IN MEMORIAM 
 FLORIAN CAJOR1 
 
bi 
 
 
RECOMMENDATIONS. 
 
 From Rev. C H. Alden, Principal of the Philadelphia High School for 
 
 Young Ladies. 
 Mr. Green : 
 
 Dear sir, — I am greatly pleased to find, that, in your " Gradations in 
 Algebra," recently published, you have rendered the elements of that de- 
 partment of the Mathematics so attractive to the young student. It has 
 long surprised me that this interesting method of analysis has been so 
 entirely excluded from our common schools. With your valuable aid, 
 however, this neglect can no longer find a suitable apology. I anticipate 
 the introduction of your excellent work into both our private primary and 
 public schools. 
 
 Very respectfully, 
 
 C. H. ALDEN. 
 
 Richard W. Green: 
 
 Sir, — I have examined your " Gradations in Algebra" with much care, 
 and have no hesitation in saying that in my opinion you have fully attained 
 your object in forming " an easy introduction to the first principles of alge- 
 braical reasoning," and of furnishing "in the same course a popular expo- 
 sition of the most important elements of arithmetic." To write a work on 
 abstruse science, adapted to the comprehension of youth, is a work of 
 extreme difficulty. I feel that I offer high but deserved and just praise 
 when I say that in the book before me you have fully succeeded. 
 
 Yours, «fec. 
 
 JAS. RHODES, 
 Principal of N. W. Grammar School, Philad. 
 
 Dear sir, — Your " Gradations in Algebra" appears to me to supply 
 what has long been wanted by those who are commencing that study, viz. 
 an initiatory text-book at once small, clear, well arranged, and compre- 
 hensive. It meddles with nothing beyond the capacity of a school-boy. 
 It is well supplied with such examples as show the learner the use of what 
 he is studying. I would rather put your book in the hands of a beginner 
 than any other work on the same subject now in common use. 
 
 Yours very respectfully, 
 
 JNO. W. FAIRES. 
 A i 
 
D RECOMMENDATIONS. 
 
 I examined " Green's Gradations in Algebra," and was so well pleased 
 with it, that I introduced it into our school. I have made use of it one 
 term, and can bear testimony that it has equaled my expectations. I con- 
 sider it as occupying an important link between arithmetic and literal alge- 
 bia. In my judgment, we have long needed such a work. 
 
 JOHN D. POST, 
 Teacher of Mathematics in Hartford Gr xmmar School. 
 
 From Porter H. Snow, A. M., formerly Principal of the Hartford Centre 
 School, and now Principal of Brainard Academy, Haddam, Conn. 
 
 I have examined critically your "Gradations in Algebra," and find it 
 admirably adapted to fulfil its design. There is nothing so good on the 
 subject in the English language for a student to read before entering col- 
 lege. All the mystery that clothes mathematics during the early part of a 
 collegiate course of study will be solved and made plain, if the student 
 will spend a few weeks on this introduction to algebra. It is also a fine 
 work for classes in schools. And I would recommend its introduction into 
 primary and higher schools in preference to any treatise on algebra I have 
 ever seen. It is also the only thing of the kind I have met with, suitable 
 for a student to peruse without an instructor. 
 
 York, Pa., Nov. 30th, 1841. 
 After a careful examination of Mr. Green's "Algebra," we have no 
 hesitation in expressing our decided approbation of the work. The clear- 
 ness of the explanations, the judicious and systematic arrangement of the 
 parts, and the gradual manner in which the student is led on from the first 
 principles to the more difficult parts of the science, render it, in our view, 
 preferable to any we have seen for primary instruction. 
 
 DANIEL KIRKWOOD, 
 Teacher of Mathematics, York Co. Academy. 
 D. M. ETTINGER, 
 Principal of the High School, York, Pa. 
 
 The great excellence of this work is, that it brings the pupil on gradually. 
 Only one difficulty is presented at a time ; and upon this, explanations and 
 examples are multiplied i ill the pupil becomes so familiar with the subject 
 that he almost wonders where the difficulty is at which he first stumbled. 
 The examples are numerous ; the few rules (and they are enough) are ad- 
 mirably expressed, in plain, concise language. On the whole, we may say, 
 that U a vouth is to begin the study of algebra, this should be his first book 
 
 Extract from the North American. 
 
GRADATIONS 
 
 IN 
 
 ALGEBRA, 
 
 IN WHICH 
 
 THE FIRST PRINCIPLES OF ANALYSIS 
 
 ARE 
 
 INDUCTIVELY EXPLAINED. 
 
 ILLUSTRATED BY 
 COPIOUS EXERCISES, 
 
 AND MADE SUITABLE FOP FRIM^RY SCHOOLS. 
 
 BY 
 
 RICHARD W. GREEN, A.M., 
 
 II 
 
 AUTHOR OF ARITHMETICAL GUIDE, LITTLE RECKONER, ETC. 
 
 PHILADELPHIA: 
 
 PUBLISHED BY E. H. BUTLER & CO. 
 
 1850. 
 
QhiS3 
 
 £7 
 
 ( Chamber of the Controllers of Public Schools, 
 ( First School District of Pennsylvania. 
 
 Philadelphia, November 15, 1849. 
 At a meeting of the Controllers of Public Schools, First District of Penn 
 sylvania, held at the Controllers' Chamber, on Tuesday, November 13, 
 1849, the following Resolution was adopted : — 
 
 Resolved, That Green's Algebra be introduced, as a Class-Book, into 
 the Grammar Schools of the District. From the minutes, 
 
 R. J. HEMPHILL, Secretary. 
 
 Secretary's Office, Harrisburg, Feb. 4, 1843. 
 Mr. Richard W. Green : 
 Sir, — I have examined the work prepared by you for the use of primary 
 schools, entitled "Gradations in Algebra." It meets with my approba- 
 tion, and I am much pleased with its design and arrangement. I consider 
 it admirably calculated to aid the pupil at the commencement of the 
 science, also to give him a general knowledge of it. 
 
 It is, in my opinion, well adapted to the use of common schools, where 
 there are students who have not the time or means of consulting more ex- 
 tended treatises upon the subject. I hope it may be introduced generally 
 into our schools throughout the state. 
 
 I am yours respectfully, 
 
 A. V. Parsons, 
 \ .' tSfyjptrintqndent Common Schools. 
 
 Entered according la the Act of Congress, in the year 1^39, by 
 
 Richard W. Green, 
 
 in the Clerk's Office of the District Court of ihe Eastern District of 
 
 Pennsylvania. 
 
 CAJORI 
 
 $3" For the convenience of those teachers who may wish to adopt the 
 author's plan of using this book, a key has been prepared containing the 
 solutions of all the examples and problems. It will also save the teacher 
 much time when he wishes to find for his pupil the errors in his v» ork. 
 
CONTENTS 
 
 Preface p age 7 
 
 NUMERAL ALGEBRA. 
 
 Preliminary Remarks 11 
 
 Addition and Subtraction of Simple Quantities 18 
 
 General Rule for uniting Terms 19 
 
 Multiplication and Division of Simple Quantities 23 
 
 Simple Equations 16 
 
 I. Equations Solved by uniting Terms 24 
 
 II. Addition of Compound Quantities 22 
 
 Transposition by Subtraction 30 
 
 Equations Solved by Transposition 31 
 
 III. Transposition by Addition 34 
 
 Equations 34 
 
 IV. Transposition of the Unknown Quantity 37 
 
 Equations 38 
 
 V. Multiplication of Compound Quantities by Simple Quantities. ... 42 
 Equations 44 
 
 VI. Fractions 48 
 
 Equations 52 
 
 VII. Fractions of Compound Quantities 57 
 
 Equatl >ns 58 
 
 VIII. Division of Fractions and Fractions of Fractions 61 
 
 Equations 64 
 
 IX. Subtraction of Compound Quantities 67 
 
 Equations 73 
 
 X. Uniting Fractions of different denominators 69 
 
 Ratio and Proportion 75 
 
 Equations 78 
 
 9185-13 
 
VI CONTENTS. 
 
 XI. Equations with two Unknown Quantities Page 82 
 
 First method of Extermination ." 83 
 
 Equations 88 
 
 XII. Second method of Extermination 93 
 
 Equations 94 
 
 XIII. Third method of Extermination 97 
 
 Equations 98 
 
 XIV. Equations with several Unknown Quantities 101 
 
 LITERAL ALGEBRA. 
 
 General Principles 103 
 
 Addition and Subtraction of Algebraical quantities 112 
 
 Multiplication of Algebraical Quantities 117 
 
 General Properties of N umbers 125 
 
 Division of Algebraical Quantities 128 
 
 Division by Compound Divisors 136 
 
 Reduction of Fractions to lower Terms .• 143 
 
 Multiplication where one factor is a fraction 146 
 
 Reducing Complex Fractions to Simple ones 148 
 
 Division of Fractions 152 
 
 Fractions of Fractions 153 
 
 Uniting Fractions of different denominators 156 
 
 Division by Fractions 159 
 
 General Theory of Equations with two unknown quantities 162 
 
 Involution and powers 1 65 
 
 Evolution 172 
 
 Extractioti of the Second Root of Numbers 177 
 
 XV. Pure Quadratic Equations 181 
 
 Equations 182 
 
 XVI. Affected Quadratic Equations 185 
 
 Equations 1 90 
 
PREFACE 
 
 The object of the author, in composing this treatise, was 
 to form an easy introduction to the first principles of alge- 
 braical reasoning ; and also to embrace, in the same course, 
 a popular exposition of the most important elements of 
 arithmetic. And he believes that he has been enabled to 
 combine the rudiments of both, in such a manner as to make 
 the operations of one illustrate the principles of the other. 
 
 In order that this method of treating the subject might 
 preserve its chief advantage, especially in the initiatory course 
 of the study ; the work has been divided into two parts— 
 Numeral Algebra and Literal Algebra. 
 
 In Numeral Algebra I have treated of the several primary 
 arithmetical operations; first making them intelligible to very 
 youngf pupils, and then exhibiting them under the algebraical 
 notation. By this means, as every lesson in algebra is imme- 
 diately preceded by corresponding numerical exercises, the 
 transition from one to the other has been made so trifling, 
 that the pupil will feel at each step that he has met with 
 nothing more than what he has already made himself familiar 
 with in a different dress. 
 
 Besides, as algebraical operations require the exercise of 
 abstraction in a greater degree than the pupil is supposed to 
 be accustomed to, I have taken care that the exercise on each 
 of the fundamental rules, shall be followed by a selection of 
 problems to be solved by equations. 
 
 7 
 
8 PREFACE. 
 
 As mathematical questions of this kind are always pleas- 
 ing to young pupils, this arrangement will serve to impart 
 an interest to the study at the commencement, and also to 
 preserve a taste for it through the whole course. 
 
 Indeed, this part of the work is the most important for 
 those pupils who do not intend to pursue the mathematical 
 sciences. For, it is in such exercises, that the mind is 
 trained to investigate the relations of one thing to another, and 
 to conduct its reasonings in a clear and forcible manner. 
 
 Under the head of Literal Algebra, I have repeated, in 
 a more strictly algebraical form, the principles which have 
 been explained in the preceding part of the work ; and have 
 shown some of their uses by applying them in the deduc- 
 tion and demonstration of several abstruse operations on 
 numbers. 
 
 But the great peculiarity of the book is, that it habituates 
 the speech and the ear to mathematical language. In any 
 study, it is necessary for beginners to receive such a course 
 of training as will imprint upon their minds each new idea, 
 as soon as it is apprehended. Learners in the mathematics, 
 especially, are accustomed to forget soon, both the names 
 and the use of the signs; and also the arrangement of the 
 several steps in the solution of their problems. On this 
 account I have required the pupil always to repeat verbally 
 the operation that he has performed ; taking care to omit no 
 part of the work that would hinder an auditor from under- 
 standing the reason for the several steps, and consenting to 
 the just conclusions of the answer which has been obtained. 
 It has been found by experience that this simple device 
 enables the young pupil to acquire the science very easily; 
 
PREFACE. 9 
 
 and while it impresses his lessons indelibly upon his 
 memory, it also developes his genius, rectifies his inventive 
 faculties, and imparts, as it were, a mathematical form to his 
 mind ; so much so, that he is generally capable of pursuing 
 the subject afterwards by himself. 
 
 In order to accomplish this end more perfectly, I have 
 swelled the number of examples beyond the ordinary limits. 
 These should be thoroughly mastered as the pupil proceeds. 
 There must be no smattering in the beginning of a science 
 if the learner is to continue the study. The author has 
 found by long experience, that a book is sooner finished 
 when each part has been made familiar to the mind, than 
 when it has been superficially attended to. 
 
 With regard to the arrangement of the several divisions, 
 I have been careful to introduce first those principles that 
 will be the most easily apprehended ; and afterwards such 
 others as would most naturally arise from the former if the 
 study were entirely new. This method appears to be the 
 best adapted for teaching the rudiments of a science ; 
 although in a succeeding text book, it is necessary that the 
 arrangement of the several parts should be more systematic. 
 On this account the advanced scholar must not be surprised 
 to find in the middle of the book, what he has been accus- 
 tomed to see near the beginning of other treatises. How- 
 ever, so much regard to a regularity of arrangement has 
 been attended to, that the pupil will be assisted by the asso- 
 ciations of method, both to understand and to remember. 
 
 As the author wishes to bring the study of Algebra withir 
 the reach of common schools, he has endeavored to pre 
 pare this work, so that it may be studied by pupils who are 
 
10 PREFACE. 
 
 not already adepts in arithmetic. And it is believed lhat 
 
 such learners will not fail of obtaining, by a perusal of it, 
 
 a full understanding of vulgar fractions, roots and powers, 
 
 proportion, progression, and other numerical operations that 
 
 are generally embraced in arithmetical treatises. 
 
 R. W. G. 
 
 ADVERTISEMENT. 
 
 The foregoing is the Preface of the author's " Inductive 
 Algebra." The first 192 pages of that book have been 
 published in this form, in order to afford a cheap manual 
 for those classes that do not wish to study beyond Quadratic 
 Equations. In the present state of education, so much of 
 Algebra should be studied by every pupil in our common 
 schools 
 
PART L 
 
 NUMERAL ALGEBRA. 
 
 PRELIMINARY REMARKS. 
 
 § 1. Algebra is a method of arithmetical computation, 
 in which the calculations are performed by means of letters 
 and signs. 
 
 § 2. When the answer of an arithmetical question is to 
 be found by algebra, we first represent that answer by a 
 letter; because we can use that letter in our calculations, 
 in the same manner as if it were the true answer. 
 
 We will explain by a few examples. 
 
 Example 1. I have 27 apples which I wish to give to two 
 children, John and Mary, in such a manner that Mary shall 
 have twice as many as John. The following will be our 
 method of rinding how many each shall have. 
 
 I wish to give to John a certain number, and to Mary 
 twice that number. Therefore, instead of saying, I will 
 give the 27 apples to John and Mary, I may say, I will give 
 away a certain number, and then again twice that number. 
 
 Now, a certain number, and then twice that number, 
 make, in all, three times that certain number. 
 
 But, I give away 27 apples ; and hence I know that 
 
 Three times that certain number is just as much as 27. 
 
 Therefore, that number itself is \ of 27, which is 9. 
 
 Hence, John has 9 apples ; and Mary has twice as many, 
 which is 18. 
 
 11 
 
12 ALGEBRA. 
 
 § 3. If, in the operation which we have just per- 
 formed, we use the letter iV, instead of the words, a cer- 
 tain number, our work will be somewhat abridged. Thus : 
 
 John will have a certain number, which is N. 
 
 Mary will have twice as much, which is twice N. 
 
 And both together will have three times N. 
 
 But, in the question, we said both shall have 27 
 
 Therefore, three times N is the same as 27: and once 
 N is j of 27, which is 9; which is John's share. 
 
 § 4. About the year 1554, Stifelius, a German, in- 
 troduced the sign -f for the words added to; and called 
 it plus. Ever since that time, -f has been used to sig- 
 nify that the quantity after it has been added to the 
 quantity before it. Thus, 2 -f 6, is read 2 plus 6, and 
 signifies 2 with 6 added to it. 
 
 Example 2. A brother told his sister that he was 4 years 
 older than she; and that his age and her age, when put 
 together, made 18 years. What was the age of each? 
 
 By the statement, his age was the same as her age with 
 4 years put with it ; because it was 4 years more than hers. 
 Therefore we will represent the sister's age by A 
 
 And his age, which is 4 years more, will be A + 4 
 
 Both ages together, will be A and A + 4, or A + A + 4 
 
 Which, because the A's can be put together, is twice A -f- 4 
 But both their ages together made 18 
 
 Therefore twice A -f 4 is the same as 18, 
 
 And twice A + 3 is the same as 17, 
 
 And twice A + 2 is the same as 16, 
 
 And twice A -J- 1 is the same as 15, 
 
 And twice A alone is the same as 14, 
 
 And A, or the sister's age, is | of 14, which is 7. 
 
 §5. About the year 1550, John Scheubelius, a Ger- 
 man, introduced the following practice. Instead of writ- 
 ing 2 times A, or 3 times A, or 4 times A, &c, he 
 wrote 2 A, 3 A, 4 A, &c. 
 
PRELIMINARY REMARKS. 18 
 
 £xample 3. Two men, A and B, owe me $270 ; and B's 
 
 cusbt is twice as much as A's. How much does each of them 
 owe me ? 
 
 We will represent A's debt by A. 
 
 Then, B's debt will be twice as much, or 2^. 
 
 And both debts put together, make A + 2 A. 
 
 But both, when put together, are equal to $270. 
 
 Therefore, A -f- 2 A is the same as $270. 
 
 Putting the A's together, 3 A is the same as $270. 
 
 Then, one A is \ of $270, which is $90. 
 
 A's debt is $90; and B's debt is twice as much, or $180. 
 
 §6. In 1557, Dr. Recorde, an Englishman, introduced 
 the sign =, which we call equals. It is used instead of 
 the expression, is the same as, or, is as much as, or, is 
 equal to. Thus, 2-j-6 = 8, is read, 2 plus 6 equals 8. 
 
 Example 4. B's age is three times A's ; and C's 
 age is double of B's ; and the sum of all their ages is 
 70 years. What is the age of each ? 
 
 Let us represent A's age by the capital A, 
 
 Then, B's age will be three times as much; that is, 3 A. 
 And C's age will be twice 3 A, which is 6 A. 
 
 Then, all their ages put together will be A + 3 A -f 6 A. 
 But all their ages put together is 70 years. 
 
 Therefore, A + 3A + QA = 70. 
 
 Putting them's together, 10 ,# = 70. 
 
 Dividing by 10, A = 7, which is A's age. 
 
 A's age =7 years. 
 
 B's is three times as much; or 21 years. 
 
 C's is twice B's ; or, 42 years. 
 
 Proof, "TO" 
 
 § 7. We very often wish to state that we have made a 
 quantity to be less. This was formerly done by using the 
 word minus. Thus, if I wished to say that your age is 6 
 years less than mine ; I may say, Your age is equal to mine 
 
 2 
 
14 ALGEBRA. 
 
 when 6 years are subtracted from mine ; or, in fewer words, 
 your age is equal to my age, minus 6 years. 
 
 § 8. Stifelius introduced the sign — for minus; so that 
 20 — 6, is read, 20 minus 6; and signifies, 20 with 6 sub- 
 tracted from it. Thus, 20 — 6 = 14. 
 
 i 
 
 Example 5. At a certain election, 548 persons voted ; and 
 the successful candidate had a majority of 130 votes. How 
 many voted for each ? 
 
 Let us represent the number of votes received by the sue- 
 cessful candidate, by the letter A. Then, as the unsuccess- 
 ful candidate received 130 votes less, his number may be 
 represented by A — 130. 
 
 Both candidates have A + A — 130 
 
 Then, .# + .#-.130 = 548 
 
 Putting the A's together, 2 A — 130 = 548, which 
 
 means, 2 A, after 130 is subtracted from it, is equal to 548 
 Now, if we had a barrel of water, minus a quart, and wanted 
 a full barrel, we should add the quart. In the same manner . 
 as we have 2 A minus 130, we must add the 130 to get th< 
 complete 2 A. 
 
 Therefore, adding the 130 to both sides, 2 A = 678 
 
 Dividing by 2, A = 339. 
 
 The successful candidate had 339. 
 
 The unsuccessful one, 130 less = 209. 
 
 Proof, 548. 
 
 § 9. Des Cartes,* a Frenchman, who wrote about 1637, 
 used the last letters of the alphabet; namely, x, y, z, w, &c, 
 to denote the unknown quantities. And this is now the 
 practice of all mathematicians. 
 
 Example 6. It is required to divide $300 among A, B, 
 and C ; so that B may have twice as much as A, and C may 
 hi*ve as much as A and B together. 
 
 • Pronounced Da-Cart. 
 
PRELIMINARY REMARKS. 15 
 
 Let us represent A's share by x. Then, B's share will be 
 
 2 a*; and C will have as much as both put together, which is 
 
 3 x. Then, 
 
 x + 2x + 3x = 300. 
 Putting the x's together, 6x = 300. 
 
 Dividing by 6, x= 50. 
 
 Therefore, A's share is $50. 
 
 B's is twice as much, or $100. 
 
 C's = as much as both, or $150. 
 Proof, $300. 
 
 From what we have shown thus far, it may be seen that 
 Algebra is a kind of language, made up with letters and signs 
 in such a manner that all the reasonings which are necessary 
 for the solution of a question, may be contained in a very 
 small space, and be perceived with great facility. 
 
 § 10. As we sometimes wish to speak of particular parts 
 of our calculations, mathematicians have given the name term 
 to any quantity that is separated from others by one of the 
 signs -j- or — . Thus, in the last example, and first line of 
 the operation, the first x is the first term, the 2 x is the next 
 term, and the 3a" is the next term; and the 300 is the last 
 term. 
 
 §11. When a figure is put before a letter, to denote how 
 many times we take the quantity which that letter stands for, 
 the figure is called a co-efficient.* Thus, in the term 2 a?, 2 
 is a co-efficient of a:; in the term 3*#, 3 is a co-efficient of .#. 
 
 § 12. It must also be understood, that a letter without any 
 number before it, has 1 for its co-efficient. Thus, x repre- 
 sents la: 1 ; a is the same as la, &c. The 1 is omitted be- 
 cause it is plainly to be understood. 
 
 § 13. When any algebraic term stands by itself, it is called 
 a simple quantity. Thus, i is a simple quantity; 2a* is a 
 simple quantity; 12 is a simple quantity, &c. 
 
 * Thin name was given by Franciscus Vieta, about 1573. 
 
16 ALGEBRA. 
 
 § 14. Quantities that consist of more than one term, 
 are called compound quantities. Thus, a + b is a com 
 pound quantity; so is also a — 6, and a?-f 7, and x — 7 
 and a + x — 7 — 6, &c. 
 
 § 15. When any algebraic quantity begins with the sign 
 4-, it is called a positive quantity; as, -fa, -f 3 a. 
 
 § 16. When any algebraic quantity begins with the sign 
 — , it is called a negative quantity ; as, — 6, — 5 a. 
 
 § 17. In algebra, the perfect representation of any simple 
 quantity requires both the specified sum, and either the sign 
 -f , or the sign — ; as, -f 5, —40, -f x, — 3x. 
 
 § 18. But, when a positive quantity stands by itself, or 
 when it is the first term of a compound quantity, the sign 
 that belongs to it is generally omitted on paper, and also in 
 our reading ; as, J*, 2, a -f 6, x — y. 
 
 § 19. Therefore, when a simple quantity, or the first term 
 of a compound quantity, does not begin with a sign, we say 
 that the sign -f is understood. That is, we think of the 
 quantity the same as if -f was before it. 
 
 § 20. In reading compound quantities, the pupil must be 
 careful to join the sign to the term that is immediately after 
 it. Thus, the following quantity must be read a; plus b; 
 plus 8 ; minus 4 ; plus 2 a ; plus 6 ; minus 36: 
 a + & + 8 — 4 + 2a-f 6 — 3 6. 
 
 § 21. When a quantity is expressed by figures, it is called 
 a numeral quantity. When it is represented by a letter, it is 
 called a literal quantity. 
 
 SIMPLE EQUATIONS. 
 
 § 22.' The most general and useful application of algebra, 
 is that which investigates the values of unknown quantities 
 by means of equations. 
 
SIMPLE EQUATIONS. 17 
 
 § 23. An equation is an expression which declares one 
 quantity to be equal to another quantity, by means of the sign 
 = being placed between them. 
 
 Thus, 5 -f 3 = 8, is an equation, denoting that 5 with 3 
 added to it, equals 8. Also, 4 — 1=3, and 3 + 2 — 1=4, 
 and 8 — 2 = 5 + 1, are equations, each denoting that the 
 quantity on one side of the =, is equal to that on the 
 other side. 
 
 § 24. The whole quantity on the left of = is called the 
 first member of the equation; and all on the right of = is 
 called the last member of the equation. 
 
 In order to be a member of the equation, it is of no import- 
 ance whether the quantity is simple or compound. Thus, in 
 the equation x = 4 -f- a — b — 18, a? is the first member, and 
 4 -f a — b — 18 is the last member. And in this case, the first 
 member represents just as great a quantity as the last. 
 
 § 25. In order that an equation may be such that it will 
 enable us to find the value of an unknown quantity by it, it 
 must contain some quantity that is already known. And then 
 we can find the value of the unknown quantity, by making 
 it stand by itself on one side of =, and all the known quan 
 tities on the other side. 
 
 § 26. But it is generally the case that the known and un 
 known quantities are mingled together. And then we are to 
 alter the arrangement of the terms, so as to bring the unknown 
 quantity by itself; and this must be done without destroying 
 the equality of the sides. This operation is called solving or 
 reducing the equation. 
 
 § 27. When the known quantities are represented by num- 
 bers, the equation is called a numerical equation. It is to 
 this kind of equations that the first part of this treatise relates. 
 
 § 28. In solving equations, the different arrangement of the 
 terms is brought about by addition, subtraction, multiplication, 
 division, &c, as the case may require. 
 B 2* 
 
18 ALGEBRA. [SECTION I. 
 
 SECTION I. 
 SIMPLE ALGEBRAICAL .OPERATIONS. 
 
 I. ADDITION AND SUBTRACTION OF SIMPLE QUANTITIES. 
 
 § 29. In algebra, simple quantities are added by writing 
 them down one after another; being careful to retain the signs 
 that are expressed or understood. Thus, we add 9 to x, by- 
 writing them x + 9 ; because, when those quantities were 
 alone, -f was understood before 9. We add — 9 to #, by- 
 writing them x — 9. 
 
 89" The pupil must understand that x stands for some 
 number; but it is often the case that we do not know what 
 that number is. For it may stand sometimes for one number, 
 and sometimes for another. 
 
 § 30. It will readily be seen, that it is of no consequence 
 which quantity is put first ; for 9 -j- 7 is the same amount as 
 7-f 9. 
 
 §31. One positive simple quantity is subtracted from 
 another simple quantity, by writing down both quantities 
 one after another, and changing from + to — the sign be- 
 fore the quantity which we subtract. Thus, we subtract 4 
 from x, by writing them x — 4; which is read x minus 4; 
 and not 4 from x. 
 
 Questions. What is algebra? See § 1. In what manner do we 
 use letters in our calculations'? 2. What is said of the sign -+- ? 4. 
 — ? 7. = ? 6. What does term signify in algebra? 10. What is 
 a simple quantity ? 13. How are simple quantities added in alge- 
 bra? 29. What is requisite for the perfect representation of an 
 algebraic term? 17. Are all terms thus represented? What is 
 understood in such cases? is the sign — ever understood? In 
 addition, which quantity must be put first? 30. What particular 
 number does a letter stand for in algebra? Why do we use letters 
 in algebra? 2. How is one simple quantity subtracted from an- 
 other? 31. Which quantity must have — before it? 
 
§ 36. J SIMPLE ALGEBRAICAL OPERATIONS. 19 
 
 § 32. When two simple quantities have been added toge- 
 ther, or one simple quantity subtracted from another, the 
 answer will consist of more than one term, and become a 
 compound quantity. 
 
 §33. It sometimes happens, that in those compound 
 quantities^ which are made by adding or subtracting, there are 
 two or more lerms of the same kind ; that is, such terms as 
 do not differ at all, or differ only in their numeral co-effi- 
 cients ; such as a? + 2a?; 4a— 2a; 5a? — 3a? + a?. Such 
 quantities are called similar quantities. 
 
 § 34. When, in any compound quantity, there are two or 
 more terms of the same kind, they may be united by per- 
 forming arithmetically the operation which is expressed by 
 the signs belonging to them. Thus, a? + 2a? is united into 
 3a?; 5a — 3a is united into 2a; 5a? — 3a? -fa? is united into 3a?. 
 
 § 35. Numeral quantities may be united in the same man- 
 ner. Thus, 4 + 3 may be united into 7 ; 9 — 4 may be united 
 into 5 ; 6 — 2 + 5 may be united into 9. 
 
 Examples. The pupil may unite the following quantities : 
 
 1. a + 3a+5a. 
 
 2. x + 5a? + x. 
 
 3. 10a? — 4a? — a?. 
 
 4. a+lla -2a. 
 
 5. b + 6b—3b. 
 
 6. la — 4a-fa. 
 
 7. 4 + 8—3. 
 
 8. 5 — 2+7. 
 
 9. 3a?+2.r+8— 4. 
 
 10. 10a— 3a+7— 2. 
 
 11. 6a+17— 3a— 7. 
 
 12. 14 + 5a?— 3a;— 5. 
 
 2. GENERAL RULE FOR UNITING TERMS. 
 
 § 36. 1st. When similar quantities have the same sign, 
 select one kind, and find the sum of the numeral co-efficients ; 
 
 Questions. What is a compound quantity ? 14. What operation 
 will produce compound quantities? What are similar quantities'? 
 What is a co-efficient] 11. Have all terms a co-efficient express- 
 ed? 12. What may be done with similar quantities? When a 
 sign is between two quantities, to which quantity does it be- 
 long? 20. How does a literal quantity differ from a numeral ? 21. 
 What is a positive quantity ? 15. What is a negative quantity ? 16 
 What is the first part of the rule for uniting terms ? What is the 
 second part of the rule for uniting terms ? 
 
20 ALGEBRA. ^SECTION I. 
 
 and, preserving the sign, annex the common letter. Then 
 select another kind, and proceed in the same manner with that. 
 
 EXAMPLES. 
 
 1. Unite the following terms. 5a? — 3?/ -f- 4a: — ly. 
 
 Ans. The x's, when united, equal -f- 9a? » and the y's, 
 when united, equal — l(h/. Ans. 9x — lOy. 
 
 2. 2x + 3y + Sx -f Ay + Ix -f y -f x -f 9y. Ans. 13a? 
 
 *$&sT As fast as the pupil adds the co-efficients, he should 
 slightly cross the term, so that he may know when all the 
 terms have been united. Thus : 
 
 2p+3y + 3$+4y+7v+y+.V+9y=lZx+l7y. 
 
 3. 4a— 3x+2a— x+a + 3a— Sx — x. Ans. 10a — 8a:. 
 
 4. — 2y+7— a?— y— 3a? + ll— 2t/. Ans. 18— 5y— 4a\ 
 
 5. — 2a— a?+3— a— 3a: + 4— 4a. Ans. — 7a— 4a?+7. 
 
 § 36. 2d. When similar quantities have different signs, 
 select one kind, and add into one sum all the p isitive co-effi- 
 cients that belong to them. Then add into another sum all 
 the negative co-efficients that belong to them. Subtract the 
 less sum from the greater, and prefix the sign of the greater 
 to the difference ; annex the common letter. Then select 
 another kind, and proceed in the same manner. 
 
 EXAMPLES. 
 
 Unite the terms in the following. 
 
 6. 3a + 46 + 26— 3c + 3c + a— 6— 3b. 
 
 Ans. The a's, when united, equal -\-4a; the 6's, when 
 united, equal +66 — 4b, which equals 4-26. Then, — 3c + 3c 
 balance each other, so as to be equal to 0. The answer is 
 4a +26. 
 
 7. 2a-|-6+4a + 46 + 8-|-6— 3a— 36— 4. Ans. 3a +26 + 10. 
 
 8. 3x+y+z + 3x + 3y—4x—2y—z. Ans. 2x + 2y. 
 
 9. 18 + 8a + 4a: + 7.r + 5a— 6x— 7a. Ans. 18 + 6a + 5a\ 
 10. 4y+7z — 4z — y + 2z + 2y — z. Ans. 5y + 4z. 
 
§ 36.] UNITING TERMS. 21 
 
 Note. — It sometimes happens that the negative quantity is 
 greater than the positive quantity. In such cases, the differ- 
 ence will have the sign — . 
 
 11. la— 3a+4z+a— 2z— 4a— 6a. Ans. — 5a+2*. 
 
 12. a+6 + 3a— c+4a— 3c + 6. Ans. 8a+26— 4c. 
 
 13. 5x + 5y + 3x—2y. Ans. 8x+3y. 
 
 14. 6+a+3a— 56+4cc— a+3a— x. Ans. 6a— 46 + 3.T 
 
 15. 4a— 56— \0 + 2y— x + 4. Ans. 4a— 56+ 2y— x— 6. 
 
 16. 2a?+ 3?/— 3x+5y— x— z. Ans. — 2x+8y— z.~ 
 
 17. #+z+;c— z+#+3?/. Ans. 3x+3y. 
 
 18. 3a— 5 — 46 + 6— 2a— 3 + 66. Ans. a+26— 2. 
 
 19. 4;r+4 + 5?/+6— 2+3a?— 2y. Ans. 7a?+3y + 8. 
 
 20. 4c + 3a— 6 — 3a+c— 6a— 6. Ans. 5c— 2a— 26. 
 
 21. 26— c+3a — 6+4c + 2tf— 6 — 5a\ -= U ^V « * 
 
 22. a— 6 + c+a*+3a+6 — 2c— 4a+10+a. 
 
 23. a+6 + 36+#— 7+4:r— 6+3a— y+2. 
 
 24. a?— 4+10— 7.r + a— 6— 10 + 2a— 1. v 
 
 25. 3a— 6+4c+2:c— c + 4a+5c+76. 
 
 26. 10— a + 6 — 6 — a?+7 + 36+2a— 40. 
 
 27. 8a— 16 — z + 91+2y — 87— 3z + 14. 
 
 28. 29 + 46+27+y— 32— 43+y. r 
 
 29. 73+4x— 36 — 3^—41+7^— y+2.r. . 
 
 30. a— 6— a-6-67— 42 + 7a+36./ 
 
 31. 4 X _37/+2a— lx+6y — 7a+5x— 7y. h2 ^ 
 
 32. 4x—3x-\-4-\-x— 2x— 5 + l + 3a?— 5a?. "-P-vX- 
 
 33. 2Z+1/+9— a:— i/— 9 + 3#+10a?y. It ^fc-^'^Z 
 
 34. 5a+36— 4c+2a— 56 + 6c+a— 46— 2c. 9 VL.-C4 
 
 35. 3a+;r+10— 5a+2x— 15— 4a— 10z+21 - i^"^ +lv 
 
 36. 5a— 36+4a— 76+7a+36— 5a— 96. H. ' 
 
 37. 6a+2;r— 10+2a— 3a?+10— 2y— 3a+2x. 
 
22 ALGEBRA. [ SECTION *• 
 
 III. ADDITION OF COMPOUND QUANTITIES. 
 
 § 37. When two or more expressions that consist of seve- 
 ral terms, are to be added together, the operation is represented 
 by writing them one after another ; taking care, in all except 
 the first, to insert the signs that are understood. Thus, a — x 
 is added to y-\-l in the following manner: 
 
 a — a? -f ?/4-7, or y-\-l-\-a — x. 
 For to a — x we add first y, and then 7 ; or to y-f 7 we add 
 a, and then because we ought to have added a — a?, we see 
 that we have added x too much, and therefore subtract it. 
 
 §38. The above example shows that it is of no conse- 
 quence in what order we write the terms. Their place may- 
 be changed at pleasure, provided their signs be preserved. 
 
 EXAMPLES. 
 
 1. Add the following compound quantities. 2« — 8a?, 
 x — 3a, — 4a — 2a?, 4x — a. Ans. — 6a — 5a?. 
 
 2. Add 2 — a? + 4?/, 3 -f 3a?— y, — 30 — x — 2y, and 
 1 — 2a? + 3y — 1 Oz together. Ans. — 24 — a? -f 4y — 1 Oz. 
 
 3. Add 3x+5y— 6z, — 2a?— 8y— 9z, 20a? -f 2y— 3z, and 
 x—y+z—4 together. Ans. 22a?— 2y— 17 z— 4. 
 
 4. Add 3— 2y+z, 4y— 2z-f5, 2—z—y, and 2z— y— 10. 
 Ans. Nothing. 
 
 5. Add 3a?— 6, 4a?+2, 64-3a?, 7— 2a?, ar-f-1. £->+ / J 
 
 6. Add 4a?— a, 3a— a?, 4a?+a, 7a— 3a?. / p &- *+ */" 
 
 7. Add2+2*4-3z, 3z + 8, 5— 2z, 4z. If +? 
 
 8. Add 4— a?, —a?— 3, 6+2a?, a?+2a?— 1. 
 
 9. Adda?— 21, 3+4a?+a?, 2a?+3a?+36. / / %~t ' ^ 
 
 10. Add a?, a?— 3, a?+a?— 3, a?+2a?— 6. 
 
 11. Add 2a?— 3z-4, 4z— 3a?, 5a?— 5, 3— 10a?+3. - kT^ 
 
 12. Add3?/-4, 6-3a?, 7x—5y, x—y—2, 3y + 5x—3. 
 
 Questions. — How are compound quantities added ? Why is the 
 sign — preserved in addition 1 Which expression must be put first! 
 
§ 43. J MULTIPLICATION AND DIVISION. 23 
 
 IV. MULTIPLICATION AND DIVISION OF SIMPLE QUANTITIES. 
 
 "§39. Were we to add a to itself four times, we should 
 write the sum thus, a + a + a+a-, which, when united, be- 
 comes 4a. Whence we have the 
 
 Rule. — Any literal quantity is multiplied by a number, 
 by putting that number before it as a co-efficient; as, 7 times 
 x is 7a?. 6 times a is 6a. 
 
 § 40. If the quantity to be multiplied has already a co- 
 efficient, that co-efficient only is to be multiplied. Thus, 3x 
 taken four times is 3a?-f 3a?-j-3a?-f 3a?, which when united 
 = 12a?. The co-efficient is multiplied by 4. Thus, 4 times 
 3a? = 12a\ 
 
 § 41. It is evident that if 2 times 3a? is 6a?, then one-half 
 of 6a? is 3a?. Whence we learn that a quantity with a nu- 
 meral co-efficient, may be divided, by merely dividing that 
 co-efficient. Thus, 8a? divided by 2 =4a?. 12a? divided by 
 4 == 3a?, &c. 
 
 § 42. In 1661, Rev. William Oughtred of England, pub- 
 lished a work, in which he introduced the sign x to represent 
 multiplication. Thus, 4x3=12, is read 4 multiplied by 3 
 equals 12 ; or, 4 into 3 equals 12. 
 
 §43. In 1668, Mr. Brancker invented the sign -r- for di- 
 vision. The sign is always put before the divisor; as, 
 20 h- 4 = 5 ; read 20 divided by 4 equals 5 ; or, 20 by 4, 
 equals 5. 
 
 But in algebra, division is frequently performed by writ- 
 ing the divisor under the dividend, so as to make a fraction ; 
 thus, 20 divided by 4, is \°. See page 48. 
 
 Questions. — How do we multiply a literal quantity? Multiply x 
 by 3, 4, 5, 6, &c. to 20. How do we multiply a quantity that has 
 already a co-efficient? Multiply 2x by 3, 4, 5, &c. to 12. Multiply 
 3x by the same numbers. Multiply ix by the same. How do we 
 divide a quantity that has a numeral co-efficient] Describe the sign 
 for multiplication. Describe the sign for division. 
 
24 ALGEBRA. [SECTION 1 
 
 EQUATIONS.— SECTION 1. 
 Equations which are solved by merely uniting terms. 
 
 In each of the following equations, ihe object is to find the 
 value of x. 
 
 Example 1. a?+2a? = 45— 15. 
 Uniting terms, Sx = 30. 
 
 Now, as we have found that three x's = 30, it is evident 
 that one x will be one-third of 30. 
 
 Therefore, dividing by 3, x = 10. 
 
 2. Find x in 8a: — 4a? — # = 7+26+51 — 15 
 
 Uniting terms, 3a? = 69 
 Dividing by 3, a? = 23. 
 
 3. Find a: in 10a?— 5a?+4a: = 56+75 + 32 — 1 
 
 Uniting terms, 9a: = 162 
 Dividing by 9, x = 18. 
 
 4. Find x in a?+2a:+3a:+4a? = 12+35+74— 11 
 
 Uniting terms, 10a? =110 
 Dividing by 10, a? = ll. 
 
 5. 8a:— 3a?+2a? = 46 + 54 + 37— 4. Ans. a?=19. 
 
 6. 4a?— 3a? + 4a? = 29—36+48+ 14. Ans. a? = 11. 
 
 7. 6a?— 8a-+ 14a? =12 + 36 + 14+22. Ans. a? = 7. 
 
 8. 5a?+4a:+3a' = 49 + 14'+22 + ll. Ans. a? = 8. 
 
 9. 7a?+a*=14— 22— 11+41— 6. Ans. a? = 2. 
 
 10. 4a:— 2a? = 96— 7 + 8— 15— 10. Ans. a? = 36. 
 
 11. 5a— a? = 2 + 3— 15— 10+72. Ans. a? = 13. 
 
 12. 6a? = 7 + 4+72— 51 — 16— 10. Ans. a?=l. 
 
 Questions. What is the most useful application of algebra ? § 22. 
 What is an equation? What is the part on the left of = called"? 
 What the part on the right 1 ? What if there are more terms on one 
 side than on the other? What is a numerical equation? How do 
 we find the value of the unknown quantity? What is this opera- 
 tion called ? 
 
§ 47.] EQUATIONS SOLVED BY UNITING TERMS. 25 
 
 13. Sx— 7x+5x— 4:X+3x=27— 12. Ans. x=3. 
 
 14. 5#— 4a?-f2a?— 3#-fa:=39— 13. Ans. a?=26 
 
 15. 17a:— 4a? — 2a;— 5a?— x =57— 32. Ans. #=5. 
 
 16. 14a?— 35a?-f29a:+47a?-f-a:=504. Ans. a?=9. 
 
 EQUATIONS FORMED FROM ARITHMETICAL QUESTIONS. 
 
 § 44. An algebraic problem is a proposition which sup 
 poses that an unknown quantity has certain relations with 
 other quantities that are known ; and requires the disco- 
 very of some arithmetical operation by which the unknown 
 quantity may be ascertained. 
 
 § 45. Problems properly belong to literal algebra ; but 
 because questions in numeral algebra are solved in a simi- 
 lar manner, they also are called problems in this work. 
 
 § 46. In the solution of problems, the first thing to be 
 done, is to represent the required answer by x, y, or some 
 other final letters of the alphabet. Then, make in alge- 
 braical language, with its explanation, a statement of each 
 of the conditions in the question, in the same manner as if 
 the letter were the true answer, and you were required to 
 prove it. 
 
 § 47- When the question has been fairly stated, it will 
 be found that two different algebraical expressions have the 
 same explanation. These two expressions must be put in 
 the same line, with the sign = between them, so as to form 
 an equation. And then, by reducing the equation, the 
 required result will be found. See § 25 and 26. 
 
 I. The sum of $660 was subscribed for a certain pur 
 pose, by two persons, A and B ; of which, B gave twice 
 as much as A. What did each of them subscribe ? 
 
 Questions. What is an algebraic problem ? What is the first step 
 in the solution of problems? How must you continue the statement? 
 How is it known when the statement is perfect? What is the second 
 part of the solution ? 
 
 3 
 
26 ALGEBRA. ^SECTION I. 
 
 Stating the question, x dollars = what A gave. 
 
 2x dollars = what B gave. 
 X -x-2x dollars = what both gave. 
 And also, 660 dollars = what both gave. 
 
 Therefore, putting the question into an equation, 
 x +2 x = 660 
 Uniting terms, 3x = 660 
 
 Dividing by 3, x = 220 A's share. 
 
 2x = 440 B's share. 
 660 proof. 
 
 2. Three persons in partnership, put into the stock $4800 ; 
 of which, A put in a certain sum, B twice as much, and C as 
 much as A and B both. What did each man put in ? 
 
 Stating the question, x = A's share* 
 
 2x = B's share. 
 
 x + 2x = C's share. 
 
 Adding all three shares, x -f 2x -f- x + 2x «■ the whole. 
 
 4800 = the whole. 
 Therefore, forming the equation, 
 
 x -f 2x -f x -f 2x — 4800 
 Uniting terms, Gx = 4800 
 
 Dividing by 6, x = 800 A's share. 
 
 2x = 1600 B's share. 
 800 -f 1600 = 2400 C's share. 
 4800 proof. 
 
 E^~ In all the succeeding problems, the learner should 
 prove his answers. 
 
 3. A person told his friend that he gave 108 dollars for his 
 horse and saddle ; and that the horse cost 8 times as much as 
 the saddle. What was the cost of each 1 
 
 Stating the question, x = price of the saddle. 
 
 Sx = " horse. 
 
 x + 8x= " both. 
 
 108 = " do. 
 
 Forming the equation x -f Sx = 108 
 
§ 47.] EQUATIONS SOLVED BY UNITING TERMS. 27 
 
 Uniting terms, 9x = 108 
 
 Dividing by 9, x = 12 = price of saddle. 
 
 96 = " horse. 
 It is advisable for the pupil, while performing his sums, to 
 write them on his slate in a manner similar to the three ques- 
 tions above; beginning the statement by making x the answer 
 to the question, and throughout the operation keeping equals 
 under equals. And in recitation, the whole of it is to be 
 recited. 
 
 4. A father once said, that his age was six times that of 
 his son; and that both of their ages put together, would 
 amount to 49 years. What was the age of each ? 
 
 Ans. Son's age 7 years ; father's 42. 
 
 5. A farmer said that he had four times as many cows as 
 horses, and five times as many sheep as cows ; and that the 
 number of them all was 100. How many had he of each 
 sort? Ans. 4 horses ; 16 cows; and 80 sheep. 
 
 6. A boy told his sister that he had ten times as many 
 chestnuts as apples, and six times as many walnuts as chest- 
 nuts. How many had he of each sort, supposing there were 
 639 in all ? Ans. 9 apples ; 90 chestnuts ; and 540 walnuts. 
 
 7. A school-girl said that she had 120 pins and needles ; 
 and that she Had seven times as many pins as needles. How 
 many had she of each sort? Ans. 15 needles, and 105 pins. 
 
 8. A teacher said that her school consisted of 64 scholars ; 
 and that there were three times as many in arithmetic as in 
 algebra, and four times as many in grammar as in arithmetic. 
 How many were there in each study? 
 
 Ans. 4 in algebra; 12 in arithmetic ; and 48 in grammar. 
 
 9. A teacher had four arithmeticians who performed 80 
 sums in a day. The second did as many as the first, the 
 third twice as many, and the fourth as much as all the other 
 three. How many did each perform ? Ans. The first and 
 second, each 10 ; the third, 20 ; and the fourth, 40. 
 
28 ALGEBRA. ^SECTION I. 
 
 10. A person said that he was $450 in debt. That he 
 owed A a certain sum, B twice as much, and C twice as much 
 as to A and B. How much did he owe each ? 
 
 Ans. To A $50, to B $100, and to C $300. 
 
 11. A person said that he was owing to A a certain sum; 
 to B four times as much ; and to C eight times as much ; and 
 to D six times as much ; so that $570 would make him even 
 with the world. What was his debt to A ? Ans. $30. 
 
 12. A boy bought some oranges and some lemons for 54 
 cents. The price of the oranges was twice the price of the 
 lemons. How much money did he spend for each sort? 
 
 Ans. 18 cents for lemons ; and 36 cents for oranges. 
 
 13. A boy bought some apples, some pears, and some 
 peaches, an equal number of each sort, for 72 cents. The 
 price of a pear was twice that of an apple, and the price of a 
 peach was 3 times that of an apple. How much money did 
 he give for each kind? 
 
 Ans. 12 cents for apples ; 24 for pears ; 36 for peaches. 
 
 14. A man bought 3 sheep and 2 cows for $60. For each 
 cow, he gave 6 times as much as for a sheep. How much 
 did he give for each ? 
 
 {y If x = price of a sheep, all the sheep will cost three 
 times as much, or 3a\ In the same manner both cows will 
 cost twice as much as one cow. One cow will cost 6a?, and 
 2 cows will cost Ylx. 
 
 Ans. $4, price of a sheep ; and $24 price of a cow. 
 
 15. A gentleman hired 3 men and 2 boys. He gave five 
 times as much to a man as he gave to a boy ; and for all of 
 ihem he gave $6.80. What was the wages of each ? 
 
 Ans. A boy's wages was 40 cents, and a man's wages, $2 
 
 16. Two men, who are 560 miles apart, start to meet each 
 other. One goes 30, and the other goes 40 miles a day. In 
 how many days will they meet? 
 
 Each will travel x days. The first will go x times 
 
§ 47.] EQUATIONS SOLVED BY UNITING TERMS. 29 
 
 30 miles, and the second will go x times 40 miles ; and both 
 together will go the whole distance. It is also evident that x 
 times 30 is the same as 30 times x ; &c. Ans. 8 days. 
 
 17. A farmer hired three labourers for $50.00 ; giving to 
 the first $2.00 a day, to the second $1.50, and to the third 
 $1.00. The second worked three times as many days as the 
 first ; and the third twice as many days as the second. How 
 many days did each work ? 
 
 Ans. The first, 4; second, 12 ; and third, 24 days. 
 
 18. A gentleman bought some tea, coffee, and sugar, for 
 $7.04 ; giving twice as much a pound for coffee as for sugar, 
 and five times as much for tea as for coffee ; and there were 
 20 pounds of sugar, 12 pounds of coffee, and 2 pounds of tea. 
 What was the price of each? Ans. 11 cents for sugar; 22 
 cents for coffee ; and 110 cents for tea. 
 
 19. A bookseller sold to a teacher at one time 10 books, 
 and afterwards 15 more at the same rate. Now the difference 
 between the whole sum received at the latter time, and the 
 whole sum received at the former time, was 60 cents. What 
 was the price of each book? Ans. 12 cents. 
 
 20. In fencing a side of a field, whose length was 450 
 yards, two workmen were employed ; one of whom fenced 9 
 yards per day, and the other 6 yards per day. How many 
 days did they work to make the whole fence ? Ans. 30 days. 
 
 8» 
 
30 ALGEBRA. ^SECTION II. 
 
 SECTION II. 
 TRANSPOSITION. 
 
 TRANSPOSITION BY SUBTRACTION. 
 
 § 48- It often happens that in the first member of the 
 equation, some number has been added to the x*u in order to 
 make them equal to the last member. Thus, in the equation 
 
 07+16 = 46 
 we see that 16 has been added to #, to make it equal to 46. 
 
 § 49. Now if x with 16 added, is equal to 46; then x 
 alone must be 16 less than 46; that is, 46 — 16. So, that if 
 we find, that #+16 = 46, we may know that a? = 46— 16; 
 or what is the same, x = 30. 
 
 § 50. But this may be proved another way. It is very 
 plain that if we subtract a quantity from one member of an 
 equation, and then subtract the same quantity from the other 
 member of the equation ; it will still be the fact that the two 
 members are equal to one another. Thus, a half-dollar = 50 
 cents. Subtract 2 cents from each member. Then a half- 
 dollar — 2 cents = 50 cents — 2 cents ; for each of them is 
 equal to 48 cents. 
 
 § 51. Thus, with the equation that we had above, 
 
 #+16 = 46 
 Subtracting 16 from both members, a;+16 — 16 = 46 — 16. 
 
 Now, in the first member of the equation, we have +16 
 — 16, which is of no value at ail, for +16 and —16 balance 
 each other, as has been seen in Ex. 6 under § 36. 
 Therefore the equation is reduced to x = 46 — 16 
 
 Uniting terms in the last member, x = 30. 
 
 Questions. Is the first member always without numeral quanti- 
 ties ? When x with another number added, equals a certain number 
 what is x itself equal to ? How can this be proved t 
 
§ 51-3 TRANSPOSITION BY SUBTRACTION. 31 
 
 EQUATIONS. — SECTION 2. 
 
 1. Suppose a?+8+3a? = 56; w hat is the value of a?? 
 Uniting terms, 4a?-f-8 = 56 
 Subtracting 8 from both sides, 4a?+8— 8 = 56—8 
 Which is the same as 4a? = 56 — 8 
 Uniting terms in the last member, 4a? = 48 
 Dividing by 4, a? = 12. 
 
 2. Suppose 2a?-fl4— a?— 7 =41+2—8; to find a?. 
 Uniting terms, a?-f-7 = 35 
 
 Subtracting 7 from £ a?-f 7— 7 = 35 — 7 or 
 
 both sides, 5 a? = 35 — 7 
 
 Uniting terms, x = 28. 
 
 3. Given a?-f5— 2a?— 3-f4a? = 26, to find x. 
 
 Uniting terms, 3a?+2 = 26 
 
 Subtracting 2 from both sides, 3a? = 24 
 Dividing by 3, a? = 8. 
 
 4. Given 5a?-f22— 2a? = 31, to find a?. Ans. a? = 3. 
 
 5. Given 4a?+20— 6 = 34, to find a?. Ans. a? = 5. 
 
 6. Given 3a? -f 12-f 7a? = 102, to find a?. Ans. a? = 9. 
 
 7. Given 10a?— 6a? -f 14 = 62, to find a?. Ans. a? = 12. 
 
 8. If 7a?— 14+5a?+20 = 246, then a? =20. For, &c. 
 
 9. If 8a?+17— 5a?+3 = 100+10, then a?=30. 
 10. If 7a?— 14+3a?+35 = 450— 29, then a? = 40. 
 
 PROBLEMS. 
 
 WtT For putting questions into equations, see page 25. 
 1. What number is that, which, with 5 added to it, will 
 be equal to 40 ? 
 
 Stating the question, a? = the number 
 
 a? +5, = after adding 
 Forming the equation, * a?+5 = 40 
 
 Subtracting 5, from both, a? = 35. 
 
32 ALGEBRA. ^SECTION II. 
 
 2. A man being asked how many shillings he had, answered, 
 Add 15 to their number, and then subtract 1, and the remain- 
 der will be 64. How many shillings had he ? 
 
 Stating the question, # = number of shillings. 
 
 a?+ 1*5 = after adding. 
 #+15 — 1 = after subtracting. 
 64 = remainder. 
 Forming the equation, #+ 15 — 1 = 64 
 Uniting terms, #+14 = 64- 
 
 Subtracting 14 from both sides, x = 50. 
 
 3. What number is that, which with 9 added to it, will 
 equal 23 ? Ans. 14. 
 
 4. Divide 17 dollars between two persons, so that one may 
 have 4 dollars more than the other. 
 
 Stating the question, x = the less share. 
 
 #+ 4 = the greater. 
 #+#-f-4 = both shares. 
 17 = both shares. 
 Forming the equation, # + #+4=17 
 
 Uniting terms, 2#+4 = 17 
 
 Subtracting 4 from both sides, 2#= 13 
 Dividing by 2, # = 6<£ share of one. 
 
 17— 6£ = 10| •• the other. 
 
 5. The sum of the ages of a certain man and his wife is 55 
 years ; and his age exceeds hers by 7 years. What is the 
 age of each ? fS9^ Let # = age of the wife. 
 
 Ans. 24 the wife's; 31 the man's. 
 
 6. A is 5 years older than B, and B is 4 years older than 
 C ; and the sum of their ages is 73 years. What is the age 
 of each ? 
 
 Stating the question, # = C's age. 
 
 #+4 = B's age. 
 #+4-f 5 = A's age. 
 #+# + 4+# + 4 + 5 = sum of all of them. 
 Forming the equation, # + # + 4 + #+4 + 5 = 73. 
 
 Ans. C 20 years, B 24, A 29. 
 
§ 51.] TRANSPOSITION BY SUBTRACTION. 33 
 
 7. Two persons were candidates for a certain office, where 
 there were 329 voters. The successful candidate gained his 
 election by a majority of 53. How many voted for each 1 
 
 Ans. 191 for one, and 138 for the other. 
 
 8. A, B, and C, would divide $200 among themselves, so 
 that B may have $6 more than A ; and C $8 more than B. 
 How much must each have ? 
 
 Ans. A must have $60, B $66, and C $74. 
 
 9. Divide $1000 between A, B, and C ; so that A shall 
 have $72 more than B, and C $100 more than A. 
 
 Ans. Give B $252, A $324, and C $424. 
 
 10. At a certain election 1296 persons voted, and the suc- 
 cessful candidate had a majority of 120. How many voted 
 for each ? Ans. 588 for one, and 708 for the other. 
 
 11. A father, who has three sons, leaves them $8000, spe- 
 cifying in his will that the second shall have $500 more than 
 the youngest, and that the eldest shall have $1000 more than 
 the second. What is the share of each ? Ans. The eldest 
 had $3500, the second $2500, the youngest $2000. 
 
 12. A bin, which held 74 bushels, was filled with a mixture 
 of corn, rye, and oats. In it there were 15 bushels of rye 
 more than of corn ; and as much oats as both corn and rye. 
 What was the quantity of each ? 
 
 Ans. 1 1 bushels of corn, 26 of rye, and 37 of oats. 
 
 13. A draper bought three pieces of cloth, which together 
 measured 159 yirds. The second piece was 15 yards longer 
 than the first, and ihe third 24 yards longer than the second. 
 Whal was the length of each? Ans. The first, 35 yards; 
 the second, 50 yards ; the third, 74 yards. 
 
 C 
 
34 ' ALGEBRA. [SECTION III 
 
 SECTION III. 
 
 TRANSPOSITION BY ADDITION. 
 
 § 52. It is frequently found that some quantity has been 
 subtracted from the a?'s ; as in the equation 5a? — 44 = 76. 
 
 § 53. Now if we had a dollar minus 5 cents, and wished 
 to make the sum just a dollar, we should add the five cents 
 that are lacking. So in the above equation, we have 5a? — 44; 
 and as we wish to make it just 5a?, we must add the 44 to it; 
 and 5a? — 44-|-44 is the same as 5a?. But if we add 44 to 
 one member of the equation, we must also add as much to 
 the other member of the equation. 
 
 So that adding 44 to both sides, 5a?— 44 + 44 = 76+44 
 Or, 5a? = 76+44 
 
 Uniting terms, 5a? = 120 
 
 Dividing by 5, a? = 24. 
 
 EQUATIONS.— SECTION 3. 
 
 1. Given a?+ 14 + 3a?— 27 = 51, to find x. Ans. a? =16. 
 
 2. Given 3a?— 30— 2a? = 46 — 7, to find x. Ans. a? = 69. 
 
 3. Given 9x— 41 + 6 = 88 — 6, to find x. Ans. x— 13. 
 
 4. Given 20 + 3a?— 46 = 35— 4, to find x. Ans. a? =19. 
 
 5. Given 4a: 39 — 2a? = 47, to find x. Ans. a: = 43. 
 
 6. Given 7a?+27— 46 = 65, to find a?. Ans. a? = 12. 
 
 7. Given 14a?— 55 — 8a?+ 14 = 85, to find x. Ans. a? = 21. 
 
 PROBLEMS. 
 
 19* For putting questions into equations, see page 25. 
 1. What number is that, from which 8 being subtracted, 
 the remainder is 45 ? 
 
 Stating the question, a? = the number. 
 
 a? — 8 = when 8 is subtracted 
 Forming the equation, a? — 8 = 45 
 Adding 8 to both sides, x = 53 
 
$ 53."] TRANSPOSITION BY ADDITION. 35 
 
 2. What number is that, from which 27 being subtracted, 
 the remainder is 41 ? Ans. 68. 
 
 3. A person bought two geese for $1.40; and gave 16 cents 
 more for one than he did for the other. What did each cost 
 him ? 
 
 B^~ In this and the four following questions, x must stand 
 for the first mentioned quantity. 
 
 Stating the question, x = the dearest. 
 
 a?— 16 = the cheapest. 
 x+x— 16 = cost of both. 
 140 = cost of both. 
 Forming the equation, x-\- x— 16= 140 
 Uniting terms, 2x— 16 = 140 
 
 Adding 16 to both sides, 2x= 156 
 
 Dividing by 2, x = 78. 
 
 Ans. 78 cents, and 62 cents. 
 
 4. Three men, who are engaged in trade, put in $2600 as 
 follows : A put in a certain sum ; B, $60 less than A ; and 
 C, as much as A and B, lacking $100. What was each 
 man's share ? Ans. A's $705 ; B's $645 ; C's $1250. 
 
 5. A purse of $8000 is to be divided among A, B, and C ; 
 bo that B may receive $276 less than A, and C $1112 less 
 than A and B together. What is each man's share ? 
 
 Ans. A's $2416; B's $2140; C's $3444. 
 
 6. A gentleman gave to two beggars 49 cents ; giving to 
 the first 15 cents less than to the second. How many cents 
 did each receive? Ans. 32 and 17. 
 
 7. A man leaves $16000 to be divided to his widow, son, 
 and daughter, in such a manner that the son is to have $2000 
 less than the widow, and the daughter $1000 less than the 
 son. What is the share of each ? 
 
 Ans. Widow, $7000; son, $5000; daughter, $4000. 
 
 8. Divide the number 60 into three such parts, that the 
 first may exceed the second by 8, and the third by 16. 
 
 Ans. 28 ; 20 ; and 12 
 
36 ALGEBRA. [SECTION III. 
 
 9. Three men having found a purse of $160, quarreled 
 about the distribution of it. After the quarrel, it was found 
 that A had got a certain sura, and that B had $30 more than 
 A, but C $50 less than A. How much did each obtain ? 
 
 Ans. A $60; B $90 ; C $10. 
 
 10. Three men, A, B, and C, trade in company, with a 
 stock of $3130; of which B puts in $350 more than A, and 
 C $220 less than A. What was the capital of each ? 
 
 Ans. A's$1000; B's$1350; C's $780. 
 
 11. How can an estate of $9931 be divided between a 
 widow, son, and daughter, in such a manner that the son 
 shall have $592 less than the widow, and $522 more than 
 the daughter ? 
 
 WtBT" After knowing the son's share, how can the daugh- 
 ter's be found ? 
 
 Ans. Widow, $3879 ; son, $3287 ; daughter, $2765. 
 
 12. A father divided $12000 among his three sons, giving 
 to the second $1500 less than to the eldest, and $750 more 
 than to the youngest. What was the share of each I 
 
 Ans. $5250 ; $3750 ; and $3000. 
 
 13. A father has willed to his four sons $25200, as follows: 
 To D a certain sum ; to C as much as to D, lacking $550; to 
 B as much as to C, together with $1550; and to A twice as 
 much as to B, lacking $10000. How much does each of 
 
 them receive ? 
 
 Ans. A $5100; B $7550 ; C$6000; D $6550 
 
§ 57.] TRANSPOSITION OF THE UNKNOWN QUANTITY. 37 
 
 SECTION IV. 
 
 TRANSPOSITION OF THE UNKNOWN QUANTITY. 
 
 § 54. We have found that when any term has the sign -}- 
 it may be removed from one member of the equation to the 
 other, if we take care to change the sign to — ; for this has 
 been done every time we have subtracted a term from both 
 sides. Thus, in the equation 
 
 #4-5 = 20; 
 if we subtract 5 from both sides, it is plain that the first 
 member becomes #, and the last member becomes 20 — 5 ; so 
 that the equation would become 
 
 # = 20 — 5. 
 § 55. So also any term that has the sign — may be re- 
 moved from one member to the other, if we take care to 
 change the sign to -}-. Because this is the same as adding 
 that term to both sides. Thus, in the equation 
 
 #—5 = 20, 
 if we add 5 to both sides, the first member becomes x, and 
 the last member becomes 20 + 5. So that the equation be- 
 comes 
 
 a? = 20 + 5. 
 
 § 56. When we remove a term from one member of an 
 equation to the other member, we say that we transpose that 
 term ; and the operation of doing it is called transposition. 
 
 § 57. Any term may be transposed from one member of 
 an equation to the other, care being taken to change the sign 
 when we change the side. 
 
 § 58. It was stated in § 25, that an equation must be 
 
 Questions. How can a positive quantity be removed from one 
 member of an equation to the other] Why? How may a negative 
 quantity be removed from one side to the other ? Why ? What do 
 we call this method of removing a term i What care is required in 
 transposing? Why are we ever obliged to transpose 1 
 
 4 
 
38 ALGEBRA. [SECTION IV 
 
 brought so that the unknown quantity will occ ipy one mem 
 ber of the equation, and the known quantities embrace the 
 other member. And, as it frequently happens that the un- 
 known quantities are on both sides, we are obliged to resort to 
 transposition in order to make one side -free from them. And 
 likewise, it is often necessary to transpose known quantities 
 from the member which contains the unknown quantity. 
 
 RULE FOR TRANSPOSING. 
 
 § 59. In transposing, it is generally best to write first the 
 unknown quantify that is already on the left ; and then bring 
 over all those which are on the right, if there are any there. 
 And in transposing the known quantifies to the right hand 
 member, write those that are already there, and then trans- 
 pose after them what known quantities there are in the left. 
 
 EQUATIONS.— SECTION 4. 
 
 1. Reduce the equation \x — 14 = 3#-f-12. 
 Transposing 3*, > ^_^ _ ^ 
 Transposing 14,3 
 
 Uniting terms, x = 26. 
 
 2. Given 21 — 7x = 40 — 11#, to find x. Ans. a? = 4f. 
 
 3. Given 40 — 6z == 136— 14*, to find z. Ans. z = 12. 
 
 4. Given Sy — 4 = y+ 12, to find y. Ans. y = 8. 
 
 5. Given bx — 15 mi 2#+6, to find x. Ans. x = 7. 
 
 6. Given 40 — 6x — 16 =•- 120 — 14r, to find x. Ans. 12. 
 
 7. Given 4 — 9y = 14 — lly, to find y. Ans. y = 5. 
 
 8. Given x+ 18 = Sx — 5, to find x. 
 Transposing, x — 3a? = — 5 — 18 
 Uniting terms, — 2x = — 23 
 Dividing by 2, — a?= — 11£ 
 
 Question. In what order do we write the terms when we are 
 ••ansposing ? 
 
 * The pupil will recite the left hand member of this line, for trans- 
 posing 3x. 
 
§ 63.] TRANSPOSITION OF THE UNKNOWN QUANTITY. 39 
 
 § 60. It is of no consequence what sign accompanies the 
 final result ; as the magnitude of the quantity is not affected 
 by the sign. If we remember that + is understood and may 
 be written with every positive quantity, it will be very evident 
 that the equation — x = — 1 1£ is just as good as the equation 
 + x = +11 5. In both cases, the quantity x is equal to the 
 number 11$. 
 
 § 61. In the result of the last question, 11| may be trans- 
 posed to the first member ; and x may be transposed to the 
 last member. Of course, this will change the signs ; and the 
 equation will become 1 \\ = x. And if 1 15 = a?, it is evident 
 that x = 1 1£. This coincides with what was shown in § 60. 
 
 § 62. From what has just been said, we see that all the 
 terms of each member may be transposed, so that the sign of 
 each term may be changed ; and still the equation shall retain 
 the same members as at nrst, though differently placed. 
 Hence, it is immaterial which member is written first. And 
 also, in any equation the signs of all the terms may be 
 changed without affecting the equality. 
 
 § 63. It is evident that all the terms of one member may 
 be transposed to the other member. When this has been 
 done, the member fr?m which the terms have been trans- 
 posed becomes, 0. Thus the equation x=ll£, may be 
 made x — ll£=0; where — ll£ balances x. 
 
 PROBLEMS. 
 
 WS r After the equation has been formed by § 47, it 
 must be transposed by § 59. 
 
 1. A man has six sons, whose successive ages differ by 
 4 years ; and the eldest is three times as old as the young- 
 est. What are their ages ? 
 
 Questions. What sign must accompany the answer? Explain. 
 Explain by transposition. To what extent may the signs be chvinged 1 
 Why? To what extent may the terms be transposed 1 Why? 
 
40 
 
 ALGEBRA. 
 
 [SECTION IV 
 
 Stating the question, 
 
 x = age of the youngest 
 
 # + 4 = " " next 
 
 # + 4 + 4 == " " next 
 
 # + 4 + 4 + 4 = " " next. 
 
 #+4 + 4-1-4 + 4-= " " next. 
 
 #+4 + 4 + 4 + 4 + 4 = " " eldest. 
 
 3x = " " eldest. 
 Forming the equation, #+4+4+4+4+4 = 3a? 
 
 Uniting terms, 
 Transposing the 3# and 20, 
 Uniting terms, 
 Dividing by 2, 
 or 
 
 #+20 = 3# 
 # — 3# = —20 
 — 2* = — 20 
 — # = — 10 
 
 x = 1 age of the 
 
 [youngest. 
 
 2. A person bought two horses, and also a hundred dollar 
 harness. The first horse, witl* the harness, was of equal 
 value with the second horse. But the second horse with the 
 harness cost twice as much as the first. What was the price 
 of each horse ? 
 
 Stating the question, x = price of the first. 
 
 #+ 100 = price of the second. 
 #+100+100 = 2d horse harnessed. 
 2x = twice price of first. 
 Forming the equation, x + 1 00 + 1 00 = 2x 
 Transpos. from both members, x — 2# = — 100 — 100 
 Uniting terms, — x = — 200 
 
 Or x = 200 &c. 
 
 3. A privateer, running at the rate of 10 miles an hour, 
 discovers a ship 18 miles off, sailing at the rate of 8 miles an 
 hour. How many hours can the ship run before she will be 
 overtaken by the privateer? 
 
 H©^ In x hours, the privateer will go 10a? miles, which 
 is the whole distance. In the same time, the ship will go 
 Sx miles. But the ship had already gone 18 miles, which, 
 added to the Sx, will make the whole distance. 
 
5 63.] TRANSPOSITION. 41 
 
 4. A gentleman distributing money among some poor peo- 
 ple, found that if he undertook to give 5s. to each, he would 
 lack 10s. Therefore, he gave only 4s. to each, and finds that 
 he has 5s. left. How many persons were there ? 
 
 W^~ It will be found that 5a? — 10= his money by the 
 first supposition, and 4x+5 = the money by the last suppo- 
 sition. Ans. 15. 
 
 5. I once had $84 in my possession ; and I gave away so 
 much of it, that what I have now equals three times as much 
 as I gave away. How much did I give away ? 
 
 ffl&T If I gave away $#, then $84 — x will be what re- 
 mains. Ans. $21. 
 
 6. A certain sum of money was shared among five persons, 
 A, B, C, D, and E. Now B received $10 less than A ; C, 
 $16 more than B ; D, $5 less than C ; E, $15 more than D. 
 And it was found that the sum of the shares of the first three 
 put together, were equal to the sum of the shares of the other 
 two. How much did each man receive ? 
 
 Ans. A $21 ; B $11 ; C $27 ; D $22 ; E $37. 
 
 7. A person wishes to give 3 cents apiece to some beggars, 
 but finds that he has not money enough by 8 cents. He 
 gives them 2 cents apiece and has 3 cents left. How many 
 beggars were there ? Ans. 11. 
 
 8. A courier who had started from a certain place 10 hours 
 ago, is pursued by another from the same place, and on the 
 same road. The first goes 4 miles an hour, and the second 9 
 In how many hours will the pursuer overtake the first? 
 
 fgy If the pursuer goes x hours, the first must go 
 x-f 10 hours. But, as both go from the same place, tha 
 distance that each goes must be the same. 
 
 In generalization see page 113. 
 
42 ALGEBRA. [SECTION V. 
 
 SECTION V. 
 
 MULTIPLICATION OF COMPOUND QUANTITIES 
 
 BY SIMPLE QUANTITIES. 
 
 v 
 
 § 64. Suppose you purchase 8 melons at 7 cents apiece, 
 and afterwards find that you must give 5 cents apiece more 
 for them. In this case you pay 8 times 7 cents, and also 8 
 times 5 cents ; that is, first, 56 cents, and afterwards 40 cents. 
 
 § 65. Let us apply this principle to Algebra. You pay in 
 all, 8 times (7 + 5,) which =56+10. Which shows that in 
 multiplying a compound quantity, you multiply each term 
 by the multiplier. 
 
 We can easily see that this operation will give the right 
 answer; for in the case of the melons, they cost 12 cents 
 apiece, and therefore their whole cost was 8 times 12 cents 
 which =96 cents. But the answer just obtained, 56+40 
 = 96. 
 
 § 66. But suppose that after you had paid 7 cents apiece, 
 a deduction of 5 cents apiece was made. The whole cost 
 would then be 8 times (7 — 5,) which =56 — 40. And this 
 agrees with the truth ; for you first paid 56 cents, and after- 
 wards 40 cents were deducted. 
 
 W£T — 5 multiplied by 8, signifies that — 5 is to be added 
 8 times. Therefore retaining the sign, (§29), we add — 40. 
 
 § 67. This shows that + multiplied by +, produces + ; 
 and — multiplied by +, produces — . 
 
 EXAMPLES. 
 
 1. Multiply a; +4, by 3. 
 
 Operation. a? +4 
 
 Ans. 'Sx +12. 
 
 Questions. How do we multiply a compound quantity] Explain 
 why. How do the signs of the answer correspond with the quantity 
 that is multiplied 1 Explain for the — , 
 
5 69.1 MULTIPLICATION OF COMPOUND QUANTITIES. 43 
 
 2. Multiply 12 + a?, by 5. Ans. 60 + 5a?. 
 
 3. Multiply x— 10, by 8. Ans. 8a? — 80. 
 
 4. Multiply 126— a-, by 4. Ans. 504— 4a:. 
 
 5. Multiply a? + 8, by 6. 
 
 6. Multiply 40 + a*, by 10. , 
 
 7. Multiply a?— 32, by 9. 
 
 8. Multiply 52— x, by 12. 
 
 9. Multiply 2a? + 14, by 7. 
 
 10. Multiply 27 + 3a?, by 14. 
 
 11. Multiply 3a?— 62, by 15. 
 
 12. Multiply 97 — 4a?, by 12. 
 
 13. Multiply a?+7— y, by 7. 
 
 14. Multiply Sx+y— 12, by 8. 
 
 15. Multiply 2a?— 3y— 6, by 6. 
 
 16. Multiply 3a?— 12+y, by 5. 
 
 § 68. Franciscus Vieta, a Frenchman, introduced about the 
 year 1600, the vinculum or a straight line drawn over the top 
 of two or more quantities when it is wished to connect them 
 together. Thus, a + 4x3, signifies that both a? and 4 are to 
 be multiplied by 3. 
 
 § 69. In 1629, Albert Girard, a Dutchman, introduced the 
 parenthesis as a convenient substitute, in many cases, for the 
 vinculum. Thus, (a?+4)x3, is the same as a* + 4x3 ; and is 
 read, a? +4, both x3. If there are more than two terms under 
 the vinculum, we say, after repeating those terms, all, &c. 
 Thus, (a?+y) x(« — b+c), is read x-\-y both into a—b + c 
 all. See also § 103. 
 
 Questions. What is a vinculum 1 How is the parenthesis used 1 
 How is a compound quantity read when embraced by a vinculum or 
 parenthesis 1. 
 
44 ALGEBRA. l SECTI0N T » 
 
 EQUATIONS. — SECTION 5. 
 
 1. Given x — 9x11 = 121, to find a?. 
 
 Writing the equation, x— 9 X 1 1 = 121 
 
 Performing the multiplication, \\x — 99 =121 
 Transposing and uniting, llx = 220 
 
 Dividing by 11, x = 20. 
 
 2. Given (#+7) x6 = 54, to find a?. Ans. x = 2. 
 
 3. Given 12 + xx5 = 100, to find x. Ans. 8. 
 
 4. Given x— 9 x8 = 96, to find x. Ans. 21, 
 
 5. Given 367— 3#x5 = 920, to find x. Ans. 61. 
 
 6. Given(8+a:)x2+14 = 72, to find a?. 
 
 10° The pupil must understand that 14 is not a part of 
 the multiplier, because there is the sign -f between it and the 
 ultiplier. Ans. 21. 
 
 7. Given (15+*) x3— 27 = 48, to find x. Ans. 10. 
 
 8. (112— 2*)x3 = (2x— 7)x4, to find x. Ans. 26. 
 
 9. (3z+14)x4 = (78— a*)x5, to find ic. Ans. 19}|. 
 
 10. 2x + 8x5 = (32 +.r)x3, to find a\ Ans. 8. 
 
 11. (3x— 14)x7 = (17— x)x6, to find x. Ans. 7Jf 
 
 12. 120— 3xx2 = (4x— 6)x9, to find a?. Ans. 7. 
 
 PROBLEMS. 
 
 1. Two persons, A and B, lay out equal sums of money in 
 trade ; A gains $126, and B loses $87 ; and now A's money 
 is double of B's. What did each lay out? 
 
 Stating the question, x, = the sum for each. 
 
 a?+126 = A's sum now. 
 x — 87 = B's sum now. 
 2x— 174 = the double of B's 
 Forming the equation, #-r-126 = 2:r— 174. 
 
 Transposing and uniting, — a?= — 300. 
 
 Changing signs, x = 300 the answei 
 
§ 69. j MULTIPLICATION OF COMPOUND QUANTITIES. 45 
 
 2. A person, at the time he was married, was 3 times as 
 old as his wife; but after they had lived together 15 years, 
 he was only twice as old. What were their ages on their 
 wedding day? 
 
 Stating the question, x = the wife's age. 
 
 Sx = the man's age* 
 a?+ 15= the wife's after 15 years 
 3a?+15 = the man's after 15 years. 
 2a?+30 = twice the wife's age. 
 Forming the equation, 3#+ 1 5 = 2.T+30 
 Transposing and uniting, x= 15 the wife's age. 
 
 Sx = 45 the man's age. 
 
 3. A man having some calves and some sheep, and being 
 asked how many he had of each sort; answered that he had 
 twenty more sheep than calves, and that seven times the num- 
 ber of calves was equal to three times the number of sheep. 
 How many were there of each ? 
 
 Ir^T 3 * If x= number of calves, then x + 20= number of 
 sheep. Ans. 15 calves, and 35 sheep. 
 
 4. Two persons, A and B, having received equal sums of 
 money, A paid away $25, and B paid away $60 ; and then 
 it appeared that A had just twice as much money left as B. 
 What was the sum that each received ? Ans. $95. 
 
 5. Divide the number 75 into two such parts, so that 
 three times the greater may exceed 7 times the less by 15. 
 
 I'lT* If x= the greater then 75— x = the less; and 3a? 
 will = (7 times the less) + 15. Ans. 54 and 21. 
 
 6. The garrison of a certain town consists of 125 men, 
 partly cavalry and partly infantry. The monthly pay of a 
 horse soldier is $20, and that of a foot soldier is $15; and 
 the whole garrison receives $2050 a month. What is the 
 number of cavalry, and what of infantry ? 
 
 If j = number of cavalry, then 20# = the whole pay 
 of cavalry, &c. Are. 35 cavalry, and 90 infantry 
 
46 ALGEBRA. [SECTION V. 
 
 7. A grocer sold his brandy for 25 cents a gallon more than 
 he asked for his wine ; and 37 gallons of his wine came to 
 as much as 32 gallons of his brandy. What was each per 
 gallon? Ans. $1.60 for wine; and $1.85 for brandy. 
 
 8. A wine merchant has two kinds of wine ; the one costs 
 9 shillings a gallon, the other 5. He wishes to mix both 
 wines together, so that he may have 50 gallons that may be 
 sold without profit or loss for 8 shillings a gallon. How much 
 must he take of each sort ? 
 
 SZW There are 50 gallons of both kinds, and after rina- 
 ing the cost by the kinds, the whole mixture will be worth 
 50 times 8 shillings. 
 
 Ans. 37 d gallons of the best; and 125 of the poorer. 
 
 9. A gentleman is now 40 years old, and his son is 9 years 
 old. In how many years, if they both live, will the father be 
 only twice as old as his son ? 
 
 JO 1 * In x years he will be 40-)-#> and his son 9-\-x. 
 
 Ans. 22 years. 
 
 10. A man bought 20 oranges and 25 lemons for $1.95. 
 For each of the oranges he gave 3 cents more than for a 
 lemon. What did he give apiece for each ? 
 
 Ans. 3 cents for lemons ; 6 cents for oranges. 
 
 11. A man sold 45 barrels of flour for $279 ; some at $5 a 
 barrel, and some at $8. . How many barrels were there of 
 each sort? Ans. 27 at $5 ; and 18 at $8. 
 
 12. 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? 
 
 !Cr" Let x = William's, and 3a? = John's. Then after 
 the change, #+20= William's, and 3x — 20 = John's. 
 
 Ans. John 24 ; William 8. 
 
 13. A person bought a chaise, horse, and harness, for $440. 
 The horse cost him the price of the harness, and $80 more ; 
 and the chaise cost twice the price of the horse. What did 
 he give for each ? 
 
 Ans. For the harness $50 ; horse $130 ; chaise $260 
 
§ 69.] MULTIPLICATION OF COMPOUND QUANTITIES. 47 
 
 14. Two men talking of their ages, the first says, Your age 
 is 18 years more than mine, and twice your age is equal to 
 three times mine. What is the age of each ? 
 
 Ans. Youngest, 36 years ; eldest, 54 years. 
 
 15. A boy had 41 apples which he wished to divide among 
 three companions as follows : to the second, twice as many 
 as to the first, and 3 apples more ; and to the third, three 
 times as many as to the second, and 2 apples more. How 
 many did he give to each ? 
 
 . Ans. To the first, 3 ; second, 9; third, 29^, 
 
 ^l 16. How many gallons of wine, at 9 shillings a gallon, 
 must be mixed with 20 gallons at 13 shillings, so that the 
 mixture may be worth 10 shillings a gallon ? Ans. 60 gallons. 
 
 17. 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 together save a sum equal to the 
 yearly income of either. What do they spend annually ? 
 
 Ans. A, $370 ; B, $330. 
 
 18. A farmer wishes to mix rye worth 72 cents a bushel, 
 with oats worth 45 cents a bushel, so that he may have 100 
 bushels worth 54 cents a bushel. How many bushels of each 
 sort must he take ? Ans. 33 £ of rye, and 66| of oats. 
 
 In generalization see page 136. 
 
48 ALGEBRA. [SECTION TI. 
 
 SECTION VI. 
 FRACTIONS. 
 
 § 70. All the division which the pupil has as yet performed, 
 has been the division either of numeral quantities, or of the 
 numeral co-efficients. But in Algebra, it is frequently neces- 
 sary to divide literal quantities. For example, after having 
 made x to stand for an unknown quantity, we may wish to 
 find the half of x, or the third of x, or the fourth of x, &c. 
 
 § 71. In common arithmetic, if we wish to divide 1 by 2, 
 we do it by writing 2 under the 1 ; thus, 5. So if we wish 
 to divide 2 by 3, we write 3 under the 2; thus, I. In the 
 same manner, 2 divided by 5 is written -f ; 3^-4 is written | ; 
 6-i-7 is written ^. The quantities that are obtained by di- 
 viding in t is manner, are called J ract ion*. 
 
 § 72. In Algebra, we most generally make use of this 
 method of dividing ; especially when we divide literal quan- 
 tities. Or, in other words, we divide a literal quantity by 
 writing the divisor under the dividend, with a straight line 
 between them; thus, x divided by 2, is written £; and is 
 read j-ha/f x-r-S, is written * ; and is read, x-third ; x-r-4, 
 is written J, and is read x-fourth; ^x—A = 3 ~, and is read, 
 Sx-fourth. 
 
 § 73. The two separate numbers that we employ in writing 
 a fraction, are called terms. The upper term is called the 
 numerator, and the lower term is called the den mi/iat >r. 
 Thus, in the fraction f , we call x the numerator, and 3 the 
 denominator. 
 
 § 74. If the one-third of x is f , it is evident that f of x, 
 is two times as much; that is I*. If J of a? is |, then | of 
 
 Questions. Can literal quantities be divided 1 In what manner 1 
 Which part of a fraction is the denominator 1 Which is the nume- 
 rator 1 What are they both called? 
 
§ 77.] FRACTIONS. 49 
 
 x is 3 |. Whence the rule to multiply a whole number by a 
 fraction, is, to multiply the whole number by the numera- 
 tor, and divide by the denominator; as £ of x is 4 £ ; f of y 
 is?*; | of a is 2 f 
 
 Examples. The pupil may multiply a, a;, and y, each of 
 them by f ; and then by £ ; and then by f , |, £ , f , $,.|, suc- 
 cessively. 
 
 § 75. As we can multiply a number of parts as well as a 
 number of wholes, and as the denominator is nothing more 
 than the name of the parts ; it is plain that to multiply a frac- 
 tion, we multiply the numerator, and retain the denomina- 
 tor without alteration. Thus, 2 times | is f ; 3 times |- is 
 £ ; 2 times f is f ; 4 times § is f ; &c. 
 
 Examples. Multiply each of the following fractions by 2, 
 then by 3, and then by 4. f, |, f , 4, f , f , f , £, f , §, |, * 
 
 3x 4x 
 
 T' 1' 
 
 § 76. As we know that 2 halves = a whole, we readily 
 conclude that 4 halves = 2 wholes ; and that 6 halves = 3 
 wholes, &c. Likewise, because 3 thirds = a whole, 6 thirds 
 must equal 2 wholes ; and 9 thirds must equal 3 wholes. In 
 the same manner 8 fourths = 2 wholes ; 20 fifths = four 
 wholes; 18 thirds = 6 wholes, &c. Such fractions as {, 
 2 £, ^, &c, are called improper fractions, 
 
 § 77. Hence, in order to find how many whole ones there 
 are in any number of halves, we have only to see how many 
 times two halves are contained in that number. Thus, in 10 
 halves there are as many whole ones as there are 2 halves 
 contained in 10 halves ; which is 5. In the same manner, in 
 12 thirds there are as many whole ones as there are 3 thirds 
 contained in 12 thirds ; which is 4. 
 
 Questions. How do we multiply a whole number by a fraction? 
 Give the answers to the examples. How do we multiply a fraction? 
 Examples. What are improper fractions? How can an improper 
 "raction be changed to a whole number? 
 D 5 
 
50 ALGEBRA. [SECTION VI. 
 
 § 78. Thus we have the rule, to change an improper 
 fraction to a whole number, divide the numerator by the 
 denominator. 
 
 When the answer consists of an integer and a fraction, it is 
 called a mixed number. 
 
 Examples. 1. How many whole ones in -|? 
 
 Ans. 8-~3 = 2$. 
 
 2. How many whole ones in } ? %1 ^\ J? f? 
 
 3. How many whole ones in' »? ?? g? *1 {?? 
 
 4. How many whole ar's in Jt $5? 8p? ^? 
 
 5. How many whole x's in if? ?| x ? Jp! 5|«? 
 
 6. How many whole a^'s in 3 times *?? 
 
 7. How many whole x's in 4 times \ ? 
 
 8. How many whole a*'s in 5 times J ?v^ 
 
 § 79. If we have the quantity f , we know that, as it takes 
 5 fifths to make a whole one, it will take 5 times this quan- 
 tity to make a whole x. Therefore, if we multiply * by 5, 
 we shall obtain ^, or exactly x. If we multiply ^ by 3, we 
 shall obtain ^, or, which is the same, x. If we multiply f by 
 4, we shall obtain 4 |, or x. 
 
 § 80. As 4 times f is equal to x ; then 4 times 2 | must be 
 equal to twice as much, or 2x ; and 4 times j must be three 
 times as much, or 3a?. As 3 times £ is equal to x; so 3 times 
 5? must be twice as much, or 2x. 
 
 § 81. Any fraction when multiplied by the number which 
 is the same as the denominator, will produce a quantity 
 which is the same as the numerator. Thus, 
 5 | X 4 = 5.t; ^xS=7x. 
 
 We shall be able to make use of this principle in the solu- 
 tion of many equations, if we operate in accordance with the 
 following axiom or self-evident truth. 
 
 Questions. What is a mixed number 1 ? What is obtained by mul 
 tiplying a fraction by a number that is the same as the denominator * 
 
§ 86.] FRACTIONS. 51 
 
 §32. Jf equals be multiplied by the same, their products 
 will be equal. Thus, if x = 10, then 2x = 20 ; 4x = 40, &c. 
 
 § 83. It is evident [§76] that each of the following frac- 
 tions, §, |, £, f , |, }, f , |, &c, is equal to 1. Therefore, 
 they must be equal to one another. Also, each of the follow- 
 ing fractions, f , J, f , |, ^ 2 , &c, is equal to 2 ; and conse- 
 quently they are all equal to one another. In the same man- 
 ner, we may make many fractions that will equal 3 ; and so 
 of any other number. 
 
 Fractions that are equal to one another but have different 
 terms, are called equivalent fractions. 
 
 § 84. Let us take from the first set of the above fractions, 
 § and * which are equal to one another. We see that both 
 the numerator and denominator in the last fraction are twice 
 as much as in the first. We find the same, by taking from the 
 second set, the equal fractions \ , | ; and also -| and £ ; and 
 also f and j 2 . 
 
 Again, in the equal fractions, § and f , we find each term in 
 the last fraction three times as great as the correspondent 
 term in the first fraction. The same may be observed in the 
 fractions | and -§■ ; and also in f and £ and also in f and f . 
 
 § 85. By pursuing this investigation, we shall find that 
 whenever we multiply both the numerator and the denomi- 
 nator by the same number, no matter what that number may- 
 be, the fraction made by that multiplication will be equal in 
 value to the first fraction. Hence, there is an equality between 
 the value of the following fractions, ^, f , £, f , T 5 „, T \, <fcc. ; 
 and also between the following, {, }, f, T \, -fa T \, &c. 
 
 § 86. The principle just explained, leads to another which 
 is of much importance. Suppose we multiply by 8, both 
 terms of the fraction f ; and obtain \\. It is plain that both 
 terms of the fraction *£• can be divided by 8, to bring the frac- 
 
 Queslions. How can we multiply without destroying the equa- 
 tion 1 What are equivalent fractions ? What will be the effect of 
 multiplying both terms of a fraction by any number? 
 
52 ALGEBRA. [SECTION VI. 
 
 tion back to I. So also, if both terms in | be multiplied by 
 6, the fraction will be -*£, which means just as much as ? ; 
 and, of course, if both terms in ±% be divided by 6, the frac- 
 tion will be brought back to £, which is equal to J-f • So, in 
 general, if we divide both the numerator and the denomina- 
 tor of a fraction by the same number, we have a new frac- 
 tion which will be equal to the first. Thus, T 8 ^ may be 
 changed to \ ; if to J ; ^ to f 
 
 § 87. It is evident, that of several fractions of equal value, 
 that which has the least denominator is the most easily un- 
 derstood. Thus, | of an apple is much better known at first 
 sight, than ff of an apple. And when a fraction is brought 
 to as small a denominator as it can be changed to, we say 
 it is reduced to its lowest terms. 
 
 §88. In order to reduce a fraction to its lowest terms, di- 
 vide both the numerator and the denominator by any num- 
 ber that will divide each without a remainder. Thus, in 
 the fraction 1 ^ y , both terms may be divided by 5, by which 
 we obtain £-f ; and both terms of this last fraction may be di- 
 vided by 3, by which we obtain £. 
 
 EQUATIONS. — SECTION 6. 
 
 PROBLEMS. 
 
 1. In an orchard, | of the trees bear apples, ^ of them bear 
 pears, -^ bear plums, and 81 bear cherries. How many trees 
 are there in the orchard ; and how many of each sort ? 
 
 Stating the question, x = number of trees. 
 
 f = apple trees. 
 £ = pear trees, 
 •fy = plum tree. 
 81 = cherry trees. 
 All these trees together = the whole number 
 Forming the equation, x = £-r-f-f-f^ + 81. 
 
 Questions. Suppose we divide both by any number ? How do 
 we reduce a fraction to its lowest terms? 
 
§ 88.J FRACTIONS. 53 
 
 Now, we know that if we multiply | by 4 we obtain a? 
 alone ; that is, we destroy the fraction, and make it a whole 
 number. And we know, that if we multiply the first member 
 by 4, and also the last member by 4, we shall not destroy the 
 equation. See § 82. We will therefore multiply both mem- 
 bers by 4, for the purpose of destroying the fraction in the 
 first term. It will then become 
 
 . 4a? = a?+f+-f£+324. 
 Next, we will multiply both members by 5, to destroy the 
 fraction in the second term. This will make 
 20a? = 5a? + 4a? + ^f + 1620. 
 Then we will multiply by 11, to destroy the remaining 
 'fraction, which will make 
 
 220a? = 55a? -j- 44a? + 40a? +17820. 
 Transposing and uniting, 81a? = 17820. 
 
 Dividing by 8 1 , a? = 220. the Ans. 
 
 2. In a certain school, £ of the boys learn mathematics, | 
 of them study Latin and Greek, and 6 study English gram- 
 mar. What is the whole number of scholars ? 
 
 ICT* After the question is stated, the equation will be- 
 come a: = £4-54- 6 
 Multiplying by 5, to de- j 5a? = a? 44? 4- 30 
 
 stroy the first fraction, 5 
 Multiplying by 4, to destroy, &c. 20a? = 4a? 4- 15a? 4- 120 
 Transposing and uniting, a; =120. Ans. 
 
 3. A gentleman has an estate, £ of which is woodland, | of 
 it pasture, and 105 acres embrace the pleasure grounds, gar 
 dens, and orchards. How many acres does it contain ? 
 
 ICT 3 " It is generally best to multiply by the greatest deno- 
 minator first, as that will sometimes destroy more than one 
 fraction. This is the case in this question ; for 6 times ^ is 
 if? = 4a?. Ans. 630 acres. 
 
 4. After paying away i and ^ of my money , I find $22 yet 
 in my purse. How much had I at first ? Ans. $40. 
 
 5* 
 
54 ALGEBRA. [SECTION VI. 
 
 5. A man bought a lot of ground, for which he agreed to 
 pay as follows : A of the mpney on taking possession, £ of it 
 in 6 months, and $250 at the end of the year. How much 
 did he pay in all ? . Ans. $600. 
 
 6. A post is one-fourth of its length in the mud, one-third 
 of it in the water, and 10 feet above the water. What is its 
 whole length ? Ans. 24 feet. 
 
 7. In a Christmas pudding, i is flour, ^ milk, £ eggs, \ 
 suet and fruit, and £ of a pound of spices and other ingredi- 
 ents. What is the weight of the pudding ? 
 
 \CT The equation will be # = £ -ff-f £+f +}. 
 
 y Ans. 15 pounds^ 
 
 Jf8. A lady being asked what her age was, replied, If you 
 add -j-, I, and £ of my age together, the sum will be 18. How 
 old was she ? 
 
 IC7* After the question has been stated, the equation will 
 be, £-f*-f£. = 18. Ans. 24 years. 
 
 9. What sum of money is that, whose ^ part, -£ part, and 
 i part, added together, amount to 94 dollars ? Ans. $120. 
 
 10. A person found upon beginning the study of his pro- 
 fession, that he had passed \ of his life before he commenced 
 his education, \ of it under a private teacher, the same time 
 at a public school, and four years at the university. What 
 was his age? Ans. 21 years. 
 
 11. How much money have I in my pocket, when the 
 fourth and fifth part of it together, amount to $2.25? Ans. $5. 
 
 12. The 3d part of my income, said a person, I expend ir 
 board and lodging, the 8th part of it in clothes and washing, 
 the 10th part of it in incidental expenses, and yet I save $318 
 a year. What was his yearly income ? Ans. $720. 
 
 13. A gentleman bequeaths, in his will, the half of his pro- 
 perty to his wife, one-sixth part to each of his two sons, the 
 twelfth part to his sister, and the remaining $600 to his servant. 
 What was the amount of his property? Ans. $7200 : 
 
§ 88.J FRACTIONS. 55 
 
 14. Of a piece of metal, -| plus 24 ounces is brass, and |. 
 minus 42 ounces is copper. What is the weight of the piece ? 
 
 jCT* The equation, when formed, will be 
 
 # = *-}- 24+ J— 42. Ans. 216 oz. 
 
 15. A farmer mixes a quantity of grain, so that 20 bushels 
 less than £ of it is barley, and 36 bushels more than £ of it is 
 oats. How many bushels are there in the whole ; and how 
 many of each sort ? 
 
 |C?" In stating the question; | — 20= barley, and f + 36 
 r= oats. Ans. 96 bushels in all; 28 of barley, and 68 of oats. 
 
 16. A teacher being asked how many scholars he had, re* 
 plied, If I had as many more, half as many more, and quarter 
 as many more, I should have 88. How many had he ? 
 
 %CT° In stating the question, he has x ; and as many more 
 is another x ; &c. • Ans. 32. 
 
 17. In a mixture of copper, tin, and lead; 16 pounds less 
 than \ was copper, 12 pounds less than \ was tin, and 4 pounds 
 more than \ was lead. What was the weight of the whole 
 mixture ; and also of each kind ? 
 
 Ans. 2881b.; and also 1281b., 841b., and 761b 
 
 18. What is that number whose \ part exceeds its \ part, 
 by 12? 
 
 10° To find what \ of it exceeds \ of it, subtract \ of it 
 from \ of it. The remainder is 12. Ans. 144. 
 
 19. What number is that whose \ part exceeds its \ part 
 by 72? Ans. 540. 
 
 20. A certain sum of money is to be divided amongst three 
 persons, A, B, and C, as follows : A is to receive $3000 less 
 than half of it, B $1000 less than the third of it, and C $800 
 more than the fourth of it. What is the sum to be divided; 
 and what does each receive ? 
 
 Ans. $38400; and also, $16200, $11800, $10400. 
 
 21. A man driving his geese to market, was met by an- 
 other, who said, Good morrow, master, with your hundred 
 
56 ALGEBRA. [SECTION VI. 
 
 geese. He replied, 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 ? Ans. 39 geese. 
 
 22. A shepherd being asked how many sheep he had, re- 
 plied, If I had as many more, half as many more, and 7 sheep 
 and a half, I should have just 500. How many sheep had he? 
 
 Ans. 197 sheep. 
 
 23. If the half, third, and fourth parts of my number, be 
 added together, the sum will be one more than my number. 
 Now, what is my number? Ans. 12. 
 
 24. 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 ? 
 
 Ans. A's, 15 years; B's, 39 years. 
 
 25. A young gentleman having received a fortune, spent i 
 of it the first year, i of it the second, and \ of it the third, 
 when he had $2600 left. What was his whole fortune ? 
 
 IC7" In stating the question, what was spent, = the whole 
 minus $2600. Ans. $12000. 
 
 26. A father leaves four sons, who share his property in 
 the following manner : the first takes half, minus $3000 ; the 
 second takes a third, minus $1000 ; the third takes exactly a 
 fourth; and the fourth son takes a fifth and $600. What was 
 the whole fortune, and what did each son receive ? 
 
 Ans. The whole fortune was $12000 ; and each son 
 received $3000. 
 In generalization, see page 151. 
 
§ 91.] FRACTIONS OF COMPOUND QUANTITIES. 57 
 
 SECTION VII. 
 
 FRACTIONS OF COMPOUND QUANTITIES 
 
 § 89. We have found, § 72, that the algebraical method of 
 dividing, is to write the divisor under the dividend, with a 
 straight line between them. It is plain that compound quan- 
 tities may be divided in this way, as well as simple quantities. 
 Thus, 14 +a? is divided by 3 as follows : 
 14-fa? 
 3 
 § 90. With the same reason, we find the fraction of a com- 
 pound number, by multiplying it by the numerator, and writing 
 the denominator under the product. Thus, 
 
 2a?— 10 
 
 I of a? — 5 is 
 
 :* 
 
 which is read 2a? — 10, both divided by 3. 
 
 EXAMPLES. 
 
 1. What is | of 8 -fa? ? Ans * 2±±!£. 
 
 4 
 
 2. What is f of #—27 ? 
 
 3. What is f of 3a?— 14? 
 
 4. What is 4 of 9 + 5a?? 
 
 5. What is 4- of 7a?— 19? 
 
 6. What is $. of 9a?— 27 ? 
 
 § 91. This may be changed into whole numbers by §78. 
 
 45.r— 135 
 Thus, q = 5a?— 15. 
 
 7. What is ^ of 86— 2a? ? 
 
 8. What is 3 of 14— 5a?? 
 
 9. What is ^ T of 2a?— 7 ? 
 
68 ALGEBRA. [SECTION VII. 
 
 EQUATIONS. — SECTION 7. 
 
 PROBLEMS. 
 
 1. A young gentleman being asked -his age, said, It is such 
 that if you add 8 years to it, and then divide by 3, the quo- 
 tient will be 9. How old was he ? 
 
 Stating the question, x = his age. 
 
 x+8 = with 8 added. 
 
 x+8 
 
 — -— = quotient. 
 
 M 
 
 Forming the equation, _ J_ — 9 
 
 o 
 
 Multiplying by 3, x+8 = 27 
 
 Transposing and uniting, x= 19 the Ans. 
 
 2. A man being asked what he gave for his horse, replied, 
 that if he had given $12 more, a of the sum would be 84. 
 What was the price of the horse ? 
 
 Stating tke question, X— the price. 
 
 rr-f-12 = when increased. 
 
 3 *+ 36 3 t* 
 
 — = I of the sum. 
 
 4 
 
 „ . 3x+36 m 
 
 Forming the equation, — - — = 84 
 
 Multiplying by 4, 3^ + 36 = 336 
 
 Transposing and uniting, Sx = 300 
 
 Dividing by 3, x = 100 the Ans. 
 
 3. What sum of money is that, from which $5 being sub- 
 tracted, two-thirds of the remainder shall be $40 ? Ans. $65. 
 
 4. It is required to divide a line that is 15 inches long into 
 two such parts, that one of them may be | of the other. 
 
 |C7* In stating the question, the parts are a?, and 15— a?. 
 
 Ans. 6f , and 8f. 
 
 5. It is required to find a number, such that if 15 be sub- 
 tracted from it, I of the remainder shall be 100 ? Ans. 140. 
 
§ 91.] DIVISION OF COMPOUND QUANTITIES. 59 
 
 6. Divide the number 46 into two parts, so that when one 
 is divided by 7, and the other by 3, the quotients together 
 may amount to 10. Ans. 28 and 18. 
 
 7 A person being askedthe time of day, answered that the 
 time past from noon was equal to £ of the time to midnight. 
 What was the hour ? 
 
 Id* From noon to midnight is 12 hours; the part of it 
 already past is x. Ans. 20 minutes after 5. 
 
 8. Two men talking of their horses, A says to B, My horso 
 is worth $25 more than yours ; and | of the value of your 
 horse is equal to f of the value of mine. What is the value 
 of each? lO™ The values will be x, and #-j-25. 
 
 Ans. A's, $125; B's, $100. 
 
 9. Two persons have equal sums of money. One having 
 spent $39, and the other $93, the last has but half as much as 
 the first. How much had each? Ans. $147. 
 
 10. A man being asked the value of his horse and chaise, 
 answered that the chaise was worth $50 more than the horse ; 
 and that one-half the value of the horse was equal to one-third 
 the value of the chaise. What was the value of each ? 
 
 Ans. Horse, $100; chaise, $150. 
 
 11. What number is that, to which if I add 13, and from 
 fa of the sum subtract 13, the remainder shall be 13 ? 
 
 Ans. 325. 
 
 12. A man being asked the value of his horse and saddle, 
 answered that his horse was worth $114 more than his sad- 
 dle, and that -| of the value of his horse was 7 times the value 
 of his saddle. What was the value of each ? 
 
 Ans. Saddle, $12; horse, $126. 
 
 13. A legacy of $1200 was left between A and B, in such 
 a manner, that | of A's share was equal to \ of B's. What 
 sum did each receive ? Ans. A, $640 ; B, $560. 
 
(JO ALGEBRA. [SECTION VII 
 
 14. A person rented a house on a lease of 21 years, and 
 agreed to do the repairs when f of that part of the lease which 
 had elapsed, should equal -|- of the part to come. How much 
 time will elapse before he repairs ? Ans. 12 years. 
 
 15. What number is that, to which if I add 20, and from 
 J of this sum subtract 12, the remainder shall be 10 ? 
 
 Ans. 13. 
 
 16. A person has a lease for 99 years; and being asked 
 how much of it was already expired, he answered that § of 
 the time past was equal to f of the time to come. What time 
 had already past ? -J Ans. 54 years. 
 
 17. Divide $183 between two men, so that ^ of what the 
 first receives, shall be equal to T \ of what the second receives. 
 What will be the share of each? Ans. $63, and $120. 
 
 18. Bought sheep for $300, calves for $100, and pigs for 
 $25; and then laid out £ of the rest of my money, which was 
 $15, in getting them home. How much had I at first? 
 
 |C7* 15 is one member of the equation. Ans. $443. 
 
 19. A gentleman paid four laborers $136. To the first he 
 paid three times as much as to the second, wanting $4; to the 
 third one-half as much as to the first, and $6 more ; and tc 
 the fourth four times as much as to the third, and $5 more. 
 How much did he pay to each ? 
 
 Ans. To the first, $26 ; second, $10 ; third, $19 ; 
 fourth, $81. 
 
 In generalization, see page 151. 
 
§ 95.] DIVISION OF FRACTIONS. 61 
 
 SECTION VIII. 
 
 DIVISION OF FRACTIONS, AND FRACTIONS OF 
 FRACTIONS. 
 
 § 92. It was shown in § 75, that in multiplying a fraction, 
 
 we multiply the numerator only, and retain the denominator. 
 
 On the same principle, a fraction is divided by dividing its 
 
 numerator, and retaining its denominator. 
 
 mL 6 2 8x i 2* c 
 
 Thus, --^3 = -. T ^4 = -,&c. 
 
 § 93. But, supposing we wish to divide |- by 3. In this 
 case, we cannot divide the numerator 2 by 3 without a re- 
 mainder; and therefore we must look for some other princi- 
 ple to assist us. We shall find it in § 85, where it was shown 
 that a fraction may be changed to one with different terms, 
 without altering the value. 
 
 § 94. It is evident, then, that we have only to change the 
 fraction which is to be divided, to some equivalent fraction, 
 whose numerator can be divided by the divisor without a re- 
 mainder. Thus, \ can be changed ^ r , if, £J, &c. ; each of 
 which can be divided by 3, giving for the quotient either ^ 2 T , 
 
 § 95. The most convenient equivalent fraction will be ob- 
 tained by multiplying both tefms of the fraction by the num- 
 ber which is to be the divisor. Because it is certain that after 
 the numerator has been multiplied by a number, the product 
 can in return be divided by that number. 
 
 Tl 4 20 _ ,20 _ 4 
 
 Thus,-^5 = --^5;and-^5, = _. 
 
 Questions. How do we multiply a fraction? How then must we 
 divide a fraction? How can we do if the numerator cannot be di- 
 vided without a remainder? How can we obtain the most convenient 
 equivalent fraction ? 
 
 6 
 
62 ALGEBRA. [SECTION VIII. 
 
 § 96. But, by examining the example just given, we find 
 the numerator of the answer to be the same as the numerator 
 of the first fraction ; for the first numerator has been multi- 
 plied and then divided again by the same number. The de» 
 nominator only is changed ; and that has been done by mul- 
 tiplying the first denominator by the number that was to be 
 the divisor. 
 
 § 97. Hence the rule for dividing a fraction. Divide its 
 numerator when it can be done without a remainder. But 
 if there would be a remainder; instead of dividing the nu- 
 merator, multiply the denominator by the divisor for a new 
 denominator ; and leave the numerator as it is. 
 
 2 2 
 Thus, 9-5 = -; 
 
 7 
 11 
 
 *•** 
 
 EXAMPLES. 
 
 
 
 3a: 
 1. Divide — by 5. 
 
 
 3a: 
 A,,S - 2D" 
 
 2a: 
 2. Divide — by 3. 
 
 • • 
 
 2a? 
 Ans. T . 
 
 4x 
 3. Divide y by 2. 
 
 
 2a: 
 
 Ans. •— - - 
 7 
 
 5x 
 4. Divide — by 5. 
 o 
 
 
 x 
 Ans. -. 
 
 ^. ., 12a: ; 
 5. Divide — - by 3. 
 5 
 
 
 4a: 
 Ans. — -. 
 5 
 
 6. Divide £t? by 2. 
 
 
 Ans. — 
 
 7. Divide — — - by 4. 
 5 
 
 
 x—6 
 ADS - "20- * 
 
 Questions. By this method, how is the numerator of the answer 
 obtained 1 What effect has this method upon the numerator] What 
 then is the rule for dividing a fraction? 
 
Ans. 
 
 3a?- 
 
 -47+2/ 
 
 
 64 ' 
 
 Ans. 
 
 7a?+4 
 
 5a?- 
 
 -5 
 
 
 21- 
 
 -3a? 
 
 J 98.] DIVISION OF FRACTIONS. 
 
 8 . what is J of ^Z_J±^ ? 
 
 9. What is 4 of ^4? 
 
 5 07—1 
 
 10. What is | of—-? . lta+63 - 
 
 fCT° In this example, although we can divide the first term 
 in the numerator, we do not do it because we cannot divide 
 the last term. 
 
 11. What is 1 of "±3?t Ans.^41-. 
 
 a Sx—2y 18a?— 1 2i/ 
 
 27— 6y 1 27— 6y 
 
 12. What is 1 of — ^-? Ans. , Q , Q * . . 
 
 ■ 31+a?— y 279 +9a?— 9y 
 
 «n - , <,32a?+16„ 4a? + 2 
 
 13. What is 4 of — -J, ? Ans. -f--. 
 
 7 a?— 6+y #— 6+2/ 
 
 § 98. We are now enabled to find a fraction of a fraction 
 by the rule in § 74 ; which is to multiply by the numerator, 
 and divide by the denominator. To multiply by the nume- 
 rator is to multiply the numerators together. And to divide 
 by the denominator, we have just shown, is to multiply the 
 denominators together. 
 
 EXAMPLES. 
 
 14. 
 
 What is 4 of ^? 
 3 6 
 
 Ans. 
 
 10a? 5a? 
 T8" = "9 # 
 
 15. 
 
 7a? 
 What is | of t± ? 
 8 
 
 Ans. 
 
 21a? 
 ~32~* 
 
 16. 
 
 What is 4 of ^- ? 
 
 5 7 
 
 Ans. 
 
 12a? 
 85* 
 
 17. 
 
 K r 1 
 
 What is j of ? 
 
 Ans. 
 
 10a?— 2 
 21 
 
 18. 
 
 What is | of ^4"—? 
 5 4 
 
 Ans. 
 
 213+6a? 
 20 
 
 Question. How do we find a fraction of a fraction 1 
 
64 
 
 ALGEBRA. 
 
 [[SECTION VIII. 
 
 19. What 
 
 20. What 
 
 21. What 
 
 22. What 
 
 23. What 
 
 24. What 
 
 25. What 
 
 26. What 
 
 27. What 
 
 28. What 
 
 9y 
 
 Ans. 
 
 24a? 
 
 8a? 
 
 207 + 9*/ 69 + 3y 
 
 is -^j- of 
 
 a?+61 
 
 Ans. 
 
 lSy 
 
 lla?+671 
 
 _jl f 4a?— 20 , 36a?— 180_ 18a?~ 90 
 
 18 iv ° 41 + 6a?' AnS * 410+^~205 + 30aV 
 
 2a?— lit/ 
 
 Ans. 
 
 12y+28a? 
 14a? — 77y 
 
 6a?+7y-10, 18a?+21y-30 
 
 - * of 4+3y=§S ' AnS ' 40 + 8^=20? 
 
 is|o f£^_Z_ 6 ? 
 9 6+a?-y 
 
 Ans *£±*£=?i 
 ' 54+ 9a?— 9y 
 
 . 4 3a?+5y— 8 6a?+10v— 16 
 
 is f of -5=-*. - ? Ans. •J/gL, 
 
 27— 4a? 
 . r 3a?-18, 
 "*° f 2^+32 ? 
 
 is s of JL+ to "^ t 
 T * 10-5i/+20a? 
 
 is ^ nf 16 *- 8 + 32 y* 
 
 135— 20a? 
 
 EQUATIONS. — SECTION 8. 
 
 1. A farmer wishes to mix 116 bushels of provender, con- 
 sisting of rye, barley, and oats, so that it may contain f as 
 much barley as oats, and i as much rye as barley. How 
 much of each must there be in the mixture ? 
 
 Stating the question, a? = oats ; and S? = barley. 
 
 Then, £ of ^ is ff- = rye. 
 Forming the equation, a? + 5?+£l = 116 
 
 Multiplying by 14, 14a , + 10a?+5a?== 1624 
 
 Uniting terms, 29a? =1624 
 
 Dividing by 29, a? = 56 the Ans. 
 
§98.] DIVISION OF FRACTIONS. 65 
 
 2. I paid away a fourth of my money, and then a fifth of 
 the remainder, which was $72. How much money had I at 
 first? 
 
 Stating the question, x = what I have. , 
 
 •£ = paid first. 
 x — | = the remainder. 
 ■j — fif = paid afterwards. 
 Forming the equation, f — ~-u = 72 
 Multiplying by 20, 4a?— # = 1440 
 
 Uniting terms, 3a? = 1440 
 
 Dividing by 3, x = 480 the Ans. 
 
 3. After paying away i of my money, and then \ of the 
 remainder, I had $72 left. How much had I at first ? 
 
 ICT" In stating the question, the remainder after the first 
 payment was x — f ; and } of that is | — ^. 
 
 Then, |-|-|— 2^4-72= all my money. Ans. $120. 
 
 4. A clerk spends J of his salary for his board, and § of the 
 remainder in clothes, and yet saves $150 a year. What is 
 his yearly salary ? Ans. $1350. 
 
 5. Of a detachment of soldiers, 1 are on actual duty, \ of 
 them sick, | of the remainder absent on leave, and the rest, 
 which is 380, have deserted. What was the number of men 
 in the detachment? Ans. 2280 men. 
 
 6. A young man, who had just received a fortune, spent | 
 of it the first year, and .£ of the remainder the next year;, when 
 he had $1420 left. What was his fortune ? Ans. $1 1360. 
 
 7. If from i of my height in inches, 1 2 be subtracted, £ of 
 the remainder will be 2. What is my height? Ans 5ft. 6 in. 
 
 8. A Christmas cake was mixed as follows : \ was sugar, 
 i butter and fruit, T J 2 eggs, and 3 pounds more than half of all 
 these was flour. How much did the cake weigh ? 
 
 Ans. 12 pounds. 
 
 9. A, B, and C, own together a field of 36 acres. B has \ 
 
 E 8» 
 
66 ALGEBRA. ^SECTION VIII. 
 
 more than A, and C has £ more than B. What is each man's 
 share ? ICT* $ more, signifies once and | as much. 
 
 Ans. A, 9 acres; B, 12: C, 15 
 
 10. A gentleman gave to three persons 98 dollars. The 
 second received |. of the sum given to the first; and the third, 
 A of what the second had. What did each receive ? 
 
 Ans. $56, $35, and $7. 
 
 11. A gentleman invested f of his property in a canal. 
 When he sold out, he lost f of the sum invested, receiving 
 only $1446. What was the value of his property when he 
 began? JO " His investment minus his loss, equals $1446. 
 
 Ans. 5511568. 
 
 12. A boy spent ^ of his money for fruit ; giving f of 
 what he spent for oranges, and 21 cents for lemons. How 
 much money had he ? Ans. 70 cents. 
 
 13. A grocer put 490 gallons of beer into three casks ; of 
 which the second held 1£ times as much as the first, and the 
 third held J as much as both the others. What did each 
 hold? Ans. 120, .160, and 210 gallons. 
 
 14. A gentleman leaves $315 to be divided among four ser- 
 vants in the following manner: B is to receive as much as A, 
 and I as much more ; C is to receive as much as A and B, 
 and J as much more ; D is to receive as much as the other 
 three, and \ as much more. What is the share of each ? 
 
 Ans. A, $24; B, $36 ; C, $80 ; D, $175 
 In generalization, see page 155. 
 
§ 102.] SUBTRACTION OF COMPOUND QUANTITIES. 67 
 
 SECTION IX. 
 SUBTRACTION OF COMPOUND QUANTITIES. 
 
 § 99. Suppose we wish to subtract the expression a?-f-6 
 from y. It is evident that we may first subtract x ; which 
 will give us, y — x. But we wish to subtract not only x, but 
 6 also. Well, after we have subtracted x we will subtract 6 
 also; and then the answer will be, y— x — 6. Therefore, 
 whenever we wish to subtract a compound quantity whose 
 terms are all positive, we write them after the other quantity 
 with -\- changed to — . 
 
 § 100. Again, suppose we wish to subtract the expression 
 x — 6 from y ; in which the number to be subtracted has — 
 instead of +• As before, we will first subtract x, by which 
 we obtain y — x. But the quantity to be subtracted was 6 less 
 than x; and we have therefore subtracted 6 too much. We 
 will therefore add 6 to our last answer for the true remainder, 
 which will give us y — x + 6. Here, we have changed the 
 positive x to — x, and the —6, to -f 6. 
 
 RULE FOR SUBTRACTING COMPOUND QUANTITIES. 
 
 § 101. Change all the signs of the expression which is 
 to be subtracted, the sign -\- to — , and the sign — to 4- ; 
 and then write the term* offer the other quantity. It is to 
 be recollected that in the quantity from which we subtract, 
 the signs are not altered. 
 
 § 102. Although the subtraction is performed the moment 
 the quantities are written according to the above rule, yet, 
 after that operation, it is always expedient to unite the term* 
 if possible. 
 
 Questions. How do we subtract a compound quantity whose terms 
 are all positive? Explain why. How do we subtract a negative 
 quantity'? Explain why. What is the Rule for subtracting compound 
 quantities? What should be done after the quantity is subtracted * 
 
68 ALGEBRA. [SECTION IX. 
 
 EXAiMPLES. 
 
 1. Subtract 7x-\-G+y from Gy — 17. 
 
 Ans. Changing the signs of the quantity we subtract, we 
 have, 6y— 17— 7a? — 6 — y; which =. 5y— 23 — 7 a?. 
 
 2. Subtract 4?/ + 3a?— 10 from 74— x. 
 
 Ans. 74 — x — 4y — 3a?-|-10; which =84 — 4a? — 4y. 
 
 3. Subtract 6— a?— 3y from 7.r-f 6i/. 
 
 Ans. 1x-\-6y — 6-\-x-\-3y, which equals 8a? + 9y — 6. 
 
 4. From 4a?— 3i/+27, subtract 6t/— 12 + x. 
 
 Ans. 3a?— 9y + 39. 
 
 5. From 6-f-a? — y, subtract 13— 9y — x. Ans. 2x + 8y—7 
 
 6. From 8a'— 2 + 3y, subtract 3y-\- 4— 3a?. Ans. 11a?— 6. 
 
 7. From 5-f-4a?, subtract 2 — 5a?-f-4y — z. 
 
 Ans. 9a?— 4y+z + 3 
 
 8. From 6a? — 8y, subtract — x— y + oQ. Ans. 7a? — 7y— 50. 
 
 2a? 2a? 
 
 9. From 14 — , subtract 3a?— 12. Ans. 26— 3a?— — • 
 
 3 3 
 
 Ji _ 8+7a? . . 3a? 8 + 7a? . 3a? 
 
 10. From — -— , subtract 5 -+--—. Ans. — 5 -• 
 
 3 5 3 5 
 
 ,, « m *!l L lOy— 4 A *- 2y lOy— 4 
 
 11. From 7 £, subtract — *- — . Ans. 7 — J £■ 
 
 3 o o b 
 
 §103. It has been shown, §68 and §69, that any com- 
 pound quantity may be , considered and operated upon as a 
 simple quantity, by merely drawing a vinculum above it, or 
 enclosing it in a parenthesis. Whenever that compound 
 quantity is a fraction, the line between the numerator and de- 
 nominator serves a> a vinculum. Thus, in the eleventh ex- 
 ample, above, — &■- — is subtracted as a simple quantity, and 
 6 
 
 Questions. What effect has ^ vinculum upon a compound quan- 
 tity 1 W T hat if the oomponnd quantity is a fraction? If a fraction 
 is to be subtracted, what sign is to be changed 1 
 
§ 104.] SUBTRACTION OF COMPOUND QUANTITIES. 69 
 
 is read, minus the fraction, <fcc. Therefore the sign of each 
 term in it is not changed ; but — is put before the whole frac- 
 tion. 
 
 ,o t> 4x ~ 6 u, 3a?+7 . 4x—6 3a?+7 
 
 12. From — - — , subtract — - — . Ans. 
 
 4 ' 5 4 5 
 
 ._ „ 3a?+8 . 51— a? 3a?+8 51— a? 
 
 13. From — - — , subtract — - — . Ans. — — . 
 
 2 3 2 3 
 
 .. ' 7— 2x . 21a? — 4 
 
 14. from — - — , subtract -— x. i 
 
 3 10 , 7 — 2x 21a?— 4 , 
 
 Ans. — — ~ W -+X. 
 
 15. From x +4— 2 a?, subtract 3a?— 6. 
 
 16. From 4a? — 6y, subtract — x-\-y— 7. 
 
 17. From 8+2a?, subtract — t/-+-3a?. 
 
 , Q n 3a?— 7 . 2— 6a? 
 
 18. From — -— , subtract — -— . 
 
 2 3 
 
 7# g 
 
 19. From 8a?, subtract 3a? . 
 
 5 
 
 20. From a?, subtract -~^- x. 
 
 3 
 
 2a?— *v 
 
 21. From 2a?— y, subtract — ;r-^+y» 
 
 3-4- a? 
 
 22. From 3+a?, subtract — !— +a?. 
 
 UNITING FRACTIONS OF DIFFERENT DENOMINATORS. 
 
 § 104. By looking at the answers to the last three exam- 
 ples in § 102, and also the next three in § 103, it will appear 
 that we ought to have some rule for uniting their terms. We 
 can easily find one by applying the principle explained in 
 § 85. For we have only to change each of the fractions to 
 an equivalent fraction, so that they will all have one with an- 
 other the same common denominator. 
 
70 ALGEBRA. [SECTION IX. 
 
 Thus, the answer to the twelfth example under § 103, is 
 
 4x-6 3a?+7 , „ ■ 
 
 — — . JNow each ot these two fractions maybe 
 
 changed to 20ths, by multiplying the first by 5, and the last 
 
 fpu ... , '. 20a;—* 30 12ar+28 , . . 
 by 4. They will then become ■■ — I which = 
 
 (20a:— 30) — (12a: -J- 28) _ 20ar— 30 — 12a; — 28 __ 8x — 58 
 90 20 20~~ 
 
 4x — 29 
 
 7 io - "* 
 
 § 105. Thus we have the rule for uniting fractions of dif- 
 ferent denominators. Multiply all the denominators toge- 
 ther for a new denominator; and each numerator by all the 
 denominators except its own, for new numerators : remem- 
 bering that if a compound numerator follows minus — , all 
 the signs in it must be changed the m ment one denomina- 
 tor is used for the whole quantity ; that is, when the short 
 vinculums are destroyed for the purpose of making the longer 
 one. 
 
 examples. 
 
 1. Unite the terms in the answer to the 13th sum in § 103. 
 Operation. 
 /3a- -f 8 51— a^ _ 9.r+24_ 102— 2x\ _ 
 V2 3~~ '™V 6 (5 / 
 
 9a--f24 — 102-f2a? _ lla;— 78 
 ~~ 6 6 * 
 
 Note. — In this operation the first minus has reference to 
 
 the whole quantity / — — j ; the second minus to the whole 
 
 . \ no 2a?\ 
 
 quantity ( ). In this last quantity, 102 has no sign 
 
 before itself and is therefore positive. Now, when the line 
 
 Questions. How can fractional terms be united 1 Give the rule. 
 What if integers are multiplied with the fractions 1 
 
§106.] DIFFERENT DENOMINATORS. 71 
 
 between the numerator and denominator is carried through 
 the ic'ole quantity, the vinculum of 102—2.7; is destroyed; 
 and then is the time for changing the signs for subtracting. 
 
 § 106. Whenever there are integers to be united with frac- 
 tions, they may be changed to fractions, by putting the number 
 
 6 x 
 
 1 under them for the denominator. Thus, 6=—; # = -. 
 
 2. Unite the terms in the quantity 7 - — - . 
 
 5 6 
 
 Operation. — The quantity is — — ~ — . 
 
 15 6 
 
 7 210 
 
 Both terms of - multiplied by 5 and by 6, equals — — . 
 
 Both terms of f- multiplied by 6, equals ^. 
 
 Both terms of ^ — multiplied by 5, equals =^~ — . 
 
 Therefore, 
 
 /7 2y lOy— 4\ _ ,2\0 _12y ^50y — 20\ 
 \l~~5~ 6 / \~30T "30 30 ~) 
 
 210— 12y— 50yf20 _ 230-r62y 115— 31y 
 30 30 15 
 
 « it • u -i 8-J-7.T " Sx 
 
 3. Unite the terms in the quantity, 5-- . 
 
 3 5 
 
 40 + 35^— 75— 9x 26.r— 35 
 AnS * IS" 15— 
 
 i 2iP 
 
 4. Unite the terms in the quantity, 26 — 3#-f-— -. 
 
 o 
 
 ^ m §x-\-2x n „ 7x 
 
 Ans. 26 ~ — = 26 -. 
 
 o o 
 
 2x 21 x 4 
 
 5. Unite the terms in the quantity, 7 — — ln -fa:. 
 
 Ans. 
 
72 ALGEBRA. [SECTION IX 
 
 6 + 2x 4x — 3 
 
 6. Unite the terms in the quantity, 12 — — — 
 
 J& Ans. *"=£?. 
 
 8 l0 
 
 x x x 
 
 7. Unite the terms in the quantity, -+- +-- 
 
 13# . # 
 
 Ans . -g— •+ ff 
 
 8. Unite the terms in the quantity, 4+ 3 — * — ~g* 
 ^J* The fractions may be united } Ang j^* 
 
 by themselves. S 
 
 3* 4z 
 
 9. Unite the terms in the quantity, ±z ——+z — jr. 
 
 43z 
 
 Ans. 5Z rr-« 
 
 35 
 
 Sy— 4 
 
 10. Unite the terms in the quantity, y - h2y— 
 
 3 15 15 
 
 6x+7 r 3*— 4 
 
 11. Subtract — - — from — — . 
 
 4 / 
 
 5x— 6 r S—x 
 
 12. Subtract — — - from — - — 
 
 x x 
 
 13. Subtract 3—- from 7-. 
 
 2 x 
 
 14. Subtract Sx -— from 5a?. 
 
 4 
 
 15. Subtract -? — - from - -J-. 
 
 16. Subtract £^+VS£! from 4*-f. 
 
§106.] SUBTRACTION OF COMPOUND QUANTITIES. 73 
 
 EQUATIONS. — SECTION 9. 
 
 1. There are two numbers, whose sum is 140; and if 4 
 times the less be subtracted from 3 times the greater, the re- 
 mainder will be 70. What are the numbers ? 
 
 Stating the question, x — the greater. 
 
 140 — x== the less. 
 
 Sx = 3 times the greater. 
 560— 4a: = 4 times the less. 
 Forming the equation, 3a?— 560 + 4a? = 70 
 Transposing, uniting, and dividing, x mm 90 
 
 Ans. Greater number, 90 ; less, 50. 
 
 2. A person, after spending $100 more than a third of his 
 yearly income, found that the remainder was $150 more thai 
 half of it. What was his income? Ans. $1500 
 
 3. Two men, A and B, commenced trade. A had twice a* 
 much money as B ; he has since gained $50, and B has losf 
 $90 ; and now the difference between A's and B's money, is 
 equal to three times what B has. How much had each when 
 they commenced trade? Ans. A, $410; B, $205. 
 
 4. A man bought a horse and chaise for $341. If $ of the 
 price of the horse be subtracted from twice the price of the 
 chaise, the remainder will be the same as if f of the price of 
 the chaise be subtracted from three times the price of the 
 horse. What was the price of each ? 
 
 IC If the price of the chaise be a?, and the price of the horse 
 
 be 341— a:; then the first remainder will be 2a? = . 
 
 But when the fraction is destroyed, the vinculum is taken 
 away, and therefore the last sign must be changed from — 
 to -j-. Ans. Chaise, $189; horse, $152. 
 
 5. A gentleman bought a watch and chain for $160. If | 
 of the price of the watch be subtracted from six times the price 
 of the chain, the remainder will be the same as if T 5 f of the 
 
 7 
 
74 ALGEBRA. ^SECTION IX. 
 
 price df the chain were subtracted from twice the price of the 
 watch. What was the price of each ? 
 
 Ans. Watch, $112; chain, $48. 
 
 6. Divide the number 204 into two such parts, that if T of 
 the less were subtracted from the greater, the remainder will 
 be equal to ^ of the greater subtracted from four times the 
 less. Ans. Greater, 154; less, 50. 
 
 7. Two travellers, A and B, found a purse of money. A 
 first takes out $2 and * of what remains ; and then B takes 
 $3 and j- of what remains ; and it is found that each has the 
 same sum. How much money was in the purse ? Ans. $20. 
 
 8. A shepherd was met by a band of robbers, who plun- 
 dered him of half of his flock and half a sheep over. After- 
 wards a second party met him, and took half of what he had 
 left, and half a sheep over; and soon after this, a third party 
 met him, and treated him in the like manner; and then he had 
 5 sheep left. How many had he at first? Ans. 47 sheep. 
 
 9. A gentleman hired a laborer for 20 days, on condition 
 that for every day he worked he should receive 14 shillings; 
 but for every day he was idle he should forfeit 6 shillings. 
 At the end of the 20 days he received 160 shillings. How 
 many days did he work, and how many days was he idle ? 
 
 Ans. He worked 14 days, and was idle 6. 
 
 10. Divide the number 48 into two such parts, that the ex- 
 cess of one of them above 20, shall be three times as much as 
 the other wants of 20. 
 
 f£J* The excess of a number above 20 is obtained by sub- 
 tracting 20 from it. Ans. 32 and 16. 
 
 11. A person in play lost a fourth of his money, and then 
 won back 3*. ; after which he lost a third of what he now 
 had, and then won back 2s. ; lastly, he lost a seventh of 
 what he then had, and then found he had but 12s. remaining. 
 What had he at first? Ans. 20s. 
 
 In generalization, see page 158. 
 
§ 109.] RATIO AND PROPORTION. 75 
 
 SECTION X. 
 RATIO AND PROPORTION. 
 
 § 107. When an unknown quantity is not, either by itself, 
 or in some connexion with others, known to be equal to some 
 known quantity or set of quantities ; we may sometimes find 
 that there is a comparison between it and some known quan- 
 tity, which is the same as the comparison between two known 
 quantities. 
 
 Thus, suppose I buy 27 yards of cloth for $72, and wish 
 to sell for $16 so much of it as cost me $16. In this case 
 the number of yards to be sold is not equal to any other 
 quantity that is mentioned. But we suppose that it must 
 compare with the number of yards bought, in the same man- 
 ner that $16 compares with $72. By knowing this compari- 
 son, we can find the number of yards ; because, as $16 is | 
 of $72, so the number of yards to be sold must be | of the 
 number of yards bought. It is 6 yards. 
 
 § 108. It will be seen that the comparison in this example 
 consists in observing how many times one of the numbers is 
 contained in the other. 72 is contained in 16, two-ninths of 
 a time. When a comparison of this kind is made, the result 
 that is obtained is called their ratio. Thus, in comparing 
 the numbers 3 and 4, we find that 4 is contained in 3, three- 
 fourths of a time ; and therefore we say the ratio of 3 to 4 is -J. 
 
 § 109. The pupil must remember that the ratio of one 
 number to another, always signifies how the first number 
 compares with the last. Thus, the ratio of 8 to 5, is } ; that is, 
 8 is | of 5. Hence the ratio is expressed by making the first 
 term to be a numerator, and the last to be the denominator. 
 
 Questions. Do we ever make use of comparison in algebra? Ex- 
 plain how. In what does the comparison consist] What is that 
 kind of comparison called in mathematical language ? How can a 
 ratio be expressed ? Why 1 
 
76 ALGEBRA. [SECTION X. 
 
 § 1 10. In the example just furnished relative to the cloth; 
 the ratio of the money paid, to the money obtained for a part 
 of the cloth; (that is, the ratio of $72 to $i6,) is T *«*§. And 
 so also the ratio of the cloth bought, to the cloth sold; (that is, 
 the ratio of 27 yards to 6 yards,) is 2 ? which equals 2 . Here 
 we see, that although the ratios are differently expressed, they 
 are, notwithstanding, equal to one another. 
 
 §111. When the ratio of two quantities is equal to the 
 ratio of other two quantities, there is said to be a proportion 
 between them; that is, an equality f ratios is called a pro- 
 portion. 
 
 § 1 12. Our chief business with ratios at present, is to learn 
 when they form a proportion ; that is, when they are equal to 
 one another. Now, as they may be expressed in the form of 
 a fraction; it is evident, that when they are brought to a com- 
 mon denominator, if the fractions are equal their numerators 
 will be the same, and if they are not equal their numerators 
 will not be the same. 
 
 For example, is 1 1 to 21 = 33 to 63 ? We pursue our in- 
 quiry as follows : 11 to 21 is the same as ^j, and 33 to 63 is 
 the same as j4« We bring the fractions to a common deno- 
 minator by § 105. 
 
 11 ,33 693 „ 693 
 
 — and — = and . 
 
 21 63 1323 1323 
 
 We find they are equal, and the four terms 11 to 21 =33 to 
 63 are proportional. 
 
 § 113. Although ratios are sometimes expressed fraction- 
 ally, they are generally expressed as follows: 11 : 21 and 
 33 : 63 ; that is, 11 divided by 21, 33 divided by 63. The 
 pupil will see that we employ the same sign that expresses 
 division, with the exception of the — between the two dots 
 
 Question. ■ Can a ratio be equal to another, and yet be differently 
 expressed ? Give an example. What do we call an equality of 
 ratios] How may we determine whether ratios are equal? Give 
 an example. Are ratios always expressed by fractions ? How else 1 
 
§117.J RATIO AND PROPORTION. 77 
 
 The sign : is read is to, and the foregoing examples are 
 read 11 is to 21 and 33 is to 63. 
 
 §114. When four quantities are proportional, they are 
 written thus, 11 : 21 : : 33 : 63. The sign :: is read as; 
 and the whole expression is read, 11 is to 21 as 33 is to 63. 
 
 §115. In a proportion, the first and the last terms are 
 called extremes, and the two middle terms are called means. 
 In the above proportion, 11 and 63 are the extremes, and 21 
 and 33 are the means. 
 
 § 116. In order to derive any important use from a pro- 
 portion, we wish the pupil to recollect the method employed 
 to find whether four quantities are proportional. We multi- 
 plied (see §112; ihe first numerator by the last denominator, 
 to find one new numerator. These were the two extremes. 
 We also multiplied the last numerator by the first denomina- 
 tor, to find the ther new numerator. These were the two 
 means. And hence we learn, that if four quantities are pro- 
 portional, the product of the two extremes is equal to the 
 product of the means. 
 
 RULE. 
 
 § 117. «/? proportion may be reduced to an equation by 
 multiplying the extremes together for one member ; and 
 multiplying the means t gether for the other member. 
 Thus, 2:7:: 8 : x, becomes in an equation 2x = 56 ; 
 whence x = 28. 
 
 Or, the fourth term may befund by multiplying the two 
 means t> gether, and dividing their product by the first 
 extreme. 
 
 Questions. How is a proportion written ? Which terms are the 
 extremes ? Which terms are the means ? In hrinjjintr the n.ti s to 
 a common denominator, what did we do with the extremes] Y\ hat 
 did we do with the two means? What principle does this show 1 
 What then can we do with a proportion ? 
 
 7* 
 
78 ALGEBRA. [SECTION X. 
 
 EQUATIONS.— SECTION 10. 
 
 1. If you divide $75 between two men in the proportion of 
 3 to 2, what will each man receive ? 
 
 Stating the question, x = the share of one. 
 
 75 — x = the share of the other. 
 Making the proportion. x : 75 — x : : 3 : 2 
 
 2x = 225— 3a? 
 
 Multiplying ext. and means 
 to reduce to an equation, 
 Transposing and uniting, 5x = 225 
 
 Dividing, x = 45 
 
 Ans. $45; and $30. 
 
 2. Divide $150 into two parts, so that the smaller may 
 be to the greater as 7 to 8. Ans. 70 ; and 80. 
 
 3. Divide $1235 between A and B, so that A's share may 
 be to B's as 3 to 2. Ans. A's share $741 ; B's $494. 
 
 4. Two persons buy a ship for $8640. Now, the sum 
 paid by A is to that paid by B, as 9 to 7. What sum did 
 each contribute ? Ans. A paid $4860 ; B $3780. 
 
 5. A prize of $2000 was divided between two persons, 
 whose shares were in proportion as 7 to 9. What was the 
 share of each? Ans. $875; and $1125. 
 
 6. A gentleman is now 30 years old, and his youngest 
 brother 20. In how many years will their ages be as 5 to 4? 
 
 iO " After stating the question, the proportion will be 
 30+a?:20+#:: 5:4. Ans. 20 years. 
 
 7. What numbei is that, which, when added to 24, and 
 also to 36, will produce sums that will be to each other as 7 
 to 9? Ans. 18. 
 
 8. Two men commenced trade together. The first put in 
 
§ 118.] RATIO AND PROPORTION. 79 
 
 $40 more than the second ; and the stock of tne first was to 
 that of the second as 5 to 4. What was the stock of each ? 
 
 Ans. $200; and' $160. 
 
 9. A gentleman hired a servant for $100 a year, together 
 with a suit of clothes which he was to have immediately. At 
 the end of 8 months, the servant went away, and received 
 $60 and kept the suit of clothes. What was the value of the 
 suit of clothes ? Ans. $20. 
 
 10. A ship and a boat are descending a river at the same 
 time ; and when the ship is opposite a certain fort, the boat 
 is 13 miles ahead. The ship is sailing at the rate of 5 miles, 
 while the boat is going 3. At what distance below the fort 
 will they be together? 
 
 The ship sails x miles from the fort ; the boat will 
 sail 13 miles less. Ans. 32£ miles. 
 
 §118. It is very often the case that a problem is easily 
 solved by using simply the ratio, instead of a proportion. 
 
 Operation by Ratio. 
 
 %C3* In these questions the pupil must not use any pro- 
 portions. 
 
 11. Divide 40 apples between two boys in the proportion 
 of 3 to 2. 
 
 Stating the question, x = the share of one. 
 
 Now, as the ratio of the first to the second is | ; then the 
 ratio of the second to the first is J. Therefore, 
 
 Org « 
 
 — = the share of the second. 
 o 
 
 2<£ 
 
 Forming the equation, a?-f-— = 40 
 
 o 
 
 Multiplying by 3, 3#-f 2x = 120 
 
 Uniting terms, x = 24 
 
 Ans. 24, and 16. 
 
80 ALGEBRA. [SECTION X. 
 
 12. Three men trading in company, gain $780. As often 
 as A put in $2, B put in $3, and C put in $5. What part of 
 the gain must each of them receive ? 
 
 Stating the question, x = A's share. 
 
 — = B's share. 
 2 
 
 — = C's share. 
 z 
 
 Sx 5x 
 
 Forming the equation, x-\- — -+- -7- = 780. 
 
 Ans. A, $156; B, $234 ; C, $390. 
 
 13. Two butchers bought a calf for 40 shillings, of which 
 the part paid by A, was to the part paid by B, as 3 to 5. 
 What sum did each pay ? Ans. A paid 15*. ; B, 25s. 
 
 14. Divide 560 into two such parts, that one part may be 
 to the other as 5 to 2. Ans. 400, and 160. 
 
 15. A field of 864 acres is to be divided among three farmers, 
 A, B, and C ; so that A's part shall be to B's as 5 to 1 1, and 
 C may receive as much as A and B together. How much 
 must each receive? Ans. A, 135; B, 297; C, 432 acres. 
 
 16. Three men trading in company, put in money in the 
 following proportion ; the first 3 dollars as often as the second 
 7, and the third 5. They gain $960. What is each man's 
 share of the gain ? Ans. $192; $448; $320. 
 
 17. Find two numbers in the proportion of 2 to 1, so that 
 if 4 be added to each, the two sums will be in proportion of 
 3 to 2. 
 
 IC7* The last expression means that the greatest is | of the 
 smallest ; or the smallest is -| of the greatest. Ans. 8 and 4. 
 
 18. Two numbers are to each other as 2 to 3 ; but if 50 
 be subtracted from each, one will be one-half of the other. 
 What are the numbers? Ans. 100 and 150. 
 
 19. A sum of money is to be divided between two persons, 
 A and B ; so that as often as A takes $9, B takes $4. Now 
 
§ 118.] RATIO AND PROPORTION. 81 
 
 it happens that A receives $15 more than B. What is the 
 share of each ? Ans. A, $27 ; B, $12. 
 
 20. There are two numbers in proportion of 3 to 4 ; but if 
 24 be added to each of them, the two sums will be in the pro- 
 portion of 4 to 5. What are the numbers ? Ans. 72 and 96. 
 
 21. A man's age when he was married was to that of his 
 wife as 3 to 2 ; and when they had lived together 4 years, his 
 age was to hers as 7 to 5. What were their ages when they 
 were married? Ans. His age, 24; hers, 16 years. 
 
 22. A certain man found when he married, that his age was 
 to that of his wife as 7 to 5. If they had been married 8 years, 
 sooner, his age would have been to hers as 3 to 2. What 
 were their ages at the time of their marriage ? 
 
 Ans. His age, 56 years ; hers, 40. 
 
 23. A man's age, when he was married, was to that of his 
 wife as 6 to 5 ; and after they had been married 8 years, her 
 age was to his as 7 to 8. What were their ages when they 
 were married ? Ans. Man, 24 ; wife, 20 years. 
 
 24. A bankrupt leaves $8400 to be divided among four 
 creditors, A, B, C, and D, in proportion to their claims. 
 Now, A's claim is to B's as 2 to 3 ; B's claim to C's as 4 to 
 5 ; and C's claim to D's as 6 to 7. How much must each 
 creditor receive ? 
 
 Ans. A, $1280; B, $1920 ; C, $2400; D, $2800. 
 
 25. A sum of money was divided between two persons, A 
 and B, so that the share of A was to that of B as 5 to 3. Now, 
 A's share exceeded f of the whole sum by $50. What was 
 the share of each ? Ans. $450, and $270. 
 
 In generalization, see page 154 
 
82 ALGEBRA. [SECTION XI 
 
 SECTION XI. 
 
 EQUATIONS WITH TWO UNKNOWN 
 QUANTITIES. 
 
 § 119. It frequently happens, that several unknown quan- 
 tities are introduced into a problem. But when this is the 
 case, if the conditions will give rise to as many equations, 
 independent of each other, as there are unknown quantities, 
 there is no difficulty in finding the value of each quantity. 
 
 § 120, An equation is said to be independent of another 
 when it cannot, either by multiplication or by division, be 
 changed into that other, Thus, Ix — i/=47, is independent of 
 the equation 10?/-f 4x = 50 ; because one of them cannot be so 
 altered as to make the other. But, Ix — 1/=47, is not inde- 
 pendent of the equation 2\x — 3i/=141 ; because the last is 
 made by multiplying the first by 3. 
 
 § 121. At present we will attend to those equations only 
 that include two unknown quantities, each represented by a 
 different letter from the other. 
 
 § 122. In equations that contain two unknown quantities, 
 our first object must be to find the value of one of them ; and 
 in order to do this, the preliminary step is to derive from the 
 equations that are given, another equation which shall have 
 but one unknown quantity. This operation is called elimi- 
 nating, or exterminating the other unknown quantities. 
 
 § 123. There are three different methods of forming one 
 equation with one unknown quantity from two equations 
 containing two unknown quantities. With each of these, the 
 learner should become familiar; as it is sometimes convenient 
 to use one of them, and sometimes another. 
 
 Questions. When is one equation said to be independent of an- 
 other 1 ? Explain. When there are several unknown quantities, how 
 many independent equations are necessary to solve the question? 
 What is it to eliminate or exterminate an unknown quantity] 
 
$ 125. J EQUATIONS WITH TWO UNKNOWN QUANTITIES. 83 
 
 FIRST METHOD OF EXTERMINATION. 
 
 § 124. It is necessary here to recollect what" was stated in 
 $ 50 and § 53, that when equals are added to equals, their 
 sums will be equal ; and also, when equals are st'btracted 
 from equals, the remainders are equal. 
 
 Thus, suppose we have the equation a?-j-14 = 36, 
 and suppose also that we know that y = 8 ; 
 
 then if we will add the two first members together, and also 
 the two last, the members will still be equal to one another, as 
 follows: a?+ 14+2/= 36 -f8. 
 
 And also if we subtract y from the first member, and 8 from 
 the second, the members will still be equal to one another ; 
 thus, #-f 14— ?/ = 36— 8. 
 
 § 125. This principle can be easily applied for the exter- 
 mination of unknown quantities. For, if in both of two equa- 
 tions, one of the unknown quantities has the same co-efficient, 
 but after different signs; it is evident that if we add both equa- 
 tions together, viz. the first member to the first member, and 
 the last member to the last member; and then unite temia, 
 we shall cancel the two quantities that are alike with different 
 signs; a new equation will be formed, in which that unknot* 
 quantity will disappear. 
 
 EXAMPLES. 
 
 1. Given the two C 3#-f-2i/=26 > to find the values of 
 equations, £ 5x — 2y =38 } x and y. 
 
 Adding together the two right hand members, and also th# 
 two left ; we have the equation 
 
 3.r + 2y -f- 5x— 2y = 26 -f 38 
 Uniting terms and canceling 2y, Sx = 64 
 Dividing, x = S 
 
 Questions. What is the effect of adding two equations together, 
 (the first member to the first member, and the last member to the 
 last member) 1 In what case can an unknown quantity be extermi- 
 nated by adding two equations together? 
 
84 ALGEBRA. [SECTION XI 
 
 Now, if x = 8 ; then in the first equation, how much will 
 3x equal ? 
 
 The first equation will then become 24-J-2y = 26, 
 from which we may find the value y = l. 
 
 _. . . C7a?-r-4u = 58 > to find the* values 
 
 2. Given the equations, 1 9x _ 4 * = 38 $ of x and y . 
 
 Ans. a? = 6. 
 Question. Why do you add the equations ? 
 If a? = 6, what does 7x in the first equation equal ? Then 
 what does y equal ? Ans. y = 4. 
 
 _ ~. , C 5a?-l-6v = 58 ? to find the values 
 
 3. Given the equations, J ^ 6 * = 34 $ of * and y, 
 
 fCT" In this example, it is plain that we cannot destroy the 
 y's by adding them together. But we have before seen, § 62, 
 that if all the signs are changed, the equation will not be af- 
 fected. Let us then change the signs of the second equation, 
 so that the y's may have different signs. The two equations 
 
 will then become j 2a?— 61/ ==—34 \ wn * cn ' 
 
 when added together, become 5x+6y— 2a?— 6y = 58— 34 
 Uniting terms, - 3a? = 24 
 
 Therefore, x = 8. 
 
 Question. What operation is performed by changing signs ? 
 
 § 126. In the last example, if we take the equations before 
 
 the alteration of the second, thus, 2„ Ta —q^c an( ^ sud " 
 
 tract the second from the first, the -result will be the same as 
 it was by changing the signs and adding. As follows : 
 5x+6y—2x—6y = 58—34. 
 
 Whence we learn that, if in both equations one of the un- 
 known quantities has the same co-efficient and also the same 
 sign ; and we subtract one equation from the other, (viz. the 
 first member from the first member, and the second member 
 
 Questions. In what case can an unknown quantity be extermi- 
 nated by subtracting :>ne equation from another? Explain the reason. 
 
5a?+6t/=64 
 
 2x +6?/== 58 
 __ _ 
 
 § 127.] EQUATIONS WITH TWO UNKNOWN QUANTITIES. 85 
 
 from the second member,) and unite the terms; we shall form 
 a new equation in which that unknown quantity will disappear. 
 
 § 1 2 7. It is evident that we may suppose the signs changed ; 
 and so unite the terms immediately, without actually writing 
 the whole work. This must be done hereafter. 
 
 . ~. A , C6# + 7w = 79? to find the values 
 
 4. Given the equations, £ te ^, B1 ^ of x and y . 
 
 Subtracting second from ) . 2ft 
 
 first, and unite. 3 
 
 Ans. a? = 5; y = 7. 
 
 5. Given J 2x+§l58 J t0 find * and tf 
 ICT 3 " In this example, the y's are alike in 
 
 both equations ; and are therefore called iden- 
 tical terms. As they have the same sign, in 
 order to cancel them we subtract the second 
 from the first, and unite. Ans. x = 2 ; y = 9. 
 
 6 - Given lltZZlll] "> find ^ and y. 
 Which are the identical terms in this example ? 
 ICT* To cancel the #'s, subtract the upper from the lower. 
 
 Ans. x = 40; i/==ll. 
 
 _, ~; C 12#-f 8y=92 } 
 
 7. Given | V2x Z 2 ly = 63 $ t0 find * and ^ 
 
 Ans. x = 7; y = 1. 
 
 8. Given J 3^g=^ J to find X and y. 
 
 ICT 3 * As the signs of the identical terms are 2a?-{-2i/=18 
 unlike, to cancel them we add the second to 33? — 2y= 7 
 the first, and unite. 5a? =25 
 
 Ans. x = 5; y = 4. 
 
 9. Given J _^+:g= -12 \ t0 find * and * 
 
 Ans. a? = 4; i/== 2. 
 
 Questions. Is it necessary to write out the whole work ? What 
 are identical terms ? 
 
 8 
 
86 ALGEBRA. [SECTION XI 
 
 10. Given 5 To?" %a £ to find # and ty. 
 
 I x-\-'Zy = 14 ^ J 
 
 |C7^ In this example, neither of the unknown quantities 
 has the same co-efficient in both equations. But both mem- 
 bers of the last equation can be multiplied by 3, without de- 
 stroying the equality, § 82 ; and then the co-efficients of the 
 
 a?'s will be alike in both equations. Thus, J _ Ifi^— 4-2 C 
 
 Ans. x = 10; y = 2. 
 
 11. Given ^ 3^^^ l 8g \ to find x and y. 
 
 |C7" Multiply the second by 2. Ans. x = 8 ; y — 16. 
 
 12. Given j 3™ _ ^— 2 C to ** nd x and &' 
 
 IC7* Make t/'s identical. Ans. x = 3 ; y = 7. 
 
 13. Given ^ 4^ + 3y = 22 5 t0 find x and ^ 
 
 |C7* Multiply the first by 2 and the ;r's will be identical. 
 
 Ans. x = 4 ; 1/ =b 2 
 
 14. Given ^+5zls2 £ t0 find * and *< 
 
 Ans. rr = 7; 2- = 8 
 
 15. Given \f x ^ y Z^\ to find z and y. 
 
 Ans. a? = 4; y = 5. 
 
 16. Given S^lg^i}? to find x and y. 
 
 Ans. a: = 8 ; y = 5. 
 
 17. Given J 4 £ + * " ^ £ to 6nd y and * 
 
 Ans. y = 24; 2- = 6. 
 
 18 Given \ £ +^Z^ to find » and y. 
 
 Ans. a? = 2 ; y = 1. 
 
 19. Given J §+£ = »»£ to find y and *. 
 
 |C7* In this example, we cannot obtain identical terms by one 
 
$ 128.] EQUATIONS WITH TWO UNKNOWN QUANTITIES. 87 
 
 multiplication. But we may apply the same principle, §105, 
 that is used for finding a common denominator. For, if we 
 multiply the co-efficient of the first equation by the co-efficient 
 of the second, the product will be the same as if we multiply 
 the co-efficient of the second by the co-efficient of the first. 
 Thus, 
 
 Multiplying the first by 4, 20?/ + I2x — 372 
 
 Multiplying the second by 3, 9y + 12a? = 240 
 
 Subtracting the second from ~) 
 
 i c i r liv == i«* 
 
 the first, and unite, 3 u 
 
 Ans. y = 12; x = 11. 
 20. Given j J^~a ~ 7 ( to find y and z. 
 
 Multiplying the first by 5, ■ 20y— 25z = 10 
 
 Multiplying the second by 4, 20?/— 162: = 28 
 
 Subtracting the first from the £ 
 second, and unite, 3 
 
 Ans. ?/ = 3 ; z = 2. 
 
 § 128. From the foregoing, we derive the following: 
 
 Rule I. to exterminate an unknown quantity. 
 
 First, Transpose, so as to bring both of the unknown 
 quantities to the left ; x's under a?'s ; y's under ?/'s, &c. 
 
 Determine which of the unknown quantities you will ex- 
 terminate; and then, if it is necessary, multiply or divide 
 one or both of the equations so as to make the term which 
 contains that unknown quantify to be the same in both. 
 
 Then if the identical terms have like signs in both equa- 
 tions, subtract one equation from the other; but if they 
 have unlike signs, add one equation to the other. And the 
 result will be an equation containing only one unknown 
 quantity. 
 
 Questions. What is the first operation for exterminating an un- 
 known quantity ] Repeat the whole rule. What are identical terms'? 
 Why do we add them when the signs are unlike 1 Why do we sub- 
 ract them when the signs are alike ? 
 
ALGEBRA. [SECTION XI. 
 
 EQUATIONS.— SECTION 11. 
 
 1. What two numbers are those whose sum is 20 and dif 
 ference 12? 
 
 Stating the question, x = greater number. 
 
 y =■ the less. 
 
 Then forming the equations, x +y = 20 
 
 x—y = 12 
 
 As the signs of the identical terms ) 
 
 ?.. .... . >2x =32.-.a?=16 
 
 are unlike, ado the equations, 3 
 
 Substituting 16 for x in the first, 16-f y = 20 
 
 Transposing and uniting, y = 4. 
 
 Ans. 16 and 4. 
 
 2. A market woman sells to. one person, 3 quinces and 4 
 melons for 25 cents; and to another, 4 quinces and 2 melons, 
 at the same rate, for 20 cents. How much are the quinces 
 and melons apiece ? 
 
 After the statement, forming the 3 3a?-j-4?/ = 25 
 equations, 3 4a?-f2i/ = 20 
 
 Multiplying the second by 2, 8#-f 4y = 40 
 
 Subtracting first from third, 5x =15 
 
 Ans. Quinces, 3 cents apiece ; melons, 4. 
 In our solutions after this, we shall number the lines, so 
 
 that any reference to them will be easily understood. 
 
 3. A man bought 3 bushels of wheat and 5 bushels of rye 
 for 38 shillings ; and at another time, 6 bushels of wheat and 
 3 bushels of rye for 48 shillings. What was the price for a 
 bushel of each ? 
 
 Let x = price of wheat, and y = price of rye. 
 
 1. By the first condition, 3x + by = 38 
 
 2. By the second, 6x + Sy = 48 
 
 3. Multiply the 1st by 2, 6x+\0y =76 
 
 4. Subtracting the 2d from the 3d," - 7y = 28.-.y — 4 
 
 5. Substituting 4 for y in the 1st, 3x +20 = 38. 
 
 Ans. Wheat for 6s. ; rye for 4* 
 
§ 128.] EQUATIONS WITH TWO UNKNOWN QUANTITIES. 89 
 
 4. Two purses together contain $400. If you take $40 
 out of the first and put them into the second, then there is 
 the same in each. How many dollars does each contain ? 
 
 Let x = the number in the first. 
 y = the number in the second. 
 gCT 3 * Although in these examples we have omitted the state- 
 ment, it is expected the pupil will state them as usual. 
 
 1. By the first condition, x + y = 400 
 
 2. By the second, x — 40=y+40 
 
 3. Transposing the 2d, x — y = 80 
 
 4. Adding the 1st to the 3d, ~2x =480 .-. x = 240 
 
 Ans. The first, $240; the second, $160. 
 
 5. A gentleman being asked the age of his two sons, re- 
 plied, that if to the sum of their ages 25 be added, this sum 
 will be double the age of the eldest ; but if 8 be taken from 
 the difference of their ages, the remainder will be the age of 
 the youngest. What is the age of each ? 
 
 Let x = the age of the eldest, y = the age of the youngest. 
 
 1. By the first condition, x+y-\-25 = 2x 
 
 2. By the second, x — y — 8=y 
 
 3. Transposing and uniting 1st, — x-j-y = — 25 
 
 4. Transposing and uniting 2d, x — 2y= 8 
 
 5. Adding the 3d and 4th, — y—— 17 
 
 6. Substituting 17 for y in the 3d, — a?-f 17 = —25 
 
 7. Transposing, — x = — 42 
 
 Ans. Eldest, 42 ; youngest, 17. 
 
 6. A gentleman paid for 6 pair of boots and 4 pair of shoes 
 $44 ; and afterwards for 3 pair of boots and 7 pair of shoes, 
 $32. What was the price of each per pair ? 
 
 Ans. Boots, $6 ; shoes, $2. 
 
 7. A man spends 30 cents for apples and pears, buying his 
 apples at the rate of 4 for a cent, and his 4 pears at the rate of 
 5 for a cent. He afterwards let his friend have half of his 
 apples and one-third of his pears for 13 cents, at the same 
 rate. How many did he buy of each sort ? 
 
 8* 
 
90 
 
 ALGEBRA. 
 
 [SECTION XI. 
 
 Let x = number of apples. 
 y = number of pears. 
 
 - cent = price of 1 apple. 
 
 - cent = price of 1 pear. 
 
 x 
 
 - cents = price of all the apples 
 
 .'/ 
 
 1. By the first condition, 
 
 2. By the second, 
 
 3. Dividing the 1st by 3, 
 
 4. Subtracting 3d from 2d, 
 
 5. Multiplying by 24, 
 
 - cents = price of all the pears. 
 
 x y 
 4+1=30 
 
 8+15- 13 
 x v 
 
 12+T5= 10 
 
 X X 
 
 8~l2 = 3 
 3x— 2x=72.-.a:=:72. 
 Ans. 72 apples ; 60 pears. 
 
 8. One day a gentleman employs 4 men and 8 boys to 
 labor for him, and pays them 40s. ; the next day he hires at 
 the same rate, 7 men and 6 boys, for 50s. What are the daily 
 wages of each ? Ans. Man's, 5s. ; boy's, 2s. 6c?. 
 
 9. It is required to find two numbers with the following 
 properties : J- of the first with -£ of the second shall make 16, 
 and ± of the first with 1 of the second shall make 9. 
 
 $C7* Performed as Problem 7. Ans. 12 and 30 
 
 10. Says A to B, Give me 5s. of your money, and I shall 
 have twice as much as you will have left. Says B to A, 
 Give me 5s. of your money, and I shall have three times as 
 much as you will have left. What had each ? 
 
 ^7» In the equations, first transpose so that x shall be 
 under x, and y under y Ans. A, lis.; B, 13s 
 
§ 128.] EQUATIONS WITH TWO UNKNOWN QUANTITIES. 91 
 
 11. Two men agree to buy a house for $1200. Says A to 
 B, Give me § of your money, and I shall be able to pay for it 
 all ; No, says B, give me | of yours, and then I can pay for 
 it. How much money had each ? Ans. A, $800 ; B, $600. 
 
 12. Find two numbers with the following properties: The 
 products of the first by 2, and the second by 5, when added, 
 are equal to 31 ; also, the products of the first by 7, and the 
 second by 4, when added, are equal to 68. Ans. 8 and 3. 
 
 13. A paid B 20 guineas, and then B had twice as much 
 money as A had left ; but if B had paid A 20 guineas, A would 
 have had three times as much as B had left. What sum did 
 each possess at first? Ans. A, 52 guineas ; B, 44. 
 
 14. A person has a saddle worth £50, and two horses. 
 When he saddles the poorest horse, the horse and saddle are 
 worth twice as much as the best horse ; but when he saddles 
 the best, he with the saddle is worth three times the poorest. 
 What is the value of each horse ? 
 
 Ans. Best, £40 ; poorest, £30. 
 
 15. A merchant sold a yard of broadcloth and 3 yards of 
 velvet for $25 ; and, at another time, 4 yards of broadcloth 
 and 5 yards of velvet for $65. What was the price of each 
 per yard? Ans. Broadcloth, $10 ; velvet, $5. 
 
 16. A person has 500 coins, consisting of eagles and dimes ; 
 and their value amounts to $1931. How many has he of each 
 coin ? JC7 3 * The solution must be in cents. 
 
 Ans. 190 eagles; 310 dimes. 
 
 17. In the year 1299, three fat oxen and six sheep together 
 cost 79 shillings ; and the price of an ox exceeded the price 
 of 12 sheep by 10 shillings. What was the value of each ? 
 
 Ans. An ox, 24s.; a sheep, Is. 2d, 
 
 . 18. Two persons talking of their ages, A says to B, 8 year* 
 ago I was three times as old as you were ; and 4 years hence, 
 I shall be only twice as old as you will be. What are their 
 present ages ? Ans. A, 44 ; B, 20 years. 
 
92 ALGEBRA. [SECTION XI. 
 
 19. A farmer sold to one man 30 bushels of wheat and 40 
 of barley for 270 shillings ; and to another, 50 bushels of 
 wheat and 30 of barley for 340 shillings. What was the price 
 per bushel of each ? Ans. Wheat, 5s. ; barley, 3s. 
 
 20. A man and his wife and child dine together at an inn. 
 The landlord charged 15 cents for the child, and for the woman 
 he charged as much as for the child and -} as much as for the 
 man ; but for the man he charged as much as for the woman 
 and child together. What did he charge for each ? 
 
 Ans. 45 cents for the man ; and 30 cents for the woman. 
 
 21. A gentleman has two horses, and also a chaise worth 
 $250. If the first horse be harnessed, he and the chaise will 
 be worth twice as much as the second horse ; but if the second 
 be harnessed, he and the chaise will be worth three times as 
 much as the first horse. What is the value of each horse ? 
 
 Ans. First, $150; second, $200. 
 
 22. A is in debt $1200, and B owes $2500; but neither 
 has enough to pay his debts. A says to B, Lend me the i 
 of your fortune, and then I can pay my debts. But B an- 
 swered, Lend me the £ of your fortune, and I can pay my 
 debts. What was the fortune of each ? 
 
 Ans. A, $900; B, $2400. 
 
 23. A wine merchant has two kinds of wine, one at 5s. a 
 gallon, and the other at 12s. ; of which he wishes to make a 
 mixture of 20 gallons that shall be worth 8s. a gallon. How 
 many gallons of each sort must he use 1 
 
 Ans. 8f gallons of that at 12s. ; 1 1-f of that at 5s. 
 
§ 131.] SECOND METHOD OF EXTERMINATION. 93 
 
 SECTION XII. 
 SECOND METHOD OF EXTERMINATION. 
 
 § 129. In each of the preceding questions, we first found 
 the value of one of the unknown quantities ; and then substi- 
 tuted that value for that unknown quantity in one of the equa 
 tions, in order to find the value of the other unknown quantity. 
 This mode of operating furnishes a hint that leads us to an- 
 other method of extermination. 
 
 Let us take the first question in the last section, [p. 88,] 
 
 in which we have the equations, \ ~i 9 £ 
 
 The last part of our operation was to substitute the value 
 of x for x itself, in one of the equations. It is evident that 
 we could make this substitution just as well if the value of x 
 was a literal quantity, instead of 16. Thus, supposing x to 
 
 y 
 
 be equal to £ ; then substituting it for x, the first equation 
 
 would be -^+i/=20. 
 
 I 
 
 § 130. Let us therefore transpose the first equation to find 
 what x will equal, just as if we knew the value of y. We 
 shall find that x = 20— y. And then in the second equation, 
 we shall use the value of x instead of x itself. 
 
 Thus, 20— y— y=12. 
 Transposing and uniting, — 2y = — 8 .*. y = 4 ; 
 
 which was our answer by the first method. Then x will be 
 found by substituting 4 for y. Whence we derive 
 
 Rule II. to exterminate an unknown quantity. 
 
 § 13 1. Select the most simple term of the unknown quan- 
 tities, and by the equation that contains it, find the value 
 of that unknown quantity, as if the other were known; and 
 
94 ALGEBRA. [SECTION XII. 
 
 then, in the other equation, substitute this value for the un- 
 known quantity itself. 
 
 We shall then have an equation with only one unknown 
 quantity ; which may be solved as usual. 
 
 EQUATIONS.— SECTION 12. 
 
 1. There are two numbers whose sum is 100 ; and three 
 times the less taken from twice the greater, leaves 150 re- 
 mainder. What are those numbers ? 
 
 Let x = greater. 
 y = less. 
 2x — 3y = the required subtraction. 
 Forming the f 1. By the first condition, x+y=l00 
 equations, c 2. By the second, 2x — 3*/= 150 
 
 3. Transposing the 1st, a? =100— y 
 
 4. Multiplying the 3d by 2, 2a: = 200— 2y 
 6. Substituting 200-2y for 2* > 800 _ 8 3 160 
 
 in the 2d, 5 J n 
 
 6. Transposing and uniting, — 5t/= — 50.\i/=10 
 
 7. Substituting 10 for y in the 1st, x -j- 10 = 100 
 
 8. Transposing and uniting, x = 90 
 
 Ans. Greater, 90; less, 10. 
 
 2. The ages of a father and his son amounted to 140 years; 
 and the age of the father was to the age of the son as 3 to 2. 
 What were their ages ? 
 
 Let x = age of the father. 
 y = age of the son. 
 
 1. By the first condition, x+y-=. 140 
 
 2x 
 
 2. By the second, y = — 
 
 o 
 
 2x 2x 
 
 3. Substituting — for y in the 1st, a?+-— = 140 
 
 3 3 
 
 4. Multiplying by 3, 3#+2.r=420 
 
 5. Uniting and dividing, x = 84 
 & Substituting 84 for x in 1st, 84 -f-y = 140 
 
 Ans. Father, 84 years ; son, 56 
 
§ 131.] SECOND METHOD OF EXTERMINATION 95 
 
 3. Find two numbers, such that -£ of the first and A of the 
 second shall be 87 ; and \ of the first and \ of the second 
 shall be 55. Ans. 135, and- 168. 
 
 4. A says to B, Give me 100 of your dollars, and I shall 
 have as much as you. B replies, Give me 100 of your dol- 
 lars, and I shall have twice as much as you. How many 
 dollars has each? Ans. A, $500; B, $700. 
 
 5. There are two numbers, such that | of the first and f 
 of the second added together, will make 12; and if the first 
 be divided by 2, and the second multiplied by 3, f of the sum 
 of these results will be 26. Ans. 15, and 10£. 
 
 6. Find two numbers in the proportion of 2 to 1, so that if 
 4 be added to each, their two sums shall be in proportion of 
 3 to 2. Ans. 8, and 4. 
 
 7. A and B owned 9800 acres of western land. A sells J 
 of his, and B sells \ of his ; and they then have just as much 
 as each other. How many acres had each 1 
 
 Ans. A, 4800; B, 5000. 
 
 8. A son asking his father how old he was, received the 
 following reply : My age, says the father, 7 years ago, was 
 four times as great as yours at that time ; but 7 years hence, 
 if you and I live, my age will be only double of yours. 
 What was the age of each ? 
 
 Ans. Father's, 35 years; son's, 14 years. 
 
 8. The weight of the head of Goliath's spear was less by 
 one pound than ■}- the weight of his coat of mail ; and both 
 together weighed 17 pounds less than ten times the spear's 
 head. What was the weight of each ? 
 
 Ans. Coat, 208 pounds ; spear's head, 25 pounds. 
 
 10. A market woman bought eggs, some at the rate of 2 for 
 a cent, and some at the rate of 3 for 2 cents, to the amount of 
 65 cents. She afterwards sold them all for 120 cents, thereby 
 gaining half a cent on each egg. How many of each kind did 
 she buy ? Ans. 50 of the first kind ; 60 of the other kind 
 
96 ALGEBRA [SECTION XII. 
 
 11. Says A to B, | of the difference of our money is equal 
 to yours ; and if you give me $2, I shall have five times as 
 much as you. How much has each ? Ans. A, $48 ; B, $12. 
 
 12. A and B possess together property to the amount of 
 $5700. If A's property were worth three times as much as 
 it is, and B's five times as much as it is, then they both would 
 be worth $23,500. What is the worth of each ? 
 
 Ans. A, $2500 ; B, $3200. 
 
 13. A gentleman has two silver cups, and a cover adapted 
 to each which is worth $20. 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 ? 
 
 Ans. First cup, $12; second, $16. 
 
 14. Two men driving their sheep to market, A says to B, 
 Give me one of your sheep, and I shall have as many as you. 
 B says to A, Give me one of your sheep, and I shall have 
 twice as many as you. How many had each ? 
 
 Ans. A, 5 sheep ; B, 7. 
 
 15. What two numbers are those, whose difference is 4, 
 and 5 times the greater is to 6 times the less, as 5 to 4 ? 
 
 Ans. 8 and 12. 
 
 16. There are two numbers such that £ of the greater added 
 to | of the less, will equal 13 ; and if i of the less be taken 
 from i of the greater, the remainder is nothing. What are 
 the numbers ? Ans. 18 and 12. 
 
§ 133.] THIRD METHOD OF EXTERMINATION. 97 
 
 SECTION XIII. 
 
 THIRD METHOD OF EXTERMINATION. 
 
 § 132. The method of substitution as explained in the last 
 section, may be modified a little. We will show how, by 
 using question 1st, in the last section of equations. 
 
 lhe two equations were < „ __q 1*0 
 
 We transpose the 1st; thus, x = 100 — y. 
 
 Now, before we substitute the value of x for x itself in the 
 second equation, we will transpose the second equation so as 
 to make x stand alone ; thus, 2x — 1 50 + 3y. 
 
 Then substitute the value of x as found before by the first 
 equation, 200— 2y = 150-j-3t/ with which we may 
 
 proceed as before. 
 
 § 133. Before we make the substitution after transposing, 
 it is generally best to find the value of x alone in the second 
 equation. Thus, 
 
 Given \ C—2y = 10 \ t0 find X and y- 
 
 Transposing and dividing the 1st, Xwm ~ ' ■-&- 
 
 2 
 
 Transposing and dividing the 2d, x = — -— - • 
 
 Now, as it is evident that things which are equal to the 
 same, are equal to one another ; one value of x is equal to 
 the other value of x ; thus, 
 
 23— 3y__ IO+2.7 
 2 " ~ 5^" 
 Destroying the fractions, 115 — 15^ = 20 + 42/ 
 
 Transposing, uniting, and dividing, y = 5 
 
 By substituting the value of y in one of the equations, we 
 find x = 4. Whence we derive 
 G 9 
 
08 ALGEBRA. [SECTION XIII. 
 
 Rule III. to exterminate an unknown quantity. 
 § 134. Find by each of the equations, the value of that 
 unknown quantity wlrich is the least involved; and then 
 form a new equatin by making one of these values equal 
 to the other. 
 
 EQUATIONS. — SECTION 13. 
 
 1. Divide $60 between A and B, so that the difference be- 
 tween A's share and 31, may be to the difference between 31 
 and B's share, as 6 to 7. 
 
 Let x = A's share; and?/=B's. 
 
 1. By the firs^ condition, x+y — 60 
 
 2. By the second, x— 31 : 31— y ::.6 : 7 
 
 3. Multiplying extremes and means, 7x — 217 m 186— 6y 
 
 4. Transposing the 1st, x = 60— y 
 
 5. Transposing and uniting the 3d, 7# = 403— fiy 
 
 6. Multiplying the 4th, 7a: = 420— ly 
 
 7. Equating 5th and C.th, 403— 63/ = 420— ly 
 
 8. Transposing and uniting, y = 17 
 
 0. Substituting 17 in the 4th, x = 60—17 = 43 
 
 Ans. A's share, $43; B's, $17. 
 
 2. There is a fraction, such that if 1 is added to the nume- 
 rator, its value will be -£ ; but if 1 be added to the denomina- 
 tor, its value will be \. What is that fraction ? 
 
 Let x = numerator ; and y = denominator. 
 
 x 
 The fraction will be, 
 
 y 
 
 x+\ 1 
 
 1. By the first condition, = — 
 
 y 3 
 x 1 
 
 2. Bv the second, — rr—T 
 
 y + l 4 
 
 3. Multiplying the 1st by y % and by 3, 3# + 3 —y 
 
 4. Multiplying the 2d by y + l, and by 4, ±x = y + \ 
 
 5. Transposing the 4th, 4x — 1 =y 
 
 6. Equating 3d and 5th, 3a? + 3 = 4.r— 1 
 
 7. Transposing and uniting, — x = — 4 
 
 8. Substituting the value of 3x in the 4th, 15 =y 
 
 Ans. £. 
 
§ 134.] THIRD METHOD OF EXTERMINATION. 99 
 
 3. There is a certain number, consisting of two places of 
 figures, which is equal to 4 times the sum of its digits ; and 
 if 18 be added to it, the digits will be inverted. What is thaJ 
 number ? 
 
 Let x = first digit or tens; and y = the units 
 10a? -fi/ = the number. 
 4x-\-4y = four times the sum of digits. 
 1037+1/+ 18, = when 18 is added. 
 
 \0y-\-x,= when the digits are inverted 
 
 1. By the first condition, \0x+y — 4x-\- 4y 
 
 2. By the second, l0x-\-y+ 18 = lOy+x 
 
 3. Transposing and uniting the 1st, 6a? = dy 
 
 4. Transposing and uniting the 2d, 9x = 9y — 18 
 
 5. Multiplying the 4th by f , 6a? = 6y— 12 
 
 6. Equating 3d and 4th, Sy = 6y— 12 
 
 Ans. 24. 
 
 4. There is a certain number consisting of two figures ; and 
 if 2 be added to the sum of its digits, the amount will be three 
 times the first digit; and if 18 be added to the number, the 
 digits will be inverted. What is the number? Ans. 46. 
 
 5. A person has two snuff-boxes and $8. If he puts the 8 
 dollars into the first, then it is half as valuable as the other. 
 But if he puts the 8 dollars into the second, then the second 
 is worth three times as much as the first. What is the value 
 of each ? Ans. First, $24 ; second, $64. 
 
 6. A gentleman has two horses and a chaise. The first 
 horse is worth $180. If the first horse be harnessed to the 
 chaise, they will together be worth twice as much as the se- 
 cond horse ; but if the second horse be harnessed, the horse 
 and chaise will be worth twice and one-half the value of the 
 first. What is the value of the second horse, and of the 
 chaise ? Ans. Horse, $210; chaise, $240. 
 
 7. There is a certain number consisting of two digits. The 
 sum of these digits is 5 ; and if 9 be added to the number itself, 
 the digits will be inverted. What is the number? Ans. 23. 
 
100 ALGEBRA. [SECTION XIII. 
 
 8. There is a number consisting of two figures. If the 
 number be divided by the sum of the figures, the quotient will 
 be 4 ; but if the number made by inverting the figures be di- 
 vided by 1 more than their sum, the quotient will be 6. What 
 is the number? 
 
 |C?* In the operation, 4 multiplied by x+y, is the same as 
 4 times x+y. Ans. 24. 
 
 9. There are two numbers such that the less is to the 
 greater as 2 to 5 ; and the product made by multiplying the 
 two numbers together, is equal to ten times their sum. What 
 are the numbers ? 
 
 Let x = the less ; and y = the greater. 
 
 2y 
 
 1. By the first condition, x —* 
 
 o 
 
 Note. — If we wish to multiply y by 4, we put 4 immedi- 
 ately before the y as a co-efficient; and in the same way, if 
 we multiply y by x, we make x the co-efficient of y. 
 
 2. By the second, xy=lf)x+l0y 
 
 3. Multiplying the 1st by 10, I0x = 4y 
 
 4. Transposing the 2d, I0x = xy — lOy 
 
 5. Equaling 4th and 3d, xy—l0y = 4y 
 
 Note. — When we divide 4x by 4, we do it by taking away 
 4 when we divide 10a; by 10, we do it by taking away the 
 U Tn the same manner we divide yx by y, in taking away 
 they 
 
 6. Divid.i a by y, x— 10 = 4 .-. 37=14. 
 
 7. Substituui.e 14 for x in the 3d, 140 = 4y .-. y = 35. 
 
 Ans. 14 and 35. 
 
 10. There are *wo numbers, whose sum is the | part of 
 their product; and ti.e greater is to the less as 3 to 2. What 
 are those numbers? Ans. 15 and 10 
 
§135.] EQUATIONS WITH SEVERAL UNKNOWN' QUANTITIES. 101 
 
 SECTION XIV. 
 
 EQUATIONS WITH SEVERAL UNKNOWN 
 QUANTITIES. 
 
 § 135. When there are three or more unknown quantities; 
 first, transpose all the unknown quantities to the left, and write 
 them so that letters of the same kind shall he under each other. 
 Then, combine successively one of the equations with each of 
 the others, so as to exterminate 'the same unknown quantity 
 from each. By this means there will be obtained a number 
 of equations one less than the original number. With these 
 perform the same process as before ; and proceed in this 
 manner till there is but one equation containing only one un- 
 known quantity; which may be solved by the usual rule. 
 Then by substitution, the value of the other unknown quan- 
 tities may be found, in the reverse order in which they were 
 exterminated. . 
 
 EXAMPLES. 
 
 in- a ♦• I x tJ f t* = ?a 1 to find *« y» 
 
 1. Given the equations < x-f-2?/ + 3z = 16 \- J 
 
 jx — y—2 z = —3j an / 
 
 4. Subtracting 1st from 2d, y-\-2z = 7 
 
 5. Subtracting 3d from 1st, 2y + 3z = 12 
 
 6. Multiplying 4th by 2, 2i/ + 4z = 14 
 
 7. Subtracting 5th from 6th, z = 2 
 
 8. Substituting value of z in 4th, y = 3 
 
 9. Substituting in the 1st, # = 4 
 
 Questions. In solving equations with several unknown quantities, 
 what must be done first] Then which unknown quantity must be 
 exterminated * 
 
 9* 
 
/02 ALGEBRA. [SECTION XIV. 
 
 rx + y + z = 29 \ 
 }ar+2y + 3* = 62/ 
 
 x + y + 2 = 29 
 8. Given }*+2y+3z = 62 £ to find x, 
 
 y, and 2. 
 
 4. Subtracting 1st from 2d, 1/4-22 = 33 
 
 5. Destroying fractions in 3d, Qx + 4y + 3z = 120 
 
 6. Multiplying 1st by 6, 6x4-61/4-62 = 174 
 
 7. Subtracting 5th from 6th, 2?/ + 3z = 54~ 
 
 8. Multiplying 4th by 2, 2y +'4z = 66 
 
 9. Subtracting 7th from 8th, z = 12 
 Whence by substitution, y = 9; and x = 8. 
 
 3. Given x+y+ ar — 7j 2x — y— 32 = 3; and 5x—3y 
 
 4-52=19; to find x, y, z. Ans. x = 4, i/ = 2, 2 = 1. 
 
 4. Given a?— y— 2 = 5; 3x + 4i/4-52 = 52 ; and 5x—4y 
 — 3z = 32 ; to find x, y, and 2. Ans. x = 10, y = 3, z = 2. 
 
 5. Given 7x-f-5y+2z = 79 ; 8x4-71/4-92=122; and 
 a?+4y -j-52 = 55 ; to find the values of x, 1/, and 2. 
 
 Ans. x = 4, i/ = 9, 2 = 3. 
 
 6. Given a: 4-y 4-2 =13; x+y-f u = 17; x+2-f-tt = 18; 
 and y+z+u = 21 ; to find the values of x, y, 2, and w. 
 
 Ans. x = 2, y=5, 2 = 6, m=10. 
 
PART II. 
 
 LITERAL ALGEBRA. 
 
 GENERAL PRINCIPLES. 
 
 § 136. The algebraical operations which we have hitherto 
 treated of, belong to that part of the science which was known 
 to the ancients, and which was in use till about A. D. 1600. 
 About that time, Franciscus Vieta, a Frenchman, introduced 
 the general use of letters into Algebra, (denoting the known 
 quantities in a problem by consonants, and the unknown ones 
 by vowels.) 
 
 § 137. This improvement gave anew aspect to the science. 
 So that now algebra is rather the representation of arithmeti- 
 cal results, than the results themselves. And therefore,' its 
 most general object is to afford means for investigating the 
 laws of calculation for every description of numerical questions. 
 
 § 138. Operations with numbers cannot furnish general 
 rules, for two reasons. In the first place, we cannot, by 
 mere inspection of the results, determine how they were ob- 
 tained. Thus, 12 may be the result, either of multiplying 3 
 by 4, or adding 5 to 7, or subtracting 8 from 20, or dividing 
 48 by 4, &c. &c. 
 
 And in the second place, every figure in an arithmetical 
 result has a determinate value which is peculiar to itself; and 
 therefore cannot be applied to any other question. 
 
 Questions. How does modern algebra differ from the ancient 
 methods 1 Who introduced the modern method 1 What is the 
 modern use of algebra] What are the two reasons why we cannot 
 obtain general rules by arithmetical operations 1 
 
 103 
 
104 ALGEBRA. 
 
 § 139. But in algebra, the letters which represent thfl 
 quantities, retain their identity throughout the whole reason- 
 ing ; and as the operations! of addition, subtraction, multioli- 
 cation, &c, are only represented by signs, we readily see the 
 depend- nee which the several quantities have upon one an- 
 othe r And as the result is represented by letters, each of 
 w' .ch may stand for whatever number we choose, its value 
 .8 entirely indeterminate; and shows merely what operations 
 it is necessary to perform upon the numbers when a particular 
 value may be assigned to them. 
 
 § 140. Hence, in Literal Algebra, the result does not de- 
 pend upon the particular values of the quantities which we 
 operate with, but rather upon the nature of /he question; 
 and it will always be the same for every question of the same 
 kind. The result is therefore a general rule. 
 
 § 141. The principal signs that are used in algebra are the 
 following. 
 
 The sign -f {plus) represents addition. Thus, a-f b de- 
 notes that b is added to a. a-{-b is the sum of two numbers. 
 
 The sign — (minus) denotes that the quantity following it 
 is subtracted. Thus, a—b is the remainder obtained by sub- 
 tracting b from a. 
 
 The sign x (multiplied by) denotes that the quantity be- 
 fore it, is multiplied by the quantity that follows it. Thus, 
 axb is the product of a and b. 
 
 The sign . is sometimes put between two literal quantities 
 instead of X ; as a,b. But more generally, in algebra, the 
 letters are joined together, to represent their product. Thus, 
 ab is the product of two numbers. 
 
 The sign -r- or : (divided by) represent the division of the 
 
 Questions. How does algebra differ in these particulars? In lite- 
 ral algebra, what is the use of the answers? Upon what does an 
 algebraical result depend ? In what cases are the results the same ? 
 Define the signs -j-, — , X, •, -4-. What is the sum of two num- 
 bers, say « and b ? What is their difference ? What is their pro- 
 iuct ? What is their quotient ? 
 
$141.] GENERAL PRINCIPLES. 105 
 
 quantity before it, by the quantity which follows it. Thus, 
 a—b is a divided by b. 
 
 Division is more generally denoted in algebra by writing 
 the divisor under the dividend, in the form of a fraction. 
 
 Thus, 7- is the quotient of a divided by b. 
 
 The sign = (equals) denotes that the whole quantity on 
 the left of it, is equal to the whole quantity on the right of it. 
 Thus, 4 + 8 = 16—4. 
 
 The sign ± or =p {plus or minus and minus or plus) 
 shows that either by addition or by subtraction, the effect will 
 be the same. Thus, there are circumstances when x = ± a ; 
 that is, x is equal to plus a or minus a. 
 
 The sign ^ or </d (the difference of) shows that it is not 
 known whether the quantity before it is subtracted from the 
 quantity after it, or the latter is subtracted from the former. 
 Thus, a \n b is the difference between a and 6, without speci- 
 fying which is the greatest. 
 
 The sign > or < (greater than or less than) denotes that 
 the quantity towards which it opens is greater than the other. 
 Thus, in a > 6, a is greater than b. 
 
 The sign (a vinculum) denotes that all which is 
 
 put under it, is to be used as one term. Thus, a-j-6 — ex 3, 
 signifies that the whole quantity under the vinculum, is to be 
 multiplied by 3. 
 
 The line which separates the terms of a fraction is also a 
 
 vinculum. Thus, — signifies that the whole quantity 
 
 above the line is to be divided by the whole quantity under it. 
 The ( ) parenthesis is frequently used instead of the vin 
 culum ; thus, (a+b). (c-j-e?) signifies the product of (a-f-6) 
 multiplied by (c + d). 
 
 Questions. Define the sign =, ±» ^f, <*> , > <> » 
 
 ( ). Is the line above a quantity the only vinculum! What is tho 
 sign of multiplication between vinculums 1 
 
106 ALGEBRA. 
 
 The sign oo {infinity) denotes a quantity that is infinitely 
 large, or a quantity so great that it may be considered larger 
 than any supposable quantity. 
 
 The cipher is sometimes used to represent a quantity 
 that is less than any quantity that may be mentioned. This 
 is always the case when the cipher is used as a denominator 
 of a fraction. 
 
 The radical sign </ denotes the root of the following quan- 
 tity. When unaccompanied by a figure, it represents the 
 square root. But when a figure is put over it, that figure ex- 
 presses the root that is designed. Thus, $/a denotes the 3d 
 root of a. The number over the radical sign is called the 
 index of the root. 
 
 A co-efficient is a number put immediately before a letter, 
 as, 2b ; and denotes how many times the quantity is to be 
 taken. 
 
 In such a case, the co-efficient is called a numeral co-effi-' 
 cient. Sometimes when one letter has been multiplied into 
 another, the first written one is called a literal co-efficient; 
 as, in ab, a is the literal co-efficient. 
 
 An exponent or index is a small figure placed a little over 
 and a little to the right of a quantity; as, a 2 . It denotes that 
 the quantity is multiplied by itself; and shows how many 
 times the quantity is to be taken as a factor. 
 
 An exponent may be either positive, negative, or fractional; 
 as, a 2 , a -2 , a*. This will be explained in another place. 
 
 The sign : : represents proportion, and : denotes the ratio 
 of two numbers. 
 
 The sign oc denotes a general proportion. 
 
 A term is any quantity that is not separated into parts by 
 either of the signs. 
 
 Questions. Define the signs oo, 0, >/, <§/ > What is the figure 
 over the radical sign called ? What is a co-efficient? Is it numeral 
 or literal? What is an exponent or index? What is the sign 
 for proportion ? and for ratio ? What is the sign for a general pro- 
 portion? What is a term ? 
 
§ 141.] GENERAL PRINCIPLES. 107 
 
 The sign .*. is used tor the word therefore 
 
 The hyphen - is used for the word which; as, 2# = ^ 4 
 - = 8 ; read, ivhich equals 8. 
 
 A simple quantity is that which is represented by one 
 term ; as, o, 26, — 3, Smnrs, &c. 
 
 A compound quantity is that which is represented by two 
 or more terms; as, a-\-2ab—x. 
 
 Similar quantities are those which consist of the same 
 letters, or combinations of letters ; as, a and 2« ; Qbx and 4bx. 
 
 Dissimilar quantities are those which consist of different 
 letters, or different combinations of them ; as, a and a 2 ; 2ax 
 and 2ab. 
 
 Identical terms are those which are not only similar, but 
 also have the same co-efficient ; as, 2ax and 2ax. 
 
 Identical expressions are sometimes made up of the same 
 letters, but differently combined in their terms ; as, (3a — 26) 
 and (2a+a+ 56— 76). 
 
 Positive quantities are those which have the sign -f- before 
 them, either expressed or understood ; as, ab, -\-ax. 
 
 Negative quantities are those which have the sign — be- 
 fore them ; as*, — 3, —2x. 
 
 Given quantities are such as have known values ; and are 
 generally represented by the first letters of the alphabet. 
 They are sometimes represented by the initial of the names 
 that stand for them ; as s for the sum; d for the difference. 
 
 Unknown quantities are those which are to be discovered ; 
 and are generally represented by some of the final letters of 
 the alphabet. 
 
 A quantity, when represented by one term, is sometimes 
 called a nomial. When it has two terms, it is called a bi- 
 
 Quesiions. What is the sign for therefore ] For which ? What 
 is a simple quantity] What is a compound quantity? 'What are 
 similar quantities'? Dissimilar quantities ] Identical terms ? Posi- 
 tive quantities'? Negative quantities'? Given quantities] Un- 
 known quantities ? What is a nomial ] Binomial 1 
 
108 ALGEBRA. [EQ. SEC. 1 & 2. 
 
 nomial; when it has three, it is called a trinomial $ when 
 it has many, it is called a multinomial or polynomial. 
 
 Each of the literal factors which compose a term, is called 
 a dimension of that term ; and the number of these dimensions 
 or factors is called the degree of the -term. The co-efficient 
 is not counted as a dimension. Thus, 2a is a term of one 
 dimension or of the first degree ; 6ax is a term of two dimen- 
 sions or of the second degree ; 5a 2 ; r 2 is a term of four dimen- 
 sions or of the fourth degree. And, generally, the degree of 
 the term is the sum of the exponents which belong to its 
 letters. 
 
 A polynomial is called homogeneous when all its terms 
 are of the same degree. Thus, 3abc — ax 2 -f-c 3 , is homoge- 
 neous ; but 8a 3 — 4ab+c is not homogeneous. 
 
 § 142. The use of the foregoing signs makes algebra a 
 species of language which brings our reasonings into a very 
 small space ; so that in solving a problem, or demonstrating 
 the existence of a numerical relation, the connection of the 
 several ideas is perceived with great facility. 
 
 Example \. The sum of $660 was subscribed for a certain 
 purpose, by two persons, A and B ; of which B gave twice 
 as much as A. What did each of them subscribe ? 
 
 Now, a question similar to this, and with the same numbers, 
 was solved in the First section of Equations, on page 25. 
 We will solve this in the same manner, with the exception 
 of using a instead of 660. 
 
 Stating the question, x a what A gave. 
 
 2x = what B gave. 
 
 Both together gave x-\-2x; also they gave a dollars. 
 Forming the equation, x+2x = a 
 
 Uniting the terms, Sx = a 
 
 Dividing by 3, x = - 
 
 3 
 
 Questions. What is a trinomial ? Polynomial 1 What is the 
 dimension of a term 1 What is the degree of a term % When is a 
 quantity homogeneous? What is the use of algebraic signs 1 
 
§ 142.] GENERAL PRINCIPLES. 109 
 
 Here we find that A subscribed ± of a, which at this time 
 stands for $660. 
 
 But it is very plain that we would solve the question in the 
 very same manner, if the sum were $240. And in that case 
 a would stand for $240 ; and A's share of it would be $80, 
 and B's share, $160. 
 
 In the same manner, if the sum were $360 ; then a would 
 stand for $360, and A's share would be } of $360; that 
 is, $120. And in the same manner we may make a repre- 
 sent any sum; and still A's share of it would be i of it. 
 Hence, this substitution of a letter for a number, is called 
 generalizing the operation. 
 
 We see that this result has given us a general ride for di- 
 viding any sum between two, so that one of them shall have 
 twice as much as the other. The rule is, The least share 
 shall be one-third of the sum ; and the greatest share, two- 
 thirds of it. ^ 
 
 Problems. — In the same manner generalize all the pro- 
 blems in the First section of Equations } page 25. 
 
 Example 2. What number is that, which, with 5 added to 
 it, will be equal to 40 ? 
 
 This is the first problem in section 2, which we will gene- 
 ralize ; using a for 40, and b for 5. 
 
 Stating the question, x = the number. 
 
 x-\-b = after adding. 
 Forming the equation, a?-f b = a 
 
 Transposing /», x = a — b. 
 
 We see that the answer is found by subtracting the 5 from 
 the 40. Thus, 40—5 = 35. 
 
 Example 3. Generalize problem 3 of section 2. It will 
 be found that the literal answer is the same as in example 2 ; 
 because the questions are similar. In this example, a repre- 
 sents 23, and /; represents 9. Whence x = 23 — 9 - = 14. 
 
 Question. What do we call the generalization of an operation 1 
 10 
 
x+x+b= 
 
 both shares. 
 
 
 x+x+b = 
 
 a 
 
 
 
 x+x = 
 
 a-b 
 
 
 
 2x = 
 
 a—b 
 
 
 
 x — 
 
 a — b 
 ~~2"-' 
 
 a 
 or,-- 
 
 h 
 _ 2 
 
 110 ALGEBRA. [EQ. SEC. 1. & 2 
 
 Example 4. Divide 17 dollars between two persons, so 
 that one may have $4 more than the other. [Prob. 4, sec. 2.] 
 Let IT be represented by a, and 4 by b. 
 
 Stating the question, , I x = the least share. 
 
 x+ b = the greater. 
 
 Forming the equation, 
 Transposing 6, 
 Uniting terms, 
 
 Dividing by 2 
 
 The answer is found by subtracting the difference or 4, 
 from the whole sum, and then dividing by 2. 
 
 And this is the rule for all similar sums. 
 
 The 5th question, on page 32, is similar to the one just 
 performed ; and in that, a represents 55, and /; represents the 
 difference or 7. The numerical answer is found by the rule 
 just shown. 
 
 ^ a—b 55—7 48 
 Thus, _,_.»_. =24. 
 
 § 143. As this rule is of some importance, it will be well 
 to remember it. If, from a number to be divided into two 
 parts, we subtract the difference of those parts, half the re- 
 mainder will be equal to the smaller part. 
 
 Applicati m. Perform the 7th and 10th by this rule. 
 
 Example 5. In the same questions, let us take x for the 
 greatest share. Then x — b = the less. 
 Forming the equation, a?-f x — b = a 
 
 Transposing and uniting, 2x = a+b 
 
 Dividing by 2, * = aJ r , ° r '2 i + 2 : - 
 
 § 144. Here we have another rule. If, to a number to 
 be divided into two parts, we add the difference between 
 those parts, half the sum will be equal to the greater part 
 
§ 145.] GENERAL PRINCIPLES. Ill 
 
 Application. Find the greater part in questions 4, 5, 7, 
 and 10, on page 32, by this rule ; without algebra. 
 
 § 145. The mere letters in the answer of an algebraical 
 operation, form what is called a formula ; because they are 
 the form of the solutions of all similar questions. And this 
 is the advantage of representing the quantities by letters. 
 For, as arithmetical operations on them can only be indicated, 
 the result also must be merely an indication ; and this indica- 
 tion will apply to any question, in the enunciation of which 
 the only things which vary, are the numerical values of the 
 
 s-i-d 
 quantities. Thus, the formula x = , denotes that the 
 
 z 
 
 greater share is found by adding the difference to the sum, 
 
 and dividing the amount by 2. 
 
 Example 6. The learner must now generalize problem 6, 
 
 on page 32 ; using b and c for the two differences. The 
 
 formula that he obtains will be the answer for questions 8, 9, 
 
 1 1, and 13. And in each of the five problems the pupil must 
 
 verify the answer by substituting the given quantities for the 
 
 i ♦♦ m, a— 2b— c 73-8—5 
 
 letters. Thus, x = = = 20. 
 
 3 u 
 
 We have said that in algebra the arithmetical operations on 
 numbers are only represented by different methods of com- 
 bining the signs that stand for those quantities. And now, 
 although we have shown in our progress thus far, what some 
 of those methods are, it may be well to review them a little. 
 
 Questions. What two important rules have we found by gene- 
 ralizing 1 ? W T hat do we call a formula in algebra? Why] What 
 advantage do we derive from formulas? What then is the real 
 effect of an algebraical result ? 
 
112 ALGEBRA. [EQ. SEC. 3 & 4. 
 
 I. 
 
 ADDITION AND SUBTRACTION OF ALGEBRAICAL 
 QUANTITIES. 
 
 § 146. One algebraical quantity is added to another by 
 writing one quantity after the other, taking care to preserve 
 to each term its respective sign. Thus, a+f— c is added to 
 d—e+b, so as to make d— e-f 6-f-a+/— c. Or, as it is 
 easier to read the letters in their alphabetical order, their sum 
 may be written a + b — c + d— e-f/. Again, when —a is 
 added to b, we preserve the sign ; thus, b — a. 
 
 § 147. One algebraical quantity is subtracted from, an- 
 other, oy changing the sign or signs of the quantity which 
 is to be subtracted, and then writing that quantify after the 
 other. Thus, a + h — y is subtracted from b — x-\-c, by first 
 making it —a — h+y, and then writing the whole quantity, 
 b— x+c— a— h+y; or, b— a+c— h-\-y— x. 
 
 § 148. After the addition or subtraction has been performed, 
 if there are any similar quantities in the result, they may be 
 united by adding the co-efficients f all the positive similar 
 terms, and affixing their literal part j then adding all the 
 negative similar terms in the same manner ; and then sub- 
 tracting the less sum from the greater, and retaini >g, in 
 the result, the sign of the greater. 
 
 § 149. In uniting the terms of compound numbers, we 
 consider the literal part of the term as a unit; thus, 2a "and 
 3a, are regarded as 2 nnits and 3 units of a particular kind 
 which when put together, make five units of that kind. Now 
 we have seen, § 141, that the co-efficient of a quantity may 
 
 Questions. How is addition performed in algebra 1 ? How is a 
 negative quantity addpd 1 How is subtraction perform? d ? How 
 are algebraical quantities united 1 In determining whether the 
 quantities are similar, what part of the quantity do we examine 7 
 
§152.] ADDITION AND SUBTRACTION. 113 
 
 also be literal ; as in ba, ca, &c. In such cases, the whole 
 term ba or ca becomes a unit, each of a different kind ; and 
 of course are not similar quantities, and cannot be united. 
 
 § 150. But if there are several similar units of this kind, 
 they may be united by the general rule. Thus, ba — ca-\-ba 
 ca-\-ba-\-ca, can be united into, 3ba + ca. ax — bx -f ax -f 2bx 
 — 3ax+bx, are equal to, — ax-\-2bx; or 2bx^-ax. 
 
 §151. Again, we have seen, § 68, that several quantities 
 are sometimes united by a vinculum. In such cases, all that 
 is embraced by the vinculum, is regarded as a unit of that 
 kind ; and may have a co-efficient. Thus, in the expressions, 
 3X0— b+x, and 5(x+ax— y); a—b-\-x is a quantity taken 
 3 times, and x+ax—y is a quantity taken 5 times. Like 
 quantities of this kind can be united; thus, 2[ay— bx+x) 
 -\-D(ay—bx+x)—7(ay—bx+x). 
 
 § 152. In uniting terms, great care must be taken that the 
 literal part be entirely alike, both in signs and letters. Thus, 
 2bx-\-3cx, cannot be united. Neither can 3y—2ay; nor, 
 6(a+bx)+2(ax+bx); nor, 3. ay— by -f 2.ay 2 —by ; nor, 
 4{ax — bx) — 2(ax~\- bx) ; neither in any other case where 
 there is the least difference in any part but the leading co- 
 efficient. 
 
 EXAMPLES. 
 
 Unite the following quantities. 
 
 1. 3ax— 2y+4ax— 5?/-f ax— 3y. Ans. Sax—lOy. 
 
 2. 3x-\-ay — 2a? — ay+4x + 3ay — 2x+4ay. 
 
 Ans. 3x+lay. 
 
 3. 4ax—y+3ay—2 — 2ax-\-ay—7y+8+2ay+y. 
 
 Ans. 2ax-\- Gay— 7y-f6. 
 
 4. ax — ay* — 3ay-\- Sax — 2ay+7ay — 4ax — Say 2 . 
 
 Ans. 2ax+2ay — 9ay 3 . 
 
 Questions. Give an example in which literal quantities are not 
 similar. What is said of quantities in a vinculum] In uniting, 
 what particular care is necessary 1 
 H l0 * 
 
114 ALGEBRA. [EQ. SEC. 3 & 4 
 
 5 . 8(a-y)+4(fl-y)+2(a-y)+7(a-y). 
 
 Ans. 16(a— y) 
 
 6. — 4(a+6)+3(a+6) — 2(a+6)+7(a+6). 
 
 Ans. 4. a-f 6. 
 
 7. 2(a6+a?)-f3(aa?+6) — 4(a?— #) — 2(ab+x). 
 
 Ans. 3(aa?+6) — 4(a?«— y). 
 
 8. 7y- 4(a + b) + 6y + 2y + 2(a + b) + (a + 5) + y 
 -3(a + 5). Ans. 16y— 4(a+6). 
 
 9. a? 3 + aa?-— ab+ab — a? 3 -f a?y-faa? + a?y — 4ab + x a +x* 
 — x+xy+xy+ax. Ans. 2x*+3ax — 4ab+4xy — x. 
 
 § 153. Sometimes the subtraction is expressed by enclosing 
 the quantity to be subtracted in a parenthesis, and prefixing 
 the sign — . Thus, 4a — 2x+3ax — (4x-\-3ay — 2ax). In 
 such cases, when the vinculum is destroyed, the signs must 
 be changed. Thus, 4a — 2x+3ax — 4a? — 3ay-\-2ax. But 
 if there is a co-efficient immediately before the vinculum, the 
 vinculum cannot be destroyed, nor the signs changed ; as in 
 ax — 2(ab — 3a?). Because such quantities are considered as 
 only one unit. § 151. 
 
 10. #+12 — ax+y — (48 — x — ax+3y). 
 
 IC^First subtract. See page 68. Ans. 2a? — 36 — 2y. 
 
 11. a b — 4xy — a— a? 3 — {2xy — 6 + 14a? -fa: 8 ). 
 
 Ans. a6 — 6a?y — a +6 — 14a? — 2a? 9 . 
 
 12. 3(a?+y) + (4.a7^/). _ Ans. 7(*?+y). 
 
 13. 2(a+b) — x — (3.a+6-a? 3 ). [See §153.] 
 
 Ans. a? 3 — (a-j-b) — a?. 
 
 14. From 4.a+6, take a-f& — 3.x — y. 
 
 Ans. 3.a-f6 + 3.a? — y. 
 
 15. a+b — (2a — 3b) - (5a+7&) — ( — 13a-f2&). 
 
 Ans. la — 56. 
 
 Questions. How may subtraction be expressed? What if the 
 vinculum in such expressions is destroyed 1 When cannot the vin- 
 culum be destroved 1 Why 1 
 
§ 155.] ADDITION AND SUBTRACTION 115 
 
 16. 37a — 5x — (3a — 2b — 5c) — (6a— 4& + 3A). 
 
 Ans. 28a-J-6& — 5x-f5c — Sh. 
 
 § 1 54. After the subtraction of a quantity has been per- 
 formed, we may transform the expression by changing the 
 signs to their original form and resupplying the parenthesis. 
 Thus, 2x — (3a-f2y — x), becomes, when subtracted, 2x 
 
 — 3a — 2y+x; and this latter expression may be restored 
 back to 2a? — (3a-\-2y — x). 
 
 By the same principle, in any quantity, we may suppose 
 there has been a subtraction, and therefore transform the ex- 
 pression to what it may have been. 
 
 Thus, 2x— 3y+la+ax— 1, 
 may be changed to 2x — (3y — la — ax+\). 
 
 ab — 3x —4y — 2ax -f 3a, 
 may become either ab — 3x—4y — (2ax — 3a) 
 
 or ab — 3x— (4y-j-2ax — 3a) 
 
 or ab — (3x+4y+2ax— 3a). 
 
 § 155- When similar quantities have literal co-efficients; 
 as, mx + nx, ay 2 — by 2 , <fcc. ; a compound quantity may be 
 expressed by placing the co-efficients of the similar quantities 
 one after another, (with their proper signs,) in a parenthesis ; 
 and then annexing their common letter or letters. Thus, 
 mx+nx maybe expressed (m-\- n)x; ay*— by* by (a — b)y 2 ; 
 axy-\-byx — abxy by (a -\-b — ab)xy, &c. 
 
 17. From ax — rx + ny — y 8 , take (a — n)y -f (a -f r)x 
 
 — (l-f-a)i/ 3 . Ans. 2ny — 2rx — ay-\-ay 3 . 
 
 18. From a(x-\-y) -f b (x -fy), take c(x -f y) — d(x — y) 
 +x — y. Ans. (a+b — c).(x+y) + (d— l).(x — y). 
 
 Questions. When a quantity has been subtracted, how can it be 
 added again so as to preserve all the terms ? Give an example. 
 How will this principle enable us to transform an algebraical ex- 
 pression 1 Give an example. What may be done with literal co- 
 efficients i 
 
116 ALGEBRA. [EQ. SEC. 3 & 4. 
 
 §156. By §154, we determined that 1 — rx — tx is the 
 same as 1 — (rx-\-tx), which by §155 is 1 — (r-j-£)x; and 
 universally, when we wish to enclose literal co-efficients in a 
 parenthesis ; if the first of them has the sign — , that sign 
 may be put before the parenthesis, and the signs of all the 
 enclosed quantities changed. Thus, 1 — mx-\-px — qx+arx 
 may be written 1 — (m — p-\- q — ar)x. 
 
 § 157. By § 146, — b is added to a, so as to make a — b. 
 Whence it will be seen that an algebraical sum is different 
 from an arithmetical sum ; and signifies, not that it is greater 
 than one of its parts, but that it is the result of putting to- 
 gether two or more different values, either of the same or of a 
 contrary signification. The same method is employed in 
 common language, when, in order to show what a man's for- 
 tune is, we declare what is owed to him and what he owes. 
 In both an algebraical sum, and an inventory of a man's 
 effects, the real positive value may be much smaller than any 
 one item in the account, and even less than nothing. 
 
 §158. Hence, it may be that, after uniting terms, we 
 shall have a negative quantity, — a. This shows that, having 
 a quantity to be subtracted, we did subtract all that we could, 
 and have yet more to be subtracted if we had anything from 
 which to subtract. 
 
 EXERCISES IN EQUATIONS. 
 
 § 159. The learner must now generalize the problems in 
 sections 3 and 4 of Equations, pages 34 and 39. This he 
 can easily do, if he takes care to make one of the first letters 
 of the alphabet stand for each numeral quantity; say a for 
 the first mentioned, b for the second, &c. When the same 
 numeral quantity occurs more than once in the same question, 
 the same letter must stand for it each time. 
 
 Questions. How, in such cases, can we change the signs 1 How 
 
 does an algebraical sum differ from an arithmetical sum? How 
 
 may an algebraical sum be illustrated in common language 1 What 
 does a negative quantity standing by itself denote ? 
 
§161.] MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 117 
 
 n. 
 
 MULTIPLICATION OF ALGEBRAICAL 
 QUANTITIES. 
 
 § 160. We have shown, §39, that a simple literal quan- 
 tity may be multiplied by writing the multiplier before that 
 quantity. This is the case whether the multiplier is numeral 
 or literal. Thus, a times x, is written ax; bxc = bc. In 
 the same manner, a times be becomes abc ; and f times abc 
 becomes fabc. As fxabc is the same as abexf we see that 
 it is of no consequence what order we make of the letters in 
 the product, abed — aedb = cadb, &c. But it is generally 
 more convenient to follow the order of the alphabet. 
 
 Case 1. 
 
 § 161. Therefore, to multiply one simple quantity by an- 
 other, write the quantities one after another, without any 
 sign between them. Thus, abx times cfy = abxcfy; 5ax 
 times cdf—5acdfx. But, if there are more than one nume- 
 ral co-efficient, those co-efficients must be multiplied as in 
 arithmetic, and placed before the product of the literal quan- 
 tities. Thus, da x 2x = 3.2ax = Qax. 2bc X 5rs = 1 Obcrs. 
 
 EXAMPLES. 
 
 1. Multiply a by b. Ans. ab. 
 
 2. Multiply ab by c. Ans. abc, 
 
 3. Multiply ab by cd. Ans. abed. 
 
 4. Multiply 2acx by by. ' Ans. 2abcyx. 
 
 5. Multiply 3brs by 2mnx. Ans. Qbmnrsx. 
 
 Questions. How is a simple literal quantity multiplied 1 In what 
 order should the factors be written 1 What is the operation when 
 there are numeral co-efficients 1 
 
118 ALGEBRA. [EQ. SEC. 5. 
 
 6. Multiply 2adn by 5cmx. 
 
 7. Multiply Srsyop by 4antx. 
 
 8. Multiply 2abcx by Sabrx. 
 
 § 162- By the foregoing principle, axa = aa. Now, as 
 in algebra the same factor is often found two or more times, 
 Stifelius (A. D. 1554) adopted a method for shortening such 
 expressions, in which he has ever since been followed. The 
 method is this : when the same letter enters as a factor two 
 or more times into any quantity, we write the factor but 
 once, and put at the right of it and a little raised, a figure 
 denoting how many times it has been multiplied. Thus, 
 aa is written a 8 ; bbb is written b 3 ; xxxx is written x*; 
 aabbbyyyy is written a 9 6 s y*. 
 
 §163. Mathematicians are accustomed to call aa or a 3 , 
 the second power of a, or, a-second power; a 3 , is called 
 a-third power, &c. 
 
 § 164, The figure that denotes the power of any quantity 
 is called the exponent or index of that quantity. 
 
 § 165. All Quantities are said to have an exponent, either 
 expressed or understood. Thus, a is the same as a 1 ; b=b 1 ; 
 &c. The written exponent affects no letter except the one 
 over which it is written ; unless it is denoted by a vinculum. 
 
 § 166. Great care must be taken by the pupil not to con- 
 found the co-efficient with the exponent; as their effects are 
 entirely different. The co-efficient shows addition, the ex 
 ponent denotes multiplication. For example, if a = 5, then 
 3a = 5 + 5+5- = 15; but a 3 = 5x5x5 - = 125. 
 
 Questions. About what time did Stifelius write ? What if two or 
 more factors are represented by the same letter? How are such 
 quantities read ? What effect has the exponent when it is on the 
 right of several letters'? What if there is no exponent written over a 
 letter? Explain the difference between an exponent and a co-effi- 
 cient, i 
 
$169.] multiplication of algebraical quantities. 119 
 
 Case 2. 
 
 § 167. We have seen that a 2 = aa; and that a 3 =aaa; 
 now aaxaaa = aaaaa. So we see that a 2 xa 3 = a 5 . Hence 
 we establish the rule that when both multiplier and multi- 
 plicand are denoted by the same letter, their product is found 
 by adding their exponents, x 2 x# 4 = x G , y 3 xy 2 = y 5 ; &c. 
 
 § 168. In order to facilitate the practice of multiplication, 
 it is best to observe the following method : First, determine 
 the sign, then the co-efficient, then the letters in their order, 
 and then the exponents. 
 
 EXAMPLES. 
 
 9. Multiply 2amn by a. Ans. 2a 2 mn. 
 
 10. Multiply dabcx by 4ax. Ans. 12a 2 bcx». 
 
 11. Multiply 5bcmn by 36c. Ans. \5b 2 c 2 mn. 
 
 12. Multiply 6a 2 xy by 4ax 2 y. Ans. 24a 3 x 3 y*. 
 
 13. Multiply 4a 2 bcx by a*b 2 c. Ans. 4a e b 3 c 2 x. 
 
 14. Multiply 3a 3 m 2 by a*m 3 n. 
 
 15. Multiply 5a 7 x 6 by 4a 3 x 4 . 
 
 16. Multiply 2m 5 r 2 a 2 x by 9a~x 4 rn*. 
 
 17. Multiply Sb*c 2 n 5 by Sa 3 d 2 mn 3 . 
 
 18. Multiply 7a 8 y*x 4 by I2a 5 b 2 x. 
 
 19. Multiply 9m 7 x 2 a 3 by la 2 b 3 m. 
 
 20. Multiply 12« 2 m 6 na? 5 by 18b 3 cn s y 
 
 Case 3. 
 
 § 169. When one of the factors is a compound quantity, 
 (§ 65 and 67,) we multiply each term by the other factor, and 
 
 Questions. How is multiplication performed when there are ex- 
 ponents? What is a good method in multiplication? What if one 
 of the factors is a compound quantity % 
 
120 ALGEBRA. [EQ. SEC. 5. 
 
 set down their products, each with their proper sign. In doing 
 this, we generally begin at the left ; thus, 
 
 a+bc ah — c. 
 
 Multiplied by x Multiplied by a 
 
 ax-\-bcx - a~b — ac. 
 
 21. Multiply ax-{-bx by xy. Ans. ax 2 y-\-bx s y. 
 
 22. Multiply 2a 2 bc — rxy* by 2ay. Ans. 4a*bcy — 2arxy 4 . 
 
 23. Multiply lay -fl by 3r. Ans. 21an/-f.3r. 
 
 24. Multiply 2xy + ab+c* by 2ax. 
 
 25. Multiply 5 — 7x+2a 3 b by 4ac*y. 
 
 26. Multiply 3m 3 n — 3rjo+2sa? 3 by 6a*bd. 
 
 27. Multiply 2en — 4an+ 5 by 12a 9 a? 5 . 
 
 28. Multiply 4y+y* by 2rcy. 
 
 29. Multiply 3a — 46+4 by 5y. 
 
 30. Multiply 6aa? 8 — a 3 — 1 by Sax 9 . 
 
 31. Multiply 4 -f 3a — a? 9 by ay. 
 
 32. Multiply 2a+56 9 -f 3c — 5e by 3a 8 . 
 
 33. Multiply 76 9 — 4+a 8 # — a: 9 by '4a»x. 
 
 34. Multiply 6x+7a—axy—2y by 3bx*. 
 
 Case 4. 
 § 170. As (a+6) x# = aa:+for; we know that x x(a+b) 
 also = ax+bx. Whence we see that if we wish to multiply 
 x by a+b ; we first find the product of a times x, and then 
 add to it the product of b times x. And generally, when the 
 multiplier and the multiplicand is each composed of several 
 terms, the product is made up of the sum of the products of 
 the multiplicand by each term of the multiplier. 
 Thus, (x+y)x{a+b) 
 Cx+y -J Cx+y ") C x+y 
 
 = < a y and < b V or < a-f-o 
 
 Lax+ayJ Lbx+byj Lax+ay+bx+by. 
 
 Question. How is multiplication performed when there are seve- 
 ral terms in the multiplier? 
 
§ 171.] MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 1<U 
 
 EXAMPLES. 
 
 35. Multiply 2a+3b by 4a+5b. 
 
 2a + 36 
 4a + 56 
 8a 2 +\2ab 
 
 + 10a6 + 15& 3 
 Uniting terms, 8a a + 22a& + 156 5 . 
 
 36. Multiply a+6 by a+6. Ans. a3+2a6+6» 
 
 37. Multiply a +6 — c by a 2 +6. 
 
 Ans. a 3 +a 2 b — a 2 c+ab+b a — be 
 
 38. Multiply x 2 +2xy+y 2 by a?+i/. 
 
 Ans. x 3 +3x 2 y+3xy 2 +y* 
 
 39. Multiply a— a? by 2a+3x. Ans. 2a 2 +ax — 3x 2 
 
 40. Multiply a? — 4 by x+8. Ans. x 2 +4x — 32 
 
 41. Multiply a? — y by a?+y. Ans. x 2 — y 2 
 
 42 . Multiply 4a 2 — 3xy 2 +lax — ay by 2a 3 + ar 2 y. 
 
 43. Multiply a 2 bc 3 + 3a 3 c 2 — ob 2 c by ax 2 -\-yx 2 . 
 
 44. Multiply m 8 a 2 +2rc 2 & 3 — 1— m 4 n by 2m 8 6 2 +4a 8 n*. 
 
 45. Multiply 8a 2 x 3 y*—2a 4 x 3 y 2 +a 5 y 5 by 3a 2 m+2ny*. 
 
 Case 5. 
 
 §171. As (a — a?) xb = ab — a$; we also know that b 
 x(a — x)=zab — xb. Therefore, to find a — x times b, we 
 first find the product of a times b, and then subtract from it x 
 times b. This principle is very plain in figures. Thus, sup- 
 pose we have 7 times 8, and wished only 4 times 8 ; if we 
 take 3 times 8 away from the 7 times 8, we should obtain 4 
 times 8 ; as 56 — 24 = 32. Whence we have the general rule, 
 if there is a negative quantity in the multiplier, the product 
 of that quantity and the multiplicand must be subtracted 
 from the product of the positive quantities and the multi 
 plicand. H^"This explains the rule in § 173. 
 
 Question. What if there is a negative term in the multiplier? 
 11 
 
122 ALGEBRA. [EQ. SEC. 5. 
 
 Thus (a + b) x(c—d) 
 a+b "J a+b 
 minus -Id lor c — d 
 
 Lac 
 
 id+bd) ac+bc—ad—bd. 
 Here we see that the product oft/ -into a-\-b is subtracted 
 from the product of c into a+b; and therefore the signs of 
 ad+bd are changed to — ad — bd. 
 
 § 17.2- By examining the answer of this last example, we 
 shall observe a principle which will enable us to be more 
 rapid in the multiplying operation. It is this : when we mul- 
 tiply a + term by a + term, the product in the answer is 
 a + term ; and when we multiply a + term by a — term, 
 the product in the final answer is a — term. It will be well 
 for the pupil to explain this. 
 
 By understanding this principle, we are able to set the final 
 answer down at first. 
 
 EXAMPLES. 
 
 46. Multiply a+b by a — 6. 
 a+b 
 a—b 
 a a +ab 
 —ab—b a 
 
 a* — b* Ans. a*— 6 s . 
 
 47. Multiply x +Sz by 3# 9 — 7xz. 
 
 Ans. 3x*+l7x»z— 56xz\ 
 
 48. Multiply x*+xy+y* by x—y, Ans. x s —y 3 . 
 
 49. Multiply 2x+3y by 3x—4y. Ans. 6x 2 +xy— 12y*. 
 
 50. Multiply b 3 + b*x+bx*+x* by b—x. Ans. b*—x*, 
 
 51. Multiply a — b by c — d. 
 
 a— b ~\ fa—b ~) a—b 
 
 c I minus < d I or c — d 
 
 ac — be J Lad — bd) ac — be — ad+bd. 
 
 Question. How is the subtraction performed 1 By what principle 
 «an we multiply and subtract at the same time 1 
 
§ 173.] MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 123 
 
 Here we see that in subtracting d times a — b, we change 
 the signs of ad — bd to — ad-j-bd. 
 
 §173. Whence we learn, that when the signs are alike, 
 the product is positive; when the signs are unlike, the pro- 
 duct is negative ; or, in abbreviated language, 
 -f multiplied by -f-, produces -f- 
 4- multiplied by — , produces — 
 — multiplied by -f , produces — 
 • — multiplied by — , produces +• 
 And by remembering this we can always set down the 
 final answer at first. 
 
 52. Multiply a—x by a— x. 
 
 a —x 
 a — x 
 a 2 — ax 
 — ax -f- a? 8 
 
 a 3 — 2ax+x*. 
 
 53. Multiply 2x— 3a by 4x— 5a. 
 
 Ans. 8x*— 22a#+15a 3 , 
 
 54. Multiply 2a — 5y by a—2y. Ans. 2a 3 — 9ay+l0y 9 . 
 
 55. Multiply a s -{-ac — c 3 by a — c. Ans. a 3 — 2ac 3 -f c 8 . 
 
 56. Multiply a +b — d by a— b. Ans. a 3 — ad— b*+bd. 
 
 57. Multiply Ax— 5a— 26 by 3#— 2a+56. 
 
 Ans. 12x 3 — 23a# -f 146#-f- 10a 3 — 21 ab — lOb*. 
 
 58. Multiply x 3 — y 3 — z 3 by x — y — z. 
 
 59. Multiply 6+ xy— a 3 — my 3 by a 3 — 3# 3 -fi/*. 
 
 60. Multiply 2a 3 6 — 3ac 3 -f 46 3 c 3 — 1 
 
 by 2a 3 c 3 -56 3 c — 8a 3 . 
 
 61. Multiply 3a 3 x 3 — 4x 5 +26 3 c 3 — abc— 1 
 
 by 2#6 3 +6c 3 # — 6c— 1. 
 
 62. Multiply Amx — nr — 2 w 2 r+ 3 nr 3 — 1 
 
 by 4ra 3 +3n 3 — r 3 — 1 
 
 Question. What is the general rule for the signs ? 
 
124 ALGEBRA. [EQ. SEC. 5 
 
 § 1 74. The rule for the signs which has been established 
 for compound quantities is also extended to simple quantities. 
 It may seem strange to the young student to hear of mul- 
 tiplying -fa by — x, or —a by -}-#, or —a by — x. But, 
 as it was intimated, §157, 158, a negative quantity implies 
 merely a contrary value from a positive quantity ; and there- 
 fore, though in algebra, the operation seems to be upon ima- 
 ginary quantities, yet in the application of the operation to 
 arithmetic and geometry, the mind readily understands their 
 signification. The extension of this rule to simple quantities, 
 will interpret the peculiar results to which algebraical opera- 
 tions sometimes lead. 
 
 § 175. It is often the case that it is better to denote the 
 multiplication of compound quantities, than to perform it, on 
 account of operations that follow. Thus, a — b times x-f y, 
 may be written (a — b) . (x+y). No general rule can be given 
 to determine when one method is preferable to the other. 
 Experience is the best teacher in this particular. But it was 
 thought best to mention it in this place. 
 
 When, after the multiplication has been denoted, the seve- 
 ral terms are actually multiplied, the expression is said to be 
 expanded. 
 
 § 176. In uniting the terms, if there are literal co-efficients 
 of similar quantities, they may be enclosed in a parenthesis, 
 as was shown in addition. § 154 and 155. 
 
 63. Multiply ma? 3 — nx — r by nx — r. 
 
 Ans. mnx 3 — (n 9 -f mr)x 2 -\-r*. 
 
 64. Multiply px z +qx — y by mx — n. 
 
 Ans. mpx?+(mq — np)x 3 — (my-\-nq)x-\-ny. 
 
 65. Multiply ax 2 — bx-\-c by x 2 — ca?+l. 
 
 66. Multiply^ 3 — rx-{-9 by a? 3 — rx — q. 
 
 Questions. How may multiplication be denoted % In what cases 
 may literal co-efficients be enclosed in a parenthesis ? 
 
§ 179.] GENERAL PROPERTIES OF NUMBERS. 125 
 
 in. 
 
 GENERAL PROPERTIES OF NUMBERS. 
 
 § 177- We have before stated that algebraical operations, 
 (§ 139,) by reason of the quantities themselves being retained 
 in their original value, do show us, in their results, important 
 general principles. We will here make a few multiplications 
 of some quantities, whose results show us some remarkable 
 general properties of numbers. These properties the pupil 
 should remember, as they are of frequent use in the subse- 
 quent parts of this study. 
 
 § 178. Suppose we have two numbers, a and 6, of which 
 a is the greatest. Then their sum = a+6 ; and their differ- 
 ence m a — 6. Then a -f- b 
 
 a — b 
 a 2 + ab 
 — ab — b* 
 
 a* — 6 s 
 
 By this operation, we find a general property of numbers 
 which it would be difficult to find by any arithmetical opera- 
 tion. It is that, if we multiply the sum of two numbers by 
 their difference, the product will be the difference of the 
 squares of those numbers. 
 
 § 179. Again, take the same quantities, and multiply their 
 sum, by their sum. 
 
 a +b 
 a +b 
 a* + ab 
 + ab -f b* 
 
 a* -|- 2ab -f- b» 
 By this operation we find the following general property 
 
 Questions. How may be found the difference of the squares of two 
 quantities] What is the square of the sum of two numbers] 
 
 11* 
 
126 ALGEBRA. [EQ. SEC. 5. 
 
 The square of the sum of two numbers is equal to the square 
 of the first number, plus twice the product of the two num- 
 bers, plus the square of the last number. 
 
 § 180. Again, take the same quantities, and multiply their 
 difference, by their difference. 
 a — b 
 a — b 
 a» — ab 
 — ab + b* 
 
 a* — 2ab + b» 
 Therefore, the square of the difference of two numbers, is 
 equal to the square of the first number, minus twice the pro- 
 duct of the two numbers, plus the square of the second. 
 
 § 181. The only difference between the square of the sum, 
 and the square of the difference, is in the second term ; being 
 in one, positive, and in the other, negative. Let us subtract 
 one from the other. 
 
 a* + 2ab + b* 
 a * — 2ab + b* ' 
 4a~b 
 The actual difference between the square of the sum, and 
 the square of the difference, is four times the product of the 
 two numbers. 
 
 § 182. If when the sum of two quantities has been raised 
 to the second power, and the co-efficient of the second term 
 has been rejected, the quantity thus obtained be multiplied by 
 the difference of the two original quantities; the result will 
 be the difference of the third powers of the two quantities. 
 Also, if we perform the same operation with the difference of 
 
 Questions. What is the square of the difference of two numbers? 
 How do the expressions of the foregoing squares differ? What is 
 their actual difference in quantities] How may we obtain the dif- 
 ference of the third powers of two quantities? How may we obtain 
 the sum of the third powers of two quantities ? 
 
§ 185.] MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 127 
 
 the two quantities, multiplying by their sum, we shall obtain 
 the sum of the third powers of the two quantities. 
 Thus, a + b a — b 
 
 a + b a — b 
 
 a* + 2ab -f 6 3 a* — 2ab + b* 
 
 Rejecting the co-efficients, 
 
 a > + ab + b» a* — ab + b* 
 
 a — b a + b 
 
 a 3 + a 2 b + ab» a 3 — a 2 b + ab* 
 
 — a *b — ab» — b 3 + a 2 b — ab* -f b* 
 
 a' — b 3 a 3 + 6 3 
 
 FACTORS. 
 
 §183. In all simple terms consisting of more than one 
 letter, the factors are evident on inspection. The same is 
 true of polynomials that have a simple quantity for a factor. 
 Thus, I0a' 2 b 2 x + \5a 3 bx — 20a 3 b 3 y, may be decomposed 
 into 5a 2 b(2bx+3ax — 4ab 2 y). 
 
 §184. In order to ascertain the factors of any quantity, 
 first see what is the greatest quantity that will divide every 
 term of the given quantity, and set that down as one factor. 
 Then divide the given quantity by the factor set down, and 
 the quotient will be the other factor. 
 
 § 185. By §178, 179, a + b is a factor of a»+2ab+b 9 ; 
 and a — b is a factor of a 3 — 2ab+b*. By §177, either 
 a-J-6 or a — b is a factor of a 8 — 6 3 . By §182, a+b is a 
 factor of a 3 +£> 3 , and a — b is a factor of a 3 — b 3 . 
 
 Any quantity is a factor of an expression made up of the 
 same quantity and an exponent. 
 
 EXAMPLES. 
 
 Find the factors of the following. 
 
 1. 4a 3 y— 4a*x+8a*xy— 12a 7 . 
 
 2. a 3 b 2 x*+3a*b*x 3 — a 3 a? a +4a 7 # 3 i/— 5a 3 x 5 y*. 
 
 3. 2a 3 bx*—4ab 4 x + 0a 5 b 2 x—2abx. 
 
 4. a 9 +2a6+6 3 -f a 3 — 6 3 -fa 3 +6 3 . 
 
128 ALGEBRA, [EQ. SEC. 5 
 
 IV. 
 
 DIVISION OF ALGEBRAICAL QUANTITIES. 
 
 §186. Division may be represented by the sign — , as 
 a-i-by is read a, divided by b; (a-f b) -f- (c — d), is read 
 a 4- 6, divided by c — rf. But the most usual way to denote 
 division (§ 72) is to write the divisor underneath the dividend; 
 . a a+b 
 
 lhus - v z=* 
 
 §187. But it often happens (§76, 78) that the fraction 
 made by this representation is an improper fraction; and one 
 in which the numerator can be actually divided by the deno- 
 minator. In such cases, it is generally best to perform the 
 division. We have always done so in the former part of this 
 
 treatise. Thus, 4x divided by 2 equals 2x ; — - = 2x. 
 
 o 
 
 § 188. Let us first look at the case where the same quan 
 tity is in both the dividend and the divisor. 7a-i-7 = a; 
 126 -h 12 = 6. In the same manner, ab-r-a = b; dcs-d = c. 
 This may easily be proved. For, ab is the product of a into 
 b; and of course if we divide by what was the multiplier, we 
 shall obtain the old multiplicand again ; as may be seen by 
 trying the product of any two numbers. 
 
 § 189. Whence we derive the general rule, that when the 
 divisor is found as a factor in the dividend, the division is 
 performed by erasing that factor from the dividend, 
 amn-r-a = mn, because amn = a times mn. amn~m = an, 
 because amn = m times an. amn-r-n — am, because amn 
 is n times am. 
 
 Questions. How is division usually represented 1 What if the 
 same quantity is in the dividend and in the divisor 1 Prove it by 
 examples. 
 
§191.] DIVISION OF ALGEBRAICAL QUANTITIES. 129 
 
 EXAMPLES. 
 
 1. Divide 8c by 8. Ans. c. 
 
 2. Divide be by b. Ans. c. 
 
 3. Divide 7ra by 7. Ans. m. 
 
 4. Divide am by a. Ans. m. 
 
 5. Divide 6rf by b. 
 
 6. Divide 7a6 by 7. 
 
 7. Divide ca6 by c. 
 
 8. Divide 6ad by b. 
 
 9. As a&rf is the same product as the last, divide that by b. 
 
 Ans. ad. 
 
 10. Divide cde by c. 
 
 11. Divide the same product in another form, thus, dec 
 by d. Ans. ec, 
 
 12. Divide «6c by c. 
 
 13. Divide a&c by b. 
 
 § 190. As a* = «a, it is evident that if we divide a 2 by a, 
 the quotient is a; because we take away one of the written 
 a's. So, if we divide a 5 by a, .the quotient is a*; because a s 
 is the same as aaaaa, which -~ a, gives aaaa or a*. So, if 
 we divide a h by a 2 , the quotient is aaa or a 3 . And if we divide 
 a 5 by a 4 , the quotient is a; because aaaaa -h- aaaa = a. 
 Hence we see, that when there are exponents in either the 
 dividend or divisor, the division is performed by subtract- 
 ing the exponent of the divisor from the exponent of the 
 dividend. b 5 -.-b* = b 3 ; x 7 -r-x 3 = x*. 
 
 § 19.1 a As every literal quantity is understood to have the 
 number 1 for its co-efficient, it is evident that if we divide by 
 1, the quotient would be the same literal quantity. Thus, 
 a-7-\ = a. And again, if we divide a literal quantity by 
 
 Questions. How is division performed when there are exponents? 
 Why ] How when the divisor is the only quantity in the dividend? 
 Why? 
 
 I 
 
130 ALGEBRA. [EQ. SEC. 5. 
 
 itself, the quotient will be 1. Thus, la—a = 1 : the divisor 
 being a factor in the dividend. 
 
 § 192. It sometimes happens that the co-efficient contains 
 the divisor as a factor. Thus, 8a is the same as 2 times 4a, 
 or 4 times 2a ; and therefore can be divided by 2 or by 4. 
 In the same manner Sab may be divided by 2a, or 4a, or 26, 
 or 4b ; because it is 2a times 4b, or 4a times 2b. 
 
 We have only to remember to take those factors out of the 
 dividend which are equal to the divisor. And in general, 
 when there are co-efficients in both the divisor and the divi- 
 dend, divide the co-efficient of the dividend by the co-efficient 
 of the divisor; and then proceed with the literal quantities as 
 before directed. I0abc-r-5b = 2ac; I2a 3 xy-r-3a 2 y — 4ax. 
 
 EXAMPLES. 
 
 14. Divide a 3 by a 9 . Ans. a. 
 
 15. Divide x 6 by x*. Ans. x*. 
 10. Divide a 3 b*y by a*b. Ans. ab 8 y. 
 
 17. Divide d 3 c*x 7 by rfc 3 x*. Ans. d 2 c z x*. 
 
 18. Divide a*m*x by a*m*. 
 
 19. Divide a*x a y 7 by ax 4 y*. 
 
 20. Divide d 6 y 7 by dy. 
 
 21. Divide p 3 r 7 s a t by r 5 st. 
 
 22. Divide ab 3 c A d* by ab»cd 5 . 
 
 23. Divide ax 7 y a by ax?y s . 
 
 24. Divide c b r 7 sHx*y 7 by c*r 5 ty*. 
 
 25. Divide 6a*bc 7 b^ 26c 5 . Ans. 3a 9 c". 
 
 26. Divide \2ax*y* by 4a?/ 9 . Ans. 3x 2 y a . 
 
 27. Divide 2\bc 3 xy R by 3ct/ 3 . 
 
 28. Divide 42c 7 rfV } by 6c 2 d 5 x. 
 
 Questions. How is division performed when there are co-efficients ? 
 Explain. 
 
§ 195.] DIVISION OF ALGEBRAICAL QUANTITIES. 131 
 
 29. Divide 36pr*st* by 4rsf. 
 
 30. Divide 54m 5 n 2 x 3 2/ by 9m 3 ny. 
 
 31. Divide 66a*c 8 aV by 3a*c s a?. 
 
 32. Divide 48c*r 7 x 3 y 5 by 8cr 5 xy*. 
 
 33. Divide 72aV?» 4 by 18a 2 rm 3 . 
 
 Case 2. 
 
 §193. We have shown, §169, that (a+b)xc = ac+bc. 
 Of course (ac + bc)-±-c = a-\-b; where we see that ivhen we 
 divide a compound quantity by a simple quantity, we divide 
 each of the terms by that quantity. 
 
 §194. We must also recollect that, as -f multiplied by 
 -f , makes + in the product, so -f in the product divided by 
 -f » must make -f in the quotient. And that as -f- multiplied 
 by — makes — , so — in the product divided by — , will 
 bring back the -f in the quotient. So that when the signs 
 are alike in the dividend and divisor, the sign in the quo- 
 tient is +. Thus, axb = ab; both of which are +• Also 
 — «X — b = ab; and — ab-. b=+a. 
 
 §195. Again, as — multiplied by -f makes — , so in 
 the product, — divided by -f-» brings back — in the quotient. 
 Also, — multiplied by — makes -J- ; and of course, + in the 
 product divided by — , brings — in the quotient. That is, 
 when the signs In the divisor and dividend are unlike, the 
 zign in the qudient is — . 
 
 examples. 
 
 34. Divide 2ad+8a»c by 2a. Ans. d+4ac. 
 
 35. Divide 8tf* 3 m 2 — I2d 5 m? by 4dm 9 . Ans. 2d»—3d*m 
 
 36. Divide ixy + Gx* by 2a?. Ans. 2y-\-3x. 
 
 37. Divide abc — acd by ac. Ans. b — d. 
 
 38. Divide I2ax—8ab by —4a. Ans. — 3x+2b. 
 
 Questions. How is a compound quantity divided ? What are the 
 rules for the signs ? For what reason ? 
 
132 ALGEBRA. [EQ. SEC. 5. 
 
 39. Divide \0xz + \5xy by 5a?. , 
 
 40. Divide I5ax— 27a? by 3a:. 
 
 41. Divide 18a- 2 — 9a? by 9a?. 
 
 42. Divide abc—bcd—bcx by —be. 
 
 43. Divide 3a?+6a?--f3aa?— 15a? by 3a;. 
 
 44. Divide 3abc + \2abx— 9a»b by Sab. 
 
 45. Divide 40a 3 6 2 -f60a a & 8 — 17 ob by ab. 
 
 46. Divide \5a-bc — l0acx* + 5ad*c by — Sac. 
 
 47. Divide 20ax+\5ax* + \0ax— 5a by 5a. 
 
 § 196. It is evident that we may divide by either factor 
 Thus, ax+bx may be divided by x, and the quotient will be 
 a + b; or it may be divided by a + b, and the quotient will 
 be x. This may appear singular to the young pupil ; but he 
 is to recollect that division is merely separating the dividend 
 into factors, being careful to make one of them of a given 
 magnitude ; that is, to make it the same as the given divisor. 
 For illustration, ax+bx means that x is taken a times, and 
 also b times. Therefore it is taken (a + b) times; and in the 
 whole quantity, a + b is the co-efficient of a?, so that (ax+bx) 
 = (a + b)x. 
 
 §197. Now we know that x times a+b = ax + bx; and 
 also that a+b times x = ax+bx. Whence, the product 
 (ax + bx)-i-(d + b)z=x. Therefore we conclude that if the 
 divisor contains just as many terms as the dividend, with 
 corresponding signs ; and the first term of it is a factor in 
 the first term of the dividend, the second term of it in the 
 second >>f the dividend, a.i<! so on through <ach of them re- 
 spectively; and the remaining factor in every term of the 
 dividend being the same ; thai remaining factor only shall be 
 the quotient. 
 
 Questions. What is the ruin for dividing when the number of 
 terms in the dividend and divisor is the samel Explain and illus- 
 trate? 
 
§ 199.] DIVISION OF ALGEBRAICAL QUANTITIES. 133 
 
 EXAMPLES. 
 
 48. Divide ax+bx+cx by a+b-\-c. Ans. x. 
 
 i9. Divide bac+bc 2 x — bx 3 by ac+c 2 x — x 3 . Ans. b. 
 
 50. Divide c 2 ax — 2abx — 3xy-\-x by ac 2 — 2ab — 3y-\-l. 
 
 51. Divide c^x—abd' + ^x 3 —,! 2 by ex— ab + x 3 — 1. 
 
 52. Divide a 2 y — bcy-\-xy by a 2 — bc-\-x. 
 
 53. Divide 6ah?n — \iabm — 3cdm by Qah — \4ab — 3cd. 
 
 § 198. If the letters of the divisor are not found in the 
 dividend, the division is expressed, as we have before shown, 
 § 186, by writing the divisor underneath the dividend, in the 
 form of a vulgar fraction. 
 
 EXAMPLES. 
 
 4t/ -4- 73? 
 
 54. Divide 4t/-f7a? by a— b. Ans. -^— ,— . 
 
 55. Divide 3a + 2b 2 — c by a+c. 
 
 56. Divide a 3 —x 2 b + c 5 by a 3 — b*. 
 
 57. Divide 3a 2 c+26 3 +c by 2c. 
 
 §199. When the dividend is a compound quantity, the 
 divisor may be placed underneath the whole dividend if we 
 choose. It may also be placed under each term of the di- 
 vidend, which is the same as dividing each term, according to 
 § 193. By this method, the answer of the last sum would be 
 
 3ac 9 26 3 c , - n . , , , . , , 
 
 f- r- — . Answer the following by both methods. 
 
 2c 2c Zc , 
 
 58. Divide 3ab+b + 2ab by a. Ans. 1 1 . 
 
 ' a a a 
 
 59. Divide 6a + ab— 3b by 2b. 
 
 60. Divide 2x+2y-\-3ax—2a 2 y by 3ay. 
 
 61. Divide ax 2 —bx—a 2 b 2 -\-ab by 2b. 
 
 Questions. What is the operation when the same quantities are 
 not in both the dividend and divisor? Must the divisor always be 
 put under the whole dividend 1 
 
 12 
 
134 ALGEBRA. £eQ,. SEC. 5. 
 
 § 200. When we divide each term separately, we may use 
 both methods of division ; that is, we may actually divide 
 such terms as we can by § 189 ; and merely express the di- 
 vision in such terms as cannot be divided. 
 
 EXAMPLES. 
 
 ttX 
 
 62. Divide cd — ax-\-ac+bc by c. Ans. d \-a+b 
 
 63. Divide ax+bx— 2ab-\-2x by x. Ans.a+6 f-2. 
 
 x 
 
 64. Divide 2am— 3a*b+b*m— 3a*m by —a. 
 
 b n m 
 
 Ans. — 2m-\-3ab ; \-3am. 
 
 a 
 
 65. Divide 2b 9 — a*b+3b*c+a*b 3 — ac by b*. 
 
 66. Divide ay*—by*+4a 3 b»—&a*by*+aby by ay. 
 
 67. Divide abx*+a?by—3ab 2 x+ax*—la 2 y by ab. 
 
 68. Divide 2abm+6a*b + 5b*m— 4a 8 m by ab. 
 
 69. Divide by — a 3 b 9 y-\-ay — aby+b s y by — by. • 
 
 70. Divide ax — bx*-\-xy*+by — ay by — y. 
 
 EXERCISES IN EQUATIONS. 
 
 § 201. The learner must now generalize the problems in 
 section 5 of Equations, page 44. And with this section, we 
 will make our calculations more purely algebraical than in the 
 preceding sections ; as we shall take care to use no numeral 
 quantities at all in stating the questions. 
 
 §202. It is customary to represent those numbers which 
 stand for times, by the letters m, n, p, a, Sic. 
 
 1. Two persons, A and B, lay out equal sums of money in 
 trade. A gains c ($126,) and B loses d ($87 ;) and now A's 
 money is m (two) times as much as B's. What did each lay 
 out ? See page 44. 
 
 Question. When may we use both methods of dividingr? 
 
§ 202.] DIVISION OF ALGEBRAICAL QUANTITIES. 135 
 
 Let x = what each lay out. 
 
 x-\-c— A's sum now. 
 
 x — rf= B's sum now. 
 
 mx — md = m times B's. 
 
 Forming the equation, mx-—md — x+c f 
 
 Trans posing, mx — x = c -f md 
 
 _...,. , ,'•■-- c+md 
 
 Dividing by m— 1, x = -. 
 
 to f m—l 
 
 Substituting numbers £ c + md 126 + 2x87 
 
 r 1 f 7" ==s 7^. 1 " ^ = "v)U. 
 
 for letters, ) m — 1 2 — 1 
 
 In sums of this kind, the only difficulty is to determine what 
 quantity we must divide by in the last step, to leave x alone. 
 But this difficulty is easily overcome, by dividing mentally 
 the left hand member by a?, and observing the quotient. Of 
 course, dividing the same member by that quotient will pro- 
 duce x; which is our only object. 
 
 2. The 2d sum on page 45, is performed as follows : 
 
 x = the wife's age. 
 mx = the man's age. 
 x-\-a= wife's after a years. 
 m#-fa= man's after a years. 
 nx+na = n times the wife's age. 
 Forming the equation, mx+a — nx-\-na 
 
 Transposing terms, mx — nx = na — a 
 
 Dividing by m — n, x = . 
 
 Substituting numbers, x = 15, the age of his wife. 
 
 All the other questions in section 5th are to be performed 
 in the foregoing manner. 
 
136 ALGEBRA. [EQ. SEC. 5. 
 
 DIVISION BY COMPOUND DIVISORS. 
 
 §203. By examining any sum in multiplication, it will be 
 seen that any one term in the product contains both one term 
 of the multiplicand, and one term of the multiplier. See each 
 term in the answer of the following example. 
 
 x+y 
 
 a+b 
 
 ax+ay+bx+by. 
 
 Therefore, if we find any term of the product that contains 
 the first term of the multiplier, we know that the remaining 
 part of it is a term in the multiplicand. And, as either term of 
 the multiplicand may be put as the first, (for y-\-x is the same 
 as x+y), we may suppose that the term which we have found 
 by this process is the first. Thus, in the answer above, ax 
 and ay contain a of the multiplier ; and therefore either x or 
 y, whichever we choose, is the first term of the multiplicand. 
 
 Now, by change of names, ax + ay+bx+by may be the 
 dividend, and a -\-b the divisor, for the purpose of finding the 
 quotient. We see that a, the first term of the divisor, is found 
 in ax and in ay ; and we may therefore conclude that the first 
 terra in the quotient is either x or y, which are the remaining 
 parts of those terms. Likewise, {a -\-b) x x or y, will be a 
 partial dividend. 
 
 Ex. 1. Suppose it were required to divide ac+bc+ad+bd 
 by a+b. We see that the first term in the divisor is contained 
 in the first term of the dividend ; and therefore judge that the 
 other factor, which is c, must be the first term in the quotient; 
 and that c times a+b, which is ac + bc, is one of the partial 
 
 Questions. What may be said of each term in a product? What 
 then do we know of any term in the product that contains the first 
 term of the multiplier? What is the change of names for division 
 ; nstead of multiplication ? Explain the first example. To what is 
 : t similar 1 
 
§ 205.] DIVISION BY COMPOUND DIVISORS. 137 
 
 dividends. Subtracting ac-\-bc from the whole dividend, and 
 there remains ad-\-bd still to be divided by a-{-b. The work 
 is as follows : 
 
 a-\-b) ac-\-bc+ad-\-bd (c+d 
 - ac+bc 
 
 ad+bd 
 ad+bd 
 
 In the last partial dividend, we see by the first term that the 
 divisor is contained in it d times. 
 
 § 204. This is the same method that is followed in arith- 
 metic ; where, in order to find a quotient figure, we see how 
 many times the first figure of the divisor is contained in the first 
 figure of the partial dividend ; and supposing the number to be 
 the true quotient, we multiply the divisor by it. But there is 
 this difference between algebraic and arithmetical division, 
 that in algebra the quotient is always the true one ; and also 
 where the divisor is not found as a factor of one of the terms 
 of the dividend, the operation of division is impossible. 
 
 Ex. 2. Divide 5a 8 #-f x 3 -\- a 3 +5ax* by a+x. 
 
 § 205- From the last example, it seems that it is best to 
 put first in the dividend those terms that contain the letter 
 which is in the first term of the divisor. It has also been 
 found easier in practice to arrange the powers of this letter 
 both in the divisor and dividend, so that the highest should 
 stand first, the next highest next, and so on. We will adopt 
 this plan in the present example. 
 
 a+x) a 8 -|-5a 8 a?-f5aa? a +ic 3 (a*+4ax+x* 
 In subtracting, a 3 -\-a*x 
 
 change the signs 4a 3 .r-f-5aa ,3 -fa? 8 
 
 mentally, and 4a»x+4ax* ^j* We find the 
 
 unite at the ax 2 +x* terms f q UO tient 
 
 same time. ax*+x* by § 2Q4 ^^ 
 
 The first term of the divisor is contained in a 3 , a 8 times ; 
 
 Questions. To what is this method similar 1 What arrangement 
 of terms is recommended ? Explain the second example. 
 
 12* 
 
138 ALGEBRA. [EQ. SEC. 5. 
 
 wherefore we determine that a a is one term in the quotient. 
 The product of that term and the divisor is a? + a 2 x, which 
 we subtract from the dividend, and there is left 4a 2 x-\-5ax 9 
 -f-cc 3 . The first term of the divisor is contained in the first 
 term of this partial dividend 4ax time's ; wherefore we deter- 
 mine, that 4ax is one term of the quotient. 
 
 We will incorporate the foregoing remarks into the follow- 
 ing general rule. 
 
 RULE FOR DIVIDING WITH COMPOUND DIVISORS. 
 
 § 206. Arrange the terms of both dividend and divisor 
 according to the powers of the same letter, beginning with 
 the highest. Divide the first term of the dividend by the 
 first term of the divisor, and place the result with its pro- 
 per sign for the first term in the quotient. 
 
 § 207. Multiply the whole divisor by this term of the 
 quotient, setting the product under the dividend; subtract 
 this product from the dividend, and use the remainder as 
 the next partial dividend. 
 
 § 208. Divide the first term of the partial dividend by 
 the first term in the divisor, and place the result for the 
 next term in the quotient. MuViply the divisor by the term 
 last written in the quotient, subtract the product as before, 
 and proceed in this manner as long as the first term of the 
 dividend can be divided by the first term in the divisor. 
 
 §209. If there is a remainder in which there is no term 
 that contains the first term of the divisor, that remainder 
 must be made the numerator of a fraction whose denomina- 
 tor is the divisor; and the fraction must be put at the end 
 )f the quotient by addition. 
 
 Questions. What is the first step in the rule ? What is done with 
 the first term of the quotient? How do we proceed with a partial 
 dividend] What is done with the final remainder? 
 
§210.] DIVISION BY COMPOUND DIVISORS. 139 
 
 EXAMPLES. 
 
 3. Divide a*+3a 2 x+3ax 2 +x 3 by a -fa?. 
 
 a+x) a 3 +3a 9 a?-f-3aa? 2 +a? 3 (a 3 -f2aa?+a? 3 
 cf+a 2 x 
 
 2a'x+3ax 2 +x 3 
 2a 2 x+2ax a 
 
 ax*+ x 3 
 ax 2 +x z 
 
 The first term 
 of the divisor is 
 contained in the 
 first term of the 
 dividend a 2 times. 
 The whole divisor 
 multiplied by a 3 , equals a?-{-a 2 x; which subtracted from the 
 dividend leaves 2a i x-\-Sax 2 -\-x 3 . The first term of the divi- 
 sor is contained in the first term of the remainder, 2ax times. 
 The whole divisor multiplied by 2ax, equals, &c. 
 
 § 210. It may be well to mention here, that if the signs 
 of the divisor are changed, we shall obtain the same quantities 
 for the quotient as before, with the exception that their signs 
 will be changed also. The pupil may divide this last sum 
 by — a — x ; and he will obtain — a* — 2ax — a? 8 . 
 
 4. Divide a? 3 — 6a?— 9 by a?— 3. For subtracting see Ex. 2. 
 x—3) #3_6a?— 9 (a? 3 -f 3a?-f 3 
 a? 3 — 3a? s 
 
 + 3a? 3 — 6a?— 9 
 3a? 3 — 9a? 
 
 3a?— 9 
 3x— 9 
 
 5. Divide cr'-f 4aa? 9 — 3a 3 a?— 2a? 3 by a a — 2aa?-f a? 9 . 
 
 fl?j? 2 — a? 3 
 a »_ 2aa?+a? 3 ) a 3 — 3a 2 x+4ax 2 — 2a? 3 (a—x+- 
 
 a>-2a 2 x+ax 2 a 3 -2aa?+a? 9 
 
 , — a a a?-f3a2? 8 — 2a? 3 
 — a 3 a?-r-2ga- a — a? 3 
 ax 2 — 3? 
 
 6. Divide a 3 — 3a 2 y+ 3ay*— y 3 by a— y. 
 
 Ans. a 3 — 2ay+y* 
 
 7. Divide 6 3 — 10& 3 +33&— 36 by 6— 4. Ans. &*_6&-f9. 
 
 8. Divide 6a*— 96 by 3a— 6. Ans. 2a 3 +4a 8 -f 8a+16. 
 
140 ALGEBRA. [EQ. SEC. 5. 
 
 9. Divide ar 3 — 3a? 8 6 + 3xb*—b s by x—b. 
 
 Ans. a? 2 — 2xb-\-b 2 . 
 
 10. Divide x* — y* by x—y. Ans. x^+xfy+xif-j-y*, 
 
 11. Divide a* — 2a?* by a -fa?. ^ 
 
 Ans. a^a^+a* 2 -* 3 - — . 
 a-fa? 
 
 2# 3 
 
 12. Divide a 3 — 2? by a-fa?. Ans. a 2 — aa?-f a? 2 
 
 a+a? 
 
 13. Divide x*+y* by a?-fy. 2 w* 
 
 Ans. or 1 — tfy+xy 1 — ^+_iL-. 
 
 14. Divide 21/ 3 — 19y 2 +26y— 16 by y— 8. 
 
 Ans. 2y 2 —3y+2. 
 
 15. Divide 48a? 3 — 76aa? 8 — 64a 8 a?-f 105a 3 by 2a?— 3a. 
 
 Ans. 24a? 8 — 2aa?— 35a a . 
 
 16. Divide b*—3y* by b—y. „ 4 
 
 Ans. b'+b^+by'+tf-.^JL-. 
 
 17. Divide 2a 4 — 13a 3 6+31a 9 6«— 38a6 3 -f 246* 
 
 by 2a fl — 3ab+4b*. Ans. a 3 — 5a&+6& 8 . 
 
 18. Divide a? 8 +/>a?+ q by a?+ a. a— pa-f-a 8 
 
 Ans. a?-f»— a-f- — *■ . 
 
 a?-f-a 
 
 19. Divide 6a?* -f 9a? 9 — 20a? by 3a? 8 — 3ar. _ 
 
 Ans. 2ar 8 +2a?-f 5- 
 
 3a? 9 — 3a? 
 
 20. Divide 9a? 8 — 46a? 5 -f 95a? 9 -f 150a? by a? 8 — 4a?— 5. 
 
 Ans. 9a?*— 10a? 8 + 5a? 9 — 30a?. 
 
 3 77 43 33 
 
 21. Divide -X s — 4a?*+— a? 8 -a? 8 — —a?-f27 
 
 4 8 4 4 
 
 1 3 1 
 
 by -a? 9 — a?-f 3. Ans. -a? 8 — 5a? 9 +-a?-f 9. 
 
 2 Z 4 
 
 §211. It sometimes happens, that in the dividend there 
 are several terms which have the same exponen. ^f that letter 
 by which the quantity has been arranged. In such cases, 
 they should be put directly under one another ; and when they 
 become the first of a partial dividend, there should be a quo- 
 tient for each term before the subtraction is performed; as 
 follows : 
 
§212.] DIVISION BY COMPOUND DIVISORS. 141 
 
 22. Divide 1 Oa 3 + 1 1 a*b -f- 3ab 2 — 1 5a c — 5b 2 c — 1 9a6c 
 -j-l5bc* by 5a 2 -f 3a& — 56c. 
 
 * 2 , o * ^ O0a 3 +lla 2 6-f 3a& 2 — 5& 3 c-fl5&c 2 7 , 
 5a W _56c{ I 15a2c _ 19a , c }2«+6-3c 
 
 10a 3 + 6a 6— \0abc 
 
 + 5a 2 6 + 3a6 3 — 56 3 c + 1 56c 3 
 
 f -f 5a*& + 'dab 2 — bb»c+lbbc 2 l 
 1 — 15a 3 c— 9a6c * 
 
 Product by +6 = 5a 2 6 -f 3a6 2 — 56 3 c 
 
 Prod, by— 3c = — 15a 2 c— 9a6c +156c 8 
 
 
 
 23. Divide a*+4a*b+8b* by a+2&. \eb*+24b* 
 
 Ans. a 3 — 2a 2 6+4a&-|-4aZ> 2 — 8& 2 — 8& 3 -' 
 
 a +2b 
 
 24. Divide 64a 3 + 64a&-fl6& 8 — 9a* 3 — 48a*-64 
 
 by 8a + 4&-f 3rf+8. Ans. 8a + 46— 3d— 8. 
 
 25. Divide 18a 2 + 33a6-f 42ac— 1 2ae?— 30& 3 + 1246c + 86a* 
 — 16c 3 — 32cd by 6a+15&— 2 c— 4a*. Ans. 3a-26+8c. 
 
 26. Divide 1 by 1— x. 
 
 l—x) 1 (\+x+x»+x*+x*, &c. 
 
 1— X 
 X 
 
 x—x* 
 
 a? 3 — x 8 
 
 x* 
 
 a a x 9 
 
 27. Divide a 3 by l—x 8 . Ans. a 3 -f-a 3 # 3 -f a 3 a? 6 + 
 
 1— X s 
 
 §212. In the last two examples, the division may be 
 carried on forever, like a decimal fraction. Any algebraical 
 fraction also may be expanded by actually performing - the 
 division in this manner. But as in this example, so in many 
 others, a few leadin/ terms of the quotient will be sufficient 
 to indicate the rest, without continuing the operation. 
 
142 ALGEBRA. [EQ. SEC. 5 
 
 28. Expand the fraction . 
 
 r a+x 
 
 „ XX 2 X 3 X* e 
 
 «+*)* (1- -+-_-+-, to. 
 
 a+x 
 — x 
 
 — a: IC7* For — --r-a = 
 
 a la 
 
 x % x* __x» 
 
 a a ' a* 
 
 x 9 x* x s _ x* 
 
 a a* ""a s "^ a = "a" 9 
 
 It is obvious that if the division is continued, the remaining 
 terms will be alternately — and -f- ; and that they will in- 
 crease one power in every successive term. 
 
 § 213. Such operations may be carried on infinitely, so as 
 to bring the result nearer and nearer to the true quotient. 
 And on this account mathematicians have called such ex- 
 pressions Infinite series. It must be understood, however, 
 
 that it is not the quantity which is infinite, but the number 
 
 2 
 of terms. Thus, .66666666G66, &c, = -. 
 
 3 
 
 x x % x 3 x* x 5 x 9 a 
 
 1 1 1 — , &c., = . 
 
 a^a* a 3 ^a* a^ a 6 ' a+x 
 
 29. Reduce to an infinite series. 
 
 \+a 
 
 \+a 
 
 30. Reduce , to an infinite series. 
 
 I— a 
 
 Ans. 1— a + a 2 — a 3 +a 4 — a 5 +a 6 , &c 
 
 an infinite series. 
 Ans. l+2a + 2a*+2a s + 2a* + 2a 5 , &c. 
 
 Questions. What may be done with an algebraical fraction? Is 
 it always necessary to continue the operation? How far may the 
 operation be continued ? What name has been given to the answer? 
 Is the quantity infinite ? 
 
§214."] FRACTIONS. 143 
 
 V. 
 
 FRACTIONS. 
 REDUCTION OF FRACTIONS TO LOWER TERMS. 
 
 §214. We showed in §88, that a fraction may be re- 
 duced to lower terms, without any alteration in its value, by 
 simply dividing both terms by a number that will divide each 
 without a remainder. Fractions that are expressed by literal 
 quantities may frequently be reduced in the same manner. 
 
 Thus, in the fraction — ^, both the ierms may be divided by 
 
 a; and the fraction will then become -^-. 
 
 o 
 
 EXAMPLES. 
 
 4abcx 
 
 Reduce to the lowest terms the fraction 
 
 Qadcy 
 
 _ , C4abcx „ 2bx 
 
 Ans. Both terms <- — ; — i-2ac = — 1 -. 
 
 C Qadcy 3 ay 
 
 ■■ _ - a 2 m 3 v . . . my 
 
 2. Reduce —r — £- to its lowest terms. Ans. — £-. 
 
 tfom^x abx 
 
 n n , 56^ 4 v 8 . , 8v 3 
 
 3. Reduce — =-r~* t0 lts lowest terms. Ans. £_. 
 
 — lx b y* x 4, 
 
 . _ , — 4x 5 y*z . . 4x 3 z 
 
 4. Reduce — -~ — to its lowest terms. Ans. . 
 
 5x y* by 2 
 
 m ~ , —I2x 4 yz . . 3a? 
 
 5. Reduce — — to its lowest terms. Ans. — . 
 
 —4x*yz* z % 
 
 10°* These examples will remind the pupil, that, (because 
 fractions are merely expressions of division,) when each 
 
 Questions. How may fractions be reduced . to lower terms ? 
 Where must a sign be put to affect the whole fraction 1 What sign 
 in e tch term will make plus for the whole 1 What sign in each term 
 will make miniu for the whole ? § 195. 
 
i44 ALGEBRA. [EQ. SEC. 6 
 
 term has its sign, then the whole fraction will have a sign 
 according to § 194 and 195. 
 
 6. Reduce • • — to its lowest terms. 
 
 I4ab*x* 
 
 9x 3 y 5 
 
 7. Reduce — — — to its lowest terms. 
 
 45ay* , 
 
 \5(t 3 r 9 s 
 
 8. Reduce — to its lowest terms. 
 
 27amst 
 
 § 215. When we divide a compound quantity by a simple 
 quantity, we divide each term, §193. Hence, in reducing 
 fractions to lower terms, we must find for the divisor a quan- 
 tity that is a factor in every term, both of the numerator and 
 denominator. 
 
 EXAMPLES. 
 
 « t> j a*x+ay 3 , ax-\-y* 
 
 9. Reduce --*- to its lowest terms. Ans. V - • 
 
 a*b a»b 
 
 m t> i a*x»+ax 9 —3a*x . . 
 
 10. Reduce r — - - - to its lowest terms. 
 
 ax*— 6ax'+9a 8 x * o 
 
 ax 2 + x — 3a 8 
 Ans. — - — -- — — . 
 x 3 — 6x-f 9a 
 
 11. Reduce — ^ — - — — - — 2 to its lowest terms. 
 
 8ax J — 4a*x 
 
 .« n j 3a 3 b*x — 9b x -6ax a . , 
 
 12. Reduce — - — — — — z — to its lowest terms. 
 
 I2ax+6abx— \5bx 
 
 ,_ -1 , 7a*r s t:> 56«ra— I4am s . . 
 
 13. Reduce --— — — r — to its lowest terms. 
 
 35a 2 ra + 2 1 abm + 56am 
 
 , _ , 3Za 3 ry { +i8ax 2 y— 6ay . . 
 
 14. Reduce -— A J? ' 1g , q J to its lowest terms. 
 
 3a 4 i/*— 2 1 a^y*— 1 oa ?/ 2 — 9ay 
 
 _ , 112a&.x — 48aca' + 100aca? . . 
 
 15. Reduce — — r — = r- to its lowest terms. 
 
 4abca — Sacd+52acx 
 
 § 216. By this principle, we may often simplify answers 
 to questions in division. That is, we may put the divisor 
 
 Questions. What if the numerator or denominator is a compound 
 ^tantity ? What application has reduction of fractions to division? 
 
§217.] FRACTIONS. 145 
 
 underneath the dividend, so as to make a fraction, and then 
 reduce that fraction to its lowest terms. 
 
 EXAMPLES. 
 
 16. Divide x 2 —2xy+xy 2 by 4xy. Ans. X ~~ y "*"^ . 
 
 4y 
 
 17. Divide I0xy—20x—5y by —5x. 2xv—4:X—v 
 
 Ans. ^ 2. 
 
 x 
 
 18. Divide 7abx— 56a*xy + l4ax 3 by 28a*bx !l y. 
 
 19. Divide 8amy 3 +lGa 3 xy—24aby !i by 48a 2 y»—72ay. 
 
 20. Divide 35o 3 6c—14aa?+42o 3 by 21a 6 — 28a 5 #+7a 4 # 8 . 
 
 21. Divide 32arty-|-16# 3 2/ 2 -f-8:ri/ 3 by 24a#+48a 3 x 3 . 
 
 22. Divide 54a 3 & 3 +45a 3 & 3 — 27a 3 6 3 by 24a 5 6 3 +30a 3 & 5 . 
 
 23. Divide 48xy+\2axy— I6ax by 4x*y*+§ax— 8xy 
 
 24. Divide 28(a— b+x) by 4m(a— &+#). Ans. - 
 
 m 
 
 25. Divide 12cc?(ra— rc) by 14ac(m— n). 
 
 26. Divide 6a/*(r+/>) by 24a(r+jo). 
 
 27. Divide (a -f&) (m+ri) by (a— a;) (m+n) 
 
 28. Divide 32a6c by 32abcx—32aby. 
 
 29. Reduce the fractions K ^-\ tk^t*'> "4^:1 n* 
 
 28a*y \2d?b* hxy 5ab 
 
 12ra 3 n — 12mn 3 
 
 § 217. It must be remembered that when all the factors in 
 the numerator are contained in the denominator, the answer 
 
 will contain 1 in the numerator. Thus, - — =— . The 
 
 Sax x 
 
 same principle will also apply to the denominator. 
 
 rp , Aax x 
 Thus, — — =-- = #. 
 4a 1 
 
 Question. What, if in reduction, the common divisor is the 
 as one of the terms of the fraction ? 
 K 13 
 
146 ALGEBRA. [EQ. SEC. 6 & 7 
 
 VI. MULTIPLICATION WHERE ONE FACTOR IS A FRACTION. 
 
 § 218. This is done, (as shown § 74, 75,) by multiplying 
 
 the whole number and the numerator of the fraction together, 
 
 and dividing by the denominator. 
 
 mi 3b — x 6ab—2ax a, ab 
 Thus, 2a X = - J 7& = -r« 
 
 EXAMPLES. 
 
 1. Multiply ^ by 3a. Ans. —-^ = xy 
 
 2. Multiply - — by 8a. Ans. — 
 
 «„i- 1n - ab . . ab 
 
 3. Multiply 3ax into — — — . Ans. — — 
 
 1 'ZCIX 4 
 
 ,-«-,,, .' . 2by iab*y—6bxy' 
 
 4. MulUply 2ab-3xy by g, Ans. - ^--HI 
 
 a&v 2a 8 #v — Saby* 
 
 5. Multiply -- 2 by 2ax—3xy. Ans. ^ ^- 
 
 C3? C 
 
 3x 3 4- a 
 
 6. Multiply 3am»— 4a? by . 
 
 - »«■ i • i 5as£ — 2m . " . 
 
 7. Multiply — by am— 2a 9 . 
 
 8. Multiply Wy+tafi by g^jy 
 
 9. Multiply sr- by 2aa?-f 3a? s . 
 
 Sax — ox 
 
 10. Multiply I4abc—3cdx by . 
 
 r J 5mx— Icm 
 
 Questions. How is multiplication performed when one factor is a 
 fraction ? What is a factor ? What if the whole number is a factor 
 in the denominator of the fraction 1 
 
§ 219.] FRACTIONS. 147 
 
 ,, . . , \0ax—6xy , n _ 
 
 11. Multiply — — - by 3mx— 2am 
 
 12. Multiply - by b. Ans. -r-^a- 
 
 ab abc ab 
 
 13. Multiply — by c. Ans. — - =-j- 
 
 § 219. In the last example, we first multiplied the nume- 
 rator by c, and then divided both the numerator and denomi- 
 nator by c. Now, multiplying the numerator by c and then 
 dividing it by c, is altogether useless ; because the numerator 
 is left as it was at first. We will use then only one part of 
 the operation ; that is, dividing the denominator by c. And 
 in general, when the multiplier is a factor in the denomina- 
 tor, the multiplication is performed by canceling that factor. 
 
 EXAMPLES. 
 
 , i „ ,. , o,rnx . amx 
 
 14. Multiply by r. Ans. . 
 
 rs s 
 
 , .. »• i . i o i obc , abc 
 
 15. Multiply x 9 by — — . Ans. — . 
 
 x y xy 
 
 16. Multiply ^ by 2m. Ans. 29*. 
 
 17. Multiply 2d by |^. 
 
 18. Multiply -—I — by 2x. 
 
 19. Multiply 36* by a ^±??. 
 
 flWl 8 
 
 20 - **** ui&=i§5* by to "- 
 
 «, n/r t • , « i 6am — 4a: 
 
 21. Multiply 3a by 
 
 3am + 6ax — 12a 3 
 
 ~, », , • , abc — bx+cx 
 
 22. Multiply — - , „ by 3aa?. 
 
 VJ 9axy + l2a 3 x— Qaxc J 
 
148 ALGEBRA. [EQ. SEC. 6 & 7. 
 
 , . . . , , am-\-4ab — 2m 
 
 23. Muluply iab by Sabm+Uab _ 4abx - 
 
 24. MuLtiply - — ■ — by 5a?— b. 
 
 r ' 5a?-f ay J 
 
 VII. REDUCTION OF COMPLEX FRACTIONS TO SIMPLE ONES. 
 
 § 220. We have shown, that when we multiply a fraction 
 by its denominator, we obtain for the answer the same quan- 
 tity as the numerator. We have also shown that where both 
 terms of a fraction are multiplied by the same quantity, the 
 value of the fraction is not altered. By these two principles, 
 we obtain the following rule for reducing a complex fraction 
 to a simple one. 
 
 § 221. Multiply both terms of the fraction by the deno- 
 minator that is found either in the entire numerator or de- 
 nominator. If the fraction is still complex, multiply the 
 result in both terms by the remaining denominator that is 
 found in the entire term. Thus, 
 
 a +-* _ ca + b ; a — J ca — b ■ 4ac — 4b 
 ~1 cd ! |T« — £ + ca 3c -f4ac* 
 
 EXAMPLES. 
 
 1. Reduce ^-to a simple fraction. 
 
 * ac 
 
 Ans. Multiplying both terms by c, — r. 
 
 a — — . , /. . ay — x 
 
 2. Reduce * to a simple fraction. Ans. ° . 
 
 ax axy 
 
 Questions. How may a complex fraction be changed to a simple 
 fraction? Explain why. How may fractions be transferred from 
 the numerator to the denominator ! 
 
§ 222.] FRACTIONS. 149 
 
 X 
 
 3. Reduce — to a simple fraction. * v 
 
 «2/+V Ans. 
 
 5ay+y—z 
 
 x — — 4x—x 3 
 
 4. Reduce - to a simDle fraction. Ans. — = -^* 
 
 xy 4yx 4y 
 
 5. Reduce ^— t to a simple fraction. Ans. y - 
 
 3a — -— * a y y-r*- 
 
 x + - 
 
 6. Reduce - to a simple fraction. 
 
 Z - x 
 
 e a?+- xy + x _ cxy-\-cx 
 
 z — l. yz — \ cyz-y* 
 
 4 + - . 24* -f Gx 
 
 7. Reduce : to a simple fraction. Ans. — . 
 
 x — f r Gxz — bz 
 
 8. Reduce =J to a simple fraction. Ans. — , ' • 
 
 a + ^ acy + bxy 
 
 3 + 4 . t „ . 3a?v + 4v 
 
 9. Reduce =- to a simple fraction. Ans. — - -• 
 
 4 — - 4;ri/ — 5a? 
 
 10. Reduce - to a simple fraction. Ans. = 3-. 
 
 § 222. It sometimes happens that we wish to transfer a 
 fraction from a numerator to the denominator, or from a de- 
 nominator to the numerator. This may be done by the fore- 
 
 2 a 
 going principles. For, supposing we have — ; multiplying 
 
 <*. & 
 
 On 
 
 by the denominator of f-, we have — . Now, if we divide 
 
 IX 
 
 this fraction by 2, we have — . Thus we see that such a 
 
 fraction is transferred, without altering the value of the whole 
 quantity, if we take care to invert it when we transfer it. 
 
 13* 
 
I5C ALGEBRA. [EQ. SEC. 6 <fc 7 
 
 EXERCISES IN EQUATIONS. 
 
 § 223- Generalize the questions in section 6th, page 52. 
 
 1. In an orchard, — (\) of the trees bear apples; - (|) 
 m v ' n V5/ 
 
 of them bear pears ; - (-fc) of them, plums ; and a (81) bear 
 cherries. How many trees are there in the orchard ? 
 
 ICT* We rarely represent unity by a letter ; but generally 
 use its own character, 1. 
 
 Let x = number of trees. 
 
 — = apple-trees. 
 
 x 
 
 — = pear-trees. 
 
 px . 
 
 ' — = plum-trees. 
 
 Then, x =* + * + 1E+a. 
 
 m n r 
 
 § 224. As the equation is to be multiplied by m, and then 
 by n, and then by r ; it is plain we may multiply it at once 
 by their product, mnr. Of course, in multiplying a fraction, 
 its denominator will not appear in the product. §219. 
 Multiplying by mnr, mnrx = nrx-\-mrx-\-mnpx+mnra 
 Transposing, mnrx — nrx — mrx — mnpx = mnra 
 
 Dividing by mnr — nr — mr — mnp, 
 
 mnra 
 a? = 
 
 mnr — nr — mr — mnp 
 Substitute figures for letters, and find the answer. 
 4x5x11x81 
 
 a? = 
 
 4x5x11 —5x11 -4x11 —4x5x2 
 
 17820 
 
 220 — 55 — 44—40 ' 
 
 17820 
 _- = 220. 
 
§ 224.] FRACTIONS. 151 
 
 2d sum, page 53. In a certain school, — of the boys learn 
 
 m * 
 
 P 
 mathematics ; — of them study Latin and Greek ; and a study 
 
 grammar. What is the whole number of scholars ? 
 
 mi X PX 
 
 1 he equation is, x = — f-- — \-a. 
 
 m n 
 
 Multiplying by mn, mux = nx-{- mpx + mna. 
 
 Transposing, mnx — nx — mpx = mna. 
 
 t\-t u mna 
 
 Dividing by mn — n — mp, x = . 
 
 r mn — n — mp 
 
 3. The pupil must go through with the whole section in 
 the same manner. And as there are several instances in 
 which the same statement and the same answer will agree 
 with two or more sums, the pupil must tell which they are, 
 and why it so happens. 
 
 4. If the teacher should think his pupils need more prac- 
 tice, he may exercise them in the 7th section in the same 
 manner. 
 
 The first question may be stated thus, 
 
 x + a 
 
 — = c 
 
 Multiplying by b, x + a = bc 
 
 Transposing, x=bc — a. 
 
 The 2d will have the following equation. 
 
 mx -f- ma . 
 
 =6 
 
 n 
 
 Multiplying by n, mx + ma = nb 
 
 nb — ma 
 
 Transposing and dividing, x 
 
 m 
 The 3d and 5th will have the following equation. 
 
 nx — na , 
 = o. 
 
 The 4th will be, a — x = — . 
 
 m 
 
152 ALGEBRA. [EQ. SEC. 8. 
 
 VIII. DIVISION OF FRACTIONS. 
 
 § 225. We have shown in § 97, that we divide fractions 
 by dividing the numerator, when the divisor is a factor in 
 the numerator ; but if the divisor is not a factor in the nu- 
 merator, we multiply the denominator by the divisor. 
 
 EXAMPLES. 
 
 , TV Sa ° U A Sb 
 
 1. Divide by a. Ans. — . 
 
 x * x 
 
 M _. ., Sax , , Sax 
 
 2. Divide -j— by o. Ans. -= — . 
 
 be * b»c 
 
 3. Divide ^ ' "~ by 3a:. Ans. ^r— 
 
 4. Divide -2 by —3a. Ans. — -3L* 
 
 z Saz 
 
 5. Divide 95 by 7a 3 . 
 
 6 . Divide. ^ by 7. 
 
 4z J 
 
 «, t^- j. 3ax — Gay . r , 
 
 7. Divide — ^ by 2#y. 
 
 4a;ry 
 
 8. Divide ~^ by 3aw- 
 
 a+y J u 
 
 m _. .. 4a+27a& t - 
 
 9. Divide — by 6ax. 
 
 Zx 
 
 -, n r^-3 6abx—\6axy . n 
 
 10. Divide — — — £ by 2a. 
 
 2am-r3xy J 
 
 „. Divide 146c3 ;+ 21ac3 by 7<ft 
 
 4abx J 
 
 Question. How do we divide fractions by whole numbers ? 
 
§ 226.] FRACTIONS. 153 
 
 ,« t%- j 10a 5 +53? 5 — 15ax . m , 
 
 12. Divide — 7 — !-— - — — — by 5a 5 x*. 
 
 a 4 +&x 2 —lQax J 
 
 ,« t.. -j 12a& — 14c3? . rt .. 
 
 13. Divide — ■ , ,_, by 2ac — 4bx. 
 
 10a.r + 166c ' 
 
 14. Divide by 4ca — cax. 
 
 x — y 
 
 15. Divide -±- by 12a3?— 9a 3 . 
 
 16. Divide * " ,* by a 3 -l. 
 
 Ha 3 — 16 J 
 
 IX. FRACTIONS OF FRACTIONS. 
 
 § 226- It was shown, §98, that a fraction is multiplied 
 by a fraction, by multiplying the numerators together for a 
 new numerator, and the denominators together for a new 
 
 denominator. Thus, =- X -y = ■=-;. 
 o a oa 
 
 EXAMPLES. 
 
 , „ , . , 3« L 4i . I2ab 
 
 1. Multiply — by — . Ans. — — . 
 
 037 C OCX 
 
 n „ , . , 9ax , 2&3? . 18C&3? 3 6a? 3 
 
 2. Multiply — by — . Ans. ^ m _ 
 
 3. Find the product of ^t^ into — 2- . Ans. Xy +y 
 
 r 3a x — y ax — ay 
 
 S3? 43? 
 
 4. Multiply — and — together. 
 
 5. Multiply ~ into -^. 
 
 Sy 73? 
 
 3a 4a 
 
 6. Multiply ~r into -. 
 
 7. Multiply - into -^r". 
 
154 ALGEBRA. [EQ. SEC. 8. 
 
 8. Multiply — - into — ^. Ans. ^. 
 
 X z xz 
 
 a Sx 4y , 4a 
 
 9. Multiply -, — , -^ together. Ans. — . 
 
 x y oZ z 
 
 . , . „ 4ax 3a*y . 2 . 
 
 10. What is the product of , -~ y and - ? Ans. 12a:. 
 
 r y 2a x 
 
 0/» 2ac—~- be 
 
 11. What is the product of , into — —, — ? 
 
 „ , . , 2am 3 — 3a 2 m , 5am 3 
 
 13 . Multiply '"f'+fy by **?-** • 
 
 r # 4a:— 3y J 5xy*+l0x*y 
 
 § 227. When the numerator of one of the factors, and the 
 denominator of the other, can both be divided by the same 
 quantity, it is best to perform that division before the quanti- 
 ties are multiplied. This is done by canceling the quantity 
 that is common to the numerator of one and denominator of 
 
 the other. Thus, -rX — = = — • 
 
 cd ay cxy cy 
 
 , ' ., . . : 2x , 5a; 9 . a? 8 
 
 14. Multiply — by -— . Ans. — . 
 
 2a: Sab 5ac 3 
 
 15. What is the product of — x — X-rr-? Ans « 15aa?. 
 
 v a c 2b 
 
 16. Multiply I*"* 3 !*' by 10a6-2c. 1K 
 
 * J 10a 3 6— 2ac J I5ac+37bc 
 
 Ans. ■ . 
 
 a 
 
 ,„ ™ i.- i 13a&— 13am , 17a . 
 
 17. Multiply r—; by -. Ans. 13. 
 
 r ' 17a 3 J 6 — m 
 
 «8^_^2 q% ^3 qZ 
 
 18. What is the product of — -j-x — - Q X ?§185 
 
 r a+b ax+x 2 a — x 
 
 a*—a*b 
 Ans. . 
 
§ 229.] FRACTIONS. 155 
 
 EXERCISES IN EQUATIONS. 
 
 § 228. Generalize the questions in Equations, section 8. 
 
 1. The 1st sum on page 64 is performed as follows ; 
 
 Stating the question, x = oats. 
 
 mx 
 
 — = barley. 
 
 n J 
 
 1 „ mx mx 
 
 — of — = — = rye. 
 
 p n np 
 
 fTfV TflX 
 
 Forming the equation, x-\ — -H = a. 
 
 § 229. In this and similar equations, we may apply the 
 principle explained in §88. For, if we multiply by the 
 greatest denominator, np, it is plain that the second term 
 
 will be — — , which = mpx. 
 
 n r 
 
 Multiplying by np, npx+mpx+mx = anp 
 
 Dividing by np 4- mp -f m, x = ■ — ~ — . 
 
 & J r r np+mp+m 
 
 x x 
 
 2. The equation in the 2d sum will be = a. 
 
 n mn 
 
 Multiplying by mn, mx — x = amn 
 
 t\- -j- 1. i amn 
 Dividingbym — 1, # = -. 
 
 3. The equation in the 3d sum will be 
 
 x , x x 
 
 x= h«» 
 
 m n mn 
 
 4. The equation of the 4th sum will be 
 
 mx , mx m 2 x 
 
 x= + a. 
 
 n n n* 
 
 The pupil will now understand how to perform the rest of 
 the section. 
 
 The problems under ratio, in section 10, page 78, are very 
 good examples for this exercise. 
 
156 ALGEBRA. [EQ. SEC. 9 
 
 X. UNITING FRACTIONS OF DIFFERENT DENOMINATORS. 
 
 § 230. Before fractional terms can be united, they must be 
 brought to a common denominator, according to the principle 
 explained in § 85. This is done, as we have shown in § 104, 
 105, by multiplying each numerator by all the denominators 
 except its own, for new numerators; and all the denomina- 
 tors together for a new denominator, 
 
 EXAMPLES. 
 
 I. Unite the following terms, --{- -—-. 
 
 X *xU ox 
 
 8ab*x — abx B +4abmx Sab*— ax 2 + 4am 
 
 4b»x* 4bx 
 
 2. Add and together. § 178. Ans. 
 
 z-f-l x—l 6 x*— 1 
 
 o o u* . Sa r 4a a S2ac—2lab 
 
 3. Subtract — - from —p. Ans. 
 
 8c 76 566c 
 
 4. Unite 
 
 x+y 
 
 y 
 
 z 
 
 5. Unite X ~ V - 
 
 x+y 
 
 Ans. ^-^-y 
 
 xz+yz 
 
 ax — Say 
 
 Ans. — „ „ J 
 
 2a 3a 6a 3 
 
 §231. If, in such cases, there is a quantity that is no 
 fractional, we multiply it by all the denominators ; and then 
 putting the common denominator under that product, unite i 
 with the other fractions. 
 
 n tt • ■ ° l ca i ! ca b ac+b 
 
 6. Unite a-\ — . Ans. a = — ; and then, j — == 
 
 c c c c c 
 
 Questions. How can fractional terms be united ? How can we 
 bring them to a common denominator ? Supposing there are quan- 
 tities connected with them that are not fractional? How do we 
 reduce fractions to the least common denominator ? 
 
§232.] FRACTIONS. 157 
 
 7. Unite x -\ ■ — . Ans. -^ — ■ — 
 
 y y 
 
 ■ m TT .- a-\-b ac—a—b 
 
 8. Unite a . Ans. . 
 
 c c 
 
 § 232. It is generally best to reduce the fractions to the 
 least common denominator. This is the least common mul- 
 tiple of all the denominators ; and is found, when the same 
 quantity is in two or more denominators, by taking that quan- 
 tity but once in the required product. Thus, 
 
 n TT . ar 2c 3cd 
 
 9. Unite — — + — — — _ 
 
 5om 2 5bn 2m s n 
 
 Ans. The common denominator is 5x2x&X m 3 xn- 
 = I0bm 3 n. To obtain this, the first denominator must be 
 multiplied by 2mn; and therefore the .numerator must be 
 multiplied by the same, and becomes 2amnr. The second 
 denominator must be multiplied by 2m 3 ; and therefore its 
 numerator must be multiplied by the same, and becomes 4cm 3 . 
 The third denominator must be multiplied by 5b ; and there- 
 fore its numerator must be multiplied by the same, and be- 
 
 2amnr-\-4cm 3 — \5bcd 
 
 comes \5bcd. The answer is 
 
 . ';— : 3m 9 5 2ar 
 10. Unite -— -f- 
 
 I0bm 3 n 
 
 n 5 r s 3mn 3 r 
 
 15a be 2m 8 r 
 
 11. Unite— + _+^-. 
 
 ,« AJJ 13a 3 6— 2c . 7ab+8c 
 
 12. Add — r — and n . 7„ _ 
 
 4ab 2b + 16ab 
 
 cde 3c d 
 
 13. Unite 
 
 2lbm 3 x 2 lb 3 m 2 x 
 
 WET To know what any denominator must be multiplied 
 by, to make the least common denominator, divide all the 
 other denominators by such factors of this as will divide 
 them, and then multiply this by the several quotients. 
 
158 
 
 ALGEBRA. 
 
 2abd—3m*c 
 
 [eq. sec. 9, 
 
 .a ^ Wad i 
 
 14. From-rrj-. take — 
 
 t _ _ a+x a—x 
 
 15. From —. r take 
 
 a(a— x) "*~"° a(axx) 
 a-f-6 a — b 
 
 ,« 4JJ fl+^ a—b A 2a 9 +2& 9 ol 46 9 
 16. Add — 5-r to — — r. Ans. __-!-— -- = 2+— — r- 
 a— 6 a + 6 a 9 — o 9 a 9 — 6 9 
 
 17. From — : take 
 
 a»—z* 
 
 •8 
 
 Ans. 
 
 3a,r— a 8 — z 9 
 a 3 — * a 
 
 a-f-z 
 'a& , 6 . a s +a6 8 +6 8 
 
 18 - Umte (^"(W'+S+S' Ans - "ISP^ST 
 
 ■ , T . .a 4 . 7bx+7a—7aby—4b 
 
 19. Unite *+-r — ay — -. Ans. -^ — ^ 
 
 v 5 « 3 + 6 ,4a 9 +26 A 37a 9 -fll6 
 
 20. Add together , , and — -r Ans. ^ 
 
 Ans. 
 
 127#+17 
 
 Ans. 
 
 28 
 
 4a? fl — llx— 5 
 5a?+5 ' 
 
 2a? 9 
 
 5*4-1 f 21x+3 
 
 21. Subtract - from — - — . 
 
 3z+l - 4x 
 
 22. Subtract — — from — . 
 
 x+l o 
 
 23. Add together^, and -J-. 
 
 , _ , 2a? — 3 r 4a?+2 
 
 24. Subtract — - from — - — . 
 
 3x 3 
 
 25. Subtract ^—7 from ^t_. 
 
 a+o a — 
 
 rtrt _ 3a+26 , 5bd—2a — 3d 
 
 26. From , subtract — -, . 
 
 c 4cd 
 
 _ * '2 • ;, 6 9 c — 5aft 9 c+a 8 
 
 27. From c+2o6 — 3ac, subtract j- — ^ . 
 
 exercises in equations. 
 
 § 233. Generalize the questions in Equations, section 9, 
 page 70. 
 
 Ans. 
 Ans. 
 Ans. 
 
 a: 9 — 9 
 
 4a? 9 +3 
 3x 
 
 4ab 
 
§234.] FRACTIONS. 159 
 
 XI. DIVISION BY FRACTIONS. 
 
 § 234. Suppose we wish to know how many times ^ is 
 contained in f-. We would divide in the same manner that 
 we follow in dividing 6 pieces by 3 pieces; and say, ^ is con- 
 tained in f, two times. In the same manner, ¥ 4 r i s contained 
 in if, Jive times. 
 
 The principle is general that when the divisor and the 
 dividend have a common denominator, the division is per- 
 formed by dividing the numerator of the dividend by the 
 numerator of the divisor. 
 
 ,„, a c a a* a 2 a 3 
 Thus, T -j- T = -; — -i- _ = _ = a. 
 bo c x x a* 
 
 EXAMPLES. 
 
 | _. ., 4a . 2b Aa 2a 
 
 1. Divide — by — . Ans. ^-r = -i— 
 
 x J x 26 b 
 
 n _. ., Sab , 4bx 3ab 3a 
 
 2. Divide — r- by — r . Ans. -z— = — — 
 
 cd * cd 4bx 4x 
 
 ' ^. .„ 2x , 3xy 2 
 
 3. Divide — by — £. Ans. — . 
 
 ab J ab 3y 
 
 . — ., 3a — b . 2a» A 3a — b 
 
 4. Divide — by — j-. Ans. ■ * 
 
 aft y a6 2a 2 
 
 . _. ., 7r#+a 9 •■ 46-faa? . 7rrr-fa a 
 
 5. Divide , ' by . ' . Ans. — =— 
 
 bast f bast 4b-\-ax 
 
 „ t^. • , « , 2a6 
 
 6. Divide 8a by 
 
 Explanation. By § 231, 8a = -^. Then ~22?-|— 
 
 ^ ' » xy xy xy 
 
 Saxy 4xy 
 " 2a7T *" T~* 
 
 1? A«. *S 
 
 2/ 
 
 7. Divide 3a by — . Ans. 
 
160 ALGEBRA. [EQ. SEC. 11. 
 
 8. Divide ab by -3. Ans. . 
 
 a c 
 
 _ _. . 4a*x 2x* 2a* 
 
 9. Divide by — . Ans. — . 
 
 y y # 
 
 in TV 'A 8mnr u 6flm * a 4nr 
 
 10. Divide by . Ans. - — . 
 
 ax * ax Sam 
 
 ii tv -a 4a v* x u l0ax * %v* 
 
 11. Divide -f— by —r — . Ans. --^-. 
 
 be J be 5x 
 
 i« tv a a + 6c u 4 a 9 — «& A a+oc 
 
 12. Divide by . Ans. 
 
 ax * ax 4a*— ab 
 
 lO TV A am * U QX "** 
 
 13. Dmde g-^ by ^--. Am. -. 
 
 U. Divide ^=^ by % 
 
 ab * b 
 
 Explanation. In this example, the dividend and divisor 
 
 has not a common denominator. Our first object, then, is to 
 
 bring them to a common denominator. This is done by §230, 
 
 ax — y xy bax — by abxy 
 
 ab ~T "" ab* ~ ab* ' 
 
 . abx — by ax — y 
 
 Ans. — r ^ = £. 
 
 abxy axy 
 
 15. Divide — by -^. 
 c .a 
 
 ~ abx dy __ a*bx cdy __ a*bx 
 
 v e ' a ' ac ' ac cdy 
 
 §235. It will be seen that in sums of this kind, after we 
 have brought the terms to a common denominator, the division 
 is performed by putting the numerator of the dividend for the 
 numerator of the answer, and the numerator of the divisor for 
 the denominator of the answer, and make no use at all of the 
 denominators. Let us see then how we obtain these two 
 terms. We multiply the numerator of the dividend by the 
 
§ 238.] FRACTIONS. 161 
 
 denominator of the divisor ; and this becomes the numerator 
 of the answer. And we multiply the numerator of the divisor 
 by the denominator of the dividend ; and this becomes the 
 denominator of the answer. By looking at the last two 
 sums, it will be seen that this is the true operation. 
 
 § 236- Hence we obtain the general rule for dividing by 
 a fraction. Multiply the numerator of the dividend by the 
 denominator of the divisor, for a new numerator ; and mul- 
 tiply the denominator of the dividend by the numerator of 
 the divisor for a new denominator. 
 
 §237- When the dividend is a whole number, it is 
 
 changed into a fraction by putting 1 under it for a denom- 
 
 x 2a 
 
 inator. Thus, x — - ; 2a = — . 
 
 EXAMPLES. 
 
 16. 
 
 3x ax 
 
 Divide — - by — T . 
 4 ' 26 
 
 
 66* 36 
 
 Ans. - — = —• 
 4ax 2a 
 
 17. 
 
 Divide " + X by - a -+y 
 a — y a-\-2x 
 
 
 a 2 + Zax+2x* 
 Ans. — - 
 
 18. 
 
 Divide x 2 — 2ax-\-a 3 by 
 
 1 
 x — a' 
 
 
 
 
 Ans. x 
 
 3 — 3# B a-f 3a*x — a*. 
 
 § 238. If both of the numerators, or both of the denomi- 
 nators have the same factor, that factor may be canceled be- 
 fore the operation is performed. 
 
 ^ ax ay x y xt 
 
 Thus, 1.— = — 7-- - ■■ — • 
 
 rs rt s t sy 
 
 , _ _.. .. a . 3a 5 
 
 19. Divide - by — . Ans. — • 
 
 Questions. What are the rules for dividing by fractions? Explain 
 the rule for fractions that are not of the same denominator. Sup- 
 posing the dividend is a whole number'? In what cases can the work 
 be shortened 1 
 
 L !4* 
 
52 ALGEBRA. [EQ. SEC. 11. 
 
 « rt ™- J X — l u X + l A 4X ~ 4 
 
 20. Divide — by _. Ans. -j—. 
 
 21. Divide by —r. Ans. — . 
 
 y 46 # , 5ay 
 
 tai . , 5a& . 4a . 5% 
 
 22. Divide — - by Ans. — -£. 
 
 4 y 16 
 
 ~. . , a . 3x 4ay 
 
 23. Divide by — . Ans. f~- . 
 
 a+1 * 4y 3a#-f3a? 
 
 ^. ., . 3a x*+ax a —xy—axy 
 
 24. Divide a?-r-a;r by . Ans. — ' -—* *. 
 
 y x— y 3a 
 
 25. Divide -JL. by ^. Ans. -. 
 
 y— i 3 y— * 
 
 * _. ., , Ay— ay 16a 9 — 4a 3 y 
 
 26. Divide 4a— ay by -^- — ^. Ans. — «. 
 
 a 4a Ay — ay 
 
 27. Divide by ■- -. See § 177,185. Ans. ;r+y. 
 
 x — y x *— y* y 
 
 na r»- j a i — x i . a+a? 
 
 28. Divide — -=- by . ; ,. . 
 
 a+b J (a+b)* 
 
 XII. GENERAL THEORY OF EQUATIONS WITH TWO UNKNOWN 
 QUANTITIES. 
 
 §239. Every equation of the first degree with two un- 
 known quantities may be reduced so as to be represented by 
 
 ax+by = c, 
 a designating the algebraic, sum by which one unknown quan- 
 tity is multiplied; and b the sum by which the other unknown 
 quantity is multiplied ; and c the algebraic sum of all the 
 known quantities. 
 
 For example, the equation mr+nx-j-py-\-qy — ry= l-\-k t 
 iwv be changed into v (m+n)x -t-(p + q— r )y =(/ + &), 
 where (*n4-n) = a ; (p + q— r) = b ; and {l+k) — c. 
 
§ 242.] FRACTIONS. 163 
 
 § 240* If there are two equations of this kind, there will 
 be two sets of co-efficients and whole numbers. But for our 
 present purpose it will be best to represent both sets by the 
 same letters ; only accompanying the second set with the 
 accent, to intimate that they stand for different quantities from 
 the first set. The two equations will then become 
 
 ax -\-by =c 
 
 a'x-\-b'y = c\ 
 
 § 241. In order to make the terms of y identical, by the 
 rule for common denominators, §231, we will multiply the 
 first equation by b', and the second by b, when we shall have 
 
 ab'x-\-bb'y = b'c 
 a'bx+bb'y = be' 
 
 Subtracting one from the other, (ab' — a'b)x = b'c — be' 
 
 tv li- b'c- be' 
 
 Dlvldm S' *rac35 
 
 -^ . , , . . , cb' — be' 
 
 Or, putting the accented quantities last, x = 
 
 ab' — ba' 
 
 By the same means we shall find y — — =- — #-, 
 J v ab'—ba' 
 
 § 242. By the use of accents in the notation of co-effi- 
 cients, the pupil may readily see the law by which these two 
 formulae may be found without performing the whole opera- 
 tion of eliminating. 
 
 The common denominator is found by forming the two 
 arrangements of the co-efficients ab and ba, with the sign — 
 betiveen them; and then accenting the last letter in each 
 term. Thus, ab' — ba'. 
 
 To obtain the numerator for answering each unknown 
 quantity, take away from the denominator the letter which 
 designates the co-efficient of that unknown quantity; and 
 put in its place the letter which designates the known quan- 
 tity, and then accent the last letter in each term. 
 
 Thus, first we have for the denominator, ab — ba. For the 
 answer to x, we will take away the a's and put in c's ; and 
 
164 ALGEBRA. [EQ. SEC. 11 
 
 we shall have cb — be, which, when accented, will be cb' — be', 
 as above. 
 
 For the answer to y, we take away the b's and put in c's ; 
 and we shall have ac — ca, which, when accented, will be 
 ac' — ca'. 
 
 § 243. It must be understood that a, 6, and c, stand for 
 any algebraic quantities, whether positive or negative; and 
 whenever those quantities are replaced for a, b, and c, their 
 own signs must accompany them in connection with the 
 signs of the above formulas. Thus, let there be two equations : 
 x — 2y = 8 
 — Sx + 4y = — 58 
 then we shall have a — 1 , b = — 2; c = 8, a' = — 3, 6'=4, 
 c' = — 58 ; and by substitution in the formulas, 
 
 _ c b' — bc' _' — 8 x4 — (— 58x -2) 32 -116 
 X -ab'—ba'~~ l^4-.(- 8x-8)""' 4-6 
 
 _acj-ca' _-58 xl-( 8x— 3) — 58— (— 24) 
 y— a h'—ba'~ 4x1— (— 2x— 3)" 4 — 6 
 
 Question 1. Change the above formula for the denomi- 
 nator common to their two values into a rule. 
 
 Question 2. Change the formula for the numerator in the 
 value of x, into a rule. 
 
 Question 3. Change the formula for the numerator in the 
 value of y, into a rule. 
 
 Examples. The pupil may perform a few sums in section 
 11, page 88, by these rules. 
 
§ 248.] INVOLUTION AND POWERS. 165 
 
 xin. 
 
 INVOLUTION AND POWERS. 
 
 § 244. We have already shown that when a quantity has 
 been multiplied into itself, the product is called the second 
 power; and that when it is taken three times as a factor, the 
 product is called the third power .'* In the same manner, any 
 number of times with which a quantity is taken as a factor, 
 will give its name to the product. § 163, 164. 
 
 § 245. In other words, the exponent of any quantity repre- 
 sents its name as a power. Thus, a or a 1 is the first power 
 of a ; a 2 the second power of a; a* the eighth power of a; 
 a n the nib. power of a ; 2y\ , or 2?/ 8 , or (2y) 3 , is the third 
 power of 2y. 
 
 § 246. When the first power of any quantity is compared 
 with the higher powers, it is called the root of those higher 
 powers ; because it is from that quantity that they may be 
 said to grow. Thus, x is called the root of x 2 and of # 3 , &c. 
 
 § 247. Involution is finding any power of a quantity. // 
 is performed by multiplying the quantity into itself till it 
 is taken as a factor as many times as there are units in the 
 exponent of the required power. 
 
 § 248. When a quantity is represented by a single letter, 
 the multiplication is performed by simply annexing the re- 
 quired exponent, by § 162. But, if the quantity consists of 
 
 Questions. What rule have we for naming powers? How is the 
 name represented on paper] What is called the root of powers? 
 Why? What is involution? How is it performed? How is in- 
 volution performed on single letters? 
 
 * Sometimes, for the sake of conciseness, the second power of a 
 quantity is called the square of that quantity ; and the third power 
 is called the cube. 
 
166 ALGEBRA. [eQ. SEC. 15. 
 
 two or more factors, the exponent must be annexed to each 
 of them ; and if there is a numeral factor, the multiplication 
 must be actually performed. Thus, the second power of ax 
 is found to be a*x 2 ; the third power of 2y is 2 3 y 3 , or Sy 3 ; 
 the fourth power of 3ab is 3*a 4 b*, or 81a 4 6 4 . 
 
 EXAMPLES. 
 
 1. What is equal to (2ab) 5 ? Ans. 2 5 a 5 b 5 = 32a 5 6* 
 
 2. What is equal to (2ax) 2 ? Ans. 4a a # 2 . 
 
 3. What is equal to [aby)*1 Ans. a 3 b 3 y*. 
 
 4. What is equal to (5mnx)*l Ans. 625m 4 n A x*. 
 
 § 249. Hence we learn, that any power of the product 
 of several factors, is equal to the product of their powers. 
 Thus, the fourth power of ab is (ab)*> or (a 4 b*). The third 
 power of 16 = 4096; or it equals (2x8) 3 which equals 2 3 x8 s 
 which =8x512 - = 4096. 
 
 § 250- From the preceding remarks, we derive a principle 
 that is of much importance in mathematics. It is the follow- 
 ing : (2a) 2 = 2 2 a 9 , or 4a 3 . That is, four times the second 
 power of any quantity is the same as the second power of 
 twice that quantity; as, 2 3 (x+|) 3 = 4(;r+§) 3 . 
 
 §251. We see that when an exponent is to affect only 
 one letter, it is annexed to that letter alone ; but when it is to 
 affect a quantity which is represented by more than one letter, 
 that quantity must first be enclosed by a vinculum, or paren- 
 thesis, and then the exponent is annexed to that. Thus, the 
 second power of a + b must be represented by either (a-f &) 3 > 
 or a-f-6| 3 , or a + b . The parenthetical form is generally the 
 best. 
 
 Questions, How is involution performed if there are more letters 
 than one? What if there is a numeral factor? What does the rule, 
 together with the representation of the power show? What im- 
 portant principle is derived from this ? Supposing an exponent is to 
 effect a quantity of more than one letter? 
 
§ 252.] INVOLUTION AND POWERS. 167 
 
 § 252. In involving compound quantities, it is found best, 
 for most purposes, to give them simply the proper exponent. 
 Thus, the fifth power of 2a— x = (2a — x) 5 . But there are 
 some cases in which it is necessary to perform the multipli- 
 cation in its extent. And that operation is called expanding 
 or developing the value of the expression. 
 
 5. Thus we expand or develope (2a— x) 5 , as follows : 
 2a— x 
 2a — x 
 4a z —2ax 
 
 —2dx + x* 
 4a' i —4ax+x 2 = (2a— x)*. 
 2a —x 
 
 8a 3 — 8a 2 x + 2ax* 
 
 — 4a-x-{-4ax* — x % 
 8a 3 — i2a*x+Qax^-- x 3 = (2a— x)\ 
 2a — x 
 
 16a 4 — 24a 3 x+ I2a 2 x 2 —2ax 3 
 
 — 8a* x + 12a'.r 2 — 6ax 3 +x* 
 r6a 4 — 32a 3 x -f 24a^ a — 8aa? 3 -f- x* = (2a- 
 2a —x 
 
 32a 5 —64a*x -f 48a 3 # 2 — 1 6a V + 2ax* 
 
 — 1 6a 4 x + 32a 3 .r 2 — 24a 2 , r 5 -f 8ax 4 — x 5 
 32a 5 — 80a i x + 80a'x»— 40a 2 r J +10a;r 4 — x 5 = (2a— a:) 5 
 
 EXAMPLES. 
 
 6. Expand the binomial (a-|-2:r) 4 . 
 
 Ans. a 4 -f8a , #+24a 2 :r 2 +32ar , + in;r 4 . 
 
 7. Expand the binomial (2x—3y) s . 
 
 Ans. 8x 3 —36x»y+54xy*r-21y 3 . 
 
 8. Expand the binomial (3— x) 4 . 
 
 Ans. 81— 108»+54x 2 — 12a? 3 +^. 
 
 Questions. What are we said to do in performing the mulliplico 
 Hon in involution? Do we always do that 1 
 
168 ALGEBRA. [EQ. SEC. 15. 
 
 9. Expand the trinomial (a+b — c) 9 . 
 
 Ans. a 3 +2a&— 2ac+b*— 2&c+c a 
 
 10. Expand the trinomial (x 9 — 2a? + l) 3 . 
 
 Ans. a? 9 — 6a? 5 + 15a? 4 — 20a? 3 + 1 5a? 9 — 6a* + 1. 
 
 0C?* If the binomial has but one letter in each term, the 
 square is immediately known, from § 179 and § 180. 
 
 § 253. The powers of a fraction are found by raising both 
 numerator and denominator to the power required. 
 
 Thus, (• y - g of f of t - 2 v ; ay - 3. 
 
 EXAMPLES. 
 
 11, 
 
 What is the sixth power of j- Ans. 
 
 3a? 27a?* 
 
 12. What is the third power of — ? Ans. —— 3 . 
 
 if if 
 
 13. What is the sixth power of -^? Ans. — -£-. 
 
 r x a? 8 
 
 14. What is the fourth power of -^? Ans. n f A 
 
 r 5a . 625a* 
 
 §254. It was shown, §167, that a*xa* = a*; and that 
 fl s xa 3 xa 3 xa 3 = a ,a . In each of these cases, the exponent 
 was added as many times as the quantity was to be taken as 
 a factor; by which we see that when a quantity has already 
 an exponent, it is raised to any power by multiplying the 
 exponent by the exponent of the required power. Thus, the 
 "ourth power of a 3 is a 3X4 = a 13 . 
 
 § 255. As to the signs of the powers, by the principle of 
 multiplication, § 172, if the root is plus ( + ), all the powers 
 are plus; but if the root is minus (—-), all the even powers 
 are plus, and all the odd powers are minus. The second 
 power of —a is —ax —a, which is -fa 3 ; the third power is 
 
 Questions. How do we find the powers of fractions 1 How do 
 we involve a quantity that has an exponent? What is the rule for 
 signs of the different powers * 
 
§ 256.] INVOLUTION AND POWERS 169 
 
 -\-a a x — a, which is — a 3 ; the fourth power is — a*X — «» 
 which is -fa 4 ; the fifth power is —a 5 . 
 
 EXAMPLES. 
 
 15. What is the third power of 6V? Ans. b«x 9 . 
 
 16. What is the fourth power of a*y*xt Ans. a™y 8 x 4 . 
 
 17. What is the fifth power of bc 3 x»1 Ans. b*c i5 x i0 . 
 
 18. What is the third power of ab*x 2 y1 Ans. a s b 6 z 6 y 3 . 
 
 19. What is the third power of —3x*y 3 ? Ans. —27x 6 y 9 . 
 
 „„ . , ,. 3a*x\ 2187a 14 # 31 
 
 20. What is the 7th power of -— ? Ans. y » 
 
 3z 27z 3 
 
 22. What is the nth power of a 8 ? Ans. a 3n . 
 
 2xr 2 Sx^r 9 
 
 23. What is the third power of — — ? Ans. — — — . 
 r 3y 27t/ 3 
 
 24. What is the nth power of — - ? Ans 
 
 ax 
 21. What is the third power of — — — ? Ans. 
 
 ay 1 " a r y mn 
 
 w , . -. i n — a 3 x(d+m) , 
 
 25. What is the second power of — ; — ^-r- — '- ? 
 
 r (x+iy 
 
 (x-f l) 6 
 
 26. What is the third power of —fa? 3 ?/ 8 ? Ans. — &x B y 9 . 
 
 § 256. If we divide any power by its root, we obtain the 
 
 a 5 
 power next below. Thus, a s -r-a = a*, or— = a 4 . Again, 
 
 a* a 3 a 3 
 
 a 4 -r-a = a 8 , or — = a 8 ; — ==a 8 ; — = a. Let us proceed; 
 a a a 
 
 — =. 1 ; but if we diminish the exponent, as we did in the 
 
 preceding divisions, we have - = a . Hence we learn, that 
 a is equal to 1, however great or small the value of a 
 may be. 
 
 Questions. What power of any quantity is equal to 11 How do 
 we learn this * 
 
 15 
 
170 ALGEBRA. [EQ. SEC. 15. 
 
 This may be shown in numbers. 49 is the second power 
 of 7. Now, 49-r-7 = 7, theirs* power; and 7-7-7 = 1, the 
 no power. 
 
 § 257. But we can proceed still farther with our division. 
 
 l-r-a = -; — 7-a = — : — -a = — ; <fcc. Or, taking a'\ in- 
 a a a 3 a 9 a 3 5 
 
 stead of 1, and dividing by subtracting the exponent of the di- 
 visor, § 190, a°-7-a = a 0_1 or a' 1 ; a~ 1 -7-a = cr 1 ~ 1 or a -3 ; 
 
 § 258. Thus, by pursuing two different methods of di- 
 viding, we obtain two different sets of expressions, both of 
 which are equal to one another. We will arrange them one 
 under the other, so that any one in the upper line shall be 
 equal to the corresponding one in the lower line. 
 
 1 11 1 1 ^ 
 
 I 
 
 a* a 3 a 3 a 1 I - ~i — ; -r — 
 
 a a 3 a 3 a* a c 
 
 a 5 a* a 3 a 9 a 1 a ar x a" 3 or 3 a* a -5 ) 
 Here, the quantities on the right of 1 in the upper line, and 
 on the right of a° in the lower line, are called the reciprocal 
 powers of a. They may be read the^r^ reciprocal power, 
 the second reciprocal power, &c. 
 
 § 259. By this method of notation, we may remove the 
 denominator of any fraction. For, we have only to put it by 
 the side of the numerator, and give it the negative exponent. 
 
 IJhus, — —ax— ,-=flxr a - = ar 2 . So - = -•■flX-r« 
 ▼^ x* x* x x l x l 
 
 ssrfl.r' 1 ; where we see that if the denominator has not any 
 
 written exponent, 1 is understood, and must be written when 
 
 it becomes negative, in order to distinguish the quantity from 
 
 a positive one. 
 
 Question. Show the principle by numbers. In how many methods 
 of notation can we continue to divide by the root 1 What are the 
 corresponding quantities below 1? What are they called 1 ? What 
 are we enabled to do by these two methods of notation 1 How 1 
 
§261.] INVOLUTION AND POWERS. 171 
 
 EXAMPLES. 
 
 Remove the denominators from the following fractions 
 
 a 
 
 27. — = abn-ixr 1 . 
 nx 
 
 clx 2 
 
 28. r— - = axty-ty-*. 
 by* * 
 
 be 
 
 29. . . . -o bcar*m-*nr\ 
 
 30. r — a(a—b)-h 
 
 a — b v ' 
 
 32. J£L- = m(2— a)" 1 . 
 2 — a v ' 
 
 § 260. It is very evident that these quantities with nega- 
 tive exponents can be returned to the denominator without 
 affecting their values. And on the same principle, any re- 
 ciprocal power in the numerator may be removed to the 
 
 ax~ 2 
 denominator. For, suppose we have the quantity - — . We 
 
 know that it is equivalent to — xar* which = — x — which 
 
 by by x 2 
 
 __ a , ax~ 2 a 
 ~~ byx 2 ' by .- ■ byx 2 ' 
 
 T l_ 1 "c , x 
 
 In the same manner, ab~ x c == -p '* and a~ 2 x\r* = — * . 
 
 b . u a*y 
 
 §261. Powers with negative exponents are involved in 
 
 the same manner as if they were positive. Thus, the second 
 
 power of a~ 2 is a -2 * 3 or ar* ; because the second power of 
 
 — is — = ar 4 . The third power of b~ 2 is 6~ 8 , &c. 
 a 2 a* * 
 
 EXAMPLES. 
 
 K 
 
 a 8 
 
 33. What is the fourth power of cPb" 1 ? Ans. a 9 b~* = j-^ 
 
 x° 
 
 34. What is the third power of x*y~ 2 ? Ans. x 9 y-° = — • 
 
 if 
 
 b mn 
 
 35. What is the nth power of b m x~ m ? Ans. — ji 
 
 x 
 
 Question. What if there are reciprocal powers in the numerator? 
 How are reciprocal powers involved ] 
 
172 ALGEBRA. [EQ. SEC. 15. 
 
 XIV. 
 
 evolution: 
 
 § 262. Whenever we meet with a quantity that is a power, 
 we know that there must be some other quantity which is the 
 root of that power. See §244, 246. 
 
 The operation of finding the root of any power is called 
 Evolution. But as for our purposes, it will be best to sub- 
 divide this part of algebra, we shall at present treat of so 
 much of Evolution as relates to finding the xoot by mere in- 
 spection; and leave that part of it which is generally called 
 Extraction of Roots for a future chapter. 
 
 § 263. When a quantity is reduced from its second power 
 to its root, we say we have found its second root ; when the 
 reduction is from the third power to the root, we say we have 
 found the third root, &c. 
 
 § 264. Definition. — The root of any quantity is a fac- 
 tor which being multiplied into itself a certain number of 
 times will produce that quantity. 
 
 § 265. As the second power is found by multiplying the 
 exponent of a root by 2, (§ 254,) so the second root will be 
 found by dividing the exponent of a power by 2. The third 
 root will be found by dividing the exponent of the power by 
 3. And, in general, any root may be found by dividing the 
 exponent of the given power by the number expressing the 
 root to be found. Thus, the second root of a 6 is a 3 , because 
 c J xa 8 = a 8 . The third root of a 8 is a 3 , because a a xa 3 X« a 
 = a 6 . 
 
 Questions. What is Evolution? What is a root of a quantity! 
 How are numeral names given to roots? How do we find the se- 
 cond root of simple quantities? 
 
§ 268.3 EVOLUTION. 173 
 
 § 266. If there are several factors in the quantity, the root 
 of each factor must be taken. The third root of a 9 b 3 c B is 
 a 3 bc 2 ; for a 3 6c 3 x a 3 bc 2 x a B bc a = a 9 b 3 c 6 . 
 
 § 267. If the quantity have a numeral co-efficient, the root 
 of the co-efficient must be taken arithmetically. The second 
 root of 16a s x 6 is 4a*x a . 
 
 §268. The root of a fraction is found by taking the 
 root of both numerator and denominator. The third root of 
 8a 6 . 2a 3 2a 9 2a*_ 2a 3 _ Sa^ 
 
 27aT» 1S 3s 5 ; 3x 3 X 3^ X 3^ ~~ 27a; 8 
 
 EXAMPLES. 
 
 1. What is the second root of 9a*? 
 
 2. What is the third root of Sx B ? 
 
 3. What is the third root of 64b° 1 
 
 ■ii* 
 
 4. What is the second root of — ^-r ? 
 
 « 4 c 6 
 
 5. What is the second root of 8lx*y*1 
 
 6. What is the third root of 27a 3 6 6 ? 
 
 7. What is the third root of -^- ? 
 
 27x 9 
 
 8. What is the fourth root of 256a 4 x 8 ? 
 
 8 a 3 
 
 9. What is the third root of — -* ? 
 
 125x e 
 
 10. What is the fourth root of S\y s x A 2 ls l 
 
 1 6a*b* 
 
 11. What is the second root of --=- -_ ? 
 
 49x 6 y 2 
 
 12. What is the second root of a 2 b~*l 
 
 13. What is the third root of 8a 6 c- fl rf- 3 ? 
 
 14. What is the third root of 64ar 3 ? 
 
 Questions. Supposing the quantity has several factors ? A nu- 
 meral co-efficient 1 How for fractions ) 
 16* 
 
 Ans. 
 
 3a s . 
 
 Ans 
 
 . 2x. 
 
 Ans 
 
 . 4b 3 . 
 
 Ans. 
 
 xy» 
 a 2 c 3 
 
 Ans. 9x*y*. 
 
 Ans. 
 
 3ab\ 
 
 Ans. 
 
 2a 3 
 3x*' 
 
 Ans. 
 
 4ax». 
 
 Ans. 
 
 2a 
 5x» 
 
 Ans. 'Sy 2 xz 3 . 
 
 Ans. 
 
 4a 2 b 
 lv»y 
 
 Ans. 
 
 ab~*. 
 
 Ans. 2a 2 c 
 
 - 3 d"\ 
 
 ns. 4x x - 
 
 4 
 
 ~~ x' 
 
174 ALGEBRA. [e<*. SEC. 15. 
 
 § 269. We may express the division of exponents in the 
 same manner that is used for the division of other algebraical 
 
 quantities. Thus, the second root of a* is a 2 - = a 2 ; the 
 
 fi 4 
 
 third root of a 6 is a 7 - = a ; the fourth root of a* is a* - = a 1 
 
 or a. By the same rule, the second root of a 1 is a? ; and the 
 
 l 1 
 
 third root of a 1 is « 3 ; the second root of a 3 is a 2 ; the third 
 
 2 l 
 
 root of a 9 is a 1 ; the nth root of a is a* ; &c. 
 
 " ■ 1 
 § 270. It is on this principle that the exponent * is now- 
 used as the sign of the second root, and the exponent 7 is the 
 sign of the third root.* * 
 
 § 271. Formerly, the sign of a root was indicated by the 
 radical sign </ ; and, for some purposes, this sign is still in 
 use.t Whenever it is used, it is placed to the left of the 
 quantity ; thus, v/a. The number of the root is denoted by 
 a little figure placed over the radical sign; unless it is the 
 second root, when the figure 2 is omitted. Thus, 
 y/a is the second or square root of a. 
 $fa is the third or cube root of a. 
 Z/a is the nth root of a. 
 \/a + x is the second root of (a -fa*.) 
 IC7* It must be remembered, that if the radical sign is 
 to effect more than one factor, the vinculum must be used 
 with it ; thus, \/5a, V ab. When the radical sign is placed 
 before a fraction, it must embrace both terms, because the 
 
 line in the fraction forms a vinculum ; thus, */— or — . 
 
 a \a 
 
 y/— is not the same as — — . In the first case, it is the 
 
 Questions. How may exponents be divided? What signs have 
 been derived from this fact? What other sign for roots? Exan 
 pies ? How is it used with fractions ? 
 
 * This method was introduced by Simon Stevinus, of Holland, 
 about 1585. 
 
 | Invented by Stifelius, in 1544. 
 
5 275.] EVOLUTION. 175 
 
 square root of the whole fraction ; in the last, it is the root 
 of the numerator, divided by 7. 
 
 § 272. With regard to the signs, a positive even power is 
 formed from either a positive or negative root ; and theretore, 
 an even root of a positive quantity is either positive or nega- 
 tive. Thus, the second root of a 2 is either +« or — a ; be- 
 cause -f a x -}-« = a 2 , and — a x — a = a*. Hence the root 
 is said to be ambiguous, and is marked with both signs, ±a. 
 The second root of 2x is ± \^2x. 
 
 § 273. The odd root of any quantity will have the same 
 sign as that quantity. The third root of — a 3 is — a. 
 
 § 274. There cannot be an even root of a negative quan- 
 tity ; because no quantity can be multiplied into itself an even 
 number of times, in such a manner as to make a negative 
 power. §255. See also §289. 
 
 §275. There are many quantities whose roots cannot be 
 exactly found. Thus, the second root of 2 can never be 
 found. The root is then expressed either by the radical sign, 
 or a fractional exponent; and called a surd, or irrational 
 quantity. The second root of 6 is ^/6; the second root of 
 x is y/x or x* ; the third root of a -f- x is (a+#) 7 or x^a-fa?. 
 
 EXAMPLES. 
 
 Express the roots of the following quantities, first by the 
 radical sign, and then by the fractional exponents. 
 
 15. The second root of ax 1 Ans. V ax or (ax) 2 . 
 
 16. The third root of a—yl Ans. *&a—y, or (a— y) 3 . 
 
 17. The second root of a 2 x ? 
 
 / l i. 
 
 Ans. v a 2 x, or a^/x, or (a 2 x) 2 , or ax^. 
 
 18. The third root of (a 2 — a) ; and the third root of x*1 
 
 19. The second root of a 3 ; and the second root of ay 3 1 
 
 Questions. Are roots positive or negative 1 Supposing the name 
 of the root is an odd number % What are surds 1 
 
176 ALGEBRA. [EQ. SEC. 15. 
 
 THE SQUARE ROOT OF BINOMIALS. 
 
 § 276. We have shown, § 179, that the second power or 
 square of a+b is a' i -\-2ab-\-b 2 ; and that the second power 
 of a—b is a 3 — 2ab+b*. For distinction's sake we call a-f b 
 a binomial, and a— »6 a residual.* 
 
 §277. Now, the second power of a binomial consists of 
 three terms, the first and last of which are complete powers, 
 and the middle one is twice the product of the roots of the 
 two powers; and all connected together by the sign -f-« 
 Therefore, whenever we meet with any such quantity, we 
 know immediately the second root of it is a binomial, and 
 is found by taking the roots of the two terms which are 
 complete powers, and connecting them by the sign -f . 
 
 § 278. The second power of a residual is altogether like 
 that of a binomial, excepting the sign before the middle term 
 is — . The root of such a quantity is taken just as if it 
 were a binomial, only making the connecting sign a — mi- 
 nus instead of a plus. 
 
 EXAMPLES. 
 
 1. What is the second root of a 9 +2ax-\-x 2 i. Ans. a + .r. 
 
 2. What is the second root of x*+2x+ 1? Ans. x + l. 
 
 3. What is the second root of 4x* — 4ax+a*l Ans. 2x — a. 
 
 4. What is the second root of x* — x+± ? Ans. x — £. 
 
 5. What is the second root of x*-\- xy + — ? Ans. x-\-~. 
 
 * 4 2 
 
 6. What is the second root of a + 2y/a~b + bl Ans. a? + b*. 
 
 7. What is the second root of 16a 3 — \6ab-\-4b' 2 l 
 
 Ans. 4a — 2b. 
 
 Questions. What is the difference between a binomial and residual ? 
 How do we know when the second root of a quantity is a binomial ] 
 How is it found ? How is a residual root found ? How can you 
 tell whether a trinomial quantity is a perfect square'? 
 
 * These terms were introduced by Dr. Recorde, in 1557. 
 
§281.] EXTRACTION OF THE SECOND ROOT OF NUMBERS. 171 
 
 EXTRACTION OF THE SECOND ROOT OF NUMBERS. 
 
 § 279. From the principle last shown, we derive a rule foi 
 finding the second root of numbers. For, supposing we have 
 a number of two digits, say 54 ; it may be represented by 
 a-{-b ; that is, 50 may be expressed by a, and 4 by b, so that 
 50 and 4=a-\-b. We have then the equation a-f-^ = 54; 
 a equaling the tens, and b equaling the units. 
 Involving both members, we have a 2 -\-2ab +b 2 ■= 2916. 
 
 §280. Now, if any of my pupils should see this last 
 equation without being told what was the root we first em- 
 ployed, he would immediately know, by §277, that the 
 second root of the first member is a +b, and of course, the 
 root of 2916 will also consist of two quantities that may be 
 represented by a and b. Thus, v/2916 = a-f&. 
 
 It would be only necessary to inform him that a stands for an 
 even number of tens, and b stands for a quantity less than ten. 
 
 § 281. As it is known that the second power of 10 is 100, 
 the second power of 20 is 400, the second power of 80 is 
 6400, &c. ; it is evident that a (which stands for tens) must 
 be the root of the hundreds. Therefore, to find a, is simply 
 to find what is the greatest second power of 29 in the hun- 
 dred, and then to take the root of it. The second power is 
 25 hundred, and the root is 5; for 6 times 6 = 36, which is 
 too much. But the root of hundreds is in tens ; therefore, 
 a = 50, and a 2 = 2500. 
 
 Questions. From what principle do we derive a rule for extracting 
 the second root of numbers'? In what manner may a number be 
 similar to a binomial? Which letter represents the tens? How 
 many figures in the power are represented by b 1 How can we find 
 the second root of a number represented by a 2 1 And then the value 
 of a? Then what part of the binomial square is left 1 
 M 
 
178 ALGEBRA. £EQ. SEC. 15. 
 
 § 282. Taking a 8 from the first member of the above equa- 
 tion, and 2500 from the second, we have left the equation 
 
 2a6+6 8 = 2916— 2500 -=416. 
 As a = 50, 2a = 100 ; hence, substituting the value of 2a, 
 we have 1006+6 3 = 416. And, of course, if we divide 416 
 by 100, we shall find very nearly the value of 6. 416-7-100 
 = 4, which we will suppose to be the value of 6. 
 
 § 283. Let us see whether b does actually equal 4. Our 
 last equation was 1 006 + 6 8 = 4 1 6 
 
 But 1006+6 8 = (100 + 6) Xb. See § 196. 
 Substituting (100+6) x6, for 1006 + 6 8 , (100+6)x6 = 416. 
 If 6 = 4, then (100+4) x4 should =416. 
 
 which is the fact. And therefore, 6 does equal 4. Hence, 
 a+6 = 50 + 4 - = 54, the second root of 2916. 
 
 § 284. We have now algebraically discovered the following 
 rule for extracting the second root of any number that con- 
 sists of not more than four figures. Find the greatest second 
 power of hundreds, and put its root in the answer. Sub- 
 tract that second power from the whole number, and divide 
 the remainder by twice the root already found. Suppose the 
 quotient to be the number that completes the root; add it to 
 the root already found, and also to the last divisor. Mul- 
 tiply the divisor thus augmented, by the quotient figure last 
 found, and subtract the product from that remainder which 
 was divided. If there is nothing left, you have found the 
 true root. 
 
 § 285. It sometimes happens that the product which is 
 obtained by multiplying the augmented divisor, is greater than 
 the number which was divided. In such cases, the figure last 
 put in the root is too large ; and therefore it must be erased, 
 and a smaller figure substituted. 
 
 Questions. Of which factor can we find the value? How? 
 What can we do with this value ? What is the rule for extracting 
 the root as now explained? What little alteration is sometimes 
 required ? 
 
§287.] EXTRACTION OF THE SECOND ROOT OF NUMBERS. 179 
 
 7396(80 
 6400 
 
 160)996(6 
 166)996 
 
 Ans. 
 
 68 
 
 Ans. 
 
 72 
 
 Ans. 
 
 23 
 
 Ans. 
 
 42 
 
 Ans. 
 
 98 
 
 Ans. 
 
 48 
 
 EXAMPLES. 
 
 1. What is the second root of 7396 ? 
 Ans. a 2 must equal 6400 ; and a 1 = 80. 
 
 Therefore the remainder 2ab + b* = 996. 
 2a = 160, which is contained in 996, 6 
 times ; for b times (2a + b), or 6 times 
 160 + 6 = 996. Therefore the second root 
 of 7396 = 86. 
 
 2. What is the second root of 4624 ? 
 
 3. What is the second root of 5184? 
 
 4. What is the second root of 529 ? 
 
 5. What is the second root of 1764 ? 
 
 6. What is the second root of 9604 ? 
 
 7. What is the second root of 2304 ? 
 
 §286. It will be found by trial, that if a number is in 
 units, its second power will be in tens; for the square of 
 9 = 81. If the number is in tens, its second power will con- 
 tain more than two figures, and less than five; for it will take 
 the square of 100 to make 10000. The principle is general, 
 that every figure in the root except the left-hand one, re- 
 quires two figures in the second power; and for the left- 
 hand figure, there may be either one or two in the power. 
 
 Hence, by pointing off the number from the right, in 
 periods of two, (as in the next example,) we may learn how 
 many figures will be in the root. 
 
 § 287. When it is found that the root will consist of more 
 than two figures, we are to suppose at first that a 3 is repre- 
 sented by the first period ; and after the first figure in the 
 root is found, the first two periods represent a 2 ; and so period 
 
 Questions. How many figures in the power belong to each figure 
 in the root ? How then shall we learn the number of figures in 
 the root? How is the operation performed when the root contains 
 more than two figures 1 
 
180 
 
 ALGEBRA. 
 
 [EQ. SEC. 15 
 
 tfter period, a represents that part of the root which is found, 
 and b embraces all which remains to be found. 
 
 8. What is the second root of 55225 ? 
 
 55225(200 
 40000 
 
 400 
 _£0 
 
 4J0 
 
 460 
 
 5225(30 
 
 Cl5i 
 1 1290( 
 
 Operation. At first, a = 200, and 
 b = 30. Afterwards, a = 230, and 
 
 1 C 2325(5 
 m I 2325 
 
 § 288. The operation may be abbreviated by omitting the 
 ciphers in the roots and powers, as in 
 the margin. When we do this, how^ 
 ever, we must remember that a repre- 
 sents an order ten times as great as 
 that represented by b ; and therefore, 
 when we double it for a divisor, we 
 must annex a cipher to it, to bring it 
 to the same denomination as b. 
 
 9. What is the second root of 2125764 ? 
 
 10. What is the second root of 20736 ? 
 
 55225(235 
 4 
 
 ^$152 
 «! J 129 
 
 «° C 2325 
 m I 2325 
 
 11. What is the second root of 10342656? 
 
 12. What is the second root of 36372961 ? 
 
 Ans. 1458. 
 Ans. 144. 
 Ans. 3216. 
 Ans. 6031. 
 
 § 289. If the number to be evolved is a vulgar fraction, it 
 must be reduced to decimals, which are to be pointed off from 
 units towards the right, instead of towards the left. 
 
 Thus, v/J ? — ^.25 - = .5 v/f = ^.428571428, &c. 
 
 It will be pointed off thus, .42857142, &c. The pupil will 
 find the answer to be .65465 + . 
 
 Question. Suppose the power is a vulgar fraction 1 
 
§ 293.] PURE QUADRATIC EQUATIONS. 1R1 
 
 SECTION XV. 
 
 PURE QUADRATIC EQUATIONS. 
 
 § 290. There are several kinds of equations. We have 
 hitherto attended to that kind which contains an unknown 
 quantity in the first degree only. But there are equations 
 that contain an unknown quantity in the second power ; and 
 also in the third power, and so on indefinitely. 
 
 §291. In order to distinguish the different kinds of equa- 
 tions, we are accustomed to call those which contain only the 
 first power of the unknown quantity, simple equations, or 
 equations of the first degree. Those which contain the se- 
 cond power of the unknown quantity, are called quadratic 
 equations, or equations of the second degree. Those which 
 contain the third power of the unknown quantity, are called 
 equations of the third degree, &c. 
 
 § 292. There are two kinds of quadratic equations. Those 
 which contain only the second power of the unknown quan- 
 tity, are called pure quadratic equations ; those which contain 
 both the second and the first power of the unknown quantity, 
 are called affected quadratic equations.* 
 
 § 293. A "pure quadratic equation is reduced by simply 
 extracting the root of both members. 
 
 Questions. How do we distinguish different kinds of equations'? 
 How many kinds of quadratics'? How do we reduce a pure quad- 
 ratic equation ? 
 
 * All pure equations are called by some algebraists simple 
 equations. 
 
 16 
 
182 ALGEBRA. [SECTION XV. 
 
 EQUATIONS.— SECTION 15. 
 
 1. There is a number, such, that by adding 5 to it for one 
 factor, and subtracting 5 from it for another factor, we may 
 obtain 96 for the product. What is that number ? 
 
 Stating the question, Let x = the number. 
 
 a?+5 = one factor. 
 
 x — 5 = the other. 
 (x+5)x(#— 5) = the product 
 Forming the equation, (a?+ 5) x (a?— 5) = 96 
 
 Multiplying, § 172, a: 3 — 25 = 96 
 
 Transposing and uniting, x a = 121 
 
 Extracting the root, x = ± 1 1. 
 
 ICT" In this example, we have given 1 1 the uncertain sign, 
 because 121 is the second power of either +11 or — 11. But 
 the x* we know to be the second power of +#, because x 
 was + in the supposition. We have therefore found that 
 the required number is either 11, or 11 less than nothing. 
 Both values will satisfy the conditions of the question. If 
 11 = #, then ll-f5=16, and 11 — 5 = 6. And 16x6 = 96. 
 If — ll=;r, then —11 + 5 = 6, and— 11— 5= -16. And 
 6x — 16 = — 96. 
 
 § 294. In this manner, every quadratic equation will have 
 two answers; but the conditions of the question will generally 
 determine which answer is the one required. In pure quad- 
 ratics, we generally suppose the positive answer to be the 
 true one. 
 
 2. What two numbers are those which are to one another 
 is 3 to 5 ; and whose squares, added together, make 1666? 
 
 Questions. Why db in the answer ? How many answers then 
 iii quadratics ? 
 
$ 294.] PURE QUADRATIC EQUATIONS. 183 
 
 Stating the question, 
 
 Forming the equation, 
 
 Multiplying and uniting, 
 Dividing by 34, 
 Extracting the root, 
 
 x = 
 
 the greatest. 
 
 Sx 
 
 1F = 
 
 the least. 
 
 x» = 
 
 square of the greatest. 
 
 9x* 
 25" = 
 
 : square of the least. 
 
 . 9# a 
 
 x 2 4 = 
 
 ^25 
 
 :1666 
 
 34z 3 = 
 
 :41650 
 
 # 3 = 
 
 = 1225 
 
 x = 
 
 :±35 the greatest. 
 
 | of 35 = 21 the least. 
 
 3. The distance to a certain place is such, that if 96 be sub* 
 traded from the square of the number of miles, the remainder 
 will be 48. What is the distance ? Ans. 12 miles. 
 
 4. There is a field containing 108 square rods; and the 
 sum of the length and breadth is equal to twice their differ- 
 ence. Required the length and breadth ? 
 
 108 
 ICT* In stating the question, x= the length, and = 
 
 the breadth; because the two multiplied together make 108, 
 the area. In this operation, there will be an equation having 
 — x z for one member. But as by § 274, we cannot find the 
 second root of a negative quantity, we must first change the 
 signs of all the terms. Ans. Length, 18 ; breadth, 6. 
 
 5. There is a rectangular field, whose breadth is £ of the 
 length. After laying out £ of the whole ground for a garden, 
 it was found that there were left 625 square rods for mowing 
 Required the length and breadth of the field ? 
 
 Ans. Length, 30 rods ; breadth, 25. 
 
 6. Two men talking of their ages, one said that he was 94 
 years old. Then, replied the younger, the sum of your age 
 and mine, multiplied by the difference between our ages, will 
 produce 8512. What is the age of the younger? 
 
 Ans. 18 years* 
 
184 ALGEBRA. [SECTION XV. 
 
 7. If three times the square of a certain number be divided 
 by 4, and if the quotient be diminished by 12, the remainder 
 will be 180. What is the number ? Ans. 16. 
 
 8. A person bought a quantity of sugar at such a rate, that 
 the price of a pound was to the number of pounds as 4 to 5. 
 If the cost of the whole had been 45 cents more, the numbei 
 of pounds would have been to the price of a pound as 4 to 5. 
 How many pounds did he buy, and what was the price per 
 pound ? 
 
 Operation. — Let a? = number of pounds, and y = the price, 
 
 and xy = the whole cost. Then, by the first condition, 
 
 4x 
 y = — . If x pounds cost xy-\- 45, then one pound would cost 
 o 
 
 -^ . Therefore, by the second condition, x = — — ; 
 
 x 5a? 
 
 4a? 1 6a? 3 
 
 or. 5a: 9 = 4zi/ +180. But y = — ;.•. 4xy=-—. Whence 
 
 16a? 9 
 the equation 5a? 9 mm — - — (-180. 
 
 Ans. 10 pounds, at 8 cts. a pound. 
 
 § 295- The pupil will find that whenever both unknown 
 quantities are found in one term, the second method of ex- 
 termination (§131) is the best. 
 
 9. There are two numbers whose product is 144, and the 
 quotient of the greater by the less is 16. What are the num- 
 bers ? Ans. 48 and 3. 
 
 10. What two numbers are those, whose sum is to the 
 greater as 11 to 7; the difference of their squares being 132? 
 
 Ans. 14 and 8. 
 
§ 300.] AFFECTED QUADRATIC EQUATIONS. 185 
 
 SECTION XVI. 
 AFFECTED QUADRATIC EQUATIONS. ' 
 
 § 296. It is but very rarely that we find quadratic equa- 
 tions so simple as those in the preceding section. Most gene- 
 rally, when reduced to their simplest terms, they contain the 
 second power of the unknown quantity in one term, the first 
 power of the unknown quantity in another term, and a known 
 quantity as a third term; thus, a? 9 +2a?==24. 
 
 § 297- Such equations are called affected quadratic equa- 
 tions, because the operation on the second power is affected 
 by a different power of the same quantity.* 
 
 § 298. It is of no importance how many terms are found 
 in an equation, provided they are the first and second powers 
 of the unknown quantity, and known numbers ; because they 
 can all be united so as to form but three. Thus, x-\-&—2x* 
 -f3a?-f-10 = 5a? 3 , can be united so as to form — 7a? 2 -f-4a? = 
 — 18; or 7a? 3 — 4x = 18. 
 
 § 299. Equations of this kind would seem at first to be 
 very difficult of solution ; but by the application of a principle 
 which is found in raising a binomial to its second power, and 
 evolving it again, we manage them with very little trouble. 
 
 § 300. Our object, then, in the first place, is to form from 
 these two terms which contain the unknown quantity, a com- 
 plete second power of some binomial. Of what quantities 
 this binomial is composed, we care not; say a? 3 -j-2aa?+« 3 . 
 
 Questions. What is the most general form of quadratics ? "What 
 are such equations called ? How many terms may they contain % 
 What is the first procedure in solving such equations ? 
 
 * This name was introduced by Vieta, about the year 1600. It 
 is derived from the Latin affedo, to pester or trouble. 
 
 16* 
 
186 ALGEBRA. [SECTION XVI. 
 
 § 301. Now, the second power of a binomial, (see § 179,) 
 has in its jirst term, the second power of the first term in the 
 binomial. Thus, 
 
 Binomial, x-\-a. 
 
 Second power, # s +2ax-|-a 3 . 
 For this, we use the second power of the unknown quantity 
 in the equation, say x*. 
 
 § 302. Again, the second power of the binomial has in its 
 second term, the first power of the first term in the binomial 
 as one factor ; and twice the last term of the binomial as the 
 other factor. That is, it has 2a times x. Thus, 
 Binomial, x+a 
 
 Second power, x 9 -f2aa?-|-a 9 . 
 For this, in our equation, we use the term that has x the first 
 power, in it. And we also determine, that whatever its co- 
 efficient is, that co-efficient shall represent the value of 2a. 
 
 § 303. Finally, the last term in the second power of the 
 binomial is the second power of half that other factor which 
 accompanies x in the second term; that is, it is the second 
 power of once a. Therefore, in our equation, whatever was 
 the co-efficient in the second term, the second power of half 
 that co-efficient will be the last term which it is necessary to 
 add, in order to make up the complete second power of a 
 binomial. 
 
 EXAMPLES. 
 
 1. Suppose we have the equation x*+4x = 96. 
 
 ICT" We are first to make the left-hand member similar to 
 the formula x 3 -{-2ax+ a 3 . The co-efficient of the second term 
 in the question is 4 ; and therefore 4 represents the 2a in the 
 formula. The half of 4, which is 2, will represent a : and 
 
 Questions. In this operation, what part of the binomial square do 
 we make of x 2 ? What do we consider as represented by the co- 
 efficient of x in the second term ] Why 1 How do we obtain the 
 last term of the binomial square? Why is this correct? 
 
§ 303.] AFFECTED QUADRATIC EQUATIONS. 187 
 
 the second power of 2, which is 4, will represent a 2 . There- 
 fore, from the left-hand member of the equation we may form 
 the complete second power x 2 -{-4x-\- 4. But if we add 4 to 
 this member, we must also add 4 to the right-hand member, 
 to preserve the equation. We shall then have the equation 
 
 a? 3 +4#+4 = 96-f4 
 Or, # 3 -+4x-f4 = 100 
 
 Extracting both sides, §277, 284, #-f 2 = ±10 
 Transposing 2, x = 8 or — 12. 
 
 Proof. If x = 8, then x 2 +4x = 64 + 32 - = 96. 
 If x=— 12, then x 2 +4x = 144— 48- = 96. For, it must be 
 remembered that if x is — 12, then 4x = — 48, which is to 
 be added to 144; and adding — 48, is done by writing it im- 
 mediately after the 144 with its own sign. 
 
 2. Given x 2 -\-6x= 16, to find the value of x. 
 
 Here, 6 = 2a in the formula. .*. a = 3, and a 2 = 9. There- 
 fore, completing the square, # 9 -f-6.r+9 = 16-f 9- = 25. 
 Extracting the root, x-\- 3 = ±5 
 Transposing, x = 2 or — 8. 
 
 3. Given rr 3 +8x = 33. 
 
 Half the co-efficient of the second term is 4 ; and the second 
 power of 4 is 16. Therefore, 
 the square is completed thus, x 2 -{-Sx-\-\& = 33-f 16. 
 
 Ans. x = 'S or — 11. 
 
 4. Given # 3 -f 4# = 32, to complete the square. 
 
 Ans. Half the co-efficient of the second term is 2 ; and the 
 second power of 2 is 4. Therefore, the square, when com- 
 pleted, is x 2 +4x + 4 = 32 + 4. And x — 4 or —8. 
 
 5. Given x 2 — 4x = 32, to complete the square. . 
 Operation. — The left-hand member of this equation is a 
 
 part of the second power of a residual quantity. See § 276. 
 Now, as the second power of a residual is the same as that 
 of a binomial, with the exception of the second term being — 
 
188 ALGEBRA. [SECTION XVI. 
 
 instead of + , the square is completed in the same manner as 
 before. Thus, half of the co-efficient is 2, and the second 
 power of 2 is 4. Therefore, when the square is completed, 
 it is x*— 4x+4 = 32 + 4-=36 
 
 Extracting the root, see § 278, . x— 2 = ±6 
 
 And transposing, a? = 8 or — 4. 
 
 6. Given X s — Gx = —8, to find x. 
 Completing the square, x* — 6:r+9 = — 8-f-9 - =1 
 
 Extracting the root, x — 3 = ± 1 
 
 Transposing, x = 2 or 4 
 
 Complete the square in the following equations, and find 
 the value of x. 
 
 7. 
 
 a; a =4;r = 140. 
 
 
 Ans. 
 
 07= 10 or ■ 
 
 -14, 
 
 8. 
 
 a- s +8x = 65. 
 
 
 Ans. 
 
 a? = 5 or - 
 
 -13. 
 
 9. 
 
 a:"— 4a? = 45. 
 
 
 Ans. 
 
 x = 9 or — 
 
 -5. 
 
 10. 
 
 x*— 15a? = —54. 
 
 
 The half of 15 is — 
 2 
 
 ; anc 
 
 completing the square, 
 
 X*- 
 
 1R 225 
 -15a?+— = - 
 
 -54+—. 
 
 
 Or, 
 
 
 X*- 
 
 i* , 225 
 -15a?+— =- 
 
 216 225 
 
 -4T+— 
 
 9 
 
 ~4 
 
 Extracting the root, 
 
 
 15 
 
 x = ±- 
 
 2 ^2 
 
 
 Transposing, x = 9 or 6. 
 
 11. x*-\-x = 42. In this example, half of the co-efficient 
 is £. Ans. a? a +a?+£ = 42£. And x = 6 or —7. 
 
 12. 2a? 3 +16a? = 40. 
 
 We must remember that the first term in the second power 
 of the binomial is a? 3 . On this account, if a? 3 in the equation 
 has a co-efficient, we must divide the equation by that co- 
 efficient. This example then becomes 
 
 a? 3 +8a? = 20. Ans. x = 2 or —10. 
 
 13. 5a? 3 — 60a? = 800. Ans. x = 20 or —8 
 
§304.] AFFECTED QUADRATIC EQUATIONS. 189 
 
 14. 8a? 3 -f 8a? =16. (See Ex. 11.) Ans. a? — 1 or —2. 
 
 5 1 
 
 15. 6a? 3 — 5a? = l. When divided, it is a? 2 — — a?=— . 
 
 6 6 
 
 Half of - = — , which squared = — — . 
 6 12 144 
 
 . . 5 25 24 25 49 
 
 Completing the square, *»- - x+ — = — + — - = — . 
 
 5 7 
 
 Extracting the root, x = ± — • 
 
 6 12 12 
 
 2 
 Transposing, x = 1 , or — — . 
 
 12 
 
 t/> a; 3 a; 133 ™ i • v , • . „ 2a? 133 
 
 16. _ — - = —. Multiplied, it is a? 2 —— = —. 
 
 n . . ■ _ 2 1 133 1 400 
 
 Completing the square, a? 3 — -a?+- = -—+-- = — . 
 
 1 20 
 
 Extracting the root, x = — . 
 
 S 3 
 
 Transposing, x = 7 or — 6J. 
 
 17. 3a? 3 — 3a: = — f. Divided, it is a? 3 — a? = — f. And 
 
 18. 14a?— a? 3 = 45. Transposed, it is — a? 3 + 14a? = 45. 
 The first term of the second power of a binomial is positive. 
 
 We must, therefore, change the signs of the whole equation, 
 § 62, in order to make the first term of the equation positive. 
 It will then be a? 3 — 14a? = — 45. Ans. a? = 9 or 5. 
 
 In review of the foregoing principles, we establish the fol- 
 lowing rule. 
 
 RULE FOR THE SOLUTION OF QUADRATIC EQUATIONS. 
 
 § 304. Unite and transpose the terms of the equation in 
 such a manner that the second power of x shall occupy the 
 first term of the equation, and be positive. Let the first 
 power of Jibe in the second term, and all the known quan- 
 tities be in the right-hand member. 
 
190 ALGEBRA. [SECTION XVI 
 
 § 305. Then, if the first term has any co-efficient, divide 
 the whole equation by that co-efficient. 
 
 To both sides of the equation, add the second power of 
 half the co-efficient of the second term; and the unknown 
 side will then be a complete square. Extract the second 
 root of each side, taking care to prefix the double sign ± 
 to the right-hand member. And then, by transposition, 
 the value of the unknown quantity is easily found. 
 
 IC7 3 " In this kind of equations, it is not necessary to destroy 
 the fractions. Indeed we are sometimes obliged to make them. 
 
 EQUATIONS.— SECTION 17. 
 
 1. What two numbers are those whose sum is 12, and 
 whose product is 35 ? 
 
 Stating the question, x = the greater number. 
 
 12 — x= the less. 
 \2x — x* = the product. 
 Forming the equation, \2x— x* — 35, 
 
 Transposing, § 304, a? 3 — 12a? = —35, 
 
 Completing the square, a? 3 — 12x-f 36 = — 35-f 36- = 1, 
 Extracting the root, x — 6 = db 1 , 
 
 Transposing, x = 7 or 5. 
 
 2. Find two numbers, such, that their difference shall be *S, 
 and their product 240. 
 
 |C7» The numbers are x and x + 8, Ans. 12 and 20. 
 
 3. The difference of two numbers is 4, and their product 
 96. What are those numbers ? Ans. 12 and 8. 
 
 4. There are two numbers, whose difference is 9; and 
 whose sum, multiplied by the greater, produces 266. What 
 are those numbers? Ans. 5 and 14 ; or — 9£ and — 18*. 
 
 Questions. What is the first step in the solution of affected quad- 
 ratic equations ? What if the first term has a co-efficient T How 
 do we complete the square] What is the next step ? What are we 
 to do in case of fractions 1 
 
§ 305.] AFFECTED QUADRATIC EQUATIONS. 191 
 
 5. Divide the number 56 into two such parts, that their 
 product shall be 640. Ans. 40 and 16. 
 
 6. Divide the number 60 into two such parts, that, their 
 product shall be 864. Ans. 36 and 24. 
 
 7. The difference of two numbers is 6, and the sum of their 
 squares is 50. What are the numbers ? Ans. 7 and 1. 
 
 8. 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? Ans. Man's, 24 years ; wife's, 18 years. 
 
 9. A merchant has a piece of broadcloth and a piece of silk. 
 The number of yards in both is 110; and if the square of the 
 number of yards of silk be subtracted from 80 times the num- 
 ber of yards of broadcloth, the difference will be 400. How 
 many yards are there in each piece ? 
 
 Ans. 60 of silk ; 50 of broadcloth. 
 
 10. A is 4 years older than B; and the sum of the squares 
 of their ages is 976. What are their ages ? 
 
 Ans. A's age, 24 years ; B's, 20 years. 
 
 11. A nursery-man planted 8400 trees at equal distances, 
 in the form of a rectangle, having 50 trees more in front than 
 in depth. What was the number in front? 
 
 IC7" Let x = the length, and x— 50 = the breadth. 
 
 Ans. 120 trees. 
 
 12. Divide the number 30 into two such parts, that their 
 product may be equal to eight times their difference. 
 
 Ans. 6 and 24. 
 
 13. A person being asked his age, answered, My mother 
 was 20 years old when I was born, and her age multiplied by 
 mine exceeds our united ages by 2500. What was his age ? 
 
 Ans. 42 years. 
 
 14. The length of a certain field exceeds its width by 8 
 rods ; and its area is 768 rods. What are the dimensions of 
 the field ? Ans. Length, 32 rods ; width, 24 rods. 
 
192 ALGEBRA. [SECTION XVI. 
 
 15. Divide the number 60 into two such parts, that their 
 product shall be to the sum of their squares in ihe ratio of 
 2 to 5. Ans. 20 and 40. 
 
 16. Each of two captains distributes 1200 dollars equally 
 among his men. A has 40 men more than B, and B's men 
 receive five dollars apiece more than A's. How many men 
 had each captain ? 
 
 IQ * If x = the number of B's men, then = the share 
 
 x 
 
 of each of them. Also, x+40 = number of A's men, and 
 
 — — — ■ = the share of each of them. Ans. A had 120 ; B, 80. 
 x+40 
 
 17. A man bought a certain number of sheep for 80 dollars. 
 If he had bought four more sheep for the same money, they 
 would have come to him 1 dollar a piece cheaper. How 
 many sheep did he buy ? Ans. 16 sheep. 
 
 18. A merchant sold a quantity of cloth for $56, by which 
 he gained as much per cent, as the whole cost him. How 
 much did it cost him ? 
 
 |C7* If x = cost, then 56— x = gain. Per cent, is always 
 
 x 
 
 written as the 100th of a number. In this case, it is » 
 
 100 
 
 and the cost multiplied by it makes the gain. Ans. $40. 
 
 19. A person buys a horse for a certain sum, and afterwards 
 sells him for $144, and gains exactly as much per cent, as the 
 horse cost. What did the horse cost him 1 Ans. $80. 
 
 20. A person bought two pieces of cloth of different sorts : 
 of which the finer cost 4s. a yard more than the other. For 
 the finer he paid £18, but for the coarser, which exceeded 
 the finer in length by 2 yards, he paid only £16. How many 
 yards were there in each piece ? 
 
 Ans. 18 yards of the finer : 20 of the coarser, 
 
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