University of California • Berkeley The Theodore P. Hill Collection of Early American Mathematics Books GRADATIONS IN ALGEBRA, IN WHICH THE FIRST PRINCIPLES OF ANALYSIS INDUCTIVELY EXPLAINED. ILLUSTRATED BY COPIOUS EXERCISES, AND MADE SUITABLE FOR PRIMARY SCHOOLS. BY RICHARD W. GREEN, A. M., AUTHOR OF ARITHMETICAL GUIDE, LITTLE RECKONER, ETC. PHILADELPHIA : PRINTED BY I. A8HMEAD AND CO. 1839. Entered according to the act of Congress, in the year 1839, by RICHARD W. GREEN, in the Clerk's Office of the District Court of the Eastern District of Pennsylvania. CONTENTS Preface, NUMERAL ALGEBRA. Preliminary Remarks, - - - - 13 Addition and Subtraction of Simple Quantities, . - 19^ General Rule for uniting Terms, - - - 21 Multiplication and Division of Simple Quantities, - 23 Simple Equations, - - - - 24 I. Equations Solved by uniting Terms, - - 25 Addition of Compound Quantities,. - - - 30 Transposition, .... 32 II. Equations Solved by Transposition, - - 32 Transposition by Addition, ... 35 III. Equations, - - - - - 36 Transposition of the Unknown Quantity, - - 38 IV. Equations, - - - - -39 Multiplication of Compound Quantities by Simple Quantities, 43 V. Equations, - - - - - 44 Fractions, - - - - - 49 Equivalent Fractions, - - - - 52 VI. Equations, .... 54 Division of Compound Quantities by Simple Quantities,- 59 VIL Equations, .... 60 IV CONTEXTS. Division of Fractions and Fractions of Fractions, - 63 VIII. Equations, - - - - 66 Subtraction of Compound Quantities, - - 68 IX. Equations, - - . - 70 Uniting Fractions of different denominators, - t2 Ratio and Proportion, ... 76 X. Equations, - - - - - 78 Xr. Equations, .... 80 Equations with two Unknown Quantities, first method, 83 First method of Extermination, - - - 84 XII. Equations, - - . - 88 Second method of Extermination, - - - 93 XIII. Equations, ... - 94 Third method of Extermination, - - - 97 XIV. Equations, .... 98 Equations with several Unknown Quantities, - 102 LITERAL ALGEBRA. General Principles, - - - Addition and Subtraction of Algebraic Quantities, Multiplication of Algebraic Quantities, General Properties of Numbers, Division of Algebraic Quantities, Reduction of Fractions to lower Tei-ms, - Multiplication where one factor is a fraction. Division of Fractions, - - - Fractions of Fractions, ... Uniting Fractions of different denominators, Division by Fractions, ... Reducing Complex Fi-actipns tq Simple ones, 104 110 113 120 123 131 133 135 139 141 143 146 PREFACE. Thb 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 pre- serve 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 arithme- tical operations; first making them intelligible to very young pupils, and then exhibiting them under the algebraical nota- tion. By this means, as every lesson in algebra is immediate- ly preceded by corresponding numerical exercises, the tran- sition 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 diflierent dress. Besides, as algebraical operations re- quire 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 equa- tions. 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 work. Under the head of Literal Algebra, I have repeated, in TI PKEFACTET. 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 ta 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 ena- bles the young pupil to acquire the science very easily; 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 sub- ject 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. PREFACE. Vll With regard to the arrangement of the several divisions, I have been careful to introduce first those prindples that will be the most easily apprehended ; and afterwards such others as would most naturally arise from the former if the study vvei-e entirely new. This method appears to be tlie 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 within the reach of common schools, he has endeavored to pre- pare this work, so that it may be studied by pupils who are not already adepts in arithmetic. And it is believed that 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. ADVERTISEMENT. The foregoing is tlie Preface of the author's " Inductive Treatise on the Elementary Principles of Algebra." The first 146 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 Simple Equations. In the pre- sent state of education, so much of Algebra should be studied by every pupil in our common schools. R. W. G. « 31 « u 35 u (( 38 (( (( 41 (( ERRATA. Before the pupil begins to perform the sums in this work, he is requested to make the following corrections : Page 17 line 15 erase comma in 130. " 43 " 11 instead o^ itself j read the multiplier. « 26 " 2 for 36x, read 35a-. 6 for 173, read 17z. 6 for 76, read 66. in §54, for 32, read 37. 9, for 24 read 20. " 45 example 5, for 181, read 367. " " ** 9, for -jV, read 1|. " 48 line 1 , for 30 read 80. " 55 " 10 insert f. " 65 example 13, for 14a;— 63, read 14a;+63. « 66 " 5, for 273, read 213. «' 67 " 5, for 1000, read 2280. " 70 " 14, for 7 — ^, read 1=?^. o 3 23? 23? « 74 « 4, for — —, read +-^, " 75 « 8, for 3+?^, read ^±^. 9 9 " « « 10, for 6 + 2y, read 3 + 2i/. " 77 line 24, for 23, read 33. " 86 example 6, for — 3y and — 6y, read -{-Sy and -\-6y. " 90 example 8, in ans. for 4«, read 6«. *' 91 " 12, for 35, read 31. " 93 line 4, for ^, read ^. " 97 " 7, for 10, read 16. *' 112 example 1, for 4a:, read Aax. PART L NUMERAL ALGEBRA. PRELIMINARY REMARKS. §1. Algebra is a method of computation, in which we find the value of an unknown quantity, by using, in our calculations, that quantity itself* Example 1. A friend has told me, that in both of his vest pockets he has 28 cents ; but in the right pocket there are 8 cents more than in the left. — I will use the following method of finding how many cents there are in each pocket. By the statement, the right pocket contains as much as the left pocket and 8 cents besides. Therefore, instead of saying, 28 cents is as much as in the left pocket and also in the right pocket; we may say 28 cents is as much as in the left pocket, and again in the left pocket, and 8 cents besides. Hence, 28 cents is as much as there is in the left pockety tvnce, and 8 cents besides. And, as this is the case, 27 cents is as much as twice the left pocket and 7 cents. 26 cents is as much as twice the left pocket and 6 cents. And so on, making the cents fewer and fewer; at last, 20 cents is as much as twice the left pocket. Therefore 10 cents is as much as once the left pocket; that is, there is 10 cents in the left pocket. Of course there is 18 cents in the right pocket. 2 14 ALGEBRA. Example 2. A brother told his sister, that he was 4 years older than she ; and that his age and her age put together, made 18 years. The age of each may be found as follows. By the statement, his age is the same as her age and 4 years besides. Therefore, instead of saying 18 years is as much as her age and his age put together ; we can say 18 years is as much as her age, and also her age and 4 years more. But her age, and her age, is the same as twice her age. Therefore, 18 years is as much as twice her age and 4 years more. And by taking away the 4 years, 14 years is as much as twice her age. Therefore 7 years is once her age; and as he is 4 years older, his age is 11 years. §2. The method in which the two preceding sums were performed, was in use for many years. But in course of time, mathematicians began to abbreviate the words that were made use of in their operations. We will proceed to show a few of these abbreviations. §J^. About the year 1554, Stifelius, a German, introduced the sign + for the words added to; and called it plus. Ever since that time, -f has been used to signify that the « quantity after it, has been added to the quantity before it. Thus, 24-.6, is read 2 plus 6'; and is the same as, 2 with 6 added to it. Example 3. Divide $1000 among A, B, and C ; so that B shall have $72 more than A, and C shall have f 172 • more than A. J By the statement, B's share is equal to A's and $72 be- sides. We may write it thus, ./2's+$72. C's share is equal to A's and $172 besides. We may write his share thus, A's+$172. Then all three shares put together, will be written, A^s, PRELIMINARY REMARKS. 15 and il's+872, and A's-\-$l72; ov, A's-\-A's + $72-i-A's +8172. Therefore, A's+A's+$72+A's+8l72, is equal to $1000. Putting together the $72 and $172, A's+A'5+A's+8244, is equal to $1000. Then, 3 times A'5+S244, is equal to $1000. Taking away $244, 3 times A'5, is equal to $756. Of course, once A^s share is equal to -i of $756 ; which is $252. B's share is $252 +$72 ; which is $324. C's share is $252 +$172 ; which is $424. Proof, $1000. §4. The student will see that in the operation of the last sum, instead of writing A^s share^ we have written simply A^s. This method of abbreviation was used for for a long time before Stifelius wrote. But he simplified the operation, by employing only some capital letter for every unknown number. Example 4. My purse and money together, are worth 40 shillings ; and the money is 7 times the worth of the purse. How much money is in the purse? Now instead of writing the worth of the purse, we will represent it by the capital P; which may well stand for Purse. Then the money will be as much as 7 tunes P. Therefore, P+7 times P, is as much as 40s. Putting the P^s together, 8 times P is equal to 40s. And of course. Once P 15 |- of 40s ; which is 5s. The money is 7 times 5s ; which is 35s. Proof, 40s. §5. About the year 1550, John Scheubelius, a Grerman, 16 ALGEBRA. introduced the following practice. Instead of writing 2 times A, or 3 times A, or 4 times B, &c.; he wrote 2 A, 3 A, 4 B, &c. Example 5. Two men, A and B, owe me $270; and B's debt is twice as much as A's. How much does each of them owe me? We will represent A's debt by A ; and then B*s debt will be twice as much, or 2 A. But both, when put together, are equal to 8270. Therefore, A+2A, is equal to $270. Putting the A's together, 3A, is equal to $270. Then A's debt is ^ of $270 ; which .s $90. B's debt is twice as much; which is $180. Proof, $270. §6. In 1557, Dr. Recorde, an Englishman, introduced the sign =, which we call equals. It is used instead of the expression, is equal to, or, is as much as. Thus, 2-f6 =8, is read, 2 plus 6, equals 8. Example 6. A's age is double of B's; and B's age is three times C's ; and the sum of all their ages is 70 years. What is the age of each? Let us represent C's age by the capital C. Then B's age will be three times as much ; that is 3C. And A's age will be twice that, which is 6C. Then all of them will be, C-f3C + 6C. Therefore 04-30+60=70. Putting the O's together, 100=70. Dividing by 10, 0=7. C's age=7 years. i?'s is three times as much, or 21 years. ji''s is twice B^s; or 42 years. ^t^^f Tf?*^ Proof, 70 PRELIMINARY REMARKS. 17 §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 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- traded from it. Thus, 20—6=14. Example 7. 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 suc- cessful candidate, by the letter A. Then, as the unsuccess- ful candidate received 1,30 votes less, his number may be represented by Ji — 130. Then A,+A— 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. Then, 2A, before 130 was subtracted from it, was equal to 130 more than it is now. That is, it was equal to 548 and 130 more. Therefore we say adding the 130, 2A=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, u, &c. to denote the unknmvn quantities. And this is now the practice of all mathematicians. • Pronounced Da-Cart'. 2* 18 ALGEBRA. f Example 8. It is required to divide $300 among A, B, and C; so that *^ may have twice as much as B, and C may have as much as *i and B together. Let us represent B's share by x. Then A's share will be 2x; and C will have as much as both put together, which is Sx. Then, a:+2ar+3ar=300. Putting the x's together, 6ar=300. Dividing by 6, x= 50. Therefore, B's share is $50. A's is twice as much, or 8100. C's=.as much as both, or $150. Proof, $300. §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 + or — . Thus, in the last Example, and first line of the operation, the first x is the first term, the 2x is the next term, and the Sx 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-ejlcient.^ Thus, in the term 2ar, 2 is a co-efficient of x ; in the term 2^, 3 is a co-efli- cient of d. ^ §12. It must also be understood, that a letter without any number before it, has 1 for its co-efficient. Thus, x re- presents la:; a=la; &c. The 1 is omitted because it is plainly to be understood. • This name was given by Franciscus Vieta, about 1573. ADDITION AND SUBTRACTION. 19 §13. Any number, or letter, or any thing else, used to denote a quantity, when it is not united to any other quan- tity by either the sign + or — , is called a simple quantity. Thus, a; is a. simple quantity ; 2x is a simple quantity ; 12 is a simple quantity ; &c. ADDITION AND SUBTRACTION OF SIMPLE QUANTI- TIES. §14. In Algebra, simple quantities are added by writing them down, one after another ; being careful to put the sign + between them. Thus we add 9 to x, by writing them 9-|-a;, or, a? +9. 0:!?" 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. §15. It will readily be seen, that it is of no consequence which quantity is put first; for 9 4-7 is the same amount as 7 + 9. §16. One simple quantity is subtracted from another simple quantity, by writing down both quantities, one after another, and putting the sign — before the quantity which tee subtract. Thus, we subtract 4 from x, by writing them x — 4. §17. When two simple quantities have been added, or one simple quantity subtracted from another, they then will consist of more than one term. §18. Quantities that consist of more than one term, are 20 ALGEBRA. called compound quantities. Thus, a-f &> is a compound quantity. So is a — b, and ar-}-7, and x — 7, &c. §19. It sometimes happens, that in those compound quan- tities which are made by adding or subtracting, there are two or more terms of the same kind ,* such as x-\'2x, 4a — 2a, 5x — Sx+x. Such quantities are called like quan- tities. §20. When in any compound quantity, there are two or more terms of the same kind, they may be united by per- forming the operation which is expressed by the sign. Thus, x-{-2x is united into 3a:. 4a — 2tt is united into 2a. 5ar — 3a; -f-a? is united into Sx. §21. Numeral quantities* may be united in the same manner. Thus, 44-3 may be united into 7. 9 — 4 may be united into 5. 6 — 2-f 5 may be united into 9. EXAMPLES. Unite, as much as possible, the terms in each of the fol- lowing compound quantities. 1 . a+ fe + 8—4 + 2a + 6 -f Sb—b -f 3a. Take one letter and go through the whole quantity with that first ; and then take another letter and do the same. And then another, &c. Ans. Uniting the a's, they equal + 6a ; uniting the &'s, they equal -j- 36 ; uniting the numeral quantities, they equal + 10. Therefore the answer is, 6a +86 + 10. 2. 2a+6+4a+26+8+6— 3a— 6— 4. 8. Sx-}-y+z+3x-{-Sy + z-i-^x—2y-]-z. 4. 18— 9+5+8a+4a:+7a;+5a— 6a:— 7a. • Numeral quantities are expressed hy figures i and literal quantities, by letters. ADDITION AND SUBTRACTION. 21 5. 4y4-72j + 9a— 5a— 42J— y4-2s+3a4-2y. 6. la — a — a+5&+4«+a — 2z+a — z-\-Aa, 7. 8— 4+6+5— 24-8+a;4-3a7— 2a?+4a?. §22. When any simple quantity begins, with the sign +> it is called a positive quantity ; as +a,+3a. §23. When any simple quantity begins with the sign — , it is called a negative quantity ; as, — 6, — 5a. §24. In algebra, the perfect representation of any sim- ple quantity requires both the specified sum, and either the sign +, or the sign — ; as, +5, — 40, +ar, — 3ar. §25. 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, x, 2, a+&, x — y. §26. Therefore, when a simple quantity, or the first term of a compound quantity, does not begin with a sign, we say that the sign + is understood. That is, we think of the quantity the same as if + was before it. §27. In reading compound quantities, the pupil must be careful to join the sign to the term that is immediately after it. Thus, the first example under §21, must be read a ; plus b ; plus 8 ; minus 4 ; plus 2a ; &c. We are now ready for the following GENERAL RULE FOR UNITING TERMS. §28. Select one kind of like quantites, and add into one sum all the positive co-efficients that belong to them. Then add into another sum all the negative co-efficients that belong to them. Then subtract the less sum from the greater; and prefix the sign of the greater lo the difference, annexing the common letter. 22 ALGEBRA. Note. — It sometimes happens that the negative quantity is greater than the positive quantity. In such cases, the difference will have the sign — . Examples. Unite the terms in the following. 1. 3a + 46+26— 3c+3c— &+a. Ans. 1st, 4a+6&— & + 3c~3c. Now +66 — &= + 56; and +3c — 3c balance each other, so as to be equal to 0. The final answer is 4a +56. 2. a+6+3a— c+4a — 3c+6. Ans. 8a+26— 4c. 3. 5a? +51/ + 3a;— 2y. Ans. 8a: +3^. 4. 6+a+3a — 56+ 4a; — a + 3a — x. Ans. 6a — 46+ 3a;. 5. 4a— 56— 10 + 2?/— a;+4. Ans. 4a— 56+22/— a;— 6. 6. 2x-\-Sy — 3a?+53/ — x — z. Ans. — 2x-^8y — z. 7. x-\-z+x — z-\-x-\-2y. Ans. 3a;+3y. 8. 3a— 5— 46 + 6— 2a— 3 + 66. Ans. a +26— 2. 9. 4a;+4+5i/+6— 2 + 3a;— 21/. Ans. 7a? +32/ +8. ^ 10. 4c+3a — 6 — 3a+c — 2a— 6. Ans. 5c— 2a — 26. 11. 26— c+3a— 6+4c+2— 2y, and 1— 2a:-f Sy—lOa; together. Ans. 24— a;-f 4y-— IO2;. 3. Add 3ar+5i/— 6z, — 2a:— 8y— 9z, 20ir+2i/— 3z and X — y-j-z — 4 together. Ans. 22a; — 2y — 173 — 4. 4. Add 3—23/4-2, 4y— 2z+5, 2—2—3/, ^"^ 2z— y — 10. Ans. Nothing. 5. Add 7a? — 61/ — 5z — 8—^ S—g—Sy—x 7^— 1— 32-fi/— a; 32— 1—^-1-31/— 2a; a?+82/— 52+9+^. Ans. 4a;+5^4-3y+2. 6. Add Zb—a—c—ll^d+ee—dy Za-\-21e—d—Zc~-2h 3e— 7i/— 8c+56 17c— 6&— 7a+9rf— 5c+lly — 2«Z — 6e— 5c — 93/— 3a+ar. Ans. 37e— 8a— 109(f— 101/4- X. 7. Add 4— 3a4-a?— 7443/ 84-63/— 3a:46a432 7a433/4-4a4-12— a? X — 4483/+6a4n ' ^ 4a 4 7 4- 3a 4 iP— 23/ Ans. 8. Add a — x-\-y-\-r — 14 — n 3n— 3/4246a— 2+3a:+r 6247r — a—y—2n 3y42a4146r44» a?— 74^4^44 — «4r. Ans. ^.. 42 -Ravim? ALGEBRA. TRANSPOSITION. 1. BY SUBTRACTION. §45. It often happens that in the first member of the equation, some number has been added to the x's in order to make them equal to the last member. Thus, in the equation, ar-J-16=46, we see that 16 has been added to .r, to make it equal to 46. §46. Now if X with 1 6 added, is equal to 46 ; then x alone must be 16 less than 46 ; that is, 46 — 16. So, that if we find, that a: -f- 16=46, we may know that a:=46 — 16; or what is the same, a:=30. §47. 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. §48. Now, with the equation that we had above, ar+16=46. Subtracting 16 from both members, x-\-lQ — 16=46 — 16. Now, in the first member of the equation, we have+16 — 16, which is of no value at all, for +16 and — 16 balance each other as has been seen in Ex. 1 under §28. Therefore the equation is reduced to a; =46 — 16. Uniting terms in the last member, ar=30. EQtJATIONSi-^SECTION 2. 1. Suppose a'+8 + 3x-»56 ; what is the value of a;? Uniting terms, 4a? 4- 8 =56. iW TRANSPOSITION BY SUBTRACTION. 33 Subtracting 8 from both, 4a;4-8— 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:+14— a;— 7=41+2— 8 ; to find x. Uniting terms, a;-f7=35. Subtracting 7 from i x-\-l — 7=35 — 7 or both sides. \ a:=35 — 7. Uniting terms, ic=28. 3. Given a; +5 =6, to find x. Writing the equaiion, a;-f 5=6. Subtracting 5, a?=l. 4. Given .5a^+22— 2a:=31, to find x. Am, x=S, 5. Given 4a?+20— 6=34, to find x. Ans. x=5. 6. Given 3a?+12+7a;=102, to find x. Ans. x=9, 7. Given 10a;— 6a:+14=62, to find x, Ans, x=\2, 8. If7a;— 14+5a;+20=246, thena?=20. 9. If8a?4-17— 5ar+3=100+10, thena?=30. 10. If 7a;— 14+3a?+35=450— 29, then a;=40. PROBLEMS. 1. What number is that, which, with 5 added to it, will be equal to 40? Staling the question, a;=the number a? +5,= after adding. Forming the equation, a?4-5=40. Subtracting 5, from both, a?=35. 2. A man being asked how many shillings he had an- swered, add 15 to their number, and then subtract 1, and the remainder will be 64. How many shillings had he? Stating the question, a;=numb. of shillings. '^' ' a;+15=after adding. a;+15 — 1= after subtracting. ^ ALGEBRA. Forming the equation, a?4-15 — 1=64. Uniting terms, a? +14= 64. Subtracting 14 from both, a? =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, a:=:the less share. a; +4=: the greater. x+x-{- 4= both shares. Forming the equation, a:+a;+4=17. Uniting terms, 2a7+4=17. Subtracting 4 from both sides, 2a?=13. Dividing by 2, x=6h 5. The sum of the ages of a certain man and his wife is 55 years ; and his age exceeds her's by 7 years. What is the age of each ? 0^ Let a;=age of the wife, .dns* 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 age is 73 years. What is the age of each ? Stating the question, ;r=C's age, a?4-4=B's age. a?-f 4+5=:A's age. a:-f a?4-4+a?+4+5=sum of all of them. Forming the equation, a? -|- a? -f- 4+ a; +4 -{-5=: 7 3. Uniting terms, 3;r+13=73. Subtracting 13 from both sides, 3a;=60, Dividing by 3, «;=20. Ans, C 20 years, B 24, A 29. 7. Two persons were candidates for a certain office, where there were 329 voters. The successful candidate TRANSPOSITION BY ADDITION. 35 gained his election by a majority of 53. How many voted for each? 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 $76, 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 successful candidate had a majority of 120. How many voted for each ? Ans. 588 fo» one, and 708 for the other. 11. A father, who has three sons, leaves them $8000, specifying in his will that the eldest shall have $1000 more than the second, and that the second shall have $500 more than the youngest. What is the share of each ? Ans. The eldest had $3500, the second $2500, the youngest $2000. 12. A cask which held 74 gallons was filled with a mix- ture of brandy, wine and water. In it, there were 15 gal- lons of wine more than of brandy, and as much water as both wine and brandy. What quantity was there of each ? Ans* 11 gallons of brandy, 26 of wine, and 37 of water. 2. TRANSPOSITION BY ADDITION. §49. It is frequently found that some quantity has been subtracted from the a?'s ,* as in the equation bx — 44=76. §50. In such cases, it is very evident, that the quantity which has been subtracted from the a?'s, must be added to 35 ALGEBRA. each side. For if in the above equation, 44 has already been subtracted from the 5x ; we must add it again if we wish to find what bx is equal to. 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 it will become 5a;— 44+44=76+44. Uniting the two 44's, 5a;=76-f 44. Or, 5a?=120. A Dividing by 5, ic=24. iJ!^ EQUATIONS. SECTION 3. 1. Given a?+ 14+ 3a;— 27=51, to find x. Ans, a;=16. JV Given-3a?— 30— 2ar=:46*-7, to find x. Ans, ar=:69. 3. Given 9a;— 41 +6=88—6, to find x, Ans. a;=l3. 4. Given 20 + 3a;— 46=35— 4, to find x. Ans. a;=l9. 5. Given 4a;— 39 — 2a; =47, to find x. Ans. a;=43. 6. Given 7a;+27— 46=65, to find x. Ans. a;=12. 7. Given 14a;— 55— 8a; +14= 85, to find x, Ans, a;=21. PROBLEMS. 1. What number is that, from which 8 being subtracted, the remainder is 45 ? Stating the question x= the number. X — ^"8=: when 8 is subtracted. Forming the equation x — 8=:45 Adding 8 to both sides a;— 8+8=45+8. Or, a;=53. 2. What number is that, from which 27 being subtracted, the remainder is 41 1 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 ? J--*' TRANSPOSITION BY ADDITION. 87 0^ In this and the following questions of this section, x must stand for the greatest quantity. Stating the question, x= the dearest. X — 16= the cheapest. x-\-x — 16= cost of both. Forming the equation, x-\-x — 16=140 Uniting terms, 2x — 16=140 Adding 16 to both sides, 2ar=1404-16 Uniting terms, 2ie=156 Dividing by 2, a?=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 ; so 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 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. 7. Divide the number 60 into three such parts, that the first may exceed the second by 8, and the third by 16. Ans. 12; 20; and 28. 8. Three men having found a purse of $160, quarreled about the distribution of it. Atler the quarrel, it was found 4 38 ALGEBRA. [Eq. SeC. X that A had got a certain sum, and that B had $30 more than A, but C 850 less than A. How much did each obtain? Ans. A $60 ; B $90 ; C $10. III. TRANSPOSITION OP THE UNKNOWN QUANTITY. §51. 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. Th«s, in the equation [x-\-5=i20 ;] if we subtract 5 from both sides, it is plain that the first member becomes x, and the last member becomes 20 — 5 ; so that the equation would become ir=20— 5. §52. 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 4-» Because this is the same as adding that term to both sides. • , " : , ; Thus, in the equation x — 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 becomes a?=20+5. §53. 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. §54. It was stated in §32, that an equation must be brought so that the unknown quantity will occupy one mem- ber of the equation, and the known quantities embrace the other member. And, as it frequently happens that the TRANSPOSITION BY ADDITION. 39 unknown quantities are on both sides, we are obliged to re- sort 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 un- known quantity. §55. 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* EQUATIONS. SECTION 4. 1. Reduce the equation 4a; — 14=3a?+12 Solution. Transposing 3a?, 4a; — 3a; — 14=12 Transposing 14, 4a; — 3a;=12 + 14 Uniting terms, a;=26. §56. In transposing, it is generally best to write first the unknown quantity that is on the left ; and then bring over those which are on the right, if there are any there. Then write those known quantities that are already in the right hand member, and then transpose after them what known quantities there are in the left. 2. Given 21 — 7a;:=40 — llx, to find x, Ans. a?=4|. 3. Given 40— 6z=136— 14z, to find z. Ans. z=l2. 4. Given t/4-12=3i/ — 4, to find y. Ans. y=8. 5. Given 5a; — 15=2a;4-6, to find x, Ans. x=7, 6. Given 40— 6a;— 16=120— 14a^, to find x. Ans. x= 12. 7. Given 4 — 9i/=14 — lly, to find y. Ans. y==5. 8. Given a;-f-18=3a? — 5, to find x. Solution. Transposing 3a;, x — 3a; +18= — 5 Transposing 18, x — 3a;= — 5 — 18 Uniting terms, — 2a? = — 23 Dividing by 2, — a;= — lid* 40 ALGEBRA. [Eq. SeC. 4. §57. 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 — a: = — 1\^ is just as good as the equation -fiC=H-ll-|. In both cases, the quantity^? is equal to the number 11|. §58. In the result of the last question, 11|^ maybe 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 ll^=a?. And if ll-|=a;, it is evident that x=l\\. This coincides with what was shown in §57. §59. 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 first ; and that it is also im- material which member is written first. And hence, in any equation the signs of all the terms may he changed without affecting the equality. §60. It is evident that all the terms of one member may be transposed to the other member. When this has been done, the memberyrom vyhich the terms have been trans- posed becomes, 0. Thus, the equation x=^y — a^y, may be made x-\-xy — 3i/=0 ; where — Sy balances x+xy. PROBLEMS. 1. A man has six sons, whose successive ages difl^er by 4 years ; and the eldest is three times as old as the youngest. What are their ages 1 (( (( next u n next (( (( next (( (( next (( C( eldest. TRANSPOSITION BY ADDITION. 41 Stating the question, x= age of the youngest. ar+4= ar+4+4= ar+4 + 44-4= a?+4+4+4 + 4= ar+4+44-4+4+4= Forming the equation, 07+4 + 4+4 + 4 + 4= 3a; Uniting terms, ic+20=3a; Transposing the 3x, x — 3a? +24=0 Transposing 20, x — 3a7= — 20 Uniting terms, — 2x= — 20 Dividing by 2, — x= — 10 or a?=10 2. A person bought two horses, and also a hundred dol- lar harness. The first horse, with 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. a;+100= price of the second. 0?+ 100 + 100= 2d horse harnessed. Forming the equation, a?+100 + 100=2a? Transposing from both members, x — 2a7= — 100 — 100 Uniting terms, — x= — 200 Or a?=200&c. 3. A privateer running at the rate of 10 miles an hour, discovers a ship 18 miles ofl?* sailing at the rate of 8 miles an hour. How many hours can the ship run before she will be overtaken by the privateer ? 0^ The equation will be 10a?=8a?+18. Ans. 9 hours. 4. A gentleman distributing money among some poor 4* 48 .. . t ALGEBRA. [Eq. SeC. 4. people, found that he would lack 10 shillings if he under- took to give 5s to each. Therefore, he gave only 4s to each, and finds that he has 5s left. How niany persons were there. 0^ It will be found that his money by the first supposi- tion =bx — 10; and by the last supposition, it =4a?+5. Ans. 15. 5. I once had $84 in my possession ; and I gave away so much of it, that I have now three times as much as I gave away. How much did I give away ? 0^ If I gave away $a?, then $84--x will be what re- mains. Ans. $21. 6. A certain sum of money was shared among five per- sons, 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 shares of the last two put together, were equal to the sum of the shares of the other three. How much did each man receive 1 Ans. A $21 ; B $11 ; C $27 ; D $22 ; E $37. 7. A person wishes to give 3 cents apiece to some beg- gars, but finds that he has not money enough by 8 cents. He gives them 2 cents apiece and has Scents left. How many beggars were there 1 Ans. 1 1 . 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 second overtake the first? Oy" In the operation, it must be remembered how far the first had the start, before the equal time for both began. MULTIPLICATION OP COMPOUND QUANTITIES. 43 MULTIPLICATION OF COMPOUND QUANTITIES BY SIMPLE QUANTITIES. §61. 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. §62. Let us apply this principle to Algebra. You pay in all, 8 times 7 + 5, which =564-40. Which shows that in multiplying a compound quantity, you multiply each term by itself, ■ 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-f40 =96. §63. 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 afterwards 40 cents were deducted. §64. This shows that + multiplied by -}-, produces + j and — multiplied by +, produces — . EXAMPLES. 1. Multiply a: 4-4, by 3. Ans. Sx+U, 2. Multiply 124-ar, by 5. Ans. 604- 5a:. 3. Multiply X— 10, by 8. Ans. 8a;— 80. 4. Multiply 126— a:, by 4. Ans. 504— 4ap. 5. Multiply X+&, by 6. H.. 44 .^ ALGEBRA. [Eq. SeC. 5. t 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 3ar— 62, by 15. 12. Multiply 97— 4a^, by 12. 13. Multiply a; +7 — y, by 7. 14. Multiply 3a:+y— 12, by 8. 15. Multiply 2x—Sy—e, by 6. 16. Multiply 3x—l2+y, by 5. §65. 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 con- nect them together. Thus, a: + 4x3, signifies that both x and 4 are to be multiplied by 3. §66. In 1629, Albert Girard, a Dutchman, introduced the parenthesis as a convenient substitute, in many cases, for the vinculum. Thus, (a: 4-4) X 3, isthe sameasa:+4x3; and is read, a?+4, both x3. If there are more than two terms under the vinculum, we say, afler repeating those terms, a//, &c. Thus, (ar-f-j^) x(« — ^+c), is vead x-\-y both into a — b-^-c alL See also §100. EQUATIONS. SECTION 5. 1. i^Xll=121, to find a:. Solution, a;— 9X11=121 Performing the multiplication, 11a: — 99=121 Transposing and uniting, llar=220 Dividing by 11, a:=20. MULTIPLICATION OF COMPOUND QUANTITIES. 45 2. Given (a?+7)x6=54, to find x. Ans. x=2. 3. Given 12+a:x5=100, to find x. Ans. 8. 4. Given x — 9x8=96, to find ar. Ans. 21, 5. Given 181— 3a:x5-=920, to find x. Ans. 61. 6. Given (8+a;)x2-f 14=72, to find x. Ans. 21. 7. Given (154-a;)x3— 27=48, to find a?. Ans. 10. 8. (112— 2a:) x 3= (2a:— 7) X 4, to find x. Ans. 26. 9. (3a;+14)x4=(78— a-)x5,to find x, Ans. 19-y\. 10. 2jr-|-8x5=(324-x)x3, tofinda:. Ans. 8. 11. (3j:— 14)x7=(17— a:)x6, to find x, Ans. 14J. 12. 120- 3arX2= (4a:— 6)x9, to find x, 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 1 Staling the question, ar,= the sum for each. a:+126=A's sum now. X — 87= B's sum now. 2a:— 174= the double of B's. Forming the equation, a: + 126=2a; — 174 Transposing and uniting, — x= — 300 Changing signs, ar=300 the answer. 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 ? 46 ALGEBRA. [Eq. SeC. 5. Stating the question, x= the wife's age. 3x= the man's age. ar-f-15= the wife's after 15 years. 3a?+15= the man's after 15 years. 2j:-f-30= twice the wife's age. Forming the equation, 3a: -j-15=2a?+<30 Transposing and uniting, a; =15 the wife's age. 3a;=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 three times the number of sheep was equal to seven times the number of calves. How many were there of each? 0^ If x== number of calves, there iP+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 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. %CJ^ Ux= the greater, thus 75 — x= the less ; and Sx will =7 times the less 4-15. 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? 10*^0;= number of cavalry, then 20a:= their whole pay, &c. Ans. 35 cavalry, and 90 infantry. 7. A grocer sold his brandy for 25 cents a gallon more than he asked for his wine ; and 37 gallons of his wine MULTIPLICATION OF COMPOUND QUANTITIES. Aj 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 ? ^3* The whole mixture will be worth 50 times 8 shillings. Ans. 37-^ gallons of the best ; and 12-| 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 ? IC?" In X years he will be 40-f-a;, 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 8279 ; 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 ? ICT* Let x= William's and 3x= John's. Then after the change, a;+20= William's and 3a; — 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 48 ALGEBRA. [Eq. SeC. 5. $30 more ; and the chaise cost twice the price of the horse. What did he give for each ? Ans. For the harness $50 ; horse 8130; chaise $260. 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 1 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. 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 income of either. What do they spend annually? Ans. A $370 ,• B $330. ii- arff h FRACTIONS. 49 FRACTIONS. §65. All the division which the pupil has as yet perform- ed, has been the division either of numeral quantities, or of the numeral co-efficients. But in Algebra, it is frequently- necessary 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 re, or the fourth of X, &c. §66. In common arithmetic, if we wish to divide 1 by 2, we do it by writing 2 under the 1 ; thus, ^. So if we wish to divide 2 by 3, we write 3 under the 2 ; thus, |. In the same manner, 2 divided by 5 is written f ; 3-^-4 is writ- ten I ; 6-i-7 is written f. The quantities that are obtain- ,ed by dividing in this manner, are caWed fractions, §67. In Algebra, we most generally make use of this method of dividing ; especially when we divide literal .quantities. Or, in other words, we divide a literal quan- tity by writing the divisor under the dividend, with a straight line between them ; thus, x divided by 2, is written ~; and is read, x-half. x-r-S, is written |; and is read, c-third ; a:-i-4, is written J, and is read xfourth ; 3a:-T-4, J, and is read, Sx fourth. §68. The two separate numbers that we employ in writ- ng a fraction, are called terms. The upper term is called he numerator, and the lower term is called the denomina- or. Thus, in the fraction -j, we call x the numerator, and J thC' denominator. §69. If the one-third 'of x is |, it is evident that | of x, i two times as much ; that is y. If -}• of x is ^, then ^ o^ 5 50 ALGEBRA. [Eq. SeC 5. X is — • Whence the rule to multiply a whole number by a 5 fraction, is, to multiply the whole number by the numerator y and divide by the denominator ; as J of a: is y ; f- of i/ is %. 2 of a is-. —, 3-01 a ih 3 . Examples. The pupil may multiply a, x, and y, each of them by | ; and then by J ; and then by f , |, |, |, f, ^, successively. ^70. 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 fraction, we multiply the numerator, and retain the deno- minator without alteration. Thus, 2 times y is |^ ; 3 times 5 is 1/ ; 2 times ^ is y ; 4 times =| is ^|' &c. Examples. Multiply each of the following fractions by 2, then by 3, and then by 4. |, f , f 4^ e, |, |, |, |, 21 3i 2z 3x. 41 3'T' 5" 5 7 §71. 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 « 4 wholes ; 18 thirds = 6 wholes ; &c. Such frac- tions are called improper fractions. §72. 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. §73. Thus we have the rule, to change an improper FRACTIONS. 51 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-T-3=:*4. 2. How many whole ones in J ? ^^ ? Y ^ \^ ^ ¥ • 3. How many whole ones in ^i ? ^ ? ^^ 1 -3/ ? \o ? 4. How many whole a:'s in ^1 ^? ?^"? ^? . How many whole xs i" -^ • '^3' • T "7 6. How many whole x's in 3 times ^? 7. How many whole x's in 4 times 1 1 8. How many whole x's in 5 times — ? 5 §74. 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 quantity to make a whole x. Therefore, if we multiply ^ by 5, we shall obtain — , or exactly ;r. If we multiply * by 3, we shall obtain — , or, which is the same, x. If we multiply 5 by 4, we shall obtain —, or x. §75. As 4 times ^ is equal to x\ then 4 times -^ must be equal to twice as much, or 2x ; and 4 times -^ must be three times as much, or ^x. As 3 times f is equal to x ; so 3 times ^ must be twice as much, or 2x. So 5 times -^ must be 3 times as much as 5 times y ; and therefore is 3ar. §76. 'Any fraction when multiplied by the number which is the same as the denominator, will produce a quantity 52 ALGEBRA. [Eq. SeC. 5 which is the same as the numerator. Thus, We shall be able to make use of this principle in the solution of many equations, if we operate in accordance with the following axiom or self-evident truth. §77. If equals be multiplied by the same, their products will be equal Thus, if a:=10, then 2a:=20 ; 4a:=:40 ; &c. EaUIVALENT FRACTIONS. §78. It is evident [§71,] that each of the following fractions, f ' h J» T' h h h h *^^'» '^^ equal to 1. Therefore, they must be equal to one another. Also, each of the following fractions, f, |, 4, |, y, &c., is equal to 2 ; and conse- quently they are all equal to one another. In the same manner, we may make many fractions that will equal 3 ; and so of any other number. §79. Let us take from the first set of the above fractions, I and I 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 see the same fact in the equal fractions -J and J ; and also in the equal fractions ^ and |. We find the same, by taking from the second set, the equal fractions f and | ,• and also | and | ; and also f and y . §80. Again in the equal fractions, -| and |, we find each term in the last fraction three times as great as the corre- spondent term in the first fraction. The same may be ob- served in the fractions | and | ; and also in f and -f ,* and also in | and |. EQUIVALENT FRACTIONS. 53 §81. 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 tliat number may be, the fraction made by that multiplication, will be equal to the first fraction. Hence, there is an equality between the following fractions, |, f , f, |, -^-q^ -j-%, &;c.; and also be- tween the following, ^, |, |, y\. /y, y\, &c. §82. The principle just explained, leads to another which is of much importance. Suppose we multiply by 8, both terms of the fraction | ; and obtain -|-|. It is plain that both terms of the fraction }f can be divided by 8, to bring the fraction back tof. So also, if both terms in ^ be mul- tiplied by 6, the fraction will be H, which means just as much as | ; and, of course, if both terms in J| be divided by 6, the fraction will be brought back to |, which is equal to ^|. So, in general, if we divide both the numerator and the denominator of a fraction by the same nuTnber, we have a new fraction which will be equal to the first. Thus, j% may be changed to | ; jf to | ; ^{ to f . §83. It is evident, that of several fractions of equal value, that which has the least denominator is the most easily un- derstood. Thus, I of an apple is much better known at first sight, than || of an apple. And ivhen a fraction is brought to as small a denominator as it can be changed to j we say it is reduced to its lowest tcr?ns. §84. In order to reduce a fraction to its lowest terms, di- vide both the numerator and the denominator by any number that will divide each without a remainder* Thus, in the fraction -j^t' ^^^^ terms may be divided by 5, by which we obtain |A ; and both terms of this last fraction may be di- vided by 8, by which we obtain |-. 54 ALGEBRA. [Eft. SeC. 6* EaUATIONS. — SECTION 6. Problems. 1. In an orchard, 1 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 1 Stating the question, a;= number of trees. 1= apple trees. y = pear trees. - , ■^y= plum trees. 81= cherry trees. All these trees together = the whole orchard* Forming the equation, |4-| + ^£-f 81s=a;. Now, we know that if we multiply ^ by 4 we obtain x alone ; that is, we destroy the fraction, and make it a whole number. And we know, that if we multiply the first mem- ber by 4 ; and also the last member by 4, we shall not de- stroy the equation. See §77. We will therefore multiply both members by 4, for the purpose of destroying the/rac- tion in the first term. It will then become a:+T+^+324=4a;. Next, we will multiply both members by 5, to destroy the fraction in the second term. This will make 5a;4-4a:+^''+1620=20a:. Then we will multiply by 11, to destroy the remaining fraction, which will make 55a;+44a;+40a;+17820=220a?. Transposing and uniting, *-81a:= — 17820. Changing signs, 81a?=sl7820. Dividing by 81, a; =220. the Ans. ' «c' EaUIVALENT FRACTIONS. 55 2. In a certain school, | of the boys learn mathematics, A of them study Latin and Greek, and 6 study English grammar. What is the whole number of scholars? 0:^ After the question is stated, the equation will be- come ?+!''+ 6= a: Multiplying by 5, a:+^+30=5a: Multiplying by 4, 4a; + 15a; +120= 20a: Transposing and uniting, — a:= — 120. Ans. 3. A gentleman has an estate, -} of which is woodland? f of it pasture, and 105 acres embrace the pleasure grounds, gardens, and orchards. How many acres does it contain 1 Ans. 630 acres. 4. After paying away i and ^ of my money, I find 22 dollars yet in my purse. How much had I at first ? Ans. $A0. 5. A man bought a lot of ground, for which he agreed to pay as follows : i of the money on taking possession, -J 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 in the water, and 10 feet abov^e the water. What is its whole length ? Ans. 24 feet* 1. In a Christmas pudding, ^ is flour, \ milk, -J- eggs, \ suet and fruit, and J of a pound of spices and other ingre- dients. What is the weight of the pudding 1 {^ The equation will be l+f +J4-?+J-^-ar. Ans. 15 pounds* 8. A lady being asked what her age was ; replied, if you add ^, i, and ^ of my age together, the sum will be 18« How old \yas she ? I- 56 ^ ALGEBRA. [Eq. SeC. 6. Cfr After the question has been stated, the equation will be, |-f £_j-|=l8. Ans. 24 years. 9. What sum of money is that, whose ^ part, i part, and -J- part, added together, amount to 94 dollars 1 Ans. 8120. 10. A person found upon beginning the study of his pro- fession, that he had passed -f of his life before he com- menced his education, ^ of it under a private teacher, the same tirne at a public school, and four years at the univer- sity. 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 in board and lodging, the 8th part of it in clothes and wash- ing, the 10th part of it in incidental expenses, and yet I save 8318 a year. What was his yearly income ? Ans. 8720. 13. A gentleman bequeaths, in his will, the half of his property to his wife, one-sixth part to each of his two sons, the twelfth part to his sister, and the remaining 8600 to his servant. What was the amount of his property? Ans. 87200. 14. Of a piece of metal, -J plus 24 ounces is brass, and .| minus 42 ounces is copper. What is the weight of the piece? 0^ The equation, when formed, will be l+24-f^— 42=ar. 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 EQUIVALENT FRACTIONS. 57 than ^ of it is oats. How many bushels are there in the whole; and how many of each sort? 0^ In stating the question ; ^ — 20= barley, and ^■+ 36= oats. Ans. 96 bushels in all ; 28 of barley, and 68 of oats. 16. A teacher being asked how many scholars he had, replied. If I had as many more, half as many more, and quarter as many more, I should have 88. How many had he? |C?^ 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 2 was copper, 12 pounds less than ^- was tin, and 4 pounds more than i 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 -J part exceeds its -J part, by 12? iHj^ The statement is the same as, ^ of it equals i of it 4-12. Ans. 144. 19. What number is that whose A 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 83000 less than half of it, B 81000. less than the third of it, and C 8800 more than the fourth of it. What is the sum to be divided ; and what does each receive? Ans. 838400; and also, 816200, 8II8OO, 810400. 21. A man driving his geese to market, was met by an- other, who said. Good morrow, master, with your hundred geese. He replied, I have not a hundred ; but if I had as -'^S ALGEBRA. [Eq. SeC. 6. 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, replied. 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. 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. 24. If the half, third, and fourth parts of my number, be added together, the sum will be one more than my num- ber. Now, what is my number? Ans. 12. 25. 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. 26. A young gentleman having received a fortune, spent ■J of it the first year, ^ of it the second, and -J- of it the third, when he had $2600 left. What was his whole fpr- tune? OO" In stating the question, what was spent, = the whole minus $2600. Ans. $12000. 27. A father leaves four sons who share his property in the following manner. The first takes half, minus $3000 ; the second lakes a third, minus $1000 ; the third takes ex- actly 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. DIVISION OF COMPOUND QUANTITIES. 59 DIVISION OF COMPOUND QUANTITIES BY SIMPLE QUANTITIES. §86. We have found, §67, 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 quantities may be divided in this way, as well 'as simple quantities. Thus, 14+x is divided by 3, as follows: 14+ar. 3~ §87. With the same reason, we find the fraction of a compound number, by multiplying it by the numerator, and writing the denominator under the product. Thus, 2 ^ ^ . 2x — 10. 4 of X — 5 IS ^ 3 EXAMPLES. 1. What is J of 84-a; ? Ans. 24 + 3ar. 2. What is I of ar— 27 ? 3 What is I- of 3a:— 141 4. What is I- of 9 + 5.T? 5. What is ^ of 7ic— 19? 6. What is | of 9a!— 27 1 §88. This may be changed into whole numbers by §73. Thus, 15f=l!£=5x-15. 4 Ans. 2ar— 54. 5 Ans, . 6a:— 28. 7 Ans. 27 + 15a-. 8 Ans. 28a:-^76. 7 Ans. 45a:— 135. 60 ALGEBRA. [Eq. SeC. 7. 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 quotient would be 9. How old was he ? Stating the question, x= his at^e. a:+8= with 8 added. >;*': Forming the equation, ^ *" —9 o Muhiplying by 3, a?+8=27 Transposing and uniting, a?=19 the Ans. 2. A man being asked what he gave for his horse, re- plied, that if he had given $12. more, | of the sum would be 84. What was the price of the horse ? Stating the question, a?= the price. ic-t-12= when increased. - — ! — s=| of the sum. Forming the equation ^' =84 4 Multiplying by 4, 3a; -f 36=336 Transposing and uniting, 3x=300 Dividing by 3, a;=100 the Ans. '3. What sum of money is that, from which 85 being subtracted, 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. 0^ In slating the question, the parts are x, and lo — x. Ans. 84; and 6-f. DIVISION OF COMPOUND QUANTITIES. 61 5. It is required to find a number, such that if 15 be subtracted from it, -J of the remainder shall be 100 ? Ans. 140. 6. Divide the number 46 into two parts, so that when one is divided by 7, and the other by 3, the quotients toge- ther may amount to 10. Ans. 28 and 18. 7. A person being asked the time of day, answered that the time past from noon was equal to | of the time to mid- night. What was the hour ? }Cj^ In stating the question, the parts are x, and 12 — x, Ans. 20 minutes after 5. 8. Two men talking of their horses, A says to B, my horse is worth $25 more than yours ; and |- of the value of my horse is equal to J of the value of yours. What is the value of each ? iCj^ The values will be x^ and a?+25. Ans. A's $125 ; B's 100. 9. Two persons have equal sums of money. One hav- ing 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. AVhat was the value of each ? Ans. Horse $100 ; Chaise $150. 11. What number is that, to which if I add 13, and from Y^-j of the sum subtract 13, the remainder shall be 13? Ans. 825. 12. A man being asked the value of his horse and saddle, answered that his horse was worth $114 more than his sad- 6 62 ALGEBRA. [Eq. SeC. 7. die, 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 person rented a house on a lease of 21 years, and agreed to do the repairs when | of that part of the lease which had elapsed, should equal ^ of the part to come. How long will he hold possession before he repairs ? Ans. 12 years. 14. What number is that, to which if I add 20, and from f of this sum subtract 12, the remainder shall be 10? Ans. 13. 15. 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 | of the time to come. What time had already past ? Ans. 54 years. 16. Divide $183 between two men, so that ^ of what the first receives, shall be equal to -^^ of what the second re- ceives. What will be the share of each ? Ans. $6.3 and $120. 17. Bought sheep for. $300, calves for $100, and pigs for $25, and then laid out $15, which was f of the rest of my money, in getting them home. How much had I at first? Ans. $443. 18. A gentleman paid four labourers $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 to the fourth four times as much as to the third, and $6. more. How much did he pay to each ? Ans. To the first $26 ; second $10 ; third $19; fourth $81. DIVISION OF FRACTIONS. 63 DIVISION OF FRACTIONS, AND FRACTIONS OF FRACTIONS. §90. It was shown in §70, that in multiplying a fraction, we multiply the numerator only, and retain the denomina- tor. On the same principle, a fraction is divided by di- viding its numerator, and retaining its denominator. Thus, --^3=-. _^4=-,&c. §91. But, supposing we wish to divide f 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 prin- ciple to assist us. We shall find it in §81, where it was shown that a fraction may be changed to one with different terms, without altering the value. §92. 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 remainder. Thus, f can be changed ^^■^, if, ||, &c. ; each of which can be divided by 3, giving for the quotient either yV or ^, or /j, &c. §93. The most convenient equivalent fraction will be ob- tained by multiplying both terms of the fraction by the number 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. TK 4 ^ 20 .20 ^ 4 Thus, --^5==---r-5; and —-^5,=—. 9 45 45 45 §94. But, by examining the example just given, we find the numerator of the answer to be the same as the numera- tor of the first fraction ; for the first numerator has been multiplied and then divided again by the same number. 64 ALGEBRA. [Eq. SeC. 7. The denominator only is changed ; and that has been done by multiplying the first denominator by the number that was to be the divisor. §95. Hence the rule for dividing a fraction. Divide its numerator when it can be done without a remainder. But if there should be a remainder, multiply the denominator by the divisor for a new denominator ; and leave the nu- merator as it is. Thus, EXAMPLES. ^-4- 1. Divide >5. 3a; 2. Divide ->y- 2a; Ans.-. 3. Divide -,^y^- Ans. -. 4. Divide fb.5. Ans.f. 5. Divide -,^.,3. 4a: Ans. — . o 6. Divide 'r^y- Abs. g . i. Divide ^s,. A ^—^ ^"'- 20. • 8. What is 1 of ^t A 3a- ^"« 28- At 9. What is i of y? A 4^ Ans. -g. 10. What is i of llzf£? 41— 2x Ans. ,2 DIVISION OF FRACTIONS. 65 11. What is 1 of'" ;;-^2/, Ans. Sx—^l+y 64 12. What is i of '^-^1 ^ X — 1 Ans. 7a: 4-4 5a:— 5' 13. Whatis|of5gf? Ans. 21— 3a: 14a:— 63 14. Whatisfoff+!!^ ' Ans. 484-7y 1 S'y 1 9/11 |C7* 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. 15. What is i of -21=^1 Ans. ,r^^=^ ,« TTn • , n 32a: 4-16^ , 4a: + 2 16. What IS I of ^^^ ? Ans. —^—. ^ X — 6+2/ X — 64-1/ §96. We are now enabled to find a fraction of a fraction by the rule in §69 ; 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 di- vide by the denominator, we have just shown, i^ to multiply the denominators together. EXAMPLES. 1. What is f of - ? Ans.—- =— . ^6 18 9 2. What is f of ^"] Ans.^. 3. What is I of - ? ^"^'W* 4. What IS f of — -— ? Ans. 21 6* •^ #y 66 ALGEBRA. 5. What is I of Il±?f? 6x 6. What is 4 of — — ? ^ 23+2/ 7. What is j\ of 91/ a;+61 Ans. Ans. Ans [Ea. Sec. 8. 2734-6iC 20 24ic 2074-91/ ISy Q.WhatiSf^of^f-^? ^^ 41 +6iC 9. What is ^ of |^±i^? ^ 2ic— 112/ Ans. ;^^^^ —~"llic+671 36ar— 180 18a;~90 10. What is 3?^ of 6x+ly— 10 , 4 + 3^— 2ar 11.. What is I of ^ ^ ^ r 12. What is I- of 6 + x—y Sx- f-5y—8 ^ 27— 4a; 410 + 60a; 205 + 30ar l2y-]-2Sx 14a? — 772/* 18a:+ 2l2/— 30 40+302/— 2.0a;* 4a; — 42/ — 24 54+9a;— 9y* 6a;+102/— 16 135— 20a; Ans. Ans. Ans. Ans. EQUATIONS. SECTION 8. 1: A farmer wishes to mix 116 bushels of provender, consisting of rye, barley, and oats, so that it may contain f as much barley as oats, and 5 as much rye as barley. How . much of each must there be in the mixture ? Stating the question, a;=oats ; and *y= barley. Then,^of^=f,= rye. Forming the equation, a;+*7-+Y|=116. Multiplying by 14, 14a;+10a;+5a;=1624 Uniting terms, 29a;=1624 Dividing by 29, a:=56 the Ans. 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 ? DIVISION OF FRACTIONS. » Stating the question. ic= what I have. £^ paid first. X — J= the remainder. Forming the equation. Multiplying by 20, Uniting terms, Dividing by 3, 1 — ^^= paid afterwards. l-w-72 4a?— a:=1440 3a?=1440 a?=480 the Ans. 67 3. After paying away ^ of my money, and then | of the remainder, I had 872 left. How much had I at first ? In stating the question, the remainder after the first pay- ment was^ ; and \ of that is|J. Forming the equation, x — ~ — 1^==72 Multiplying by 20, 20a?— 5a?— 3iP=1440 Uniting terms, and dividing, a?=120 4. A clerk spends f of his salary for his board, and | of the remainder in clothes, and yet saves $150 a year. What is his yearly salary? Ans. 81350. 5. Of a detachment of soldiers, f are on actual duty, | sick, I of the remainder absent on leave, and there are 380 officers. What is the number of men in the detachment? Ans. 1000 men. 6. A young man, who had just received a fortune, spent 1^ of it the first year, and J of the remainder the next year ; when he had 81420 left. What was his fortune? Ans. 811360. 7. If from ^ of my height in inches, 12 be subtracted, ^ of the remainder will be 2. What is my height ? Ans. 5 ft. 6 in. 8. A bowl of punch was mixed as follows : \ was rum. 68 ALGEBRA. [Eq. SeC. 8. c i brandy, ^j acid and sugar, and 3 pints more than half of all these was water. How much did the bowl contain ? Ans. 6 quarts. 9. A, B, and C, own together a field of 36 acres. B has •J more than A, and C has i more than B. What is each man's share? Ans. A, 9 acres; B, 12 ; C, 15. 10. A gentleman leaves 8315 to be divided among four servants in the following manner. B is to receive as much as A, and | as much more ; C is to receive as much as A and B, and ^ as much more ; D is to receive as much as the other three, and i as much more. What is ihe share of each? Ans. A $24 ; B $36 : C $80 ; D $175. SUBTRACTION OF COMPOUND QUANTITIES. §97. Suppose we wish to subtract the expression x+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. W^ell, 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 quan- tity with -f changed to — . §98. 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 re- mainder, which will give us y — a?-J-6. Here, we have changed the positive a: to — ; and the — 6, to -f-6. SUBTRA-CTION OF COMPOUND QUANTITIES. 69 §99. Hence the propriety of the following rule for sub- tracting compound quantities. Change all the signs of the expression which is to be subtracted, the sign -}- to — , and the sign — to -{■ ; and then write the terms after the other quantity. It is to be recollected that in the quantityyrowi which we subtract, the signs are not altered. §100. Although the subtraction is performed the moment the quantities are written according to the above rule, yet it is always expedient to unite the terms if necessary after the operation. EXAMPLES. 1. Subtract 7a?+6+y from 63/— 17. Ans. 6y — 17 — 7x — 6 — y; which =5y — 23 — 7x. 2. Subtract 4y+3a:— 10 from 74— ar. Ans. 74 — X — 42/ — 3a:+10; which =84 — 4ar — 4y. 3. Subtract 6 — x — Sy from 7ic+6y. Ans. 7«+6y — d-^x-^Sy; which equals 8a; -|-9y — 6. 4. From 4a?— 3i/+27, subtract 6y — 12+ar. Ans. 3a; — 93/+ 39. 5. From 6-\-x — y, subtract 13 — 9y — x. Ans. 2a;+82/— 7. 6. From 8a;— 2 + 3i/, subtract Sy-\-^—3x. Ans. 11a; — 6. 7. From 5+4a;, subtract 2 — 5x-\-4:y — z. Ans. 9a; — 4y-f-3-f 2;. 8. From 6x — 8y, subtract — x — y-^50, Ans. 7a; — 7y — 50. 2x 9. From 14 — ;^, subtract 3a; — 12. o 2r Ans.26— 3a;— -. 70 ALGEBRA. [Eq,. SeC. 9. 10. From — ^ — , subtract 5 + — . S 5 , 8+7x ^ Sx AnS. :; 5 - 3 5 11. From 7 — -^, subtract — ^- — §101. It has been shown, §65 and §66, 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 denominator serves as a vinculum. Thus, in the eleventh example, above, — ^ is subtracted as a simple quantity ; and therefore the sign in it is not changed. 4a?— 6 Sx+7 ,„ ^ 4.r— 6 ^ ^x+7 12. From — : — , subtract — -- ,_ ^ 3a;+8 , 51 — X 13. rrom -—, subtract — -— . Ans. 3a?+8 51— a; Ans. — -: --. ,. ^ 7— 2a; , 21x— 4 14. J:^8— 42y _ 2:3--7y = Jo ; which ^g - ^ . 8+7x 3a; 3. Unite the terms in the quantity, — 5 — — . 40+35a?— 75— 9x 26a?— 35 ^°'- 15 =="^[5~' 2x 4. Unite the terms in the quantity, 26 — 3a? . 78— 9a;— 2a: 78— 7a; 78 7a; ^^ 7ar Ans. -^ =._.^_=26--. TT . , . , . ^ 52a: 21a;— 4 J 5. Unite the terms m the quantity, 7 jj: — +a;. Ans. — 2— 53a; 30 * 6. Unite the terms in the quantity, 12- 6+2a; 4 4a;— 3 5 12a; 8 ' Ans 111— 28ar 10 mi i^ii • ,'. Si^ifies therefore. r, '-: ■; ;■,' *■ -ii|U«i^.A'. DIFFERENT DNEOMINATORS. 75 X X X 7. Unite the terms in the quantity, J + 3" + 4"' 13a; X Afls. — =x+-. X 2x S. Unite the terms in the quantity, 4-1- 17 — ^+"5 * Ans.l-h|. Sx 4z 9. Unite the terms in the quantity, 42-}- — -{-z — — . 7 5 Ans.5.--. 10. Unite the terms in the quantity, y ^— — {- 2y— o 3 * 26i/— 3 , lly— 8 Ans ^— — - = 1/H — . 15 ^^ 15 §104. Whenever there are integers to be united with fractions, they may be changed to fractions, by putting the number 1 under them for the denominator. See example 2. Thus, 6= ^ ; 8= -. 76 0it*/m^LGEBTLA. RATIO AND PROPORTION. §105. When an unknown quantity is not, either by it- self, or in some connexion with others, known to be equal to some known quantity or set of quantities ; we may some- times find that there is a comparison between it and some known quantity, which is the same as the comparison be- tween two known quantities. Thus, suppose I buy 27 yards of cloth for S72, and wish to sell for 816 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 manner that $16 compares with $72. By knowing this comparison we can find the number of yards ; because, as $16 is |- of $72, so the number of yards to be sold must be I of the number of yards bought. It is 6 yards. §106. It will be seen that the comparison in this exam- ple, consists in observing how many times one of the num- bers is contained in the other. 16 is contained in 72, 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 con- tained in 3, three-fourths of a time ; and therefore we say the ratio ofSto^is^. §107. The pupil must remember that the ratio of one number to another, always signifies how the frst number compares with the last. Thus, the ratio of 8 to 6, is | ; that is, 8 is I of 5. Hence the ratio is expressed by mak- ing the Jirst term to be a numtrator, and the last to be the denominator. RATIO AND PROPORTION. T7 §108. 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 $16,) is ^|= |. And so also the ratio of the cloth bought, to the cloth sold ; (that is, the ratio of 27 yards to 6 yards,) is y which equals |. Here we see, that although the ratios are differ- ently expressed, they are, notwithstanding, equal to one another. §109. When the ratio of two quantities is equal to the ratio of ether two quantities, there is said to be a propor- tion between them ; that is, an equality of ratios is called a proportion, §110. 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 express- ed in the form of a fraction, it is evident, that when they are brought to a common denominator, if they are propor- tional their numerators will be the same, and if they are not proportional their numerators will not be the same. For 'example, is 11 to 21=33 to 63? We pursue our inquiry as follows : 11 to 21 is the same as |-J-, and 33 to 63 is the same as -^|. We bring the fractions to a common denominator by §103. 11 J 23 693 J 693 2l""^ 63 =1323 ^"^1323- We find they are equal, and the four terms 11 to 21ss33 to 63 are proportional. §111. 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 w ill see that we employ the same sign that expresses 7* division, with the exception of the — between the two dots. The sign : is read is to, and the foregoing examples are read 11 is to 21 and 33 is to 63. §112. 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. §113. In a proportion, the first and the last terms are called extremesyand the two middle terms are called means. In the above proportion, 11 and 63 are the extremes, and 21 and 33 are the means. ^ §114. In order to derive any important use from a pro- portion, we wish the pupil to recollect the method employed td find whether four quantities are proportional. We mul- tiplied, (see §110,) the first numerator by the last denomi- nator,, to find one new numerator. These were the two extremes. We also multiplied the last numerator by the first denominator, to find the other new numerator. These were the two means. And hence we learn, that if four quantities are proportional, the product of the two extremes is equal to the product of the means. §115. Therefore, whenever in our operations we find a proportion, we can easily reduce it to an equation by mul- tiplying the extremes together for one member ; and multi- plying the means together for the other member. Thus, 2 : 7 : : 8 : a:, becomes in an equation 2a7=56 ; whence :r=28. ■ . •'^ wm ^ im»- ^^ -f? f-^i,>ry^ BaUATIONS. SECTION 10. Jiff 1. Tf you divide $75 between two men in the proportion of 8 to 2, what will each man receive ? RATIO AND PROPORTION. 79 Stating the question, x= tlie share of one. 75 — a?= the share of the other. Making the proportion, x ; 75 — x : : 3 : 2 Reducing to an equation, 2x=225 — 3a; Transposing and uniting, 5a:=225 Dividing, a?=45 Ans. $45 ; and $30. 2. Divide 150 into two parts, so that the parts may be to each other 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? |C?^ After stating the question the proportion will be 30 +a;: 20 + 0" : : 5:4. Ans. 20 years. 7. What number 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 $40 more than the second ; and the stock of the 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, toge- ther with a suit of clothes which he was to have immedi- 80 ALGEBRA. [Eq. SkC 11. ately. 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? Ans. 32^ miles. §116. It is very often the case that a problem is easily solved by using simply the ratioj instead of a proportion. . '; EQUATIONS. SECTION 11. Operation by Ratio. .1. 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 J ; then the ratio of the second to the fii-st is |. Therefore, 2ic --- = the share of the second. 3 2x Forming the equation^ ^+-^ =40 o Multiplying by 3, 3a?4-2x= 120 Consequently, a?= 24. Ans. 24; and 16. 2. 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, a;= A's share. — - ■= B's share. '' > -~- =. C's share^ < • •.. .* RATIO AND PROPOKTION. 81 Forming the equation, ^+-77- -\ — 17 = '^^^' lit 4t Ans. A $156; B $234 ; C$390. 3. 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 pa'd 15s. ; B 25s. 4. Divide 560 into two such parts, that one part may be to the other as 5 to 2. Ans. 400 ; and 160. 5. A field of 864 acres is to be divided among three far- mers, A, B, and C ; so that A's part shall be to B's as 5 to 11, and C may receive as much as A and B together. How much must each receive? Ans. A 135 ; B 297 ; C 432 acres. 6. Three men trading in company, put in money in the following proportion : the first 3 dollars as often as the se- cond 7, and the third 5. They gain $960. What is each man's share of the gain? Ans. $192 ; $448 ;, $320. 7. 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. Ans. 8 and 4. 8. 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. 9. 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 it happens that A receives $15 more than B. What is the share of each ? Ans. A $27 ; B $12. 10. 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 proportion of 4 to 5. What are the numbers ? Ans. 72 and 96. 9^ 4. /. • ALGEBRA. [Eq. SeC. 11. 11. A man's age when he was married was to that of his wife as a to 2 ; and when they had Hved 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. 12. A certarn 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. 13. 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. 14. 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 10 5 ; and C's claim to D's as 6 to 7. How much must each creditor receive? Ans. A 81280 ; B $1920 ; C $2400 ; D $2800. 15. 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 J- of the whole sum by $50. What was the share of each ? Ans. $450 ; and $270. EaUATIONS WITH TWO UNKNOWN QUANTITIES. 83 EQUATIONS WITH TWO UNKNOWN QUANTITIES. §117. 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 quanti- ties, there is no difficulty in finding the value of each quan- tity. §118. An equation is said to be independent of another when it cannot be changed into that other. Thus, 7x — y =47, is independent of the equation 10y-^4x=50; be- cause one of them cannot be so altered ns to make the other. But, 7x — 1/=47, is not independent of the equation 21 ar — 8^=141 ; because the last is made by multiplying the first by 3. §119. At present we will attend only to equations that include two unknown quantities, each represented by a dif' ferent letter from the other. §120. In equations that contain two unknown quanti- ties, 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 opera- tion is called exterminating the other unknown quantities. §121. 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 con- venient to use one of them, and sometimes another. 84 ,^r^,T^I^^.^, ALGEBRA. iTr^HH^I** ^6^« First Method. §122. It is necessary here to recollect what was stated in §42, that when equals are added to equals, their sums will be equal ; and also when equals are subtracted from equals, their remainders are equal. Thus, suppose we have the equation, a?+14=36 ; and suppose also that we know that y=8 ; ihen if we will add y to the first member, and 8 to the last, the members will still be equal to one another, as follows: x+l4.-\-y—S6+8. And also if we subtract y from the first member, and 8 from the last, the members will still be equal to one another; thus aj-f-14 — ^2/==^ — 8» §12'3. This principle can be easily applied for the exter- mination of. unknown quantities. For, if in both of two equations, one of the unknown quantities has the same co- effccieni, but after different signs, it is evident that if we add both equations together, viz. the first member to the first member, and the last member to the last member ; and then unite terms, a new equation will be formed in which that unknown quantity will disappear. EXAMPLES. 1. Given the two equations, < - _.2iy=38 v ^^ ^"^ ^^^ values of x and y. Add together the two right hand members, and also the two left ; and we have the equation 3a: + 21/ +5x— 21/^26 + 38 Uniting terms, 8a:=64 Dividing, a:=8 Now, ifz^^^Xhen 3a:=24 ; and the first equation will become .-^'fij iiVs 24+2y=26, from which we may find the value of y. .1^* W** r* f^^ii W tw»* m** vi kwitn^t EQUA-TIONS WITH TWO UNKNOWN QUANTITIES. 85 2. Given the equations, < qSZ./Uj^28 i ^° ^^^ ^^® values of a: and y. Ans. a;=6 ; i/=4. 3. Given the equations, < o-cXe ---S4. i ^° ^^^ ^^® values of ic and y* OCr In this example, it is plain that the y^s will not be destroyed by adding them together. But we have before seen, §59, that if all the signs are changed, the equation will not be affected. Let us then change the signs of the second equation. The two equations will then become < 'o^_«'^Z 34 ( which, when added together, become 5x+6y--2a?— 6i/=58— 34, or 3a?=24. .•.a;=8; and the first equation becomes 40+6i/=58. Whence, y=S. §124. In the last example, if we take the equations before the alteration of the second, thus, < o^iXfi — S4 i ^"^ subtract the second from the first, the result will be the same as it was by changing the signs. As follows ; 5x-{-6y — 2a;— 6i/=58— 34. Whence we learn that, if in both equations one of the unknown 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 se- cond member from the second member,) and unite the terms ; we shall form a new equation in which that un- known quantity will disappear. EXAMPLES. 4. Given the equations, j g^tS^Sl ( ^° ^"^ ^^® values of ar and yl Ans. a;=5 ; y=7. 8 86 ALGEBRA. ^•^^^^^ I 2^-6^58 ( to find :r and y. Ans. x=2; y=9* 6. Given ^ ^ZepIoG | to find a^ and y. Ans. a?=40 ; 2/=lI' 7. Given J }2^21j,=63 J ^ A-d ^ and 3,. Ans. x=7 ; y^^l- 8. Given i 3^Z.2v= 7 ( ^° ^"^^ ^ ^"^ 2/* Ans. ar=5; y=4. 9. Given ^ _4^]}]2y=— 12 | to find a; and y. Ans. x=4: ; y=2. 10. Given J^^+2y;j4j to find ^ and y. (jfO" In this example, neither of the unknown quantities has the same co-efl5cient in both equations. But both mem- bers of the last equation can be multiplied by 3, without destroying the equality, §77 ; and then the a;'s will be alike in both equations. Thus, ) 3^T6^Z42 ( Ans. a?=10; y=2. 11. Given ^ 3^:]:4p^88 ] ^° ^"^ ^ ^^^ ^' ^d^ The second may be multiplied by 2. Ans. a;=8; y=l6» 12. Given ^ ^l^yZ^l I *° ^"^ ^ ^"^ 2/- OO' Make the j/'s alike. Ans. x=S ; y=7, 13. Given ^ 4^;;j:3p22 ^ to find x and 2^. OO" Multiply the first by 2 and the a;'s will be alike. Ans. a:=4; y=2» EaUA-TIONS WITH TWO UNKNOWN QUANTITIES. 87 '*-<^-^-\elXl:=ll\ to find. and.. Ans. x=7 ; z=8. 15. Given J ^^l|=Jo | to find :. and y. Ans. 07=4; 3/=5. 16. Given | 4^jl3pi7 \ to find x and y. Ans. a?=8; 2/= 5* 17. Given {'^iXlZlf I '° ^"^ ^ ^"^ '■ Ans. y=24; «=6. 18.Given52^+%^=7^ to find .= and j,. Ans. x=2; y=l' 19. Given lly+llZlll to find , and a.. Multiply the first by 4, 20?/ + 12ar==372 Multiply the second by 3, %+12a:=:240 Subtracting the 2d from the 1st, lly =132. Ans. 2/=12; a;=ll. 20. Given 5 l^~llZ^ I to find 2j and ^. Multiply the first by 5, 2O3/— 252=10 Multiply the second by 4, 20y— 16z=28 Subtracting the 1st from the 2d, 92=18. Ans. 2/= 3 ; z=2. §125. From the foregoing, we derive the following : Rule I. to exterminate an unknown quantity. 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 quantity to be the same in both. Then if the identical terms have like signs in both 88 ALGEBRA. [Ea. Sec. 12. equations, subtract one equation from the other ; hut if they have unlike signs, add one equation to the other. And the result will be an equation containing only one un- known quantity. . , EQUATIONS. SECTION 12. 1. What two numbers are those whose sum is 20 and difference 12 ? » Stating the question, x= greater number. y= the less. Then forming the equations, x+y=:20 X — 2/ =12 Adding the equations, 2a; =32 .•. a;=16 Substituting 16 for a; in the first, 16+2/=20 Transposing and uniting, 2/= 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? Forming the equations, 3ar-f 42/=25 4a;+2i/=20 Multiplying the second by 2, 8a?+4i/=40 Subtracting 1st from 2d, .5a? =15. Ans. Quinces 3 cents ; Melons 4. In any of our solutions after this, we shall number the lines, so that any reference to them will be easily under- stood. 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 ? EQUATIONS WITH TWO UNKNOWN QUANTITIES. 89 Let x= price of wheat, and 2/= price of rye. 1. By the first condition, 3a:4-5y=38 2. By the second, 6a;+3y=48 3. Multiplying the 1st by 2, 6a?+10//=76 4. Subtracting the 2d from the 3d, 7i/=28 .-.i/=4 5. Substituting 4 for yin the 1st, 3x+20=38 6. Transposing and dividing, 37=6. Ans. Wheat for 6s. ; Rye for 4s. 4. Two purses together contain 8400. 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 1 Let 3?= the number in the first. y = the number in the second. 1. By the first condition, a;+y=400 2. By the second, x — 40=2/4-40 3. Transposing the 2d, x — y ^80 4. Adding the 1st to the 3d, 2x =480 .-. 37=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 diflierence of their ages, the remainder will be the age of the youngest. What is the age of each 1 Let x= the age of the eldest, y^ the age of the youngest. 1. By the first condition, x-\-y+2^=z2x 2. By the second, x — y — 8 ==y 3. Transposing and uniting 1st, — x-{-y~ — 25 4. Transposing and uniting 2d, x — 2i/^ 8 5. Adding 3d and 4th, —y=—n 6. Substituting 17 for y in the 3d, —37+17 = —25 7. Transposing — 37= — 42 Ans. Eldest 42 ; Youngest 17- 8* 90 ALGEBRA. [Eq. SeC. 12. 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 832. What was the price of eacli 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 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 ? Let x= number of apples. y= number of pears. -J cent = price of 1 apple. — cent = price of 1 pear. 5 — cents = price of all the apples. ^ cents = price of all the pears. 1. By the first condition, 4+ I" = 30 4 o 2. By the second, 5-f X= 13 o lO 8. Dividing the 1st by 3, ^^ -{-^ = 10 12 15 4. Subtracting 3d from 2d, 2. —^^= 3 8 12 5. Multiplying by 24, 3a?— 2ar=:72 .-. ir=72. Ans. 72 apples ; 60 pears. 8. One day a gentleman employs 4 men and 8 boys to labour for him, and pays them 405. ; the next day he hires at the same rate, 7 men and 6 boys, for 505. What are the daily wages of each 1 Ans. Man's, 4*. ; boy's, 2s Qd. EQUATIONS WITH TWO UNKNOWN QUANTITIES. 91 9. It is required to find two numbers with the following properties. ^ of the first with -^ of the second shall make 16 : and ^ of the first with ^ of the second shall make 9. Ans. 12 and 30. 10. Says A to B, give me 55. of your money, and I shall have twice as much as you will have left. Says B to A, give me 55. of your money, and I shall have three limes as much as you will have left. What had each ? Ans. A lis; B 13s. 11. Two men agree to buy a house for $1200. Says A toB, give me f 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 8800 ; 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 35. 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 92 ALGEBRA. [Eq. SeC. 12. 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 ? ^CT" The solution must be in cents. Ans. 190 eagles; 310 dimes. 17. In the year 1299, three fat oxen and 6 sheep toge- ther 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 24 shillings ; a sheep Is. 2d. 18. Two persons talking of their ages^ A says to B, 8 years ago I was three times as old as you were ; and 4 years hence, I shall be only twice as old as you. What are their present ages ? Ans. A 44 ; B 20 years. 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 harpessed, 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. SECOND METHOD OP EXTERMINATION. 93 22. A is in debt $1200, and Bowes $2500; but nei- ther 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 answered, 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. 8^ gallons of that at 12s. ; 11^ of that at 5s. SECOND METHOD OF EXTERMINATION. §126. In each of the preceding questions, we first found the value of owe of the unknown quantities ; and then sub- stituted that value for that unknown quantity in one of the equations, in order to find the value of the other unknown quantity. This mode of operating furnishes a hint that leads us to another method of extermination. Let us take the first question in the last section [p. 88,3 in which we have the equations, < ^ io i The last part of our operation was to substitute the value of ar 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 94 ALGEBRA. [Eq. SeC. 13. X to be equal to ~ ; then substituting it for a;, the first equa- te tion would be |- +3/= 20. ^. t §127. 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 a?^20 — ?/. And then in the second equation, we will use the value of a; instead of x itself. Thus, 20— y— 1/=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 RVLE II. TO EXTERMINATE AN UNKNOWN QUANTITY. §128. By one of the equations, find the value of one of the unknown quantities, as if the other were known ; and then, in the other equation, substitute this value for the un- known quantity itself. EQUATIONS. SECTION 13. 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. 2/= less. 2x — Sy -= the required subtraction. 1. By the first condition, oc+y= 100 2. By the second, 2x — Sy=i60 3. Transposing the 1st, x=100—y 4. Multiplying the 3d by 2, 2a? 200— 2i/ 5. Substituting 200—2?/ for x in the 2d, '200— 2y— 3y= 150 6. Transposing and uniting, — 5y -= — 50 .•.y=lO SECOND METHOD OF EXTERMINATION. 95 7. Substituting 10 for y in the 1st, a? +10= 100 8. Transposing and uniting, a? =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 a?= age of the father. y= age of the son. 1. By the first condition, 37+3/= 140 2. By the second, 2/== "^J o ^3. Transposing the 1st, 2/=il40 — x 2x 4. Substituting 140— x in the 2d, l40-~a?= — 5. Multiplying by 3, 420— 3ar=2a? 6. Transposing and dividing, ar=84 7. Substituting 84 for x in 3d, 2^=140—84=56. Ans. Father 84 years ; Son 56. 3. Find two numbers, such that | of the first and i of the second shall be 87 ; and -J- of the first and -J- of the se- cond 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. Two servants went to market. A laid out as much above 4 shillings, as B did under 6 ; and the sum spent by A was to that spent by B, as 7 to 8. How much did each lay out ? Ans. A, 4s. Sd, ; B, 5«. 4d 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. 96 ' ALGEBRA. [Eq. SeC. 13. 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. S, 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 1 Ans. Father's 35 years; Son's 14 years. 9. The weight of the head of Goliah'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. 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 1 Ans. A $2500; B $3200. THIRD METHOD OF EXTERMINATION. 97 13. A gentleman has two silver cups, and a cover adapt- ed to each which is worth 820. If the cover be put upon the first cup, its value will be twice that of the second ; but i^ 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, 812 ; second, 810. 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. THIRD METHOD OF EXTERMINATION. §129. The method of substitution as explained in the last chapter, may be modified a little. We will show how, by using question 1st, in the last section of equations. riM. r ^ ^+y =100 The two equations were ^ 2x--dy=l50 We transposed the 1st ; thus, 07=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, 207=1 50 +3y. Then substitute the value of x as found before by the first equation, 200 — 2y—l50-\-Sy with which we may proceed as before. §130. Before we make the substitution after transposing, it is generally best to find the value of x alone in the se- cond equation. Thus, Given Is^lgplO^ ^^ find x and y. dS AtGEBRAr ■' [Eq. Sec. 14. JiU^' ^; Vlr^ 23— 3V Transposing and dividing the 1st, x= — ~- Transposing and dividing the 2d, x— -— ^. o 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—32/ _10+22/ 2 5~~ Destroying the fractions, 115 — 15i/=20+4y Transposing, uniting, and dividing, y=^ By substituting the value of y in one of the equations^ we find a?=4. Whence we derive Rule III. to exterminate an unknown quantity. '§131. Find by each cfthe equations, the value of that unknown quantity which is the least involved ; and then form a new equation by making one of these values equal to the other, equations. — section 14. 1. Divide $60 between A and B, so that the difference between 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 i/= B's. 1. By the first condition, x-\-y=60 2. By tlje second, x — 31 : 31— 2/ : •* 6 : 7 3. Multiplying extremes and means, 7x — 217=186 — 6y 4. Transposing the 1st, . a?=60 — y 5. Transposing and uniting the 3d, 7a?=403 — 6y 6. Multiplying the 4th, 7iP=420 — 7y 7. Making 5th and 6th equal, 403—6^=420—72^ 8. Transposing and uniting, y=17 9. Substituting 17 in the 4th, ar=60— 17=43 Ans. A's share $43 ; B's $17. THIRD METHOD OF EXTERMINATION. ^9 2. There is a fraction such that if 1 is added to the nu- merator, its value will be \ ; but if 1 be added to the de- nominator, its value will be i. What is that fraction 1 Let ar= numerator ; and y = denominator. X 1. The fraction will be. y 2. By the first condition, x-\-l 1 y 3 3. By the second, X 1 2/ + 1 4 4. Multiplying the 2d hyy, and by 3, 3:r + 3=2/ 5. Multiplying the 3d by y-^l, and by 4, ^x=y+l 6. Transposing and dividing the 5th, 4a:— 1=2/ 7. Making 4th and 6th equal. 3a;+3=4iC— 1 8. Transposing and uniting. — a:=— 4 9. Substituting the value of 3x in the 4th, 15=y. 4 Ans. -. 3. 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. 4. 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 tbat number 1 Let iP= first digit or tens ; and y= the units. 10x-\-y= the number. ^x-\-^y— four times the sum of digits. 10a;+2/+18,= when 18 is added. lOy-\-x,=: when the digits are inverted. By the first condition, 10a?+2/=4a;+4i/ By the second, 10a?-f i/-|-18=10«/+a; Ans. 24. 100 ALGEBRA. [Eq. SeC. 14. 5. 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. 6. A person has two snuflT-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. 7. 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 second 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. 8. 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. 9. There are two numbers such that ^ of the greater added to 4 of the less, will equal 13 ; and if ^ of the less be takep from ^ of the greater, the remainder is nothing. What are the numbers ? Ans. 18 and 12. 10. 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 divided by 1 more than their sum, the quotient will be 6. What is the number ? Ans. 24. 11. There are two numbers such that the less is to the THIRD METHOD OF EXTERMINATION. lOl 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=i the less ; and 2/=» the greater. 2v 1. By the first condition, ^—~R o 2. By the second, xy=10x-\-10y Note. — If we wish to multiply y by 4, we put 4 imme- diately before the f/ as a co-efficient ; and in the same way» if we multiply y by x, we make x the co-efficient of y. 3. Destroying the fraction of the 1st, 5x=:2y 4. Multiplying by 2, 10x=4:y 5. Substituting 4y for 10a? in 2d, xy=^y-\-10y 6. Dividing by y, a:=4+10 = 14. Note. — When we divide 4y by 4, we do it by taking away 4; when we divide lOy by 10, we do it by taking away the 10. In the same manner, we divide by y, in taking away the y. 7. Substituting 14 for or in the 3d, 70 ^2y .-. 2/=35. Ans. 14 and 35. 12. There are two numbers, whose sum is the ^ part of their product ; and the greater is to the less as 3 to 2. What are those numbers? Ans. 15 and lO* 102 V ^AI'GBBJIA. EQUATIONS WITH SEVERAL UNKNOWN QUANTITIES. §132. When there are three or more unknown quantities, they are exterminated one after another by the same me- thods which are used for two unknown quantities. §1 33. But it must always be the case that there are as many independent equations as there are unknown quanti- ties. See §117. EXAMPLES. 1. Given the equations or < {greater than or less than) de- notes that that quantity towards which it opens is greater than the other. The sign (a vinculum,) denotes that all which is put under it, is to be used as one term. The ( ) parenthesis is frequently used instead of the vin- culum. The sign c» {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 's always the case when the cipher is used as a denominator of a fraction. The radical sign V denotes the root of the following quantity. When unaccompanied by a figure, it represents the square root. But when a figure is put over it, that figure expresses the root that is designed. A co-efficient \s a number put immediately before a letter," as, 26. In such a case, the co-efficient is called a numeral co- efficient. Sometimes when one letter has been multiplied into another, the first written one is called a literal co-effi- cient ; as, ab. An exponent or index is a small figure placed a little over and a little to the right of a quantity ; as, a^. An exponent may be either positive, negative, or frac- tional, as, a^, a~^, a*. The sign : : represents proportion, and : denotes the ratio of two numbers. The sign oc denotes a general proportion. GENERAL PRINCIPLES. 107 A simple quantity is that which is represented by one term. A compovnd quantity is that which is represented by two or more terms. A quantity when represented by one term is sometimes called a nomial ; with two terms it is called a binomial ; with three it is called a trinomial ; with many terms it is called a multinomial, or polynomial, §141. The algebraical mode of designation is of the greatest use in universal arithmetic ; as every conclusion, and indeed every step by which it is obtained, becomes a general rule for performing every possible operation of the kind. Example 1. Suppose we wish to divide $660 between a son and a daughter, so that the daughter shall have twice as much as the son. What must we give to each ? Now, a question similar to this, and with the same numbers was solved in the First section of Equations, on page 26. We will solve this in the same manner, with the exception of using a instead of 660. Stating the question, x=. What the son has. 2x= What the daughter has* Both together have x-\-2x ; also they have a dollars. Forming the equation, x-\-2x=a Uniting terms, 3a?=a _,.-.-_ a Dividmg by 3, x= — Here we find that the son's share is | of a, which at this lime, 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 the son's share of it would be $80, and the daughter's share, $160. 108 ALGEBRA. In the same manner, if the sum were ^360 ; 'then a would stand for $360, and the son's share would be ^ of 360 ; that is, $120. And in the same manner we may make a represent any sum ; and still the son's share of it would be -J- 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 rule for di- viding any sum between two, so that one of ihem 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. §142. In the same manner, we may generalize all the questions in the First Section of Equations. That is, with each question we may find a rule by which any other ques- tion like it, may be answered with fewer figures than the Algebraic operation required. But, as such sums are not apt to occur, there will be no practical use of finding rules for them. Notwithstanding, as the exercise may be instructive, the teacher may require his pupils to go through them if he sees fit. 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, a;= the number. x-{-b= after adding. Forming the equation, x-}-b—a Transposing 6, x=a — b We see that the answer is found by subtracting the 5 from the 40. Thus, 40—5=35. The third question is similar to it ; and in that, a repre- sents 23, and b represents 9. The answer is 23 — 9=14. GENERAL PRINCIPLES. 109 Example 3. Divide $17 between two persons, so that one nnay have $4 more than the other. [Prob. 4. Sect. 2.] Represent 17 by a, and 4 by b. Forming the equation, x+x-^-b^a Transposing b, x-\-x=a — b Uniting terms, 2x=a — b Dividing by 2, x= a — b ~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 5lh question is similar to it : and in that, a represents 55, and b represents the difference or 7. It is found by the rule just shown. Thus, 523! = !?=, 4. -i lit Perform the 7th and lOih by the same rule. §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 J we subtract the difference of those parts, half the remainder will be equal to the smaller part. Example 4. In the same questions, let us take x for the greatest share. Then x — b = the less. Forming the equation, x-\-x — &=a Transposing and uniting, 2a*==a+6 Dividing by 2, x=a+5 §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, 10 110 ALGEBRA. Find the greater part in questions 4, 5, 7, and 10, by this rule. §145. The mere letters in the answer of an algebraical operation are called di formula. They are not called a rule^ until they are turned into common language. Example 5. The learner may now generalize question 6; and by the formula that he obtains, he may find the an- swers in questions 8, 9, and 11. The two differences will be h and c. We have said that in Algebra the arithmetical operations on nuqibers are only represented by different methods of combining 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. ADDITION AND SUBTRACTION OF ALGEBRAICAL QUANTITIES. §146. One algebraical quantity is added to another by writing one quantity after the other, taking care to put the sign plus + between them. Thus, a+/ — c is added to d — c+^j so as to make d — e-\-h-\-a-\-f — 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/» §147. One algebraical quantity is subtracted from an- other, by changing the sign or signs of the quantity which is to be subtracted, and then writing that quantity after the other. Thus, a+A — y is subtracted from b — x-fc, by first ALGEBRAICAL QUANTITIES. Ill making it — a — A+y, and then writing the whole quan- tity, b — x-\-c — a — h-\-y \ or, h — a-f-c — h-\-y — x. §148. When quantities that have been nnade by addition or subtraction, have like terms in them, they may be re- duced to smaller expressions^ by uniting the like terms. This is done by putting into one part, all those which have the sign +, and into another part, all those which have the sign — ; then subtract the least result from the greatest, and give to the remainder the sign that belonged to the greatest result. §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 units and 3 units of a particular kind, which when put together, make 5 units of that kind. Now, we have seen, §139, that the co-efficient of a quantity may also be literal ; as in 6a, ca, &c. In such cases, the whole term ha or ca becomes the unit, each of a different kind ; and of course are not like 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, ha — ca -^ha-\-ca-\-ha-\-ca, can be united into, Zha-\-ca. ax — hx -{■ax-\-2bx — t^ax-\-hx, are equal to, — ax-\-2hx \ ov 2hx — ax, §151. Again, we have seen, §65, 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 expres- sions, 3xa — h-\rx^ and b(x-\-ax — i/), a — h-\-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 — hx-\-x)-\-b{ay — hx-]-x) = l{ay — hx-\-x). 112 ALGEDRA. §152. In uniting terms, great care must be taken that the literal part be entirely alike. Thus, 2bx-\-ZcXj cannot be united. Neither can Sy — 2ay ; nor, 6(a-\-bx)-\-2{ax ■i-bx) ; nor, 3. ay — by + 2.ay^ — by, nor, A{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. Sax — 2y-\-4:X — oy-\-ax — Sy, Ans. 8aar — lOy 2. Sx-^ay — 2x — ay-i-^x-\-Say — 2x+Aay. Ans. ^x-^-lay 3. Aax — y+Say — 2 — 2ax-\-ay — ly-\-Q-\-2ay-\-y. Ans. 2ax-\-^ay — 7^-4-6 4. ax — ay^ — Say-\-bax — 2ay-\-lay — Aax — Say^ Ans. 2ax-\-2ay — 9«y^ 6. 3(a - y) + 4(a — y) + 2(« — y) +7(a — y). Ans. 16(a — y) 6. _4(a+&)+3(a+6)— 2(a + 6)+7(a-f&.) Ans. 4. a-\-b 7. 2{ab-\'X)-it-S{ax+b) — A{x^y) — 2{ab-\-x) Ans. S{ax-\-b) — 4(a: — y) 8. 7y — 4(fl + ft) + 62/ + 22/ + 2(a + 6) + (a + 6) -f y — 3(a + b.) Ans. 16y — 4(a+6) 9. x^ -\-ax — ab-\-ab — x^-^xy-\-ax-\-xy — A.ab-\-x^-\- x^ — x-^xy-\-xy+ax. Ans. 2x^-\-Sax — Aab-\-\xy — x 10. X+12-— aa;+2/ — (48 — a: — ax+3.y). Ans. 2x — 36 — 2i/, MULTIPLICATION OF ALGJ^RAICAL QUANTITIES. 113 11. ab — Axy — a~x^ — (2xy — b-\-lAx-\-x^) Ans. ah — 6xy — a + b — 140^ — 2x^^ 12. S{x-\-y)-\-{^.x-^y). Ans. l{x-\-7j). VS. 2{a--b) — x—{'S.a+b — x''). Ans. x''—{a+b)—x. 14. From 4.a+6, take a+b — S.x — y. Ans. S.a-\-b-{-S.x — y. 15. a-^b — {2a — .36) — {5a-\-7b) — ( — lSa+2b). Ans. 7a — 56. 16. 37a — ox — (3a — 26 — 5c) — (6a — 46 + 3A). Ans. 28a+66 — 5a: + 5c— .3/t. EXERCISES IN ECiUATIONS. §153. The learner may now generalize the problems in Sections 3 and 4 of Equations, pages 35, and 40. This he can easily do, if he takes care to make one of the first let- ters of the alphabet stand for each numeral quantity. When the same numeral quantity occurs more than once in the same question, the same letter must stand for it each time. MULTIPLICATION OF ALGEBRAICAL QUANTITIES. §154. We have shown, §29, that a literal quantity may be multiplied by writing the multiplier before that quantity. This is the case whether the multiplier is numeral or literal. Thus, a limes x, is written ax; bxc=bc. In the same manner, a times be becomes a6c ; andy times a6c becomes fabc. As fxabc is the same as abcxf, we see that it is of no consequence what order we make of the letters in th« 10* 114 ALGEBRA. product, ahcd = acdb = cadby &c. But it is generally more convenient to follow the order of the alphabet. §155. Therefore, to multiply one simple quantity by an- other, write the quantities one after another, without any sign between them. Thus, abx times c/v/ = abxcfy; bax times cdf=^ bacdfx. But, if there are more than one numeral co-efficientj those co-efficients must be multiplied as in arithmetic, and placed before the product of the literal quantities. Thus, 3a x 2a? = Z.2ax = Qax, 2bc x5rs = lObcrs. 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 "ibrs by 2mnx. Ans. Qbmnrsx. 6. Multiply 2adn by 5cmx, 7. Multiply ^rsyop by 4an/x. 8. Multiply 2abcx by 8abrx, §156. By the foregoing principle, axa — aa. Now, as in Algebra the same factor is often found two or more times, Stifelius adopted a method for shortening such ex- pressions,* 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 oncej 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^ ; bbb is written 6^ ; xxxx is written ar* ; aabbbyyyy is written a* b^ y*. • A, D. 1554. i^^i MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 115 §157. Mathematicians are accustomed to call aa or a", the second power of a, or, a-second 'power ; a*, is called a-third power, &c. §158. The figure that denotes the power of any quantity is called the exponent or index of that quantity. §159. All quantities are said to have an exponent, either expressed or understood. Thus, a is the same as a* ; 6=6* ; &c. The written exponent affects no letter except the one over which it is written ; unless it is denoted by a vincu- lum. §160. Great care must be taken by the pupil not to con- found the co-efftcient 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=54-5-f5=15; but ^3=5x5x5 = 125. §161. We have seen X\\aX a^=^aa ; and that a^^^aaa; now aaxaaa=aaaaa. So we see that a^xa^=a^. 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^Xx*=x^, y^Xy^=s y' ; &c. EXAMPLES. 9. Multiply 2amn by a. Ans. 2a^mn. 10. Multiply Sabcx by 4ax. Ans. 12a^bcx^, 11. Multiply 5bcmn by 'Sbc, Ans. 15b^c^mn, 12. Multiply 6a'^xy by ^ax^y. Ans. 2Aa^x^y^ 13. Multiply Aa^bcx by a^b^c, Ans. Aa^b*c^x. 14. Multiply 'Sa^m^ by a'^m^n. 15. Multiply ba^x^ by 4a^a?*. 16. Multiply 2771 Wic by 9aTa?*rn». 116 ALGEBRA. 17. Multiply Sb^c^'n^ by SaH^mn^. 18. Multiply 7aya;* by \2a'>¥x. 19. Multiply ^ni'x^a' by la^'h^m, 20. Multiply \2aJ^mHx^ by ISb^cn^y. §162. When one of the factors is a compound quantity, (§61 and 62,) we multiply each term by the other factor, and set down, their products, each with their proper sign. In doing this, we generally begin at the left ; thus, a-{-bc ab — c. Multiplied by x Muhiplied by a ax-\-bcx a% — ac. , 21. Multiply ax-\-bx by xy. Ans. ax^y-\-bx!^. 22. Multiply 2d^bc — rxy^ by "lay. Ans. Aa^bcy — 2arxy*. 23. Multiply 7ay+l by Sr. Ans. 2lary+3r. 24. Multiply 2xy+ab-\-c* by 2ax. 25. Multiply 5 — 7x-{-2a^b by 4acy 26. Multiply 2m"n — Srp-{-2sx^ by 6a^ba. 27. Multiply 2en — Aan+5 by 120^^x5. 28. Multiply Ay-\-y^ by 2xy. 29. Multiply 3a — 46+4 by 6y, 30. Multiply 6ax''-^a^— 1 by 2ax\ 31. Multiply 4 + 3a — ic^ by ay. 32. Multiply 2a+5i2_^3c — 5e by 3a». 33. Multiply 7b^ — 4-ha«a? — x"^ by 4a«a?. 34. Multiply 6a; -f-7a — axy — 2y by Zbx^. §163. As(a+6) y.x=^ax-\-bx\ weknow thatar x(a-|-6) also =ax+bx. Whence we see that if we wish to multi- ply X by a-\-b ; we first find the product of a times a?, and then add to it the product of h limes x. And generally^ MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 117 when the mvltiplier is composed of several terms^ the pro- duct is made up of the sum of the products of the multipli- cand hy each term of the multiplier. Thus, {x-{-y)x{a-\-b) = ^ a > and minus < or c — d \ac-\-bc y \ad-j-bdl ac-\-bc — ad — bd. Here we see that the product of fZ into a 4-^ is subtracted from the product of c intoa + 6 ; and therefore the signs of ad-\-bd are changed to — ad — bd. §165. 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 multiply 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 ; as we have only time to remem- ber that, §166. When the signs are alike, the product is +. §167. When the signs are unlike, the product is — . EXAMPLES. 46. Multiply a-\-b by a—b. a-\-b a—b —ab—b^ Ans. a^- -4i%'. MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 119 47. Multiply x+8z by dx''—7xz. Ans. 3a:3+17»^z— -56a;2^ 48. Multiply x^+xy + y^ by x — y. Ans. x^ — y^. 49. Multiply 2x-\-3y by 'Sx—^y, Ans. Gaj^-f a:^ — 12^^ .50. Multiply b^-^b^x+bx^+x^ by b — x. Ans. b* — x*, 51. Multiply a — b by c — d, a — b 1 C a — b ^ a — b c > minus or c — d ac — be J ac — be J V ad — bd J ac — be — ad-\-bd. Here we see that in subtracting d times a — b, we change the signs of ad — bd to — ad-{-bd. §168. Whence we learn, -f multiplied by +, produces + + multiplied by — , produces — — multiplied by +» produces — — multiplied by — , produces -f. And by remembering this we can always set down the Jinal answer at first. 52. Multiply a — x by a — x, a — X a — X ix-\-x^ tt« — 2ax+x'^. 53. Multiply 2x — Sa by 4a? — 5a. Ans. Sx^ — 22aa;-f 15a». 54. Multiply 2a — 5y hy a — 2y. Ans. 2a* — Qay-^-lOy^. 55. Multiply a^+ac — c^ by a — c. Ans. o* — 2ac'^-\-c', 56. Multiply a-\-b-^dhy a — b. Ans. o" — ad — b^-j-bd. k 120 ALGEBHA. 57. Multiply Ax — 5a — 2b by Sx — 2a-\-5b. Ans. l2x^ — 2Sax-\-lAbx+\0a^ — 21ab^l0b^. 58. Multiply x^ — y^ — z^ by x — y — z, 59. Multiply 6+xy — a^ — mif by a^ — Sx^-^y*, 60. Multiply 2a^b — 3ac2-f-46V — 1 by 2a^c^ -— 5b^c — 8a*. §169. 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. Gdneral Properties of Numbers. §170. We have before stated that algebraical operations, §137, by reason of the quantities themselves being retained in their original value, do show us, in their results, import- ant general principles. We will here make a few naultipli- cations of some quantities, whose results show us some re- markable general properties of numbers. These properties the pupil should remember, as they are of frequent use in the subsequent parts of this study. §171. Suppose we have two numbers, a and 6, of which a is the greatest. Then their sum = a-\-b ; and their dif- ference = a — b. Then a-i-b a — b aFTab —ab—b^ a^—b^ By this operation, we find the general property of num- bers which it would be difficult to find by any arithmetical operation. It is that, if we multiply the sum of tico num- bers by their difference, the product will be the difference of the squares of those numbers. ^n MULTIPLICATION OF ALGEBRAICAL QUANTITIES. 121 §172. Again, take the same quantities. and multiply their sum, by their sum* a+6 a-\-b a^+ub + ab-\-b^ a^+2ab-\-b^ By this operation we find the following general property. 2'he square of the sum of two numbers is equal to the square of the first number, plus twice the product of the two numbers, plus the square of the last number, §173. Again, take the same quantities, and multiply their difference, by their difTerence. 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 product of the two numbers, plus the square of the second. §174. 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+¥ Aab The actual difference between the square of the sum, and the square of the difference, is four times the product of the two numbers. 11 122 ALGEBRA* ^175. 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 quart' titles. Also, if we perform the same operation with the difference of the two quantities, multiplying by their sum, we shall obtain the sum of the third powers of the two quantities. Thus, a+h a — h a-\-h a — b Rejecting the co-efficients, a^+ab + b^ a^—ab-j-b" a — b a+b a^ + a^'b+ab^ a^—a^b+ab^ ^c^b—ab^—b^ + a^b—ab^ + b^ §176. It is often the case that it is better to denote the multiplication of compound quantities, than to perform it j on account of operations that follow. Thus, a — h times ic+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 par- ticular. 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. aiVISION OF ALGEBRAIC aUANTITIES. 128 DIVISION OF ALGEBRAIC QUANTITIES. §177. Division may be represented by the sign -r- , as « -H 6, is read a, divided by h ; (a+6) H- (c — d)y is read a-\-h, divided by c — d. But the most usual way to denote division (§67,) is to write the divisor underneath the tiivi- dend : thus, 7-, — :;. b c — a §178. But it often happens (§71, 73,) that the fraction made by this representation is an improper fraction ; and one in which the numerator can be actually divided by the denominator. In such cases, it is generally best to pe rfom the division. We have always done so in the former part lOrc of this treatise. Thus, 4a; divided by 2=2a: ; — - = 2x, o §179. Let us first look at the case where the same quan- tity is in both the dividend and the divisor. 7a-r-7=a ; 126-r-12=6. In the same manner, ab-r-a^=b\ dc-7'd=^c^ This may be easily proved. For, ab is the product of a into b ; and of course if we divide by what was the multi- plier, we shall obtain the old multiplicand again; as may be seen by trying the product of any two numbers. §180. Whence we derive the general rule, that when the divisor is found as a factor in the dividend, the divisionis performed by erasing that factor from the dividend, where we see that when 7ve divide a compound quantity ^ we divide each of the terms. §185. We must also recollect that, as + multiplied by -f J makes -f- in the product, so + in the product divided by +, must make -f in the quotient. And that as + ^v^• tiplied by — makes — , so — in the product divided by — will bring back the + in the quotient. So that when the signs are alike in the dividend and divisor, the sign in the quotient is +. Thus, axb=ab; both of which are +. Also — ax — b=ab ; and — ab-i 6=H-a. §186. Again, as — multiplied by + makes — , so in the product — divided by +, brings back — in the quotient. Also, — multiplied by — makes + ; 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 sign in the quotient is — . EXAMPLES. 34. Divide 2ad-\-8a^c by 2a. Ans. d+Aac. 35. Divide 8d^7n^^l2d'm^ by 4(Zm». Ans. 2d^—3d*m. 36. Divide 4.xy + 6x^ by 2x. Ans. 2y-^Sx. 37. Divide abc — acd by ac Ans. b — d. 38. Divide I2ax—8ab by —4a. Ans. — 3a;+26. 39. Divide l0xz-{-15xy by 5x. 40. Divide 15ax — 27iC by Sx, 41. Divide 18a^— 9a; by 9a;. DIVISION OF ALGEBRAIC QUANTITIES. 127 42. Divide abc — bed — hex by — be. 43. Divide 3a?+6a^-|-3aa? — 15a; by 3a:. 44. Divide ^abe-\-V2abx—%a^b by 3a6. 45. Divide \Qa^b^-^e,Oa'b^—\lah by ab. 46. Divide Iba^bc — \Octex^-\-^ad^c, by — 5ac. 47. Divide 2(iax-\-lbax^-\-\Qax—ba, by 5a. §187. It is evident that we may divide by either factor. Thus, ax-\-bx may be divided by a?, and the quotient will be a+h ; 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 rriake one of them of a given magnitude ; that is to make it the same as the given divisor. §188. Now we know that x times a-\-b,=ax-{-bx ; and also that a-^b times x,=ax-\-bx. Hence we know that the product {ax-^bx)-^{a-\-h)=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 fac- tor in the first term of the dividend, the second term of it in the second of the dividend, and so on through each of them respectively ; and the remaining factor in each term of the dividend being the same ; that remaining factor shall be the quotient. EXAMPLES. 48. Divide ax+bx — ex, by a+b — c. Ans. x. 49. Divide bac-{-bc^x — bx^, by ac-\-c^x — x^. Ans. b. 50. Divide c'^ax — '2abx — 3xy + x, by ac^ — 2ab — 3?/+l. 51. Divide cd^x—abd^'+d^x^—d'', by cx—ab+x^—1. 52. Divide a^y — bcy-^xy, by a^ — bc + x. 53. Divide 6ahm — 14a&m — *icdm, by 6ah — lAab — 3a/. I 128 ALGEBRA. §189. If the letters of the divisor are nol found in the dividend, the division is expressed, as we have before shown, §177, by writing the divisor underneath the dividend, in the form of a vulgar fraction. EXAMPLES. 4i/"4~7.c 64. Divide 4y+7a;, by a — b. Ans.— — —. 55. Divide 3a+262— ,c, by a+c, 56. Divide a^— a^^ft + c^, by a^— fe^. 57. Divide Sa''c+2b^ + c, by 2c, §190. 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 §184. By this method the answer of the last sum would Sac^ 2b^ c be 1- f- — . Answer the following by both me- liC 'iC 4a a?/ 8. Divide ; — - by 3ay. a+2/ ^. .^ 4a+27a6 . . 9. Divide by 6aa,'. 2x ^. ., 6a6ar — 16axy ^ ^ 10. Divide-— — — - — - by 2a, 2am-\-Sxy ^. ., 146c^a?+21ac^^ ^ - 11. Divide — f byTc^ 4a 6a; 12. Divide , /i „ 77; bySa^r*. a^4-5x^ — lOaa; „ ^. ., 12a6 — I4ca: , „ ., 13. Divide—- —7— by 2ac — Abx. 10aarH-166c '' FRACTIONS, 137 1 4. Divide ^: — by 4ca — cax. x—Sy ,_ ^. ., 9am 4-1 Sax , ,„ ^ „ 15. Divide -^ by I2ax — 9a'. 6mx 16. Divide ^^'"^^^^ by a^ — 1. 14a«— 16 ^ EXERCISES IN EQUATIONS. §201 . Generalize ihe questions in Section 6th, page 54. 1. In an orchard, — (i) of the trees bear apples i — (i) P of them bear pears ; — (^-j-) of them plums, and a (81) bear cherries. How many trees are there in the orchard? 0:!?° We rarely represent unity by a letter ; but general- ly use its own character, 1. Let X = number of trees. — = apple-trees. X — = pear-trees. n ^ J)X — = plum-trees. iheii, X — — I [- — + a. 771 71 r Multiplying by mnr, innrx=^nrx-}-mrx -\-mnpx+mnra Transposing, mnrx — rirx — mrx — mnpx = mnra Dividing by mnr — nr — mr — mnp, mnra x= mn r — n r — mr — mnp Substitute figures for letters, and find the answer. 12* 138 ALGEBRA. 2d sum, page 55. In a certain school, — of the boys learn mathematics ; — of them study Latin and Greek ; and a n •' study Grammar. What is the whole number of scholars? The equation is, ar = — [- — 4- a. Blultiplying by mn, 7nnx = na? + mpx + mna. Transposing, mnx — nx — mpx = mna. r^. .,. , mna Dividmff by mn — n — m», x = . ° "^ mn — n — mp 3. The pupil may go through with the whole section in the same manner. And as ih&xe are several instances in which the same statement and the same answer will agree with two or more sums, the pupil may tell which they are, and why it so happens. §202. If the teacher should think his pupils need more practice, he may exercise them in the 7th section in the same manner. The first question may be stated thus, X -^ a Multiplying by 6, a? + a = 6c Transposing, a? = 6c — a. The 2d will have the following equation. mx 4- «_ , n ~ Multiplying by n, mx -j- a ^ nb Transposing and dividing, _ nb — a m The 3d and 5th will have the following equation. nx — na m nx The 4th will be, a — x=—. m FRACTIONS. 139 FRACTIONS OF FRACTIONS. §203. It was shown, §96, 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, -r- X -r= r-f d bd EXAMPLES. ,,,,., 3a ^ 4& , I2ab 1. Multiply— by—. Ans. — — . ^ •' 5a; "^ c 5cx ^ ,, , . , 9ax , 2bx , leabx^ 6x'' 2. Multiply ^ by— . Ans.j^^=-. 3. Find the product of ^i^ into — ^. Ans.-^^i^. ^ Sa X — y ax — ay 3x 4 a; 3a:^ 4. Multiply — and — together. Ans. -— . . ,c , • , ^ . 6m ,2 .5. Multiply— into -^. Ans.—. ^ -^ Sy 7x - 7 ^ ,, , . , 3a . 4a , Sa^ 6. Multiply -^ into — - . Ans. — -^ . 7. Multiply linto-^. Ans. ^^±^ . ^ -^ a x+2y ax-\-2ay a . y . ay 5. Multiply into — — . Ans.—. ^ ' X Z XZ 9. Multiply — , — , -^ together. Ans. — . 4a a? 3a?'M 2 10. What is the product of , -^, and — ? Ans. I2a?. , , ^ 2a . 2ac — be ^ 11. What IS the product of . into 36 + c Sab 140 ALGEBRA. 12. Multiply -j;^7+-2— ^y 2S5;r3r5'c- 13. Multiply ^Jg/ ty g;^-^^^^- EXERCISES IN EQUATIONS. §204. Generalize the questions in Equations, Section 8. 1. The 1st sum on page 6(5, is performed as follows. Stating the question, x= oats. mx . , — = barley. n p n np mx , mx Forming the equation, *+~;^'^ "no ~ Multiplying by np, npx+mpx+mx = anp anp Dividing by np+mp-hm, x = —j-^-j-^. , X X 2. The equation in the 2d sum will be -— — = a Multiplying by mw, mx — x^ amn __ amn Dividing by m — 1, ^ ~" ^_i' 3. The equation in the 3d sum will be X XX X = «• m n mn 4. The equation in the 4th sum will be nx nx n^ __ m m m^ The pupil will now understand how to perform the rest of, the section. FRACTIO>S. ] 41 UNITING FRACTIONS OF DIFFERENT DENOMINATIONS. §205. Before fractional terms can be united, they must be brought to a common denominator, according to the principle explained in §81. This is done as we have shown in §102, 103, by multiplying each nvmerator by all the denominators except its own^for new numerators ; and all the denominators together for a new denominator, EXAMPLES. , TT •. .L rn • . 2a6 ax , am 1 . Unite the following terms, 7/ "^ A~ * Sa¥x — abx^4-Aabmx 8ab' — ax^-\-Aa7n Ans. 77- = ~j . Ab^x^ Abx X x 2x^ 2. Add — -rrand together. Ans. -— ^ . o C5 u. * 3« r 4a . 32ac— 21a6 6. bubtract -- from — r. Ans. — :-j . 8c lb ^Qbc . TT . 3? V . xz — xy — y^ 4. Unite — •^. ' Ans . . x-{-y z xz+yz _ ,^ . X — y x+y . ax^5ay 5. Unite ^ -^ ~-^. Ans. tt-o—^' 2a Sa 6a^ §206. If, in such cases, there is a quantity that is not fractional, we multiply it by all the denominators ; and then putting the common denominator under that product, unite it with the other fractions. ^ TT . . ^ A ca , , ea , b ac-\-b 6. Unite a-\ — . Ans. a=— ; and then, — = . c c c c c -r TT •. . «+^ A xy+a-^x 7. Unite X-] —. Ans. -^ . y y 8. Unite a Ans. . 142 ALGEBRA. ayz+8xz — '2x^y ^ ,^ . a , 4 ar . 0. Unite ^+2'+"-*. z c -U'>"%-f-.+S-^ a 4 2. Unite xH-^ — a^/ — ;^- Ans. Ans. 2xy2 cx4-cy4-az — bz Ans. — — Ans. cz 2ay^+Sa^x+6ax 2ay^+4y^ lbx-\-la — laby — 46 lb „ .,,2a;+l 4a?+2 , a? , , 169a: + 77 13. Add — - — , — - — , and — together. Ans. — -— — o o 7 105 ,. *.i u 5a2+6 , 4a2+26 ^ 37a2+ll6 14. Add together , , and — —r — . Ans. —r — 1^ cj u. . ^^+1 r 21a;+3 ^ 127a:+17 15. Subtract — - — from : . Ans. 28 ,« CI, 3x4-1. 4a: ^ /^x'^ — UX'-b 16. Subtract ■— from — . Ans x+l 5 5x4-5 ,- Ajj 1. 2a:— 5 ^a:— 1 ^ 4x3— .7a: — 3 1 / . Add together — - — , and —r — . Ans. 3 ' 2a; 6a: 18. Add together r, and — — . Ans. ° X — 3 a:+3 x^ — 9 io C! K* .2a: — 3. 4a:42 ^ 40:^^+3 1 9. Subtract — from — - — . Ans. ^— 3a: 3 3x 20. Add together — — j-, and — —7. Ans. a — b a + b a^ — 6^ ^^, ^ 3a426 , ^hd—^a—Zd ^1. rrom .subtract -, . -, , c 4crf r •<* 12af/4-3iJ4-2a+3rf Ans. -—: ■ — . ^cd 22. From c42a6 — 3«c, subtract j- = — ■ — . 0^ — be 2ab^ — 6c^4 Sabc" — «' Ans. FRACTIONS. 143 DIVISION BY FRACTIONS. §207. Suppose we wish to know how many times -f- is contained in ^. We would divide in the same manner that we follow in dividing 6 pieces by 3 pieces ; and say ^ is contained in -f-, two times. In the same manner, ^r '^ contained in |^, fae 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^ a' Thus, -j-H-T-= — ; — ■7-—=— —a. b b c X X a^ EXAMPLES. ^ ^. . , 4a , 26 . 4a 2a 1. Divide — by — . Ans.--, = -r . ^ ^. ., Sab , Abx , Sab Sa 2. Divide —7- by — r- Ans.— r--= -— . cd •' cd Abx Ax 3. Divide ~r by —r* Ans. •— . ab ^ ab Sy 4. Divide T— by -t~. Ans. ^ - . ab ^ ab 2a^ I ^ ^. .J 7raT + a« ^ Ab+ax lrx-{-a'' 5. Divide — - — -— by —= — -— Ans.-y- bast 5u8t 46+oa* ^ T.. .. ^ . 2a6 6. Divide 8a by . Explanation, By §206, 6a = -^. Then — ^ 8a6 _ 8axy __ Aocy xy 2ab b 144 ALGEBRA. 7. Divide Sa by —. Ans. — -. U 4a; ^ ,^. . , , , c ^ abd 8. Divide ao by —. Ans. — . a c 9. Divide by . Ans. — - . y y x^ 10. Divide by . Ans. . oa? ''ax Sam ^. ., 4ay^x , lOaa;'^ . 2i/'' 11. Divide —f- — by — r — . Ans.--^. oc be 5x 12. Divide by . Ans ax -^ ax 4a* — ab' 13. Diviae -; by-; . Ans. — . ab — c '' ab — c x , ^ T^. . , ax — ly xy 14. Divide j-^ by -^. ab '' b Explanation, In this example, the dividend and the di- visor has not a common denominator. Our first object then, is to bring them to a common denominator. This is , , {.•r^^ ^^ — V xy ^ax — by abxy done by 6^5, —^ -^ •# = rr-^ -^ — i/. ^ ^ ab b ab'^ ab^ abx — by ax — ?/ Ans. ^ — '' Id. Divide by -^. c ^ a abxy axy abx (ly a"bx cily a^bx Operation. . . — , . c a ac ae edy §208. It will be seen that in sums of this kind, after we have brought the terms to a common denominator, the divi- sion is performed by putting the numerator of the dividend for the numerator of the answer, and the numerator of the FBACTIONS. 145 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 divi- dend by the denominator of the divisor ; and this becomes the numerator of the answer. And we miultiply the nume- rator of the divisor by the denominator of the dividend ; and this becomes the denominator of the answer. By look- ing at the last two sums, it will be seen that this is the true operation. §209. 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 multiply the denominator of the dividend by the numerator of the divisor for a new denominator. §210. When the dividend is a whole number, it is chang- ed into a fraction, by putting 1 under it for a denominator. Thus,^ = ^; 2a = 5^. EXAMPLES. 16. Divide- by ^^. Ans. ^^^=2^. 17. ^. ., a+x , o+V A a^+'6ax-\-2x^ Divide -!— by —ff-- Ans. —4 — ^ — . a— y ^ a-\-2x a'^ — y^ 18. T^i'virlr ^^ "/yy-t-iT^ hv jjiviae X ■— "CuXT^tt uy • X — a Ans. a?3 — dx^a+Sa'^x — a 19. Divide -by-. Ans.-. 20. ^. ., x — l , x+l , 4a; — 4 Divide 3 by ^ . Ans. ^^ _^ 3. 13 146^ ALGEBRA. 01 r^- -A "+^ u ^" A 4a6+4&r 21. Divide by T/- Ans. — . y 4o day 22. Divide — r- by Ans. — --#. 4 -^ y 16 23. Divide 4- by ??. Ans. -i^. . a+1 "^ 41/ 3tta;+3a? 24. Divide a;+aa; by . Ans.-—!- ^ i. •'a? — y Sa 25. Divide -^— by |-. Ans. ^. 26. Divide 4a — av by -^^ -. Ans. -• ^ '' A^a 4y — ay 27. Divide by — „. Ans. x-{-y. X — y ^ x^ — y^ ^ REDUCTION OF COMPLEX FRACTIONS TO SIMPLE ONES. §211. We have shown, that when we multiply a fraction by its denominator, we obtain for the answer, the same quantity as the numerator. We have also shown that where both terms of a fraction are muhiplied 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. §212. 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+A ca+b^ a — f ca ^ b 4ac— 46 cd 3 , ^ — 3c 4+a ^ + ca 3c+4ac 4 4 • • FRACTIONS. 147 EXAMPLES. 1. Reduce r- to a simple fraction. c ac Ans. Multiplying both terms by c, ac — b' 2. Reduce a • — iTto a simple fraction. Ans. — . ax ^ X 3. Reduce to a simple fraction. ay + ^ ^ 5x Ans ^^y+y — z 4:X — X 3 4. Reduce ^ 4 to a simple fraction. Ans , 5. Reduce - — - ,,_i to a simple fraction. Ans. ^— -. . 3a — =^ ^ 3ay— 1/+1 X 6. Reduce ^ + y to a simple fraction. z — ± e Ans. y _ xy -\ - x cxy -f- ex z — JL yz — y^ ~'cyz — y^ c £. 24z + 6x 7. Reduce 4 + z to a simple fraction. Ans. — . I VXZ — OZ X — ± 6 X cxv^ "4" ex 8. Reduce ^y + ^ to a simple fraction. Ans. — ^ . , .,. ^TfTT ' acy-\-bxy 148 AL6EBRA. 9. Reduce 3 4-^ to a simple fraction. Ans. — ^^- — - 4 — ± Adcy — bx oy hcx~~~'CiC'U JO. Reduce x — 6 to a simple fraction. Ans. -, r^* ^Zf^ bcy+abx c §213. It sometimes happens that we wish to transfer a frac- tion from a numerator to the denominator, or from a denomi- nator to the numerator. This may be done by the foregoing principles. For, supposing we have ~ ; multiplying by the denominator of f , we have — . Now, if we divide this / X CL fraction by 2, we have — . Thus we see the fraction is transferred, without altering the value of the whole quantity, if we take care to invert it when we transfer it. THE END.