 
 to the amount of $288, which was reckoned at 6 
 per cent. What was the debt . ? 
 
 80. The amount, principal and rate being given, 
 to find the time. 
 
 r p = the interest for 1 year. 
 
 t r p = the interest for t years. 
 
 p -f- t rp = the amount. 
 
 Therefore, p -\~ t r p = a, 
 
 or t r p = a — p 
 
 v , a — p 
 
 and t = —r. 
 
 rp 
 
 That is, From the amount subtract the principal; 
 and divide the remainder by the product of the rate 
 and principal. The quotient xvill be the time. 
 
 81. A man lent $460 at 5 per cent, and received, 
 for principal and interest, $529. How long was the 
 money kept? 
 
 82. In what time will $780 amount to $1014, 
 interest being reckoned at 6 per cent. ? 
 
 These examples show the manner in which gen- 
 eral results, or formulas, are obtained; and also how 
 they may be used in solving particular questions. 
 Let the learner now turn back to Chapter VIII., and 
 generalize the questions marked with a star (*) ; and 
 then solve the same questions, numerically, by their 
 respective formulas. He will thus be prepared to 
 generalize some of the more difficult questions, in the 
 same chapter, which are not marked. 
 
196 TIRST LESSONS IN ALGEBRA, 
 
 CHAPTER X 
 EVOLUTION. 
 
 SECTION I. 
 Introduction. 
 
 When a quantity is multiplied by itself one or more 
 times, the product is called a Power of that quantity. 
 Thus, a 2 , being the product of a X «, is the second 
 power or square of a; and 6 3 , that is, b X b X b, is 
 the third power or cube of 6. [See Chap. VIL 
 Sec. I.] 
 
 On the contrary, the quantity which is multiplied by 
 itself to produce any power, is said to be the Root of 
 that power. Thus, a is the second or square root of 
 a s ; and b is the third or cube root of b 3 . 
 
 Powers and Roots are, therefore, correlative terms ; 
 and Evolution and Involution are the reverse of each 
 other. Involution is the method of raising a given 
 root to a proposed power ; but Evolution is the method 
 of finding the roots of given powers. 
 
 Involution is more perfect, however, than Evolu- 
 tion ; for if any proposed power of a given quantity 
 be required, it can be exactly obtained ; but there are 
 many quantities whose exact roots cannot be found, 
 
EVOLUTION. 
 
 197 
 
 It is evident, for instance, that the square root of a 
 cannot be determined ; for there is no quantity, which, 
 being multiplied by itself, will produce a. 
 
 The roots and powers of numbers have the same 
 relation to each other as those of literal quantities. 
 Thus, the second powers of 2 and 3 are 4 and 9 ; 
 and the square roots of 4 and 9 are 2 and 3. The 
 exact roots of the intermediate numbers, 5, 6, 7 and 
 8, cannot be found. 
 
 TABLE OF ROOTS AND POWERS. 
 
 Roots. 
 
 1 
 
 s 
 
 3 
 
 4 
 
 5 
 
 6 
 
 36 
 
 7 
 
 8 
 
 9 
 
 10 
 
 Squares. i 
 
 1 
 
 4 
 
 9 
 
 16 
 
 25 
 
 49 
 
 64 
 
 81 
 
 100 
 
 Cubes. 
 
 1 
 1 
 
 8 
 
 16 
 
 27 
 81 
 
 64 
 256 
 
 125 
 
 625 
 
 216 
 
 343 
 
 512 
 
 729 
 
 1000 
 
 4th powers. 
 
 1296 
 
 2401 
 
 4096 
 
 6561 
 
 10000 
 
 5th powers. 
 
 1 
 
 32 
 
 243 
 
 1024 
 
 3125 
 
 7776 
 
 16807 
 
 32768 
 
 59049 
 
 100000 
 
 The roots of quantities are indicated either by 
 means of the radical sign \/, or by a fractional index. 
 
 Thus, %/a, or \/a, is the square root of a. 
 *y/a is the cube root of a. 
 4 \/a 3 is the 4th root of a 3 . 
 ^64 is the square root of 64, which is 8. 
 3 ^/a -\- x is the cube root of a -f- x. 
 
 If the quantity affected by the radical sign be not 
 a complete power, that is, if its root cannot be ex- 
 actly found, it is called a Surd, or Irrational Quantity. 
 Thus, ^/S5, \Ar 2 , 5 \/a 3 , &c. } are surd quantities. 
 
 Express the roots of the following quantities by 
 means of the radical sign : 
 17* 
 
198 FIRST LESSONS IN ALGEBRA, 
 
 1. The square root of x. 
 
 2. The fourth root of fa. 
 
 3. The cube root of a? 2 -f- y- 
 
 4. The fifth root of 79. 
 
 5. The square root of a 2 — b -f- 14. 
 
 When the root of a quantity is expressed by means 
 of a fractional index, the numerator of the fraction 
 indicates the power of the quantity, and the denomina- 
 tor the root required. 
 
 Thus, a 2 is the square root of a 1 or a. 
 a 2 is the square root of a 3 , 
 a* is the cube root of a. 
 (a 2 -j- b)i is the fourth root of a 2 -f b. 
 a 3 is the cube root of a 2 . 
 
 2 
 
 The expression a? may be regarded either as the 
 second power of the third root of «, or as the third 
 root of the second power of a. And so with all 
 other quantities having fractional indices. 
 
 Suppose the value of a to be 27. The third root of 
 27 is 3, and the second power of 3 is 9. Again, the 
 second power of 27 is 729, and the third root of 729 
 is 9. The value is the same, whichever mode of ex- 
 pression is used. 
 
 Express the roots of the following quantities by 
 means of fractional indices : 
 
 6. The square root of x. 
 
 7. The fourth root of y 3 . 
 
 8. The cube root of (a 2 + #) 2 . 
 
EVOLUTION. 199 
 
 9. The m th root of c. 
 10. The square root of x 3 — c 2 -f 12. 
 
 If the numerator and denominator of a fractional 
 index be the same, the value of the quantity is not 
 
 2 
 
 affected by it ; for a 2 , that is, the second root of the 
 second power of a, is evidently a. 
 
 As the value of a fraction is not altered, when 
 both the numerator and denominator are either mul- 
 tiplied or divided by the same number, fractional in- 
 dices may be changed into other indices of the same 
 
 value ; as, a 2 , a*, a 5 , a ¥ , &c, which are all equal. 
 
 Suppose the value of a to be 16. Then the sec- 
 ond root of a is 4, whose first power is also 4. Again, 
 the fourth root of a, or 16, is 2 ; and the second 
 power of 2 is 4. And so with the others. 
 
 We can, therefore, reduce different fractional in- 
 dices to other indices which shaH express the same 
 root, by reducing the fractions to a common denom- 
 inator. 
 
 When a letter or figure is prefixed to a quantity 
 affected by the radical sign, it is to be regarded as a 
 coefficient, and the two quantities are supposed to 
 be multiplied together. 
 
 Thus, a \/ x implies that the square root of x is 
 multiplied by a; and 5 \/ a 3 is the product of the 
 square root of a 3 , multiplied by 5. But 5 -|- \/ a 3 , 
 or 5 — \/ a 3 , implies that the square root of a 3 is to 
 be added to, or subtracted from, 5, and not multiplied 
 by that number. 
 
200 FIRST LESSONS IN ALGEBRA. 
 
 SECTION II. 
 Roots of Simple Quantities 
 
 1. What is the square root of a 6 ? Ans. « 3 » 
 We are here required to find two equal factors, 
 
 whose product shall be a 6 ; and, as we multiply pow- 
 ers by adding their exponents, [See Chap. VII. Sec. 
 VI.] a 3 X « 3 = a 6 . Or the required root may be 
 
 expressed by a fractional index, thus, a 2 ; which, the 
 fraction being reduced, becomes a 3 . 
 
 2. What is the cube root of a 6 ? Ans. a 2 . 
 Here we are required to find three equal factors, 
 
 whose continued product shall be a 6 ; and, by the 
 rule for multiplying powers, a 2 X « 2 X a 2 == a 6 . If 
 the required root be expressed by a fractional index, 
 
 it will be cr — a 2 , as above. 
 
 3. What is the square root of 16 a 2 ? Ans. 4 a. 
 
 For 4 a X 4 a == 16 a 2 . The root of the coeffi- 
 cient is found and prefixed to the root of the literal 
 quantity, which is obtained as above. 
 
 4. What is the square root of 9 a 4 h 2 x 6 ? 
 
 Ans. 3 a?bx?. 
 For 3 a 2 b x 3 X 3 a 2 b x 3 = 9 a 4 6 2 a*. We di- 
 vide the exponent of every letter by the index of the 
 required root, and annex the result to the root of the 
 coefficient. 
 
 5. What is the cube root of a 5 ? Ans. cr. 
 As the exponent of the given power cannot be di- 
 vided by the index of the required root, without leav- 
 
EVOLUTION. 201 
 
 ing a remainder, the root must be represented by a 
 fractional index. 
 
 6. What is the fourth root of 81 a 4 c 3 ? 
 
 Ans. 3 a c 4 . 
 
 7. Required the fifth root of 1024 a 5 x 10 . 
 
 Ans. 4 a x 2 . 
 
 From these examples and observations we derive 
 the following Rule for extracting the root of a sim- 
 ple quantity, viz : 
 
 Divide the exponent of the given power by the index 
 of the root to be found, and annex the result to the root 
 of the coefficient. 
 
 8. What is the square root of 64 a 4 6 2 ? 
 
 9. What is the cube root of 27 a 3 b 6 x 9 ? 
 
 10. What is the fourth root of 81 a* x 4 y 12 ? 
 
 11. Required the fifth root of 32 x* y 10 . 
 
 12. Extract the cube root of 64 « 6 x 3 y 12 . 
 
 13. Required the square root of 5 a 2 x 4 . 
 
 14. What is the cube root of 1 x 6 y 9 } 
 
 15. Extract the fourth root of 1296 a 4 b s x™. 
 
 16. Required the fourth root of 16 a 8 b 3 . 
 
 17. What is the third root of 9 a 3 b 4 x G ? 
 
 18. Find the square root of 25 a 4 b. 
 
 19. Extract the cube root of 64 x 4 y 5 . 
 
 The root of a fraction is found in the same man- 
 ner. Extract the root of the numerator for a new nu- 
 merator, and the root of the denominator for a new de- 
 nominator. 
 
 9 a 2 
 
 20. What is the square root of — ? 
 
 21. Required the cube root of g^. 
 
202 FIRST LESSONS IN ALGEBRA. 
 
 22. What is the fourth root of ~? 
 
 23. Extract the cube root of ,,,,, . 
 
 64, A 3 c 8 
 
 16 a^ 
 
 24. Required the square root of 
 
 5x*y* 
 
 25. What is the cube root of rr-rr? 
 
 9a 6 b 
 lb 
 
 26. What is the square root of 
 
 4>x*f 
 
 To determine what sign should be prefixed to a 
 root, observe, in general, that the root, when multi- 
 plied by itself the requisite number of times, must 
 re-produce the given power. Therefore, 
 
 An odd root of any quantity must have the same 
 sign which the quantity has. The cube root of — a 3 
 is — a; for — a X — a = -}- a 2 , and -f- a 2 X — a 
 == — * a 3 . And the cube root of -f- a 3 is -f- a; for 
 |«X + a = + a 2 , and -J- a 2 X + « = + a 3 . 
 
 ^4w even roo^ o/" a positive quantity has two signs, 
 the one positive, the other negative. Such a quantity 
 is said to be ambiguous. The square root of a 2 may 
 be either -f- a, or — a ; for -f- a X -f- « = + a 2 , and 
 — a X — « = + a 2 , also. When it is not known, 
 from the nature of the question, whether the root is 
 positive or negative, it should be marked with the 
 ambiguous sign ; thus d- a. 
 
 There is no such thing as the even root of a nega- 
 tive quantity ; for neither — a X — «, nor -f aX-fo, 
 will produce — a 2 . 
 
 27. What is the square root of 25 a 2 b 4 ? 
 
 28. What is the cube root of 125 a 3 b 6 > 
 
EVOLUTION. 203 
 
 29. What is the square root of 7 x 4 y 3 ? 
 
 30. Required the fifth root of — 243 x™. 
 
 31. Extract the fourth root of 256 a* b 4 c™. 
 
 32. Required the square root of 64 a 2 x 4 . 
 
 33. Extract the cube root of — 125 m 6 n 3 . 
 
 34. What is the sixth root, of 4096 x 6 z™ ? 
 
 35. Required the cube root of 64 x 3 z 6 . 
 
 36. Extract the square root of 17 x 4 y 3 z 2 . 
 
 37. What is the cube root of — 18 a 6 y 4 z 3 } 
 
 a 10 * 5 
 
 38. Required the fifth root of 
 
 39. Find the cube root of 
 
 18c* x 
 12 a? x^ 
 
 125^*4' 
 
 SECTION III. 
 To extract the Square Root of a Compound Quantity. 
 
 1. What is the square root of 12 x 2 y 4 -{- 4 y 6 -f- 
 9 x 4 y* ? 
 
 Since the power given in this question consists of 
 three terms, it is evident that its root must contain 
 more than one term ; for the second power of a sim- 
 ple quantity is a simple quantity, and the second 
 power of a binomial consists of three terms. [See 
 Chap. VII. Sec. III.] We are, therefore, required 
 to find the binomial, whose square is the quantity pro- 
 posed in the question. 
 
 We may here remark that no binomial, as x 2 -f- y* t 
 can be a complete power ; and the root of an incom- 
 plete power can be found only by approximation. 
 
204 FIRST LESSONS IN ALGEBRA. 
 
 The second power of the binomial a -f- b, is a 2 -}- 
 2«H^ [See Chan. IX. Quest. 18.] in which, it 
 will be observed, the several terms are arranged ac- 
 cording to the powers of the letter a. 
 
 If we arrange the terms of the proposed quantity 
 according to the powers of x, it will become 
 
 9 x 4 %p + 12 x* y 4 -f- 4 if. 
 
 In the formula, (a + bf — a 2 + 2 a b + 6 2 , the 
 first term of the power, a 2 , is the square of the first 
 term of the root, a. 
 
 Hence we infer that the first term of the given 
 quantity, 9 x 4 y 2 , is the square of the first term of the 
 root sought, which is, therefore, 3 x 2 y ; for 3 # 2 y X 
 3 a? 2 y = 9 x 4 y 2 . 
 
 Again, the second term of the power expressed in 
 the formula, 2 a b, is twice the product of the two 
 terms of the binomial, a -\- b ; of course, if it be di- 
 vided by twice the first term, a, the quotient is the 
 second term, b. 
 
 To apply this principle, we must divide the second 
 term of the given power, 12 a? 2 y 4 , by twice the first 
 term of the root already obtained, 3 x 2 y ; that is, by 
 6 x 2 y. The quotient is 2 y 3 , which is the other term 
 of the root required. 
 
 Therefore, 3 x 2 y -j- 2 y 3 is the whole root sought ; 
 and (3 a: 2 y + 2 y 3 ) X (3 x* y -f 2 y 3 ) = 9 x 4 y* + 
 12 x 2 y 4 + 4 2/6. 
 
 The several steps taken in this investigation may 
 be expressed together in the following manner : — 
 
6 a? 2 y -j- 2 y 
 
 EVOLUTION. 205 
 
 9 x* y 2 -f 12 oP y 4 + 4 y 6 ( 3 x 2 y + 2 y 3 
 9 a? 4 ?/ 2 
 
 12 a; 2 # 4 + 4 y 5 
 12 a: 2 y 4 -f 4 y 6 
 
 The given quantity being arranged according to 
 the different powers of x, as in Division, the root of 
 the first term is found to be 3 x 2 y, which is put in 
 the quotient's place. 
 
 The second power of 3 a: 2 y is next subtracted from 
 the first term of the given quantity, 9 a; 4 y 2 , when 
 there is \jio remainder ; which proves that the true 
 root of this term has been found. 
 
 To ascertain the next term of the root, we must 
 divide the second term of the given power by twice 
 the root already found ; therefore, the remaining 
 terms of the power, 12 a; 2 y 4 -j- 4 y 6 , are brought 
 down for a new dividend. 
 
 That portion of the root which has been found, 
 3 a? 2 y, is then doubled, and put in the divisor's place. 
 
 The first term of the new dividend, 12 x 2 y 4 , is di- 
 vided by 6 a: 2 y, and the quotient is found to be 2 y 3 , 
 which is the second term of the root. 
 
 To ascertain whether 3 x 2 y -f- 2 y 3 be the true 
 root required, we should involve it, and subtract its 
 second power from the given quantity, when there 
 must be no remainder. But the second power of the 
 first term, 3 # 2 y, has already been subtracted. Now, 
 in the formula, (a -\- b) 2 = a 2 -j- 2 a b -f- b 2 , twice 
 the first term of the root, multiplied by the last term, 
 18 
 
206 FIRST LESSONS IN ALGEBRA. 
 
 is equal to the second term of the power ; also the 
 last term of the root, multiplied by itself, gives the 
 last term of the power. That is, if we add the last 
 term of the root to twice the first term, and multiply 
 their sum by the last term, the product will be equal 
 to the second and third terms of the power. There- 
 fore, 2 y 3 is added to 6 x 2 y, to complete the divisor. 
 
 The whole divisor, 6 x 2 y + 2 y 3 , is then multi- 
 plied by 2 y 2 ; and the product is subtracted from the 
 new dividend. There being no remainder, we are 
 certain that 3 a? 2 y -f- 2 y 3 is the true root required. 
 
 From this investigation may be derived the follow- 
 ing Rule for extracting the square root of a com- 
 pound quantity. 
 
 1. Arrange all the terms of the given quantity ac- 
 cording to the powers of one of the letters, so that the 
 highest power shall stand first, the next highest next t 
 and the rest in order, as in Division. 
 
 2. Find the root of the first term, and put it in the 
 place of a quotient. 
 
 3. Subtract the square of this root from the first 
 term of the given quantity, and bring down the remain- 
 ing terms for a dividend. 
 
 4. Double the root already found, and put it in the 
 divisor's place. 
 
 5. *See how often this divisor is contained in the first 
 term of the new dividend, and annex the quotient both 
 to the root already found and to the divisor. 
 
 6. Multiply the divisor, thus increased, by the term 
 of the root last found, and subtract the product from 
 the dividend. 
 
EVOLUTION. 
 
 207 
 
 7. Double the whole root for a new divisor, and di- 
 vide as before. 
 
 8. Proceed in this way, until the entire root of the 
 given quantity is extracted. 
 
 2. What is the square root of 9 x 4 — 12 a: 3 -f- 16 a: 2 
 
 — 8a: -f 4? 
 
 9 x * _ 12 X 3 _|_ 16 #2 _ Sx + 4( 3 a: 2 — 2a:+2. 
 9tf 4 
 
 6^- 2 
 
 — 12a: 3 -f-16a:2 
 
 — ■ T2 a: 3 -f- 4 a: 2 
 
 6^2 — 4^_j_2 
 
 12 #2 — 8 x + 4 
 12 #» _ g a; + 4 
 
 In this example, the second term of the root, 2 x, 
 is — , because it is the quotient of — 12 a 3 divided by 
 -f- 6 a: 2 . The second divisor, 6 x 2 — 4 a:, is obtained, 
 as before, by doubling the whole of the root already 
 found. 
 
 3. What is the square root of a 2 -f- 2 a b -j- 6 2 , — 
 2«c — 26c-f-c 2 ? 
 
 «2_j_2a6-|-&2 — 2ac — 2&c-fc 2 (a + £ — c. 
 a 2 
 
 2a + 6 
 
 2a6-f-&2 
 2ab + & 
 
 2a-\-2b — c 
 
 — Sac — Zbc + c* 
 
 — 2 a c — 26c + c 2 
 
208 FIRST LESSONS IN ALGEBRA. 
 
 4. What is the square root of 4 x 2 + 1%ocy -{-9y 2 ? 
 
 5. Extract the square root of 
 
 a 4 _ 4 ft 3 £ _j_ 6 a 2 x 2 _ 4 ^3 _|_ #4. 
 
 6. What is the square root of 
 
 #4 __ 4 ^3 _j_ 6 a: 2 — 4 a? -f 1 ? 
 
 7. Required the square root of 
 
 4 x 6 + 12 x 5 + 5 x* — 2 a? -f- 7 #2 — 2 a? + I. 
 
 8. What is the square root of 
 
 l_4a-|-4a 2 + 2a: — 4aa:-{-a: 2 ? 
 
 9. Extract the square root of 
 
 x e _- 2 #5 + 3 x* — 2 #3 -J- a: 2 . 
 
 10. Whaf is the square root of 
 
 a?4 _|_ 4 ^2 y _|_ 4 y 2 _ 4 ^2 _ 8 y -j_ 4 ? 
 
 11. Extract the square root of 
 
 9 <*4 £2 _ 30 a 2 6 + 12 a 4 ¥ + 25 — 20 a 2 fr* -f- 
 4 a 4 ¥. 
 
 12. What is the square root of 
 
 4 + 4 a 2 6 +.a 4 fc 2 — 12 y 2 — 6 a* 6 y» + 9 y 4 ? 
 
 13. Required the square root of 
 4 a 2_4 a 5_|_j2_|_i2ac — 6 J c + 9 c 2 . 
 
 14. What is the square root of 
 
 a* + ia* + 10 a: 4 + 20 x 3 -f- 25 a: 2 + 24 x + 16 ? 
 
 15. Find the second root of 
 
 4 06 _ 4 a A + 12 « 3 + a 2 — 6 a -f 9. 
 
 16. Extract the square root of - — m * ,; J. . 
 
 n c 2 — 2crf-f-eP 
 
 Extract the square root of the numerator and of the 
 denominator, separately, as in simple quantities. 
 
 *m t. • j a j r 16 * 2 + 48 *?/-}- 36 y 2 
 
 17. Required the second root of — — — to — ~. 
 
 1 9 a 2 — 12 a c + 4 c 2 
 
 to t> • j *u r 25« 2 — 90a+81 
 
 18. Required the square root of — — — - . 
 
 1 ■ 36 + 24 x + 4 or 2 
 
EVOLUTION. 209 
 
 SECTION IV. 
 
 To extract the Square Root of a Number. 
 
 The following Rule for the extraction of the square 
 root of a number, is derived directly from the method 
 of extracting the square root of an algebraic quantity : 
 
 1. Separate the given number into periods of two 
 figures each, beginning at the right hand. The period 
 on the left may contain either one or two figures. 
 
 2. Find the greatest second power contained in the 
 left-hand period, and write its root in the, quotient's 
 place. 
 
 3. Subtract the square of this root from the first 
 period of the given quantity, and bring down the next 
 two figures for a dividend. 
 
 4. Double the root already found, and put it in the 
 divisor's place. ± 
 
 5. See how many times the divisor is contained in 
 the new dividend, rejecting the right-hand figure ; and 
 annex the quotient both to the root already found and 
 to the divisor. 
 
 6. Multiply the divisor, thus increased, by the figure 
 of the root last found, and subtract the product from 
 the dividend. 
 
 7. Bring down the next two figures of the given 
 number for a new dividend. 
 
 8. Double the whole root for a new divisor, and di- 
 vide as before. 
 
 9. Proceed in this way, until the entire root of the 
 given quantity is extracted. 
 
 18* 
 
210 FIRST LESSONS IN ALGEBRA. 
 
 Should the divisor not be contained in the divi- 
 dend, must be annexed both to the root and the di- 
 visor ; the next two figures must then be brought 
 down, and a new trial made. 
 
 The only important point, in which this rule differs 
 from the one given for extracting the square root of 
 an algebraical quantity, is that which requires the 
 given number to be separated into periods of two 
 figures each. This is done to ascertain how many 
 figures the required root will consist of. As the 
 square root of 100 is 10, the square root of every 
 number less than 100 must be less than 10, and con- 
 sist, of course, of but one figure. And as the square 
 root of 10000 is 100, the square root of every num- 
 ber less than 10000, and more than 100, must be 
 composed of two figures. So, also, as the square root 
 of 1000000 is 1000, the square root of every num- 
 ber smaller than 1000000 must be less than 1000, 
 and consist of not more than three figures. Hence 
 it is, that the given number is separated into periods 
 of two figures each. 
 
 1. What is the square root of 106929 ? 
 
 106929 ( 327. 
 9 
 
 62 169 
 124 
 
 647 4529 
 4529 
 
EVOLUTION. 211 
 
 By separating the nrnber, given in the question, 
 into periods of two figures each, we find that the root 
 will consist of three figures. The greatest even power 
 in the first period, 10, is 9, the square root of which 
 is 3. This root is, therefore, put in the quotient's 
 place ; its second power is subtracted from the first 
 period, 1Q ; and the next period, 69, is brought down 
 and annexed to the remainder, for a dividend. This 
 dividend (the last figure, 9, being omitted) is divided 
 by twice the root already found, 6 ; and the quotient, 
 2, which is the second figure of the root, is annexed 
 to 6, to complete the divisor. The divisor, thus in- 
 creased, is multiplied by 2, and the product is sub- 
 tracted from the dividend. The next period, 29, is 
 annexed to the remainder, and the work is continued 
 as before. 
 
 2. What is the square root of 43264 ? 
 
 43264 ( 208. 
 4 
 
 408 
 
 3264 
 3264 
 
 In this example, the divisor, 4, is not contained in 
 the dividend after the right-hand figure is rejected ; 
 a zero is, therefore, added to both the root and the 
 divisor, and the next period, 64, is brought down to 
 complete the dividend. 
 
 3. What is the square root of 148225? 
 
 4. What is the square root of 1522756? 
 
 5. Required the square root of 5499025. 
 
212 FIRST LESSONS IN ALGEBRA. 
 
 6. What is the square root of 28153636 ? 
 
 7. Extract the square root of 9247681. 
 
 8. What is the square root of T 9 ^f Ans. f. 
 We find the root of a fraction by extracting the 
 
 root of the numerator and of the denominator sep- 
 arately. 
 
 9. Required the square root of f f ? 
 
 10. What is the square root of 3£§? Ans. If. 
 The mixed number, 3^f, reduced to an improper 
 
 fraction, becomes y^, the square root of which is -\\ 
 or If, as given above. 
 
 11. What is the square root of 5^-f ? 
 
 12. Extract the square root of 7£. 
 
 13. Required the square root of 30£. 
 
 14. What is the square root of 27699169? 
 
 15. Required the square root of ff. 
 
 16. Find the square root of 13ff|. 
 
 17. What is the square root of T 5 ^- ? 
 
 18. Extract the square root of 3637175481. 
 
 SECTION V. 
 Approximate Roots. 
 
 1. What is the square root of 7 ? 
 
 Since the second power of 2 is 4, and the second 
 power of 3 is 9, the square root of 7 must be more 
 than 2, and less than 3. Therefore, 7 is not a com- 
 plete power, and its exact root cannot be found ; but, 
 by annexing two zeroes to each of the successive re- 
 
EVOLUTION. 213 
 
 mainders, we can obtain its approximate root, in de- 
 cimals, to any assignable degree of accuracy. Two 
 zeroes are annexed for each new decimal of the root, 
 for the same reason that two figures are brought 
 down, when the operation is confined to whole num- 
 bers. Of course, the decimal places of the root will 
 be equal to half the number of zeroes used. 
 
 7 ( 2.645 &c. 
 4 
 
 46 
 
 524 
 
 300 
 276 
 
 2400 
 2096 
 
 5285 
 
 30400 
 26425 
 
 3975 
 
 If there were no remainder, the square root of 7 
 would be 2 T 6 (jVir ; but, as there is a remainder, it is more 
 than 2-$r&V> although less than 2 T 6 (j 4 (y 6 <j. Therefore, 
 2t«tVzt differs less than the thousandth part of one from 
 the true root required. By annexing additional ze- 
 roes, and continuing the work, we may obtain the root 
 still more accurately. 
 
 2. Required the square root of 5. 
 
 3. What is the square root of 2? 
 
 4. Extract the square root of 823. 
 
 5. Find the square root of 527. 
 
 6. What is the square root of § ? 
 
 The square root of 9 is 3 ; but, since the numera- 
 
214 FIRST LESSONS IN ALGEBRA. 
 
 tor is not a complete power, the root whose second 
 power is nearest to 5 must be taken, which is 2. The 
 difference between § and the true square root off, is 
 less than ^ of a unit. 
 
 When both the numerator and the denominator of 
 a fraction are multiplied by the same quantity, its value 
 is not altered ; and if both terms of this fraction be 
 multiplied by any perfect power, a nearer approxima- 
 tion to the true root will be obtained. Let them be 
 multiplied by 9. 
 
 5X9 _ 45 
 9X9 ~~ 81' 
 
 The square root of 81 is 9, and the nearest square 
 root of 45 is 7 ; so that £ expresses the value of y'l 
 within ^ part of a unit. 
 
 Again, let f-f- De multiplied by 144, which is the 
 second power of 12. 
 
 45X 144 _ 6480 
 81 X 144 ~~ 11G64* 
 
 The square root of ttVft i s -i 8 o°ff nearly ; conse- 
 quently, T 8 o° ¥ differs less than T £ F of a unit from the 
 value of the exact square root off. 
 
 In general, the larger the power by which the terms 
 of a fraction are multiplied, the nearer to the true 
 root will be the approximation. 
 
 7. What is the square root of 3f ? 
 Reduce this mixed number to an improper fraction, 
 and proceed as before. 
 
 1 22 , 22X7 154 
 
 3 y = T , and ^r = -. 
 
EVOLUTION. 
 
 215 
 
 The nearest square root of -^ 1S V- or If? which 
 is the root required within less than | of one. 
 
 8. Required the square root of f . 
 
 9. What is the square root of 5^ T ? 
 
 10. Find the square root of f. 
 
 11. Required the square root of 2§. 
 
 12. What is the square root of a? -f- x 2 ? 
 
 It has been shown, already, that no binomial is a 
 perfect second power. The approximate root of a 
 surd can be found by the common rule for extracting 
 the square root of a compound quantity, thus : 
 
 a 2 
 
 5x8 
 128 a> 
 
 &C 
 
 ' 9 /» 
 
 2 a 
 
 1 4 a 2 
 
 2a + 
 
 X* 
 
 8^3 
 
 4 a* 
 x* 
 
 4 a* So* "*" 64 a* 
 
 x* x< , x« 
 
 2 a H — + — - 
 
 a 4 a 6 \b a? 
 
 •" a 4a3'8a5 
 
 x« 
 8a4 
 
 — 
 
 *8 
 
 64 a« 
 
 
 
 
 x« 
 
 8«4 
 
 + 
 
 x8 
 
 16 «e 
 
 x'° 
 
 64 a8 
 
 + 
 
 X»* 
 
 256 a"> 
 
 5x8 5x« , #10 
 
 5 fi4/»6 ' fi4«8 
 
 128 a 
 
 256 a"> 
 
 13. Required the square root of 1 -{- a\ 
 
 14. What is the square root of a? 2 — z 2 } 
 
 15. Extract the square root of a 4 -f- I. 
 
 16. What is the square root of T 5 ? ? 
 
 17. Extract the square root of 7641. 
 
216 FIRST LESSONS IN ALGEBRA. 
 
 CHAPTER XI. 
 EQUATIONS OF THE SECOND DEGREE. 
 
 SECTION I. 
 
 Pure Equations. 
 
 An equation of the second degree contains the 
 second power of the unknown quantity. When the 
 unknown quantity appears only in the second power, 
 the equation is said to be pure. 
 
 1. What number is that, which, being multiplied 
 by itself, and the product doubled, will give 162 ? 
 
 Let x = the number. 
 
 Then 2 a? 2 == 162, by the question ; 
 
 and jc 3 = 81, by division. 
 
 x t± 9. 
 
 Ans. 9. 
 
 2. A farmer, being asked how many cows he had, 
 answered, that if the number were multiplied by 5 
 times itself, the product would be 720. How many 
 
 had he . ? 
 
 Let x = the number of cows. 
 
 Then 5« 2 = 720, by the question. 
 # 2 = 144, and x = 12. 
 
 Ans. 12 cows. 
 
EQUATIONS OF THE SECOND DEGREE. 217 
 
 3. A gentleman, being asked the price of his hat, 
 answered, that if it were multiplied by itself, and 26 
 were subtracted from the product, the remainder mul- 
 tiplied by 5 would be 190. What was the price of 
 the hat ? 
 
 Let x == the price of the hat. 
 
 Then 5 x* — 130 = 190, by the question. 
 
 5 # 2 = 190 -f- 130, or 320, by transposition. 
 
 # 2 = 64, and x = 8. 
 
 Ans. $8. 
 
 4. A gentleman, being asked the age of his son, 
 replied, that if from the square of his age were sub- 
 tracted his own age, which was 30 years, and the re- 
 mainder were multiplied by his son's age, the product 
 would be 6 times his age. How old was he ? 
 
 Let x t=. the son's age. 
 
 Then, by the conditions of the question, 
 
 (#2 _ 30) x , or x 3 — 30 x = 6 x. 
 
 x 2 -— 30 = 6, by dividing by x. 
 # 2 == 36, and x = 6. 
 
 Ans. 6 years. 
 
 5. What two numbers are those, which are to each 
 
 other as 3 to 4, and the difference of whose squares 
 
 is 112? 
 
 Let x = the larger number, 
 
 and — - a± the smaller. 
 
 4 
 
 Then X*— ^ = 112, 
 
 or 16 a? 2 — 9^= 1792, by multiplication. 
 
 Ans. 16 and 12. 
 19 
 
218 FIRST LESSONS IN ALGEBRA. 
 
 6. There is a certain room, the sum of whose length 
 and width is to its length as 5 to 3 ; and the same 
 sum, multiplied by the length, is equal to the width 
 multiplied by 60. What are the dimensions of the 
 
 room 
 
 Let x z± the length, 
 
 and y == the width of the room. 
 
 6 x 
 
 a. Then x -f- y — — by the question, 
 
 b. and 60 y = x 1 -\- x y, that is, (x -f- y) x. 
 
 c. y = — , by reducing equation a. 
 
 X s 
 
 jy. y = , by reducing equation b. 
 
 #* 2 x 
 
 e. = — , by comparing equations c and d. 
 
 3 a: 2 = 120 x — 2 a: 2 , 
 
 and 3 x = 120 — 2 a:, by dividing by a?. 
 
 3 a? -f 2 a? = 120, and x =. 24. 
 
 y, or — = 16. 
 
 Ans. Length, 24 feet; 
 Width, 16 feet. 
 
 From these operations may be derived the follow- 
 ing Rule for solving pure equations of the second 
 degree : 
 
 Find the value of the second power of the unknown 
 quantity, in the same manner as the value of the un- 
 known quantity is found in simple equations ; and then 
 extract the square root of each member of the equation. 
 
 Sometimes, as in the last question, the second 
 power can be made to disappear by division. 
 
 7. A boy bought a number of oranges for 36 cents ; 
 
EQUATIONS OF THE SECOND DEGREE. 219 
 
 and the price of an orange was to the number bought 
 as 1 to 4. How many oranges did he buy, and what 
 did he give apiece ? 
 
 8. A merchant sold a quantity of flour for a certain 
 sum, and at such a rate, that the price of a barrel was 
 to the number of barrels as 4 to 5 : if he had re- 
 ceived 45 dollars more for the same quantity, the price 
 of a barrel would have been to the number of barrels 
 as 5 to 4. How many barrels did he sell, and at what 
 price ? 
 
 9. A gentleman exchanges a field, 81 rods long 
 and 64 rods wide, for an equal quantity of land in the 
 form of a square. What was the side of the square ? 
 
 10. How long and wide is a rectangular field con- 
 taining 864 rods, the width of which is equal to § of 
 the length ? 
 
 11. A certain street contains 144 rods of land ; 
 and if the length of the street be divided by its width, 
 the quotient will be 16. How long and wide is the 
 street ? 
 
 12. A trader sold two pieces of broadcloth, which 
 together measured 18 yards ; and he received as many 
 dollars a yard for each piece as it contained yards. 
 Now, the sums received for the two were to each 
 other as 25 to 16. How many yards were there in 
 each piece? 
 
 Let x = the yards in the longer piece, 
 and y as the yards in the other. 
 
 Then x + y = 18 ; 
 and x = 18 — y. 
 
220 FIRST LESSONS IN ALGEBRA. 
 
 Again, x 2 rfc the whole price of one, 
 and y 2 = the price of the other. 
 
 Therefore, # 2 = — ~, by the question ; 
 
 and x == -J-, by evolution. 
 
 -£- = 18 — y, by comparing the values of a:. 
 
 Ans. 10 yards ; 8 yards. 
 
 13. A man divided 14 dollars between his son and 
 daughter in such a manner, that the quotient of the 
 daughter's part divided by the son's, was T 9 F of the 
 son's part divided by the daughter's. What was the 
 share of each ? 
 
 14. A house contains two square rooms, the areas 
 of which are to each other in the proportion of 25 to 
 9 ; and a side of the larger room exceeds a side of 
 the smaller by 10 feet. What are the dimensions of 
 the rooms? 
 
 15. In a certain orchard there are 4 more rows of 
 trees than there are trees in a row ; and if the same 
 number of trees were so arranged that there should 
 be 64 added to each row, the number of the rows 
 would be reduced to 4. How many trees are there 
 in the orchard ? 
 
 Let x = the trees in a row. 
 
 Then x -j- 4 = the number of rows, 
 
 and x 2 -{- 4 x = the number of trees. 
 
 Also (x + 64) X 4, > . , f4 
 
 , ^^ ^ — the number of trees, 
 or 4 x -\- 256, > 
 
 Then x* + 4x = 4x-{- 256. 
 
 Ans. 320 trees. 
 

 EQUATIONS OF THE SECOND DEGREE. 221 
 
 16. When an army was formed in solid column, 
 there were 9 more men in file than in rank ; but when 
 it was formed in 9 lines, each rank was increased by 
 900 men. Of how many men did the army consist? 
 
 17. A gentleman has two squares of shrubbery in 
 his grounds, the difference of whose sides is to the side 
 of the greater square as 2 to 9 ; and the difference of 
 their areas is 128 yards. What are the sides of the 
 squares ? 
 
 18. Says A to B, "Our ages are the same ; but if 
 I were 5 years older, and you were 5 years younger, 
 the product of our ages would be 96." What are 
 their ages ? 
 
 19. What number is that, which being added to 
 10 and subtracted from 10, the product of the sum, 
 multiplied by the difference, will be 51 ? 
 
 20. There is a rectangular field, whose length is to 
 its breadth in the proportion of 6 to 5. A part of this, 
 equal to £ of the whole, being an orchard, there re- 
 main for tillage 625 square rods. What are the length 
 and breadth of the field ? 
 
 21. It requires 108 square feet of carpeting to cover 
 a certain entry ; and the sum of its length and breadth 
 is equal to twice their difference. How long and 
 wide is it ? 
 
 22. Required two numbers which are to each other 
 as 1 to 3, and the sum of whose second powers is 
 equal to 5 times the sum of the numbers. 
 
 23. The area of an oblong room is 400 square feet ; 
 and if its width were equal to its length, its area woulcf 
 
 19* 
 
222 FIRST LESSONS IN ALGEBRA. 
 
 be T % greater. What are the dimensions of the 
 room ? 
 
 24. A charitable person distributed a certain sum 
 among some poor men and women, the numbers of 
 whom were in the proportion of 4 to 5. Each man 
 received ^ as many shillings as there were persons 
 relieved ; and each woman received twice as many 
 shillings as there were women more than men. The 
 men received, altogether, 18 shillings more than the 
 women. How many were there of each ? 
 
 25. A gentleman, being asked the ages of his two 
 sons, replied, that they were to each other as 3 to 4 ; 
 and that the product of their ages was 48. What 
 were their ages ? 
 
 26. A gentleman has an oblong garden of such di- 
 mensions, that if the difference of the sides be mul- 
 tiplied by the greater side, the product will be 40 
 square rods ; but if the difference be multiplied by the 
 shorter side, the product will be 15 rods. What are 
 the length and width of the garden ? 
 
 Let x r= the less side, 
 and y = the difference of the sides. 
 Then x -\- y = the greater side. 
 Therefore, x y -\- y 2 = 40, ") by the conditions of the 
 and x y — 15, ) question. 
 
 15 
 
 x = — . 
 y 
 
 By substituting the value of x in the first equation, 
 
 we shall obtain 
 
 15 -f y f = 40, 
 
 and y = 5. 
 
EQUATIONS OF THE SECOND DEGREE. 223 
 
 SECTION II. 
 
 Affected Equations. 
 
 As a. pure equation of the second degree contains 
 the unknown quantity only in the form of its second 
 power, all the terms in which it appears can be united 
 in one term, whose root, as we have seen, can be 
 readily found. 
 
 An affected equation of the second degree contains 
 not only the square of the unknown quantity in one 
 term, but also the unknown quantity itself in another 
 term. 
 
 Thus, x 2 -f- 4 x =. 11 is an affected equation of the 
 second degree, in which the unknown quantity ap- 
 pears in two terms ; for a: 2 and x cannot be actually 
 added together so as to make but one term. 
 
 When an equation of this sort is formed, it may 
 
 contain the unknown quantity in any number of terms, 
 
 provided it be only in the first and second powers ; 
 
 for, in this case, the terms may all be reduced to two. 
 
 Thus, if we have the equation 
 
 5 x * _|_ 8 x — 24 + x 2 — 3 x = 4 x* + x + 48, 
 
 by transposition we obtain 
 
 5 # 2 + # 2 — 4x* + 8x — 3 x — # := 48 + 24 ; 
 
 and, by adding the similar terms, 
 
 2 z 2 + 4 x = 72 ; 
 
 and, by dividing all the terms by 2, 
 
 x^ -f 2 x = 36. 
 
 Affected and pure equations of the second degree 
 
224 FIRST LESSONS IN ALGEBRA. 
 
 are solved in the same manner: in both, ive find the 
 value of the unknown quantity by extracting the square 
 root of each member of the equation. When the mem- 
 ber containing the unknown quantity is a complete 
 power, the process is as direct and simple in an affect- 
 ed as in a pure equation. 
 
 1. Given afl + 2 a x + a 2 = 6 2 . What is the 
 value of x 1 Ans. b — a. 
 
 In this question, a and b represent known quanti- 
 ties. Now. we know by inspection, that the first mem- 
 ber of the equation, x 2 -f- 2 a x -f- a 2 , is the complete 
 second power of the binomial quantity x -f- a ; and 
 the other member, 6 2 , is the second power of b. By 
 extracting the square root of each member, therefore, 
 we obtain the equation, 
 
 x -f a — b. 
 And x = b — «, by transposition. 
 
 If we suppose a = 4 and b = 9, by raising x -(- 4 
 and 9 to their second powers, instead of the above 
 equation, we shall have 
 
 a? _}_ 8 x + 16 sa 81, 
 
 or s/ a? 3 + 8 x + 16 = %/ 81. 
 That is, a; + 4 £= 9, 
 and # = 9 — 4 = 5, by transposition. 
 
 2. Given « 2 -f 2 o x = i. What is the value 
 
 of a?? Ans. \/b + « 2 — a- 
 
 In this question, the member of the equation con- 
 taining the unknown quantity, is not a complete power ; 
 
EQUATIONS OF THE SECOND DEGREE. 225 
 
 and, of course, while it remains in its present form, 
 its root, and consequently the value of the unknown 
 quantity, cannot be found. But it is possible to add 
 such a quantity to the first member as shall make it a 
 perfect square. The necessary quantity must be 
 added to both members of the equation, to preserve 
 its equality. 
 
 But how shall we find the quantity which must be 
 added to complete the square ? If we examine the 
 equation given above, 
 
 x* + 2 a x = b, 
 
 we ^hall be satisfied that the root of the first member, 
 whatever it may be, is not a single term ; for any 
 power of one term consists of but one term. But if 
 its root consist of two terms, one term is wanting to 
 complete the square ; for the second power of a bi- 
 nomial quantity contains three terms, [See Chap. VII. 
 Sec. III.] whereas the given quantity contains but two. 
 The terms given, # 2 -f- 2 a x, are the first two terms 
 of the second power of the binomial x -f- a ; and the 
 third term of the same quantity is a 2 . [See Chap. 
 IX. Ques. 18.] By adding this quantity, therefore, 
 to both of the given members, we make the first mem- 
 ber a complete square ; and the equation becomes 
 
 x* -f- 2 a x -f- a 2 — _|_ a 2. 
 And V x* + 2 a x + « 2 = ^H" 2 , 
 or x -J- a = s/ b -\- a 2 , by evolution, 
 and x = ^/ b + a 2 — - a, by transposition. 
 Observe that the coefficient of the second term 
 
226 FIRST LESSONS IN ALGEBRA. 
 
 given is 2 a, one half of which is 1 a or a; and that 
 the quantity added to complete the square, a 2 , is the 
 second power of a. 
 
 If, in the last example, we suppose a = 4, and 
 b =s 84, the given equation will be 
 
 #2 -f 8 x = 84. 
 
 Now, if we add, as before, the second power of the 
 value of a, (4 X 4 = 16,) to both sides of the equa- 
 tion, the first member will be a complete square ; and 
 the equation will be solved thus : 
 
 # 2 -f- 8 x + 16 = 84 + 16, or 100. 
 And +/x 2 + 8 x -f 16 as V 100, 
 or x -f 4 = 10; 
 and a? cs 10 — 4, or 6. 
 
 Observe that 16, which is added to complete the 
 square, is the second power of 4, or half of the co- 
 efficient of the second term of the given equation. 
 
 Hence, to render the first member a complete 
 second power, we add the square of half the coeffi- 
 cient of the second term to both members of the equation. 
 
 3. Given the equation 87 + 7 x* — 123 + 3 x = 
 5 a; 2 -f 118 — 5 x. What is the value of x 1 
 
 7#2— 5# 2 H-3tf+5a^ll8-fl23— 87, by transposition. 
 
 2 # 2 -f- 8 x t=. 154, by uniting terms, 
 
 and x 2 -\- 4 x r= 77, by division. 
 
 a? 2 -f-4 x-\-4 = 77+4, or 81, by completing the square. 
 
 x -f- 2 = 9, by evolution, 
 
 and x = 9 — 2, or 7. Ans. 7. 
 
EQUATIONS OF THE SECOND DEGREE. 227 
 
 4. Given the equation 3 x* -f 89 + j+ 20 x = 224 
 -f- 3 a? 2 -|- 8 a? ; to find the value of x. 
 
 3 a: 2 -f j — 3^2_j_ 20 ^— 8z=224— 89,by transposition. 
 
 X 2 
 
 — -f- 12 x =z 135, by uniting terms. 
 
 a? 2 -j- 36 a? = 405, by removing the denominator. 
 jr 2 -f-36#+324 =405+324, or 729, by completing the 
 
 x -f- 18 = 27, by evolution. [square. 
 x = 27 — 18, or 9. 
 
 5. What is the value of x in the equation 3 a: 2 -f- 
 
 2 x m 161 ? 
 
 3 a: 2 + 2x = 161. 
 
 1x 
 
 x 2 -\- — = ifi, by removing the coefficient of a? 2 . 
 
 2x 
 
 r 2 -}- — -j-|=i|i_|_|.— 4|4 ? D y completing the square. 
 In this equation, the second term has a fractional 
 
 2x 
 
 coefficient; for — is the same as f x. The half off 
 
 is £, the second power of which is £. The fractions 
 of the second member, J-f^ and £, are reduced to a 
 common denominator, and added together in the 
 usual way. 
 
 6. Given 5 a? 2 + 3 x = 344. What is the value 
 
 of a?? 
 
 5 a? 2 + 3 x = 344. 
 
 3 x 
 
 a? 2 -\- — = 3 f *-, by removing the coefficient of x 2 . 
 
 ** + T + Th> = #5 + t&it, or -6AV- 
 Half of | a? is T 3 ^ a? ; and the second power of T 3 T 
 
228 FIRST LESSONS IN ALGEBRA. 
 
 7. Required the value of x in the following equa- 
 tion; x 2 — 14 x = 51. 
 
 x*—Ux=. 51. 
 
 # 2 — 14 x -f- 49 = 100, by completing the square. 
 
 x — 7 = 10, by evolution ; 
 
 and x = 10 + 7, or 17. 
 
 In this operation, as the second term of the first 
 member, — 14 x, is a negative quantity, the second 
 term of the root must also be negative. It should 
 not be forgotten, that the square root of a positive 
 quantity may be either positive or negative. 
 
 It will be observed, that the square root of the first 
 member of the equation always consists of the wiknown 
 quantity, and half the coefficient of the second term. 
 In practice, therefore, it is not necessary to complete 
 the square of that member, and then extract the root. 
 
 8. Divide 34 into two such numbers that their 
 product shall be 225. 
 
 Let x = one number. 
 Then 34 — x = the other, 
 and x (34 — x) = their product. 
 Therefore, 34 x — x 2 = 225, by the question. 
 
 It will be remembered, that there can be no such 
 quantity as the root of — x 2 . [See Chap. X. Sec. 
 II.] Consequently, in the second power of a bino- 
 mial or residual quantity, the first and last terms must 
 always be positive, as each is the second power of 
 one term of the root. Therefore, the first member 
 of the last equation, 
 
 — a? + 34 x = 225, 
 
EQUATIONS OF THE SECOND DEGREE. 229 
 
 is not, in its present form, a part of the second power 
 of any binomial whatever. The signs of all the terms 
 may be changed, however, without affecting the equality 
 of the members; and, when this has been done, the 
 square of the first member can be completed in the 
 usual manner. 
 
 x* — 34 x = — 225, by changing all the signs. 
 
 x * _ 34 x + 289 = — 225 + 289, or 64. 
 
 x — 17 =s =F 8 ? by evolution, 
 
 and x == =p 8 -f- 17. 
 
 Ans. 25 and 9. 
 
 From these examples and observations, we derive 
 the following Rule for solving affected equations of 
 the second degree : 
 
 1. Collect all the terms containing the unknown 
 quantity in one member of the equation, and all the 
 known quantities in the other. 
 
 2. Arrange the terms that contain the unknown 
 quantity according to their powers, as in Division. 
 
 3. If the square of the unknown quantity have a 
 coefficient or a divisor, ii should be removed in the 
 usual manner. 
 
 4. If the term containing the square of the unknown 
 quantity be negative, make it positive by changing the 
 signs of all the terms. 
 
 5. Complete the square of the first member, by add- 
 ing the square of half the coefficient of the second 
 term to both sides of the equation. 
 
 6. Reduce the equation by extracting the square root 
 
 20 
 
230 FIRST LESSONS IN ALGEBRA. 
 
 of each member, and transposing the Tcnown part of the 
 binomial. 
 
 9. Required two numbers, whose difference shall 
 be 8 and their product 105. 
 
 Let x — the smaller number, 
 and x -j- 8 = tne larger. 
 Then x 2 -f- 8 * = 105, by the question. 
 
 x + 4 = s/ 105 + 16, or 11. 
 
 10. What two numbers are there whose sum is 30, 
 and whose product is equal to eight times their dif- 
 ference ? 
 
 Let x =c the greater number. 
 
 Then 30 — x = the less, 
 and 2 x — ■ 30 = their difference. 
 Therefore, x (30— a?) = 8 (2 x — 30), by the question, 
 
 or 30 x — x* = 16 x — 240 ; 
 
 and a? 2 — 30 x = — 16 x + 240, by changing the signs. 
 
 x 2 — 14 x = 240, by transposition, 
 
 and a: 2 — 14 x + 49 = 240 + 49, or 289. 
 
 Another Way to complete the Square. 
 
 11. Given a? 2 = c — b x; to find the value of x. 
 
 a. x* -f- b x = c, by transposition. 
 
 b. x* -\- b x -\- — = c -}- — , by completing the square. 
 
 Let this equation be freed from fractions, and it 
 will become 
 
 c. 4 a 2 + 4 b x -f- 6 2 = 4 c -f- 6 2 . 
 
EQUATIONS OF THE SECOND DEGREE. 231 
 
 The first member of this equation being a perfect 
 square, its root may be found in the usual manner. 
 
 y' 4 x 2 _J_ 4 & a _|_ £2 _ ^ 4 c _|_ £ 2j 
 
 or 2 a: -{- 6 = \/ 4 c -f- 6 2 , by evolution, 
 and 2x = \/4c-^-b 2 — b, by transposition. 
 
 ANS. y/ — 
 
 We might have obtained equation c directly from 
 equation a, by multiplying it by 4 and adding the 
 square of b, the coefficient of the second term, to 
 each member. 
 
 12. Required the value of x in the following equa- 
 tion ; ax* — c = — b x. 
 
 a. a a: 2 -f- b x = c, by transposition. 
 
 b. x 2 -f- — ~ — "^"j °y removing the coefficient of a: 2 . 
 
 c . x z _|_ h -l 4- £ 2 = L + 1L, by completing the square. 
 
 If we remove the denominators from this equation, 
 we shall obtain 
 
 d. 4 a 2 x* -f 4 a 6 a? + 6 2 = 4 a c -f- 6 2 ; 
 
 and V 4 a 2 * 2 + 4 a b x -f 6 2 = ^/ 4 a c + 6 2 , 
 
 or 2 a x -{- 6 = ^/ 4 a c -f- 6 2 , by evolution, 
 
 and 2ax = \/4ac-\-b^ — 6, by transposition. 
 
 . , /4ac+A s b 
 
 Ans -^^5 — «r 
 
 Compare equations a and d. It will be seen that 
 if we multiply the former by 4 a, that is, by 4 times 
 the coefficient of # 2 , and add to each member the 
 
232 FIRST LESSONS IN ALGEBRA. 
 
 second power of b, which is the coefficient of x, we 
 shall obtain the latter equation. 
 
 From these operations may be derived the follow- 
 ing Rule for completing the square, which will be 
 found more simple and convenient than that already 
 given, where the coefficient of # 2 is a small number, 
 and where the coefficient of x is an odd number ; as 
 it does not introduce fractions into the operation. 
 
 Having prepared and arranged, the equation as be- 
 fore, multiply both members by four times the coefficient 
 of the first term; that is, of the term which contains 
 the square of the unknown quantity. If it have no 
 coefficient, multiply by four. 
 
 Add the square of the coefficient of the second term 
 to both sides of the equation ; and extract the square 
 root of each member. 
 
 13. Given # 2 -{- 5 x = 126 ; to find the value of*. 
 As the first term, # 2 , has no coefficient, the equa- 
 tion must be multiplied by 4. 
 
 4 x 2 -f 20 x == 504. 
 
 Next, add the square of 5, the coefficient of the 
 second, term, to both members. 
 
 4 x* + 20 x + 25 = 504 -f 25, or 529. 
 The first member being now a complete square, 
 extract the root as before. 
 
 V 4 a? 2 + 20 x + 25 = s/ 529, 
 
 or 2 x + 5 = 23. 
 
 2 x = 23 — 5, or 18, by transposition, 
 
 and x = 9. 
 
EQUATIONS OF THE SECOND DEGREE. 233 
 
 14. Required the value of x in the following equa- 
 tion ; 3 x* + 3 x =s 216. 
 
 Multiply by 4 times the coefficient of a: 2 , that is, 
 by 4 X 3 = 12. 
 
 36 x* -f- 36 x = 2592. 
 
 Add the square of 3, the coefficient of x. 
 
 36 x 2 + 36 a; + 9 = 2592 -f 9, or 2601. 
 
 V 36 x 2 + 36 x -f 9 = V 2601, 
 
 or 6 a? -|- 3 = 51. 
 6 a? = 51 — 3, or 48 ; and x = 8. 
 
 15. What is the value of x in the following equa- 
 tion ; 3 a; 2 — 19 a: == — 6? 
 
 3 a: 2 — 19 a? = — 6. 
 
 36 a: 2 — 228 x = — 72, by multiplying by 12. 
 
 36 # 2 — 228 a: + 361 = 289, by adding 19 2 . 
 
 16. Given — 11 = -r\ to find the value of x. 
 
 3 3 
 
 3 3 
 
 4 a: 2 — 33 = a?, by removing the denominators. 
 4 a: 2 — x — 33, by transposition. 
 64 a: 2 — 16 x = 528, by multiplying by 16. 
 64 a: 2 — 16 x + 1 = 528 -f 1, or 529, by complet- 
 ing the square. 
 
 17. There are two numbers whose difference is 9, 
 and \ their product is 10 more than the square of the 
 smaller number. What are the numbers ? 
 
 18. The length of a room exceeds its width by 9 
 feet ; and its area is 400 feet. What are the dimen- 
 sions of the room ? 
 
 20* 
 
234 FIRST LESSONS IN ALGEBRA. 
 
 I 
 SECTION III. 
 
 Questions producing Affected Equations. 
 
 1. The ages of a man and his wife amount to 42 
 years, and the product of their ages is 432. What is 
 the age of each ? 
 
 2. A gentleman, being asked the ages of his son 
 and daughter, replied, that his son was 5 years older 
 than his daughter, and that the product of their ages 
 was 266. What were their ages? 
 
 3. The length of a room exceeds its width by 8 
 feet, and its area is 768 feet. What are its length and 
 width ? 
 
 4. The difference of two numbers is 6 ; and the 
 square of the greater exceeds twice the square of the 
 less by 47. Required the numbers. 
 
 5. A gentleman divided 28 dollars between his two 
 sons in such a manner, that the product of their shares 
 was 192. What was the share of each? 
 
 6. The wall which encloses a rectangular garden, 
 is 128 yards long, and the area of the garden is 1008 
 yards. What are its length and breadth ? 
 
 7. A man bought a certain number of sheep for 80 
 dollars. If he had bought 4 more for the same money, 
 they would have come to him a dollar apiece cheaper. 
 How many did he buy ? 
 
 8. It is required to find two numbers, whose sum 
 shall be 14; and such, that 18 times the greater shall 
 be equal to 4 times the square of the less. 
 
EQUATIONS OF THE SECOND DEGREE. 235 
 
 9. A man paid 120 dollars more for his watch than 
 for a chain ; and the price of the watch was to that 
 of the chain as the price of the chain was to 10. Re- 
 quiied the price of each. 
 
 10. In a parcel containing 24 coins of silver and 
 copper, each silver coin is worth as many pence as 
 there are copper coins ; and each copper coin is worth 
 as many pence as there are silver coins ; and the 
 whole is worth 18s. How many coins are there of 
 each sort ? 
 
 11. A drover bought a number of oxen for 675 
 dollars ; which he sold again for 48 dollars a head ; 
 and he gained, by the bargain, as much as he gave for 
 one ox. How many oxen did he buy ? 
 
 12. Two travellers, A and B, set off at the same 
 time to a place distant 150 miles. A travels 3 miles 
 an hour faster than B, and arrives at his journey's end 
 8 hours and 20 minutes before him. How many miles 
 did each travel per hour ? 
 
 13. What two numbers are there, whose sum is 25 
 and product 144? 
 
 14. The age of A is 12 years more than that of B ; 
 and the product of their ages is 640. What is the 
 age of each ? 
 
 15. The sum of two numbers is 30; and if 18 be 
 added to £ of their product, the sum will be equal to 
 the square of the smaller number. What are the 
 numbers ? 
 
 16. Says A to B, "The product of our ages is 
 120 ; and if I were 3 years younger, and you were 2 
 
236 FIRST LESSONS IN ALGEBRA. 
 
 years older, the product of our ages would still be 
 120." What are their ages ? 
 
 17. A farmer sold a certain number of sheep for 
 £120. If he had sold 8 more for the same money, 
 he wuuid have received 10 shillings less for each 
 sheep. How many did he sell ? 
 
 18. Two benevolent gentlemen, A and B, dis- 
 tributed each 1200 dollars among a certain number of 
 poor persons. A relieved 40 persons more than B ; 
 but B gave 5 dollars more to each person than A. 
 How many persons did each relieve ? 
 
 19. A person bought two pieces of cloth; the finer 
 of which, at 4 shillings a yard more than the other, 
 cost £18. But the coarser piece, which was 2 yards 
 longer than the finer, cost only £16. How many 
 yards were there in each piece ? and what was the 
 price of a yard of each ? 
 
 20. An officer would arrange 1200 men in a solid 
 body, so that each rank may exceed each file by 59 
 men. How many must be placed in rank and file ? 
 
 21. In an orchard containing 900 trees, the trees 
 are so planted that there are 11 more rows than there 
 are trees in a row. Required the number of rows ; 
 also the number of trees in a row. 
 
 22. The perimeter of a room is 48 feet ; and the 
 area of the floor is equal to 35 times the difference 
 of its length and breadth. What are the dimensions 
 of the room ? 
 
 23. A drover bought a number of sheep for 190 
 dollars. Having lost 8 of them, he must sell the re- 
 mainder at a profit of 8 shillings apiece, not to lose 
 
EQUATIONS OF THE SECOND DEGREE. 237 
 
 money by the bargain. How many sheep did he buy ? 
 and at what price ? 
 
 24. A merchant sold a quantity of sugar for £56, 
 by which he gained as much per cent, as the whole 
 cost him. How much did it cost ? 
 
 Let x == the cost of the sugar. 
 Then 56 — x = the gain. 
 
 Again, — ec the rate per cent. ; 
 
 and ^o — the S ain * 
 Therefore, -- = 56 — x. 
 
 25. A trader sold a quantity of flour for 39 dollars, 
 and gained as much per cent, as the flour cost him. 
 What did he give for the flour ? 
 
 26. A butcher bought a certain number of calves 
 for 200 dollars ; and, reserving 15, he sold the rest 
 for 180 dollars, by which he gained 2 shillings a head. 
 How many calves did he buy ? and at what price ? 
 
 27. A grass-plot, 18 yards long and 12 wide, is 
 surrounded by a border of flowers of a uniform width. 
 The areas of the grass-plot and border are equal. 
 What is the width of the border ? 
 
 28. A square court-yard has a gravel walk round 
 it. The side of the court wants 2 yards of being 6 
 times the breadth of the walk ; and the number of 
 square yards in the walk, exceeds the number of yards 
 in the periphery of the court by 164. What is the 
 area of the court ? 
 
 29. If the square of a certain number be taken 
 from 40, and the square root of their difference be 
 
238 FIRST LESSONS IN ALGEBRA. 
 
 increased by 10, and the sum be multiplied by 2, and 
 the product divided by the number itself, the quotient 
 will be 4. Required the number. 
 
 Let x =± the number. 
 
 If we take the square of x from 40, we shall have 
 
 40 — a; 9 , 
 
 whose square root is expressed thus, 
 
 y 40 — x*. 
 
 This square root, increased by 10, becomes 
 
 y' 40 — # 2 -f 10; 
 
 and 2 V 40 - *2 -f 20, by multiplying by 2 
 and — - — , by dividing by x. 
 
 Therefore, 2 _ y^-* 2 + 20 _ 4. by the question. 
 2 y' 40 — a; 2 _j_ 20 = 4 a?, by multiplying by x~, 
 and v'~40^^ 2 + 10 ~ 2 a?, by dividing by 2. 
 
 ^40 — . x^ = 2 a? — 10, by transposition. 
 To free this .equation from the radical quantity, 
 
 ^/ 40 — . a: 2 , we must raise both members to the sec- 
 ond power. The second power of a quantity affected 
 with the radical sign, is evidently that quantity without 
 the sign; for ^/ «, being the square root of a, the 
 product of \/ a multiplied by \/ a must be a. There- 
 fore, the second power of \/ 40 — a? 2 is 40 — a? 2 ; and 
 the whole equation, when raised to the second power, 
 becomes 
 
 40 — x* = 4 a: 2 — 40 x -f 100. 
 
 30. What number exceeds its square root by 42 ? 
 
EQUATIONS OF THE SECOND DEGREE. 239 
 
 31. A school-boy, being asked the ages of himself 
 and sister, replied, that he was 6 years older than his 
 sister ; and that twice the square of her age was 47 
 less than the square of his own. What were their 
 ages ? 
 
 32. A gentleman has two square flower-plots in his 
 grounds, which together contain 2120 yards ; and the 
 diameter of the larger plot exceeds that of the smaller 
 by 12 yards. What are the sides of each ? 
 
 33. A laborer, having built 105 rods of fence, found 
 that, had he built 2 rods less a day, he would have 
 been 6 days longer in completing the job. How many 
 rods did he build per day ? 
 
 34. A farmer sold 7 calves and 12 sheep for 50 
 dollars ; and the price received for each was such, 
 that 3 more calves were sold for $10 than sheep for 
 $6. What was the price of each? 
 
 35. A young lady, being asked her age, answered, 
 " If you add the square root of my age to half of my 
 age, the sum will be 12." What was her age? 
 
 36. A trader bought some barrels of flour for $60. 
 Had he bought 3 more barrels for the same sum, each 
 barrel would have cost him one dollar less. How 
 many barrels did he buy ? 
 
 37. A man had a field whose length exceeded its 
 breadth by 5 rods. He gave 3 dollars a rod to have 
 it fenced ; and the whole number of dollars was equal 
 to the number of square rods in the field. Required 
 the length and breadth of the field. 
 
 38. Says A to B, " I have 9 dollars more than you ; 
 and if the number of dollars we both have, be multi- 
 
240 FIRST LESSONS IN ALGEBRA. 
 
 plied by the number that I have, the product will be 
 266." How many dollars has each ? 
 
 39. A man has three children, A, B and C ; A 
 being the oldest, and C the youngest. Now, the dif- 
 ference of A and B's ages exceeds the difference of 
 B and C's by 6 years. The sum of all their ages is 
 33 years, and the sum of their squares 467. Required 
 their ages. 
 
 Let x = B's age, 
 
 and y = the difference of B and C's ages. 
 Then x — • y = C's age, 
 and x -f- y -f- 6 = A's age. 
 a? + a? — y-\-x-\-y-\-6 = 33, by the question, 
 or 3 x — 33 — 6, or 27, 
 and x = 9. 
 
 Again, according to the question, the sum of the 
 squares of all their ages is 467 ; that is, 
 
 * 2 + (* — yy + + y + <3) 2 = 467, 
 
 or, by involution and addition, 
 3 x* + 2 y 2 -f 12 x + 12 y + 36 = 467 ; 
 
 and, by substituting the value of x, we have 
 
 243 + 2 y 2 + 108 + 12 y + 36 = 467, 
 or 2 y 2 -j- 12 y = 80, by transposition. 
 
 40. A farmer sold 80 bushels of wheat and 100 
 bushels of rye for £65 ; and each at such a rate, that 
 he sold 60 bushels more of rye for £20 than of wheat 
 for £10. What was the price of each? 
 
MISCELLANEOUS QUESTIONS. 241 
 
 CHAPTER XII. 
 MISCELLANEOUS ClUESTIONS. 
 
 1. A gentleman bought three kinds of wine, of each 
 an equal quantity. For the sherry he gave $8 a 
 dozen ; for the port, $9; and for the madeira, $11. 
 The whole came to $ 168. How many dozen of each 
 kind did he buy ? 
 
 2. A steam-boat has 81 passengers; there being 
 twice as many women as children, and three times as 
 many men as women. What is the number of men, 
 women and children ? 
 
 3. Divide $9289 between A and B in such a man- 
 ner, that A's share shall be to B's as 2 to 5. 
 
 4. A man, dying, left an estate valued at $14832. 
 In his will he gave £ of his property to his wife ; and 
 directed the remainder to be so divided between his 
 son and daughter, that the daughter's portion might 
 be to the son's as 3 to 5. What was the share of 
 each ? 
 
 5. What is that number, to which if you add £ of 
 itself, and from the sum subtract f of itself, i of the 
 remainder is 3 ? 
 
 6. A farmer bought 12 sheep, 5 cows, 2 yoke of 
 
 21 
 
242 FIRST LESSONS IN ALGEBRA. 
 
 oxen and 3 horses, for 795 dollars. A cow cost as 
 much as 6 sheep, a yoke of oxen as much as 3 cows, 
 and a horse as much as 3 oxen. What did he give 
 for each ? 
 
 7. A man divided a certain sum of money equally 
 between his son and daughter ; but had he given his 
 son 33 dollars more, and his daughter 47 dollars less, 
 her share would have been but £ of his. What was 
 the sum divided ? 
 
 8. Divide 46 dollars into two such parts, that \ of 
 one and £ of the other may be 10 dollars. 
 
 9. A man divided 198 acres of land between his 
 three children in such a manner, that A's part was to 
 B's in the ratio of 3 to 8 ; and C had as many acres 
 as both his brothers. What was the share of each? 
 
 10. A man bought a certain quantity of wine for 
 94 dollars ; and after 7 gallons had leaked out, he 
 sold i of the remainder, at cost, for $20. How many 
 gallons did he buy ? 
 
 11. If a certain number be divided by the sum of 
 its digits, the quotient, will be 8; but if the digits be 
 inverted, and that number divided by 2 less than their 
 difference, the quotient will be 9. What is the 
 number f 
 
 12. Two friends bought a horse together; and 
 when one had paid § and the other f of the price 
 agreed upon, they still owed 21 dollars. What was 
 the price of the horse ? 
 
 13. In a certain university there are 384 students, 
 £ of whom belong to the academical department; 
 and in the departments of law, divinity and medi- 
 
MISCELLANEOUS QUESTIONS. 243 
 
 cine, the students are to each other as the numbers 
 1, 2 and 3. How many students are there in each de- 
 partment? 
 
 14. A man agreed to carry 47 earthen jugs to a 
 certain place. For every one he delivered safe, he 
 was to receive 6 cents ; and for every one he broke, 
 he was to pay 10 cents. He received $1,54. How 
 many jugs did he break ? 
 
 15. Divide $1170 among three persons, A, B and 
 C, in proportion to their ages. Now, B is a third part 
 older than A, and A is half as old as C. What is 
 the share of each ? 
 
 16. Three men, A, B and C, pay a tax of 594 dol- 
 lars. The property of A is to that of B as 3 to 5 ; 
 and the property of B is to that of C as 8 to 7. 
 What part of the tax is paid by each ? 
 
 17. A father gives to his six sons $2010, which 
 they are. to divide according to their ages, so that 
 each elder son shall receive $24 more than his next 
 younger brother. What is the share of the youngest 
 son? 
 
 18. If I multiply a certain number by 6, add 18 to 
 the product, and divide the sum by 9, the quotient will 
 be 20. What is the number ? 
 
 19. Divide 119 into three such parts, that the second 
 divided by the first will give 3 for a quotient, and 3 
 for a remainder; and the third divided by the second 
 will also give 3 for a quotient, and 3 for a remainder. 
 
 20. A school-master, being asked how many dollars 
 he received a month for teaching, replied, " If I add 
 9 to £ part of the number of dollars I receive, mul- 
 
244 FIRST LESSONS IN ALGEBRA 
 
 tiply the sum thus obtained by 7, subtract 15 from 
 the product, multiply the remainder by 6, and then 
 take away the cipher from the right of the number 
 last obtained, I shall have $54." What were his 
 wages ? 
 
 21. Some travellers find a purse of money, which 
 they agree to share equally. If they take 5 dollars 
 apiece, one man will receive nothing ; but if they 
 take 4 dollars, there will be seven dollars left. What 
 is the number of travellers ? What is the sum to be 
 divided ? 
 
 22. There are two numbers, the product of whose 
 sum multiplied by the greater, is 144 ; and whose 
 difference, multiplied by the less, gives 14. Required 
 the numbers. 
 
 23. A courier had been travelling 4 days, at the rate 
 of 6 miles an hour, when another was sent after him, 
 who travelled 8 miles an hour. In how many days 
 will the second courier overtake the first, if they both 
 travel 15 hours a day? 
 
 24. A gentleman has a rectangular garden 36 rods 
 in circumference ; and the square of the width is to 
 the square of the length as 16 to 25. Required its 
 dimensions. 
 
 25. Two travellers, A and B, began a journey of 
 300 miles at the same time. A travelled A mile an hour 
 faster than B, and arrived at his jojRiey's end 10 
 hours before him. How many miles an hour did each 
 travel ? 
 
 26. Two sportsmen, walking over a marsh, started 
 a flock of plover. The first one fired, and brought 
 
MISCELLANEOUS QUESTIONS. 245 
 
 down f of the whole flock. Afterwards, the second 
 one fired, and killed a number equal to the square 
 root of half the flock ; when only 2 birds were left. 
 How many birds were there in the flock ? 
 
 27. Required the side of a square field, which shall 
 contain the same quantity of land as another field, 
 which is 72 rods long and 18 rods wide. 
 
 28. Three planters, A, B and C, together possess 
 2658 acres of land. If B sell A 215 acres, then will 
 A's plantation exceed B's by 236 acres ; but if B buy 
 167£ acres of C's plantation, they two will have the 
 same quantity of land. How many acres has each ? 
 
 29. Required two numbers, whose sum, multiplied 
 by their product, shall be equal to 12 times the differ- 
 ence of their squares ; the numbers being to each 
 other in the ratio of 2 to 3. 
 
 30. It is required to form a regiment, containing 
 865 men, into two squares, one of which shall con- 
 tain 7 more men in rank and file than the other. 
 How many men must each of the squares contain ? 
 
 31. A man, having travelled 108 miles, found that 
 he could have performed the same journey in 6 hours 
 less, if he had travelled 3 more miles an hour. At 
 what rate did he travel ? 
 
 32. Two persons, A and B, set out at the same 
 time from two towns, distant 396 miles ; and, having 
 travelled as many days as A travelled miles daily more 
 than B, they met each other. It then appeared that 
 A had travelled 216 miles. How many miles did 
 each travel per day ? 
 
 33. Two merchants, A and B, trade in company, 
 
 21* 
 
246 FIRST LESSONS IN ALGEBRA. 
 
 and gain $1930,28. Of the capital employed, A fur- 
 nished $4000 and B $ 7000. What is each man's 
 share of the gain . ? 
 
 34. A farmer, being asked how many acres of land 
 he owned, answered, that the number was expressed 
 by two digits, whose sum, increased by 7, would be 
 equal to three times the left-hand digit; and he 
 added, that, if he owned 18 acres less, the digits ex- 
 pressing the number would be inverted. How many 
 acres were there in his farm ? 
 
 35. Several gentlemen made an excursion, each 
 taking the same sum of money. Each had as many 
 servants attending him as there were gentlemen ; the 
 number of dollars which each had, was double the 
 number of all the servants, and the whole sum of 
 money taken out, was 3456 dollars. How many gen- 
 tlemen were there ? 
 
 36. Four farmers, A, B, C and D, hired a pasture, 
 for which they paid 81 dollars. A put in 4 cows for 
 3 months ; B, 8 cows for 2 months ; C, 7 cows for 
 5 months ; and D, 3 cows for 6 months. How much 
 of the rent must each man pay ? 
 
 37. There is a certain number, the left-hand digit 
 of which is equal to 3 times the right-hand digit ; and 
 if 12 be subtracted from the number, the remainder 
 will be equal to the square of the left-hand digit. 
 Required the number. 
 
 38. A man has two horses and two saddles, one of 
 which is worth $40, and the other $5. When the 
 best saddle is upon the first horse, and the worst sad- 
 dle upon the second, the former is worth just twice 
 
MISCELLANEOUS QUESTIONS. 247 
 
 as much as the latter ; but when the worst saddle is 
 upon the first horse, and the best saddle upon the 
 second, the latter is worth $5 more than the former. 
 What is the value of each horse ? 
 
 39. Find three such numbers, that the first, with £ 
 the sum of the second and third, shall be 78 ; the 
 second, with £ the excess of the -third over the first, 
 shall be 60 ; and £ the sum of the three shall be 66. 
 
 40. If I had 3 shillings more in my pocket, I could 
 give 2s. 6d. to each of a certain number of beggars ; 
 but if I give them only 2s. apiece, I shall have 4s. left. 
 How much money have I in my pocket? What is 
 the number of beggars ? 
 
 41. A person had £27, 6s. in guineas and crown 
 pieces. Having paid a debt of £14, 17s., he finds 
 that he has as many guineas left, as he has paid away 
 crowns ; and as many crowns left, as he has paid away 
 guineas. How many crowns and guineas had he at 
 first ? 
 
 Remark. A guinea is 21 shillings, and a crown 5 
 shillings, sterling. 
 
 42. A bill of £27, 9s. was paid in half-guineas and 
 crowns ; and twice the number of guineas was equal 
 to half the number of crowns. How many of each 
 were paid away ? 
 
 43. A laborer agreed to work 24 days for 75 cents 
 a day, and to forfeit his wages and 25 cents every day 
 he was idle. At the end of the time, he received 
 $12. How many days was he idle ? 
 
 44. A colonel would arrange a regiment of 1152 
 men in such a manner, that each rank may exceed 
 
248 FIRST LESSONS IN ALGEBRA. 
 
 each file by 24 men. What numbers must he place 
 in rank and file ? 
 
 45. A cistern, containing 276 gallons, is emptied 
 in 21 minutes by two cocks running successively. 
 One cock discharges 16 gallons, and the other 11 
 gallons, in a minute. How many minutes is each 
 cock running ? 
 
 46. A merchant has two kinds of wine; one of which 
 is worth 95. 6d. per gallon, and the other, 135.6c?. 
 How many gallons of each must he take, to form a 
 mixture of 104 gallons which shall be worth £56? 
 
 47. A gentleman bought a quantity of broadcloth 
 for $48; and four times the number of yards were 
 equal to three times the price of a yard. How many 
 yards did he buy, and at what price ? 
 
 48. Two gentlemen, A and B, have rectangular 
 gardens contiguous to each other. A's garden is 20 
 yards wide, and f as long as B's ; and the surface of 
 B's garden is to that of A's as 5 to 3. What is the 
 width of B's garden ? 
 
 49. A miser, dying, left a certain number of eagles, 
 as many quarter-eagles, f the number of half-eagles, 
 and dollars enough to make the whole number of coins 
 equal to | of the value of the whole in dollars; and 
 the eagles and dollars together were 2 more than £ 
 the number of coins. How much money did he 
 leave ? 
 
 50. A farmer sold 120 bushels of rye and barley ; 
 receiving, for a bushel of each kind of grain, as many 
 cents as there were bushels of that kind ; and the 
 barley brought only f as much as the rye. How many 
 bushels of each kind did he sell ? 
 
MISCELLANEOUS QUESTIONS. 249 
 
 51. A gentleman distributed $47,50 among 30 
 men and women, giving the women 85. and the men 
 10s. 6d. each. How many men and how many 
 women were there ? 
 
 52. A criminal, having escaped fromprison, travel- 
 led 10 hours before his escape was known. He was 
 then pursued, so as to be gained upon 3 miles an hour. 
 After his pursuers had travelled 8 hours, they met an 
 express going at the same rate as themselves, who met 
 the criminal 2 hours and 24 minutes before. In what 
 time from the commencement of the pursuit did they 
 overtake him ? 
 
 53. A farmer has an irregular piece of land, con- 
 taining 5 acres, which he wishes to exchange for a 
 square field of the same size. Required one of the 
 sides of the square field. 
 
 Remark. An acre of land contains 160 square 
 rods. Only an approximate answer to this question can 
 be found, as the given quantity is not a perfect square. 
 
 54. I have a field containing 10 acres ; and the 
 length of the field exceeds its width by 18 rods. Re- 
 quired its dimensions. 
 
 55. A man bought a field whose length was, to its 
 breadth as 8 to 5. The number of dollars paid per 
 acre was equal to the number of rods in the length 
 of the field ; and the number of dollars paid for the 
 whole was equal to 13 times the number of rods 
 round the field. What did he give for the field? 
 
 56. A father gave to each of his children, on new 
 year's day, as many books as he had children ; for 
 each book he gave 12 times as many cents as there 
 
250 FIRST LESSONS IN ALGEBRA. 
 
 were children ; and the cost of the whole was $15. 
 How many children had he ? 
 
 57. A messenger had been gone from a certain 
 place 8 hours, when another was sent after him. The 
 first went 7 miles an hour, and the second 11. • In 
 what time did the second overtake the first? 
 
 58. The members of a lyceum, wishing to buy 
 an air-pump, found, if they paid 80 cents each, that 
 they should raise $18 more than they wanted for the 
 purpose ; but if they paid only 50 cents each, they 
 would not have enough by $12. What was the price 
 of the air-pump? 
 
 59. A servant was sent to market with a basket of 
 eggs, which he was directed to sell for 12 cents a 
 dozen. Having carelessly broken 6 dozen of the 
 eggs, he was obliged to get 15 cents a dozen for the 
 rest, that there might be no loss. How many dozen 
 of eggs did the basket contain at first? 
 
 60. A man wished to plant a certain number of 
 trees in the form of a square. At the first trial, he 
 had 39 trees left. He then determined to enlarge, 
 the square, by adding one tree to each row ; to do 
 which, he found it necessary to procure 50 trees more. 
 How many trees had he at first ? 
 
 61. A grocer, being asked the size of 3 wine-casks, 
 replied, " If I fill the first empty cask from the second 
 full cask, § of the wine will remain; if I fill the 
 second empty cask from the third full one, £ of the 
 wine will remain; and the third empty cask will con- 
 tain the contents of the first full cask and 23 gallons 
 more." Required the size of the casks. 
 
MISCELLANEOUS QUESTIONS. 251 
 
 62. A cistern, which holds 2340 gallons, is filled in 
 £ of an hour by 3 pipes; the first of 'which conveys 
 13 gallons more, and the second 6 gallons less, than 
 the third per minute. How many gallons does each 
 pipe convey in a minute f 
 
 63. Two persons, A and B, set out at the same 
 time from two towns at the distance of 672 miles. 
 B travelled 8 miles a day more than A ; and when 
 they had travelled half as many days as A went miles 
 in a day, they met. How many miles did each travel 
 daily ? 
 
 64. A farmer has a rectangular peach-orchard, with 
 unequal sides. If the difference of the sides be mul- 
 tiplied by the greater side, and the product divided 
 by the less, the quotient is 24 rods ; but if their dif- 
 ference be multiplied by the less side, and the product 
 divided by the greater, the quotient is only 6 rods. 
 What are the dimensions of the orchard ? 
 
 65. There is a school-room in Boston, whose length 
 is to its breadth as 6 to 5. If it were a square, hav- 
 ing its sides equal to the length, it would contain 891 
 feet more than it would were the sides of the square 
 equal to its width. What are the dimensions of the 
 room? 
 
 66. The fore-wheel of a carriage makes 6 revolu- 
 tions more than the hind-wheel in going 120 yards; 
 but if the circumference of each wheel be increased 
 one yard, it will make only 4 revolutions more in go- 
 ing the same distance. What is the circumference 
 of each wheel? 
 
 67. When the passengers on board of a steam-boat 
 
252 FIRST LESSONS IN ALGEBRA. 
 
 took their seats at the dinner-table, it was observed 
 that the number of the men was to that of the women 
 as 9 to 4 ; but when 6 men, with their wives, had re- 
 tired, there remained at the table 3 times as many 
 men as women. What was the number of passen- 
 gers? 
 
 68. Two clerks, A and B, sent ventures in a ship 
 bound to India. A gained $11; and, at this rate, 
 he would have gained as many dollars on a hundred 
 as B sent out. B gained $36, which was but one 
 fourth part as much per cent, as A gained. How 
 much money was sent out by each ? 
 
 69. Required two fractions, whose product is ^, and 
 the sum of whose squares is ££. 
 
 70. A company of persons spend £3 10s. at a tav- 
 ern. Four of them go away without paying ; in con- 
 sequence of which, each of the others has to pay 2s. 
 more than his proper share. How many persons were 
 there in the company ? and what was the proper share 
 of each ? 
 
 71. A gentleman bought a rectangular lot of land, 
 giving $10 for every foot in the perimeter. If the 
 same quantity of land had been in the form of a 
 square, and he had bought it in the same way, it would 
 not have cost him so much by $330 ; and if he had 
 bought a square piece of the same perimeter, he would 
 have had 12£ rods more. What were the dimensions 
 of the lot ? 
 
 THE END. 
 
 *