C^g. «/> LI BRARY OF THB UNIVERSITY OF CALIFORNIA. Received (^pi^^^ , fS^^. ^^ Accessions No.4^^fy^.Sl^ Shelf No. . ....■ ^ O^ ♦SO Digitized by the Internet Arciiive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementarytreatiOOsherrich AN ELBMENTAKY TEEATISE ALGEBRA, FOR THE USE OF STUDENTS HIGH SCHOOLS AND COLLEGES. BY THOMAS SHERWIN, A. M., Principal of the English High School, Boston. ^ 0? THB SIVERSIT??^^^'^^ EDITION. t >5^*/rO».»^ BOSTON. HALL AND A^^HITING-, 32 BEOMFIEU) STREET. ^.< ISntered according tc act of Congress, in the year 1841, Bv Thomas Sherwik, In the Ck-rk's Office of the D' strict Court of MaaMcnusetts. PREFACE. The author of this treatise has endeavored I: prepare a work which should sufficiently exercise the ability of most [earners, without becoming, at the same time, repul- sive to them by being excessively abstract. Some writers err in expecting too much, and others err, in an equal degree, by requiring too little of the student. What success has attended an attempt to attain a proper medium it is left for competent teachers to decide. This work commences in the inductive manner, be- cause that mode is most attractive to beginners. As the learner advances, and acquires strength to grapple with it, he meets with the more rigorous kind of demonstration. This course seems the most natural and effective. In- duction is excellent in its place ; but when an attempt is made to carry it into all the departments of an exact science, the result often shows, that the main object of study was misapprehended. The young frequently fail to deduce clearly the general principle from the particular instances which have engsiged their attention. Several parts of algebra, which are either omitted or not explained with sufficient distinctness in other works, liave received particular attention in this. These partb treat of principles and operations, with which students rarely become familiar, but which are essential to a clear conjprthension of the subject. Among these operations may be mentioned the separation of quantities into factors, finding the divisors of quantities, and the substitution of numbers in algebraic fo mulae. ly PREFACE Most of the problems are original ; others have teen Belected, which seemed the most appropriate. Although this treatise is designed for students in the higher grade of seminaries, it is not beyond the reach of any, who have a good knowledge of arithmetic, and who are under the guidance of Competent instructors. Should Articles 46, 58, 59, 153 and 154 be found too difficult for the beginner, on his first perusal of the book, they may be postponed for investigation in a review. The writer is unwilling to close his remarks, without expressing his obligations to others, who have done so much to introduce into our country a natural and rational mode of studying mathematics. Among these none merits greater praise than Colburn ; and his works have served as a guide in the composition of several others on the inductive plan. Day, Smyth, Davies and Peirce deserve also to be mentioned with great respect. THOMAS SHERWIN. English High School, Boston, Sept. 10, 1841. In this new edition of his work, the author would remark, that the few errors of the first edition have been carefully corrected ; that a Key to the Algebra has been published ; and that, in both the Algebra and Key, a marked distinction has been made between the full point when used as the sign of multiplication and whe*i used EU5 a decimal point; in the latter case, the type being inverted, and the sign consequently elevated. T. 8. ApRii 4, 1843. CON! ENTS. Preliminary remarks, 1 Algebraic signs, lb. Axioms, .' 2 1 Equations of the first degree, having only unknown t^rms in one member, and known terms in the other, 3 Definition of equation, member, term, coefficient, 4 II. Equations of the first degree having unknown terms in one member only, and known terms in both, 7 Transposition defined — rule tor transposition, ti — 10 III. Equations of the first degree, in which known and unknown terms may occur in each member, 13 IV. Equations of the first degree, containing fractional parts of sin- gle terms, 17 V Equations of the first degree, containing fractional parts of quantities consisting of several terins, 21 VI. Equations of the first degree, which require the subtraction of quantities containing negative terms, 25 VL Multiplication of monomials, 29 Exponents and powers defined, 31 1^11 1 . Reduction of similar terms, 33 IX. Addition, 35 X Subtraction, 37 Subtraction of polynomials represented, 39 X I Multiplication of polynomials, 40 Multiplication of polynomials indicated, 44 Product of sum and difierence — second power of a binomial, 45 Third power of a binomial, 46 & 1 1 Division of monomials, 47 Value of aO, , 4g CONTENTS. •V5CT PAOe ' Xlll Division of polynomials, ,. 49 Infinite series, 55 • Divisibility ofa'" — 6"* by a — A, 56 Conditions under which z* — y* and z* + t/'* are divisi- ble by 2 -f- y, 'tS Division of a product^by dividing one of its factors, b. XIV. Multiplication of fractions by integral quantities, o9 XV. Division of fractions by integral quantities, 64 XVI. Factors and divisors of quantities — prime quantities, 67 Separation of quantities into their factors, 6d Finding ail the divisors of quantities, 59 XVII. Greatest common divisor, 71 X VllJ Least common multiple, 74 XIX. Reduction of fractions to their lowest terms, ... 75 XX. Multiplication of fractious by fractions, 77 XXI. Addition and subtraction of fractions — common denom- inator, 79 XXll. Division of integral and fractional quantities by fractions,. 83 XX ill Literal equations, • b6 XXIV. Equations of the first degree with two unknown quantities, 87 First method of elimination, 90 Second method of elimination, 92 Third method of elimination, 93 XXV. Equations of the first degree with several unknown quan- tities, yfi XXVI. Numerical substitution of algebraic quantities, 105 XXVII. (Generalization, 107 XXVIU. Negative quantities and the interpretation of negative re- sults, 119 XXIX. Discussion of problems, 126 Signification of the symbol ^, 129 Signification of the symbol Q., iUsJ XXX Extraction of the second roots of numbers, 133 Irrational and imaginary quantities defined, 140 XXXI. Second roots of fractions — and the extraction of second roots by approximation,.. i4J XXXIl. Questions producing pure equations of the second degree, 144 XXX 111. Aftected equations of the s(?cond degree, \.4f XXX iV. Extraction of the third roots of numbers, 155 CONTENTS Vll ISCT. PAGB XXXV. Third roots of fractions — and the extraction of third roots by approximation, 163 XX XVI. Questions producing pure equations of the third degree, lb6 XXXVII. Powers of monomials, }G8 XXXVIIl. Powers of polynomials, 169 XXXIX. Binomial theorem, IT'-i XL. Roots of numbers to any degree, 180 XLI. Roots of monomials, , 18i\ Origin and signification of fractional «= jponents, ISo Separation of a quantity into any number of factors, , 1S6 XLIl. Roots of polynomials, ib XLIII. Simplification of irrational or radical quantities, 191 XLIV. Operations on irrational quantities with fractional ex- ponents, ID4 XLV. Operations on irrational quantities with radical signs,... 201 XLVl. Negative exponents, 209 XLVII. Inequalities, 213 XLVIII. Equidifierence, 217 XLIX. Ratio and proportion, 219 L. Progression by difference, 229 LI. Examples involving progression by difference, 234 Lll. Progression by quotient, 236 LIU. Examples in progression by quotient, o . . . 242 LI V. Exercises in equations of the second degree, 244 LV. Exercises in equations of the second degree with two unknown quantities, 250 LVI. Logarithms, 258 LVII. Use of the tables in finding the logarithms of given num- bers, and the reverse, 265 l»VIII. Application of logauthms to arithmetical operations,.... 278 LIX Compound mtf rest, 282 LX Annuities, 288 Misceilaneou-) qiiestions 299 OP ifffB >^ 'UHI7BKSIT71 ELEMENTS OF ALGEBRA. PRELIMINARY REMARKS. Art. 1. Arithmetic treats of numbers which have known and definite values ; but Algebra makes use of symbols, which may represent known, unknown, or indeterminate quantities. These symbols are the letters of the alphabet. Moreover, in Arithmetic, after an answer to a question has been obtained, it contains nothing in itself to show by what operations it was found. For instance, suppose the number 6 is ascertained to be the answer to a particular question ; this ex- hibits no marks to show whether it was obtained by addition, multiplication, division, or by some other process or combina- tion of processes ; but the results of pure A Igebra, that is, when both known and unknown quantities are represented by letters, always irndicate, or may be made to indicate, the means by which Miey were produced. Algebra enables us also to carry on a course of reasoning with much greater ease and rapidity than Arithmetic, and to arrive at the solution of problems, which, by the aid of Arithmetic alone, would be exceedingly difficult, if not impossible. Art. ^. We proceed to notice some of the signs, which mos» frequently occur in Algebra. The sign -\- is used to express addition^ and is called plus^ which signifies more ; thus, 6 -|- 3 is read 6 plus 3, and means I PRELIMINARY RKMARKS that 6 and 3 are to be added together, or indicates the sum of t and 3. The sign — expresses subtraction ^ and is called minus ^ uhicb signifies less ; thus 8 — 3 is read 8 minus 3, and means that 3 i? to be subtracted from 8, or indicates the difference between 8 and 3. jMoreover, quantities having before them the sign -|-, expressed or understood, are called positive ; those having before them iIip sign — a»-f» railed negative quantitirs. Multiplication is represented by the sign X ; thus, 6X4 means that 6 and 4 are to be multiplied together, or indicates the product of 6 and 4. Sometimes a point between the quanti- fies, as 6 . 4, has the same signification. Division is represented by the sign -i-, or : ; but more fre- <^uently it is expressed in the form of a fraction ; thus, 6 ~ 3, 6 : 3 and f , each signifies the division of 6 by 3, or indicates the quo- iient of that division. To express equality we use two horizontal parallel lines, thus =;z ; this is read equal to, equals, or by some words of similar import; for example, 6 + 4= 10 means that the sum of 6 and 4 is equal to 10, and is read 6 plus 4 equals 10. Accordingly, 5 X ^ -\- "t ==^ ^2^ — 3 means, that if 5 be multi- plied by 4, and 7 be added to the product, the result will be the same as if 60 be divided by 2, and 3 be subtracted from the quo- tient. The sign ^, or <^, is used to express the inequality of quan- tities ; thus, 8 > 5, or 5 <^ 8, signifies that 8 is greater than 5, or that 5 is less than 8, the open end always being placed towards the greater quantity. To represent unknown quantities, we use some of the last let- ters of the alphabet, as x, y, &c. ; and to represent known quan titles, we use some of the first letters, as a, b, c, &c. ; althouj^h, in many problems of this work, known quantities are expressed by figures. Art. 3. There are some propositions, the truth of which ia manifest, as soon as they are presented to the mind. These pro- liositions are called axioms ; the following are of this kind. AXIOMS. EQUATIONS OF THE FIRST DKGBEF. 1. If the same quantity or equal quantities be added to equa. quantities, the sums will be equal. 2. If the same quantity or equal quantities be subtracted from equal quantities, the remainders will be equal. 3. If equal quantities be midtiplicd by the same quantity or by equa. qiantities, the products will be equal. 4. If equal quantities be divided by the same quantity or by equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed. 6. If a quantity be both multiplied and divided by another, ita value will not be changed. 7. Two quantities, each of which is equal to a third, are equal to each other. 8. The whole of a quantity is greater than a part of it 9. The whole of a quantity is equal to the sum of all its par^s SECTION I. EQTTATIOIfS OF THE FIRST DEGREE, HAVING OJVLY UNKNOWN TERMS IN ONE MEMBER AND KNOWN QUANTITIES IN THE OTHER. Art. 4:. 1. An apple and an orange together cost 6 cents; but the orange cost twice as much as the apple. What was the price of each 1 In this question, if we knew the price of the apple, we should, by doubling it, obtain that of the orange. The price of the apple, then, may be considered as the unknown quantity. Suppose that X represents the number of cents given for the apple ; twice as much, or the price of the orange, would je represented by 2 x. Hence, z -j- 2 x = 6. Putting the z's together, we have 8 z = 6 ; one x will be ^ as much: therefore, x = 2 cents = tne price ol ihe apple ; and 2 x =z 4 cents = the price of the orange. 4 EQUATIONS OF THE FIRST DEGREE. DEFINITIONS, ETC. L Remark. Questions in Algebra may be proved as uell as ihose in Arithmetic. The proof of the foregoing, would consist in adding the price of the apple to that of the orange, and ascer- taining that their sum is 6 cents. Let the learner prove the cor- rectness of his answers, as he advances. A representation of the equality of quantities, is called an t'juation. Thus, x-\-2x=zfy is an equation. A member or side of an equation, signifies the quantity or quantities on the same side of the sign =, theirs/ member being <5ii the left, and the second member on the right hand side of this sign. An equation of the first degree is one, in which the unl nown quantities are neither multiplied by themselves nor by each other. The separate parts of an algebraic expression affected by the signs -\- and — , are called terms. Those terms which have no sign prefixed to them, are supposed to have the sign -|-» ^nd a quantity is said to be affected by a sign, when it is immediately preceded by that sign, either expressed or understood. When the first term of a member of an equation, or of any algebraic quantity, is affected by the sign +, it is usual to omit writing the sign before that term; but the sign — must always be written before any term affected by it. The equation, x -)- 2 a; n: (^^ consists of three terms, two in the first member and one in -ne second, and each of these terms is affected by the sign -\-. The number written immediately before a letter, showing how many times the letter is taken, is called the coefficient of that letter; thus, in the expressions, ^z, 5a:, 7a;, the coefficients of x are -J, 5 and 7. A letter which has no number written before it, is supposed to have 1 for its coefficient ; thus, x is the same as 1 a. Letters, as we shall see hereafter, may be used as coefficients. 'J'he i>, ocess by which an equation is formed from the condi lions of a (question, is called putting the question into an equa- Hon ; and the process by which the value of the unknown quan- tity is found from the equation, is called solving the equation. 2 Said A to B, my horse and saddle are worth $110 ; but mj I EQUATIONS OF THE FIRST UEf^REE. 5 horse is worth 10 times as much as my saddle Required the (vorth of each. 3. A man bought some corn and rye for 60 shillings, the corn at 4s. per bushel and the rye at 6s., and there was the same number of bushels of each. How many bushels were there of each? Let z represent the number of bushels of each ; then x hushelg of corn at 4s. per bushel, will come to 4 x shillings, and % bush- els of XY^ at 6s. per bushel, will come to 6 x shillings. Hence, 4 2 + 6 X = 60. 4. A man sold an equal number of oxen, cows and sheep; the oxen at $40 apiece, the cows at $15, and the sheep at $5; the whole came to $660. How many were there of each? 5. A woman bought some peaches, pears and melons for $110; the peaches at I cent apiece, the pears at 2, and the melons at 12 ; there were twice as many pears as melons, and three times as many peaches as pears. How many were there of each ? Let X represent the number of melons ; then 2 x will represent the number of pears, and 6x, the number of peaches. At 1 cent each, 6x peaches come to 6x cents, 2x pears at 2 cents each will come to 4 x cents, and x melons at 12 cents each will come to 1 2 / cents ; hence, 6 x -(- 4 x -|- 12 x z= 110. 6. A gentleman hired a man and a boy to work a certain number of days, the man at 8s. and the boy at 4s. per day, and paid them $30. How many days were they employed, and how much did each receive ? 7. Three numbers are in the proportion of 1, 2 and 3, and the Bum of them is 630. What are these numbers ? The proportion of 1 , 2 and 3, means that the second is twice and the third three times as much as the first. 8. Divide 100 into three parts, in the proportion of 5, 7 and 8. The proportion of 5, 7 and 8, means that the 2d is -^, and the '5d I as much as the 1st. Suppose tne 1st part z=: 5x, then the 2d will be 7x, an^ the W, 8x I* 6 EQUATIONS OF THE FIRST DEGREE 1 9. Two persons set out at the same time from two towns 150 miles apart, and travel towards each other till they meet, one at 8 miles an hour, and the other at 7. How many hours will they be on the road, and how far will each travel ? JO. Three robbers, having stolen 48 guineas, quarrelled about the division of them, and each took as nmch as he could get ; the first obtained a certain sum, the second twice as much, and the third as much as both the others. How many guineas did each obtain ? 11. A gentleman wished to divide an estate of $81000 be- tween his wife and two sons, so that his wife should have $4, as often as the elder sou had $3, and the younger $2. How much ^ould each receive? 12. A fortress has a garrison of 1200 men, a certain portion of whom are cavalry, three times as many artillerymen, and six times as many infantry. How many are there of each corps ? 13. In fencing a field, three men. A, B and C, were employed A could fence 9 rods a day, B 7, and C 5 ; B wrought twice as many days as A, and C five times as many as B. The distance round the field was 584 rods. How many days did each work, and how many rods of fence did each build ? 14. A man bought three pieces of cloth for $280. The sec- ond piece was twice as long as the first, and the third was as long as the first two. He gave $4 a yard for the first piece, $5 a yard for the second, and $7 a yard for the third. Required the num- ber of yards in each piece. 15. Four cows, 3 calves and 10 sheep cost $112. A cow cost 5 times as much as a calf, and a calf cost twice as much as a sheep. Required the price of each. 16. A cistern holding 140 gallons, was filled with water by means of two buckets, the greater of which held twice as much as the less. The greater was emptied 7 times and the less 6 times. How many gallons did each bucket holdt 17. A boy being sent to market, bought some beef at 14 cents A pound, and twice as much mutton at 9 cents a pound. He was (I EQUATIONS OF THE FIRST DEGREE. 1 intrusted with $4 and brought back 80 cents. How n.ny pounds of each kind of meat did he buy? 18. A man wished to pay $60, with dollars, halves, quarterSj and eighths, of each an equal number. How many coins of each kind would he require? 19. A man paid c£l44 in guineas at 21s. and crowns at 5s. each. There were three times as many crowns as guineas. Required the number of each. 20. A man on a journey traveled twice as far the 2d day as he did the Ist ; on the 3d day, as far as he did the first two days ; on the 4th day, as far as he did the first three days ; and on the 5th day, half as far as on the 4th. The whole distance traveled was 150 miles. How far did he go each day ? 21. A merchant exchanged rye at 7s. and wheat at 9s. a bushel, of each the same quantity, for 32 bushels of corn at 4s. a bushel. How many bushels of rye and wheat were given in exchange ? 22. A drover bartered 6 oxen and 10 cows for a farm of 50 acres at $1J per acre. He reckoned each ox worth as much as two cows. What price was assigned to an ox and a cow re- spectively ? Art. o. In the preceding questions, ar's, that is, unknown quan- tities, have been found only in the first member of the equation, and they have all been affected by the sign -|- ; and we perceive, that, after an equation was formed, the first step was to reduce or combine all the unknown quantities into one term, which is done by adding the coefficients ; after which, the value of the unknown quantity was found by dividing both members by the eoefficient of the unknown quantity. SECTION II. EQUATION'S OF THE FIRST DEGREE, HAVING UNKNOWN TERMS IM ONE MEMBER ONL.Y, AND KNOWN TERMS IN BOTH MEMBERS. Art. 6. Two brothers had together $20, but the elder had two dollars more than the younger. How much money had each? S EQUATIONS OF THE FIRST DEGREE. II Let X represent the number of dollars the younger had; then X -f- 2 = the number of dollars the elder had ; consequently, z -|- z -f- 2 =: 20 ; or, combining the z's, 2x + 2 = 20. Now as the two members are equal, we can subtract 2 from each, and the remainders will be equal (ax. 2, Art. 3); 2 sub tracted from 2 z -|- 2, leaves 2 z, and 2 subtracted from 20, leaves 18; hence, 2z= 18; zrz: 9, number of dollars the younger had, and z -|- 2 r= 11, number of dollars the elder had. Instead of actually subtracting 2 from the second member at once, we may subtract it from the first, and represent it as sub- tracted from the second; thus, 2z = 20 — 2; now performing the subtraction indicated, we have 2z=: 18, the same as before. The equation 2zzz:20 — 2 is obtained from 2z + 2z=20 merely by removing the 2 to the other side of the sign rz:, and changing its sign from -(- to — . Art. 7. Removing a term from one member of an equation to the other, is called transposing that term, or transposition. Any term^ therefore^ affected by the sign -(-, may be transposed^ if this sign be changed to — . 1. Two men, A and B, hired a house for $650, of which A paid $150 more than B. What did each pay? Let X represent the number of dollars B paid. Then z -|- 150 will represent the number A paid. Hence, z -(- z -f- 1 50 rr 650. Reducing, 2 z-j- 150 = 650; transposing 150, 2 z = 650 — 150 ; reducing the 2d member, 2z = 500, X = $250 = what B paid, z + 150 z= $400 = what A paid. 2 Two men possess together $56000, but the second has HOOOO more than the first. How much money has each? 3. Two towns are at unequal distances and in opposite direc- tions from Boston ; the distance between th ese towns is 23(' U. EQUATIONS OF THE FIRST DEGREE. 9 miles, but one is 10 miles more than twice as far from Boston aa the other. What is the distance of each from that city ? 4. The sum of the ages of A, B and C is 100 years ; but B's age is twice that of A and 5 years more, and C's age is equal to the sum of A's and B's. Required the age of each. 5. A grocer wishes to make a mixture of four kinds of tea, so that there shall be 6 lbs. more than twice as much of the 2d kind as of the 1st, as many lbs. of the 3d as there are of the first two, and as many of the 4th as there are of all the others ; the whole mixture is to contain 120 lbs. How many lbs. must there be of fiach sort? 6. A man has five sons, each of whom is two years older than his next younger brother, and the amount of their ages is 50 years What is the age of each ? 7. A merchant bought 10 pieces of cloth for $331 ; 5 pieces were blue, 3 green, and 2 black ; a piece of green cost $2 more than one of black, and a piece of blue $3 more than one of green. How much did each kind cost per piece ? 8. Says A to B, my age is 10 years more than yours, and twice my age added to three times yours, makes 120 years. Re- quired the age of each. 9. A gentleman leaves an estate of $10000, to be divided be- tween his three daughters and two sons, in the following man- ner, viz : the daughters are all to share equally, but the elder son is to have $1000 more than twice as much as the younger and the younger exactly twice as much as one of the daughters What is the share of each ? 10. A laborer undertook to reap 6 acres of wheat and 10 acres of oats for $21f , or 130 shillings ; but he was to have 3s. more an acre for the wheat than for the oats. What was the price of reaping an acre of each ? Let X shillings represent the price of reaping the wheat per acre Then x — 3 will be the price of reaping the oats per acre. Six acres of wheat will cost (i x shillings ; and ten acres of oats will cost 10 a: — 30 shillings. 10 EQUATIONS OF THE FIRST DEUKEE. U Hence, 6x+10x — 30r= 130. Reducing, 16x — 30=:U0. By adding 30 to each member (ax. 1, Art. 3), the equation becomes 16x — 30 + 30 = 130 + 30, or, 16 X = 130 + 30, since 16 a: — 30 + 30 is the same as 16 » (ax 5) ; hence, 16 a; = 160, herefore, x =. 10, and X — 3 =: 7. Ans. wheat 10s., oats 7s. per acre Most of the preceding questions in this section, may be solved »n a similar way. The equation 162 = 130 + 30 is obtained from 16 x — 30 = 130, merely by removing the 30 to the second member of the equation, and changing its sign from — to -\-. Art. 8. HencCf any term affected by the sign — , may be trans posed from one member to the other ^ if its sign be changed to -{- ; for, this is adding the same quantity to each member (ax 1). This principle, together with that established in Art. T, gives the following general RULE FOR TRANSPOSITION. Art. O. Any term may be transposed from one member oj an equation to the other, care being taken to change its sign from — to -}-) or from -j- to — . It may be remarked, that the value of every such expression as, 1 — 1, 2 — 2, 3 — 3, &-C., or x — x, 4x — 4x, a — a, 5 a — 5«, &.C,, is or nothing; that is, the plus and minus quantities equal in value cancel each other. Moreover, when quantities are connected by the signs -\- and — , it is of no importance in what order they stand, provided ihe} have their proper signs prefixed to them ; thus, 3 -J- 7 — 2 may be written 7 -|- 3 — 2, or — 2 -|- 7 + 3, the value of each expression being 8. When the first term is affected by the sign +, it is usual to v/uiit writing that sign ; but the sign — must never be omitted The learner cannot be too careful with regard to the signs^ a mistake in the sign occasions an error equal to twice the 11. EQUATIONS OF THE FIRST DEGREE. LI value ol' the term aflfected by it ; thus, 12 + 3 is equal to 15, and 12 — 3 is equal to 9 ; now the diffe ence between 15 and 9 is G or twice 3. We perceive also, that when a quantity consisting of severa' terms, as x — t^, is to be multiplied, each term must be multi- plied and the same signs retained; thus 10 times x — 3 is 10a; — 30 ; in like manner, 7 times 12 — 32; is 84 — 21 x. 1. At a certain election, two persons were voted for; but -^ he candidate chosen had a majority of 87, and the whole numjer of votes was 899. How many votes had each? 2. In a manufactory, 205 persons, men, boys and girls, are employed ; there are four times as many boys as men, and 20 less than ten times as many girls as boys. How many of each are employed 1 3. A general, on reviewing his troops, found he had in all 2300 men, of whom a certain portion were cavalry, three times as many riflemen, and 100 less than four times as many infantry as riflemen. How many were there of each? 4. Four men, A, B, C and D, enter into partnership. A con- tributes a certain sum, B three times as much, C twice as much as A and B both, and D as much as the other three wanting $1000. The whole sum invested was $65000; how much did each put in trade ? 5. Divide $491 among three persons. A, B and C, so that A shall have $270 more, and B $100 less than C. Suppose X =. C's share. Then x -\- 270 = A's share, and » — 100 = B's share. Hence, x + 2 + 270 + a; — 100 = 491 Reducing, we have 3x-|- 170 = 491 ; for adding 270 and Bubtracting 100, is the same as adding their difference. 6. A man aged 80 years, had spent a certain part of his life in France, three times as much and 30 years more in England and twice as much wanting 10 years in America. How mary years had he lived in each country ? 7. A certain town contains 2900 inhabitants, English, Irish, and French ; there are 600 fewer Irish than English, and 400 fewer French than Irish. How many are there of each? 12 EQUATIONS OF THE FIRST DEGREE. (I Let X =z the number of English ; then X — 600 = the number of Irish, and X — 600 — 400 z=. the number of French. Hence, a: + a; — 600 + z — 600 — 400 = 2900 ; by reducing we have 3x — 1600 = 2900; for all three of the numbers af- fected by the sign — , are considered as separately subtracted and it would evidently be the same thing to subtract the sum of them at once. 8. In a casket containing 390 coins, there is a certain num- ber of eagles, 10 less than twice as many half-eagles, and 20 less than three times as many dollars as there are half-eagles. How many coins are there of each kind ? 9. A merchant bought a certain number of yards of broad- cloth at $8 per yard, 6 less than three times as many yards of cassimere at $4 per yard, and twice as much silk at $1 per yard as there were yards of cassimere. The whole came to $1264 How many yards of each kind did he buy? 10. A man, engaged in trade, gained, the first year, $500 ; the second year he doubled what he then had ; but the third year he lost $2000, when it appeared that he had remaining $3000. How much money had he at first? Suppose X ■=. his money at first. Then x + 500 = his money at the end of the 1st year ; 2 x 4- 1000 = his money at the end of the 2d year, • and 22;+ 1000 — 2000 = his money at the end of the 3d year Hence, 2 x + 1000 —2000 = 3000 ; or reducing, 2ar— 1000=^3000; for 2 r + 1000 — 2000, signifies that 1000 is added to 2 x, and from the sum 2000 is subtracted, which is the same thing as subtracting 1000. 11. An inheritance of $92500 is to be divided among fi\G heirs, A, B, C, D and E, in the following manner, viz : B is to have $600 more than A ; C twice as much as B, wanting $400 ; D as much as A and B both, wanting $300 ; and E $500 more •ban A and D both. What is the share of each ? [II. EQUATIONS OP TtiE FIRST DEGREE I? Suppose X = the share of A Then x + 000 = the share of B, 2 a: 4- 1200 — 400 = the share of C, a; _j_ X + 600 -- 300 = the share of D, and X + a: + X -f 600 — 300 + 500 = the share of E. Hence, x + x + 600 + 2 x-{- 1200 — 400 + z + x + 600 — 300 _|. X _|_ x -|- z -f. 600 — 300 + 500 = 92500. Reducing, 9 1 ■\- 2500 zzz 92500 ; for the sain of the numbers affected by the sign + is 3500, and the sum of those affected by the sign — is 1000; but adding 3500 and subtracting 1000 is the same a» ndding 2500. Had the sum of the numbers, affected by the sign — , been greater than that of the numbers, affected by the sign +, the dif- ference of these two sums would have had the sign — . In the above question, the labor would have been abridged, if the expressions for the several shares had been reduced, as far as possible, previous to forming the equation. 12. A drover has a certain number of oxen; three times as many cows, wanting 25; just as many calves as cows; and 100 more sheep than he has oxen and cows together. The number of the whole is 905 ; how many of each has he ? 13. In a company of 140 persons, consisting of officers, mer- chants and students, there were 4 times as many merchants as students, wanting 25 ; and 5 more than 3 times as many officers as students. How many were there of each class ? SECTION 111. tQITATIONS OF THE FIRST DEGREE, IN WHICH BOTH KNOWN AND UNKNOWN TERMS MAY OCCUR IN EACH MEMBER. Art. lO. What number is that to which if 18 be added, thf sum will be equal to four times the number itself? Let x represent the number ; then x-|-18z=4x, or 4ar = z-f- 18 ; as it is evidently indifferent which quantity is made the firsl member. 2 14 EQUATIONS OF THE FIRST DEGREE. Ill Now it IS our object to make all the x's, or unknown quantif des, stand in one member of the equation, and the known quan tities in the other ; and, for the sake of uniformity, we gerjcrally collect the unknown quantities into the first member. In the equation 4x=:x-\- 18, by transposing the x from tlw second member to the first, that is, by subtracting x from both members, we have 4 x — x=z 18, or reducing, 3 a: =18, and xz=i6, Ans. Or we might have taken the equation z -(- 18 = 4 x j by transposing the 18, we have x = 4 x — 18 ; then, by transposing the 4 x, we have x — 4xz=z — 18 ; reducing, — 3 x = — 18. Here both members are wholly minus, but by transposing both, we have 18 = 3x, which is the same as 3x:= 18; hence, x = 6. The equation, 3x= 18, might have been obtained from — 3z = — 18, merely by changing the signs to -|-. In like manner, in the equation 3x — 5x = 20 — 46, which reduced gives — ^xz=z — 26, we might change all the signs be- fore reducing, which would give — 3x-|-5x = — 20 + 46, or 2 X = 26 and X = 13. Art. 11. Hence, the signs of all the terms in both members oj an equation may bt changed ; for this is the same as transposing^ all these terms. This change of signs should be made, whenever the first mem- ber becomes minus ; but the learner must recollect, that terms having no sign, are supposed to have -f-, and that he must change all the signs, otherwise great errors will ensue. 1. Says A to B, if to my age twice my age and 30 years more be added, the sum will be. five times my age. How old is he? Let X = his age ; then 6x = x + 2x + 30. Reducing the 2d member, 5 X = 3 X + 30 ; transposing 3 x, III. EQUATIONS OF THE FIRST DEGREE. 15 6x — 3 a; = 30; reducing, 2z=r30, and z = 15 years. Ans. 2. A merchant sells two kinds of cloth, the finer at $2 a yard more than the coarser; 12 yards of the coarser come t) as much Bs 8 yards of the finer. What is the price of each per yard? 3. Says A to B, four times my age is equal to five times yours, and the difference of our ages is 10 years. What is the age of* ftach ? 4. A man having a certain number of cows and the same number of sheep, bought 4 more cows and 16 more sheep ; he then found that three times his number of cows was equal to twice his number of sheep. How many had he of each at first ? 5. A father distributed a certain sum of money among his four sons. The third received 9d. more than the youngest ; the sec- ond, I2d. more than the third; and the eldest, 18d. more than the second. The whole sum was 6d. more than seven times what the youngest received. How much had each, and what was the whole sum distributed ? 6. A sum of money was to be divided among six poor persons, so that the second should have 3s., the third 2s., the fourth 5s., the fifth 7s., and the sixth 8s., less than the first. Now the sum divided was 7s. more than four times the share of the second. What did each receive ? 7. A person bought two casks of beer, one of which held twice as much as the other • from the larger he drew out 20, and fiom the smaller 25 gallons, he then found that there remained in the larger 4 times as much as in the smaller. What did each cask contain at first? 8. A man bought 10 bushels of wheat and 16 bushels of rye, the wheat cost 2s. more per bushel than the rye, and the whole cost of the wheat wanted 16s, to be equal to that of the rye. What was the price of each per bushel ? 9. An instructor, wishing to arrange his pupils in rows with a certain number in each row, found that there were 3 too many to make six rows, and 4 too few to make seven rows. How 16 EQUATIONS OF THE FIRST DEGREE. HI many did he wish to place in a row, and how many scholars had he? Let X z=z the number in each row ; then, 6 X -|- 3 r= the whole number of scholars ; also, 7 X — 4z=z the whole number of scholars. Hence, 7x — 4 = 6x-\-3. (ax. 7). 10. A boy being sent to buy a certain number of pounds of meat, found, that if he bought pork, which was 9 cents pei pound, he would have 5 cents left, but if he bought beef, which was 10 cents per pound, he would want 5 cents. How many pounds was he to buy, and how much money had he ? 11. Two workmen received equal wages per day; but if the first had received 2s. more, and the second 2s. less per day, the first would have earned in 8 days as much as the second would in 12. What were the daily wages of each ? 12. A and B began trade with equal stocks. In the first year A gained a sum equal to his stock and $27 over ; B gained a sum equal to his stock and $153 over. The amount of both their gains was equal to five times the stock each had at first. What was the stock with which each began 1 13. A man is 40 years old, and his son 9; in how many years will the father be only twice as old as the son ? 14. A father is 66 years old and his son 30 ; how many years ago was the father three times as old as his son ? 15. A grazier had two flocks of sheep, each containing the same number ; from one of these he sold 50, and from the other 100, ard found twice as many remaining in the one as in the other. How many did each flock originally contain ? 16. A courier, who traveled 80 miles a day, had been gone one day, when another was sent from the same place to overtake him. In what time will the second, by traveling 90 miles per day, overtake the first, and at what distance from the starting- place ? 17. A gentleman bought a horse and chaise; for the chaise ke gave $75 more than for the horse, and three times the price [V EQUATIONS OF THE FIRST DEGREE. 1 1* of the horse, diminished by $50, was equal to twice the price of the chaise. Required the price of each. 18. When wheat was worth 5s. a bushel more than oats, a farmer gave 8 bushels of oats and 8s. in money for 4 bushels of wheat. What were wheat and oats worth per bushel ? 19. A merchant, engaging in trade, during the first year doubled his stock, wanting $500 ; the second year he doubled the stock he tljen had, wanting $500; and so continued to double his stock each year, wanting $500 ; until, at the end of the fourth year, he found he had $500 more than eight times the stock with which he commenced. What was his stock at first? 20. Four towns are situated in the order of the four letters, A, B, C and D, and in the same straight line. The distance from B to C is 10 miles less than twice the distance from A to B ; and the distance from C to D is 20 miles more than that from B to C ; moreover, the distance from A to B, added to that from B to C, is equal to the distance from C to D and 5 miles more. What is the whole distance from A to D ? SECTION IV. EQUATION'S OF THE FIRST DEGREE, CONTAINING FRACTIONAIi PARTS OF SINGLE TERMS. Art. IS. 1. A merchant sold a bag of coffee for $16, which was only four fifths of what it cost him. How much did it cost t Let X z=. the number of dollars it cost. 4x Then four fifths of x may be written f x, or more properly — -, o wliich may be read either four fifths of x, four x fifths, one fifth jf four X, or four x divided by five, the last of which is preferable 4x Hence, — = 16. Dividing both members by 4, we have - = 4 ; if one fifth of x is equal to 4, the who e of o I will be five times as much, or, z = $20, Ans. 2* 18 EQUATIONS OF THE FIRST DEGREE. IV Or, we might first multiply by 5, and since a fraction is multi" 4x plied by dividing its denominator, we have — or 4 x = 80, and X =: 20, as before. This latter method is generally preferable to the former. j 2. A man said that one half and one fourth of his money amoimted to $75. How much money had he? Let X =. his money. nr nr Then o + J — '^^' Multiplying by 2, X + - = 150 ; multiplying this by 2, 2z -\-x = 300 ; reducing, 82;=: 300 ; dividing both members by 3, X = $100. Ans. 3. In a certain school, one half of the boys learn Arithmetic , one fourth, French ; one eighth. Grammar ; one sixteenth, Al- gebra; and 10, Geometry. These classes constitute the whole school. How many boys does the school contain ? Suppose x:=. the whole number of scholars. Then, x= 1+1 + 1+^ + 10. (ax. 9). -Multiply by 2, 22:= x + 1 + ^+1 + 20; multiply by 2, 4x = 2x + a: + |+^ + 40; multiply by 2, 8x=4x+2x + x+^ + 80; multiply by 2, 16x=i8x+4x + 2x + x+160; reducing, 16x= 15x + 160 ; transposing 15 x and reducing, x=160. Ans. Remark. Although it is generally safest to multiply by the denominators separately, we might, in this question, have multi- plied Aie first equation by 16, the least common multiple of the IV. EQUATIONS OF THE FIRST DEGREE. 19 denominators ; or we might have reduced all the unknown terms to a common denominator, which would have given —x- =— -r-r nr 4-10, and — = 10, consequently 2= 160. 4. A man found that he had spent one third of his life in Germany, one fourth in France, two fifths in England, and one year in the United States. How old was he, and how many years had he spent in the first three countries mentioned ? 5. A merchant, on settling his affairs, found that he owed to one man ^, to another ^, and to a third ^-^ of the money he had on hand; and that, after paying them, he should have $3018 left. How much money had he, and how much did he owe each of the three creditors ? 6. A goldsmith wished to make a mixture of gold, silver and copper, so that 2 ounces more than one third of the whole should be gold, 8 ounces more than one fourth of the whole, silver, and 2 ounces less than one sixth of the whole, copper. How many ounces in the whole mixture, and how many of each kind of metal ? 7. A man left his estate to be divided between his wife and his three sons, in the following manner, viz : the wife was to have $1000 less than one third of the whole estate; the eldest son, $2000 more than one fifth of the whole ; the second son, $2000 more than one sixth of the whole ; and the youngest son, exactly one sixth of the whole. What was the whole estate, and what were the portions of the several heirs? 8. A gentleman had spent 4 years more than one fourth of his life with his parents and at school, 12 years less than three fifths of it in the study and practice of his profession, and had lived in retirement 20 years. How old was he ? 9 A's age is to B's as 4 to 3, and if twice B's age be added to A s, the sum will be 100 years. Required the age of each. The meaning of the first condition is, that A's age is f of B's, or that B's is f of A's. 10 What is the length of a fish, whose head is 3 inches long 20 EQUATIONS OF THE F1R8T DEGREE. IV his tail f the length of his body, and his body as long as his head and tail ? Let X z=z the length of the body. 11. Three fourths of a certain number exceeds five ninths of it by 14. What is that number 1 12. A person, having spent one half of his money and one third of the remainder, had $50 left. How much had he at first? 13. Says B to C, lend me $200 ; C replies, I have not $200 on hand, but if I had as much more and half as much more as I now have, and $12^, I should have $200. How much had he'' 14. Divide 60 cents among three boys, so that the second shall have half as many as the first, and the third 10 more than one third as many as the second. 15. A man wished to distribute a certain number of apples amongst his four children, in such a manner, that the first should have one third of the whole ; the second, three fifths as many as the first ; the third, two thirds as many* as the second ; and the fourth, half as many as the third and 8 apples more. What was the whole number, and how many would each child receive? 16. A gentleman bought two horses and a chaise ; the second horse cost once and a half as much as the first; and the chaise cost three times as much as the first horse; moreover the price of both horses wanted $50 to be equal to that ^ of the chaise. What was the cost of each horse and of the chaise ? 17. A man found, that he expended one third of his yearly income for board, one eighth of it for clothes, and one twelfth of it for other purposes ; and, that he had remaining $550. What was his income, and what were his whole expenses? 18. A drover, having a certain number of sheep, sold one third of them and then bought 60, when he found he had twice as many as he had at first. What was his first number of sheep ? 19. A gentleman gave to three persons .£98. The second received five eighths of the sum given to the first, and the third, pne fifth as much as the secqnd. What did each receive ? V EQUATIONS OP THE FIRST DEGREE 21 20. A person set out on a journey, and went one se^ enth of the whole distance the first day, one fifth the second, one fourth the third, and 114 miles the fourth, at which time he completed his journey. How many miles did he travel in all, and how many each of the first three days ? SECTION V. EQl'ATIOJjrS OF THE FIRST DEGREE, CONTAINING FRACTIONAI, PARTS OF QUANTITIES CONSISTING OF SEVERAL TERMS. Art. 13. 1. A says to B, I am 6 years older than you, and two thirds of my age is equal to three fourths of yours. What is the age of each ? Let X =. B's age , then, a; -j- 6 = A^s age. According to the conditions of the question, three fourths of the former must be equal to two thirds of the latter. One third of a: -|- 6 is written T^ , and two thirds will be twice as much, o 2x + 12 „ 3x 22 + 12 ,,,.,. ^ , . or ^ . Hence, -j- = ^ . Multiplymg by 4, 8x4-48 , „ o X = ; multiplymg this by 3, 9x=:8x-|-48; transposing and reducing' X =z 48 years, B's age. x -|- 6 z=: 54 years, A's age. Remark. The division of a quantity consisting of several terms, as x -\- 6, is represented by placing the divisor under the dividend, care being taken to extend the line of separation under all the terms of the quantity to be divided, 2. A man bought a horse and saddle ; for the horse he gave 1*2:30 more than for the saddle ; and five times the price of the saddle was equal to two fifths of the price of the horse. R&< quired the price of each. 22 EQUATIONS OP THE FIRST DLGREE V Let % ■=. the price of the saddle ; then, X -f 230 = the price of the horse. Hence, according to the conditions of the question, 5 X = '- . Multiplying by 5, o 25 X = 2 2; -{- 460 ; transposing and reducing, 23x = 460; dividing by 23, X =z $20, price of the saddle, X + 230 =: $250, price of the horse. 3. A father's age is to that of his son as 5 to 2, and the dif ference of their ages is 30 years. Required their ages. The first condition signifies that the father's age is f of the son's, or that the son's is f of the father's, or that 5 times tho son's is equal to twice the father's. Suppose X =. the age of the father ; then, X — 30 = the age of the son. 5a:_150 ,,,.,. ^ ^ Hence, x = . Multiplying by 2, 2 X = 5 X — 150 ; transposing and reducing, — 3 X = — 150 ; changing the signs, 3 X == 150 ; dividing by 3, a: z= 50 years, father's age ; X — 30 = 20 years, son's age. 4. A and B traded together. A put in $100 more than B T he whole stock was to what A put in as 5 to 3. How much did each invest in trade? 5. A man's age, when he was married, was to that of his wife as 4 to 3; but after they had been married 10 years, his age wag to hers as 5 to 4. How old was each at the time of their mar- riage T 6. A man's age, at the time of his marriage, was to that of his wife as 10 to 9; but if they had been married 10 years sooner, his age would have been to hers as 8 to 7. What were their re- upoctive ages at the time of marriage? 7. 4 and B have equal sums of money; but if B gives A V. EQUATIONS OF THE FIRST DEGREE 23 ^€10, ^ of what A then has, will be equal to £ of what B has left How much money has each? & Three towns are situated on the same straight road, and in the order of the letters A, B, C. The distance from B to C is 20 miles more than the distance from A to B, and is equal to | of the whole distance from A to C. What are the distances from A to B, from B to C, and from A to C ? 9. A merchant sold three packages of cloth ; the second con- tamed 15, and the third 30 yards more than the first; moreover, the third contained f as much as the first two. How many yards were there in each? 10. A and B commence trade with equal stocks; A gains r^lO per year, and B loses «£5 per year; at the end of three years B has only f as much property as A. How much has each at first? 11. Two boys, standing with bows and arrows on the bank of a river, undertook to shoot across it ; the arrow of the first boy fell 10 yards short of the opposite bank, and that of the second fell 10 yards beyond it ; now it was found that the first boy shot only y\ as far as the second. What was the breadth of the river ? 12. Two men have equal sums of money, but if one gives the cither $40, the former will have only | as much as the latter. How much has each? 13. A, B and C counting their money, it was found that B had $50 more than A and $75 less than C, and that the sum of what A and B had, was f of the sum of what B and C had How much money had each ? 14. A farmer, having a certain number of cows and twxs aa many sheep, sold 15 cows and bought 5 sheep ; he then found that the number of cows was to the number of sheep is 3 to 13, Ilow many of each had he at first ? 15. A man engaged to work a year for $200 and a suil of clothes ; bat falling sick, he worked only 5 months, and received $60 and the suit of clothes. What was the value of the suit of tlothes? 24 eQUAlIONb OF THE FIRST DEGREE. V 16. A man engaging in trade, gained the first year 6500, but the second year he lost -^ of what he then had ; after which he found that his stock was to that with which he began as 6 to 5. What was the stock with which he commenced 1 17. A grocer bought 6 barrels of cider, and 7 barrels of beer; he gave $2 a barrel more for the beer than for the cider ; and | of the price of the cider was equal to f of the price of the beer. What was the price of each per barrel? 18. Two numbers are to each other as 9 to 10 ; but if 6 be added to each, the sums will be as 10 to 11. What are these numbers 1 19. A man built two pieces of wall, one jof which was 20 rods longer than the other ; for the shorter he was to have $S a rod, Hnd for the longer $4 a rod ; now the whole price of the former was to that of the latter as 3 to 8. What was the length of each piece 1 20. A gentleman has two horses and one chaise ; now if the first horse, which is worth $100, be harnessed, he, with the chaise, will be twice the value of the second horse ; but if the second horse be harnessed, he, with the chaise, will be four times *he value of the first horse. What is the value of the chaise and of the second horse 1 21. A man bought a horse, and afterwards paid $50 for keep jng him ; he then sold him for ^ of what he had already cost in- cluding the keeping, and received for him $20 more than he first gave. How much did he pay for him at first? 22. Two cars run on different rail-roads; the speed of the second is 2 miles an hour greater than that of the first ; and the distance passed over by the first in 8 hours, is f of that passed over by the second in 9 hours. What is the speed of each per hour 1 23. A gentleman started on a journey with a certain sum of racmey ; after having had $60 stolen from him, he expended one third of what he had left, and found that the remaining two thirds wanted $90 to be equal to the sum which he carried from home How much money had he on commencing his journe) ? VI ■ EQUATIONS OF THE FIRST DEC REE. 25 24. A man having a gold watch, paid $10 for repairirg it and then exchanged it for two silver watches of equal value, and after paying $5 for repairing one of these, he found that it had cost him $65. What was the value of the gold watch at first? 25. A shepherd, in time of war, was plundered by a party of soldierg, who took ^ of his flock and ^ of a sheep ; another party took from him ^ of what he had left and ^ of a sheep ; then a ihird party took ^ of what remained and ^ of a sheep ; after vhich he had but 34 sheep left. How many had he at first ? SECTION VI. KQl/ATIONS OF THE FIRST DEGREE, WHICH REQUIRE THE SUBTRAC- TION OF QUANTITIES COPfTAINING NEGATIVE TERMS. Art. 14:. 1. A and B commenced business, A with twice as much money as B ; A gained .£20 and B lost .£10 ; then the difference between A's and B's money was .£70. How much did each begin with? Suppose we knew, that B had .£40 and A <£80, when they be- gan. Then, after A had gained <£20, he would have 80 -j- 20, or £100; and B having lost £10, would have left 40—10 or £30 ; now" to find the difference, we must subtract 30 from 100, which leaves 70. But, as in algebra most of the operations can only be represented, let us see how we can represent the prece- ding subtraction. Instead of 100 put its equivalent 80 -(- 20, and instead of 30, its equivalent 40 — 10 ; our object is to sub tract the latter from the former. If we subtract 40 from 80 + 2'0, it will be represented thus, 80 -f- 20 — 40, which is the same as GO; but we wished to subtract only 30 or 40 — 10 ; we hav« therefore subtracted too much by 10, and the remainder is too small by 10, consequently 10 must be added to 80-|-20 — 40 which then becomes 80-|-20 — 40-j- 10 or 70. Hence we see, that, to subtract 40 — 10, we must change the -{-^0 to — 40 tnd the — 10 to + 10 3 26 EQUATIONS OF THE FIRST DEGREE. Vi Now to solve the question ; let 2; = B's money ; then 2xz= A's money. When A had gained .£20, he would have 2x-|-20 ; and B having lost .£10, he would have x — 10. , Subtracting B's from A's gives 'Jix-]-20 — 2 -j- 10. For X — 10 is less than x by 10, and if we subtract x, wt sub (i ict too much by 10; the remainder then, after x has been sub- tracted, being 10 too small, 10 must be added to correct it Hence, 2 a; +20 — a; + 10 r= 70. Reducing the first member, a; -|- 30 r= 70 ; transposing and reducnig, X = .£40, B's money ; 2x=: .£80, A's money. 2. Divide 40 into two parts such, that if three times the les* be subtracted from twice the greater, the remainder will be 5. Suppose X = the greater part ; then 40 — xz=z the less part. For if the greater were any known number as 30, for example, the less would be the remainder of 40, which is 40 — 30 or 10 ; or if the greater were 28, the less would be 40 — 28 or 12. So when the greater is represented by 2, the less will be 40 — t. Twice the greater is 2x, and three times the less is 120 — 32 which, being subtracted from 22, gives 22 — 120 -|- 3 2. Hence, 2 2 — 120 -|- 3 2 := 5. Transposing and reducing 5xz= 125 ; hence, 2 = 25, the greater part, and 40 — 25 =: 15, the less. Art. \5. It follows from the preceding questions and eTrpIana* lions, that any quantity is subtracted by changing the signs of all its terms f and writing it after the quantity from which it is to be subtracted. 1 A man has a horse and chaise, which together are worth $400. Now if the value of the chaise be subtracted from twice that of the horse, the remainder will be the same, as if tnree limes the value of the horse be subtracted from twice that oj the chaise. Required the value of each. . 2. A vintner has two equal casks full of wine ; he drawj 20 . . llAcSvOut of one and 40 out of the other, and finds the differ VL EQUATIONS OF THE FIRST DEGREE. 2? ence between the number of gallons remaining in the two casks equal to one fourth of what each cask contained at first How many gallons does each cask hold ? 3. Divide the number 60 into two parts, such that the greater subtracted from 50, shall be equal to three times the less sul> tracted from 90. 4. A poulterer had a certain number of geese and twice as many turkeys ; after having sold 10 geese and bought 30 turkeys, he found that if he subtracted f of his number of geese from his number of turkeys, the remainder would be the same, as if he subtracted y5_ of his number of turkeys from four times his num- ber of geese. How many of each had he at first 1 Let X =. the number of geese ; . then 2xz=. the number of turkeys. After selling 10 geese and buying 30 turkeys, he would have X — 10 geese and 2 z -|- 30 turkeys. Then, according to the conditions of the question, _ , __ 3x — 30 . .. 162; + 240 -_ , . . . 2 a; 4-30 =r4x — 40 ^^ . Muluplyrag o lo by 5, 10a; + 150-^3x + 30 = 20x — 200 — ^^^^i^» multiply. o ing by 3, 30Z + 450 — 9a; + 90 = 60a; — 600— 16x — 240; transpos- ing and reducing, — 23 z z=: — 1380 ; changing the signs, 23 a; = 1380 ; dividing by 23, X bz 60, the number of geese ; and 22= 120, the number of turkeys. Observe that, after the equation was formed, in multiplying by 5, 3 z was changed to — 3 2, and — 30 to + 30 ; for, the sign — precedmg the fraction, belongs to the whole fraction, and not to any particular part of it, and when the fraction is multiplied by 5, the numerator is to be subtracted ; consequently 3 a:, which is supposed to have the sign -f-> must be changed to — 3 a:, and --30 to +30. 28 EQUATIONS OF THE FIRST DEGREE. VL Also, in multiplying by 3, both the terms 16 a; and 240, being iiffected by the sign -|-, must receive the sign — . If, however, these fractions had been preceded by the sign -J-, the signs of the numerators would have remained unchanged. The same remarks are applicable to all similar cases. Care must be taken also, when fractions are preceded by the signs -(- and — , to make these signs stand e'en with the lines ecparating the numerators and denominators 5. A farmer has 60 tons of hay ; of this he sells a certain por- tion, and finds that ^ of what he sells subtracted from f of what he retains, gives the same remainder, as f of what he retains sub- tracted from f of what he sells. How many tons does he sell 1 6. Two men, A and B, set out on a journey, each with the same sum of money. A spends $40, and B $30 ; then f of A's money subtracted from f of B's, would give | of what each car- ried from home. How much money had each on commencing the journey ? 7. Divide 147 into two parts, so that ^ of the less subtracted from the greater, shall be equal to -J of the greater subtracted from the less. 8. A vintner had two casks of wine, each containing the same quantity ; from the first he drew 10 and from the second 40 gal- lons ; he then drew from the first ^ as many gallons as the sec- ond contained after the first draught, and from the second ^ as many gallons as the first contained after the first draught, and found the number of gallons remaining in the first cask to the number remaining in the second as 7 to 3. How many gallons did each cask hold? 9. A market woman having a certain number of eggs, sold 30 of them, and found that f of what she had left, subtracted from wliat she had at first, would leave f of what she had at first Uow many had she before she sold any? 10. A man having a lease for 100 years, on being asked how much of it had already transpired, answered, that f of the time past, subtracted from | of the time to come, would leave the tame remainder, as if ^j of the time to come were subtracted VTl. MULTIPLICATION OF MONOMIALS 29 from f of the time past. How many years had already trans- pired ? 11. There is a pole consisting of three parts; the middle part is 4 feet longer than the lower and 4 feet shorter than the upper part ; moreover, if fV of the upper part be subtracted from § of the lower part, the remainder will be the same, as if ^ of the middle part be subtracted from | of the upper part. What is the length of each part and of the whole pole? 12. If a certain number be successively subtracted from 30 and 52, then |- of the first remainder be taken from f of the sec- ond, the last remainder will be 10. What is that number ? SECTION VII. MULTIPLICATION OF MONOMIALS. Art. 10. It may be remarked, that the addition, subtraction multiplication and division of algebraical quantities, cannot, strictly speaking, be actually performed, in the same sense aa they are in arithmetic, but are, in general, merely represented ; these representations, however, are called by the same names aa the actual operations in arithmetic. A monomial, or simple quantiti/, consists of only one term (Art. 4) ; as a, b c m, or — . A binomial is a quantity consisting of two terms, ^ a-\-b^am a — z y , or - -j- c m. A trinomial is a quantity consisting of three terms, as a + ft 1 — cd. Polynomial is a general name for any quantity consisting of leveral terms. Moreover, any quantity containing more than one term, ii called a compound quantity. 3» 30 MULTIPLICATION OF MONOMIALS. VIl Art 17. The product of two simple quantities, such as a an.< 6, is expressed, either by writing them with the sign of multipli- cation between them, as a X 6 or -a . 6, or by writing them after each other without any sign, as ah. The last form is most usual It is evidently immaterial in what order the letters are written ; for, suppose a = 5, and 6 = 7; 5 X 7 is the same as 7X5; hence, the product of a by 6 may be written either a 6 or ha. In like manner, the product of a, h and c may be written ahc^ achjb a c, he a^ ch a Of cah. It is most convenient however to write them in the order of the alphabet. The product of 5 ah by 2c d might be written 5ah2cd; but, as the order of the factors is unimportant, we may place the numerical factors next to each other, thus, 5 X 2 a 6 c 6?, or per- forming the multiplication of 5 by 2, we have 10 ah cd. But we could not write 5 2 a 6 c 3 q — 5mnr + 76c. 6. 22x2 — y3_|_32;2_|_822 — 7x2 + 7n + 7y3_i6y3_f.3;„ \ 7»i + 2;2-f 2y3__6w». QC. ADDITION. 36' v^mp-^^amp^ — Samp^ — 12 a^mp -\- 10 a^m p. 8. abc^ + 7mn^p^r'^ + Qabc^ + 9ab^c-{-l2m^n + 2abc» a62c-[-a6c2 — 3w2»2p3^4_3a6c3--2a6c2. SECTION IX. ADDITION. Art. 33. The addition of positive monomials or simple quati" titles J consists merely in writing them after each other, and giv- ing to each of them the sign 4-» except the first, which is also supposed to have the sign -f-. Thus, to express the addition of a and 6, we write a-\-b or b-^-a. Also, to signify the addition of a, 6, c and d^ we write a -|" ^ H" ^ H~ ^* ^^ ^^^® manner, the addition of a 6, 3zy, and 4mn, is expressed thus, ab-\-^xy -\- Amn The order of the terms, as was observed in Art. O, ia unimportant. Art. 241. If it were required to add together the polynomials, a -f- 6 -|- c, and m -f- w, in which all the terms are affected with the sign -{-, the process would evidently be the same, as if it were required to add together the separate terms of which these poly- nomials are composed ; that is, we should write them after each other, giving the sign -(- to every term except the first ; and the sum would bea-|-6-f-'^~l~'"~h'** But if some of the terms in the polynomials to be added, have the sign — , they must retain the same sign in the sum. Take an example in arithmetical numbers. Let it be proposed to add 10 — 3 to .12 ; 10 — 3 is 7 ; we wish then to add 7 to 12. But, if we first add 10, which \s expressed thus, 12 -\- 10, the sum is too great by 3 ; therefore, after having added 10, we must sub- tract 3, and the true sum is 12 + 10 — 3 or 19. Let us no\^ add b — c to a. First add 6 to a, and we have a 4- 6 • this is too great by c, because the quantity b — c, which d6 ADDITION. IX was to be added, is less than h by the quantity c ; therefore, aftei having added 6, we must subtract c, and the true result is a -j- We see, in these instances, that we have merely written the terms after each other without any change in their signs ; and the reasoning used in explanation of the process, is applicable to the addition of all polynomials, in which some of the terms are af- fected with the sign — . The sum of the polynomials c^h-\-^c — 4a and Vic-\Sa^h — 3a isa26 + 3c — 4a+12c + 8a26 — 3a. But this sum contains similar terms, which may be reduced, according to the principles given in Art. 2S8. This reduction being made, the sum, in its simplified form, becomes 9 a^ 6 -|- 15 c — la. Art. 25. From what has been said above, we deduce the fol lowing BULK FOR THE ADDITION OF ALGEBRAIC QUANTITIES. Write the several quantities one after another ^ giving to each term its proper sign, and then reduce the similar terms. Observe, that those terms which have no sign, are supposed to have the sign -j- 1. Add together 4 a, 6 6, 7 c, 9 a, 3a + 6, 6c, 4 the quotient will be -\-b, because -j- a multiplied by -\-b gives -\-ab. If -|-a& be divided by — a, the quotient will be — 6, because — a multiplied by — 6 gives -\-ab. If — ab be divided by -f- a, the quotient will be — b^ because ya multiplied by — b gives — ab. Lastly, if — ab be livided by — a, the quotient will be -|- i, tiecause — a multiplied by -|- 6 gives — ab. Hence, the rule for the signs in division, is the same as that in multiplication ; that is, if the two terms , one of which is to be di" mded by the other y have the same sign, either both -f- or both — , the quotient must have the sign -j- ; but if they have different signs f that is, one -\- and the other — , the quotient must have thi nrn — . 5 50 DIVISION OF POLYNOMIALS. XlII Art. 42. Hence, we have the following RULE FOR DIVIDING A POLYNOMIAL BY A MONOMi IL. Divide each term of the dividend by the divisor ^ according to hi principles given in Art. 40 for dividing one monomial by another, observing the rule established above for the signs ; and the partial quotients taken, together ^ will form the entiic quo tient, 1. Divide a^ b -]- a m by a. 2. Divide I2x i/ — 6x^ -{-Sx y^ by 3 x. 3. Divide 15 a m^ + 30 m^ z — 45 m^ by 15 m^. 4. Divide36 62ce/— 24 63 c2— 12 64c by 662 c. 5. Divide 9 a5 66 _ 3 ^2 63 by3«2 63. 6. Divide— 21x2 3^ — 7 + 42z by— 7. 7. Divide 56 a^ 63 c — 28 a^ 63—168 a? 6^ c^ by — 28 a^ 6-. Art. 43. When the dividend and divisor are both polynomi- als, the process becomes rather more difficult ; but, if we observe (he manner in which a product is formed by multiplication, we shall readily see the course to be pursued in division. Multiply 3 a3 +2a2 6 + a 62 by 2a2 -|- 3a 6. Operation, 3a3+ 2a26 + a62 2a2-(- 3a6 6a5+ 4a46 + 2a362 -f 9an+6a362-f 3q26 3 6a5-j-13a4 6 + 8a3 62 4-3a2 63. Product. If this product be divided by the multiplicand, the quotient will De the multiplier ; or, if it be divided by the multiplier, the quotient will be the multiplicand. Since each term in the multiplicand is multiplied by each term in the multiplier, if no reduction takes place, the number of terms in the product will be equal to the number produced b) multiplying the number of terms in the two factors together Thus, if one factor have 4 terms and the other 3, the product will contain 12 terms. In most cases, however, a reduction ^J.11. DIVISION OF POLYNOMIALS. 51 takes place, by which some terms are united and others wholly disappear. But there are always two terms, which can neither be united with any others, nor cancelled by any others ; viz : one, which is produced by multiplying the term containing the highest power of any letter in the multiplicand, by the term containing the highest power of the same letter in the multiplier; and the other, which arises from the product of the terms with the lowest expo* uents of the same letter. Now, since the dividend is to be considered as the product of the divisor and quotient, it is evident, that if the term containing the highest power of a particular letter in the dividend, be divided by the term containing the highest power of the same letter in the divisor, the result will be the term of the quotient containing the highest power of that letter. J.et us reverse the process of multiplication, and divide Ga^-j- Operation. Dividend. 6q5-4-13a^6 + 8g3 62-|-3a2 63f 3a3 + 2a26 + q52 . Divisor. 6 a5-[- 4a^6-f ^«^^^ X'^a^ + ^ab, Quotient.. 9a46+^^^ + 3a2 63 9g^6-f 6«^^^ + 3flQ63 0. According to the preceding remarks, 6 a^ divided by 3 o^ must produce the term containing the highest power of a in the quotient ; 3 a^ is contained 2 a^ times in 6 a^ ; then, as the en- tire dividend is produced by multiplying the whole of the divisor by the whole of the quotient, if we multiply the whole divisor by this first term 2 a^ of the quotient, a part of the dividend will be produced. The product of the divisor by 2 a^ is QaP -\-4,af^h -|- 2 a3 ft2j which being subtracted from the dividend, leaves ^a^h + 6a362^3a253. This remainder is to be considered as a new dividend, and as produced by multiplying the divisor by the remaining part of the ijuotient. The first term, 9 a* 6, of this new dividend, must 62 DIVISION OF POLYNOMIALS. Xlli have been produced by multiplying 3 a^ by the term containing the next highest power of a in the quotient ; therefore, if it be divided by 3 a^y that term of the quotient will be obtained. Di- viding 9 a'* 6 by 3a^, we have 3a6 for the second term of the quotient ; then, multiplying the divisor by 3 a 6, the product is 9 a"* 6 + 6 a3 6^ -|- 3 a^ ^3^ which subtracted from the last divi- dend, leaves no remainder. The entire quotient therefore is 2 a^ + 3a6. We perceive, from the preceding example, that we always divide the term containing the highest power of some letter in the divi- dend, by the term containing the highest power of the same letr ter in the divisor It will, therefore, be found convenient to write the quantities in such a manner, that these two terms may stand, the one on the left of the dividend, and the other on the left of the divisor. This will be accomplished by arranging the dividend and divisor according to the powers of the same letter, beginning with the highest. A polynomial is said to be arranged according to the powers of a particular letter, when the terms are so written, that the powers of that letter go on increasing or diminishing from left to right. Thus, in the example just performed, the quantities were arranged according to the diminishing powers of a. With regard to the signs of the partial quotients, the same rule Is applicable, that was given for the division of a polynomial by a monomial. Art. 4L4. From what precedes we deduce the following RULE FOR THE DIVISION OF ONE POLYNOMIAL BY ANOTHER. 1. Arrange the dividend and divisor according to the powcs of the same letter ^ beginning with the highest. 2. Divide thejirst term of the dividend by the first term of the divisor y and place the result as the first term of the quotient; rec- ollectingy that if both terms have the same sign, the partial quo- tient must have the sign -|-, but if they have different signs, thi partial quotient must have the sign — . 3. Multiply the entire divisor by this term of the quotient, sub- XIII. DIVISION OP POLYNOMIALS. 53 tract the product from the dividend, and the remainder will form a new dividend. 4. Divide the first term of the new dividend hy the first tern- of the divisor J and the result will form the second term of the quo^ tient ; multiply the entire divisor hy this second term of the quo- tientj and subtract the product from the second dividend. The remainder will form a new dividend, from which another term of the quotient can he found. These operations are to he repeated, until all the terms of the >*rginal dividend are exhausted. N. B. The same arrangement must be preserved, in each of the partial dividends, as was made at first in the whole dividend. When the first term of any remainder cannot be divided by the first term of the divisor, the process must terminate, unless the quotient be continued in a fractional form. When the divi- sion terminates, the remainder, if there be one, may be written over the divisor, in the form of a fraction, and annexed to the en- tire part of the quotient. Let it be proposed to divide 75a^h* — 27 a h^ — 49a^b^-\- 20a^h — l9a*h^hy 4a^b + 2h^^7ahK Operation. 20a56_i9a462__49a353_^75a254__27a65 ( 4.a^h^7ah^+Sb^ 20a56_35a462^15a353 \ 5a^+^a%—9ah^ + 1 6a462— 64fl363_^75a264— 27a65 -fl6q^6Q— 28a3&3+12a^6^ _36a363^03a264_27a65 _36«363_|_63a254_27g55 0. After having arranged the two quantities according tc the powers of a, and placed the divisor on the right of the dividend, we divide 20 a^ 6 by 4 a^ 6, which gives -\-5a^ for the first term of the quotient ; we then multiply the divisor by 5 a^, write the product under the dividend, and subtract it from the dividend. The subtraction of the product is performed by changing itj signs, considering it as written after the dividend, and redu- cing similar terms ; thus, the signs being changed, it becomes 5* 54 DIVISION OF POLYNOMIALS. XIII. — *20a5 6_|_35a4 52_i5a3 53. then, by reduction, +20a5j and — 20a^b cancel each other, — 19 a^ 6^ and -{-25 a* b* make + 16a4 62, and —i9a^b^ and — 15 a^b^ make — 64^3 fts. bringing down the other terms of the given dividend, we have for a remainder IGo^i^ — 64 a^ 63 _j_ 75 ^2 M — 27^65^ which forms a new dividend We now divide 16 a^b'^ by the first term of the divisor 4 a^ 6, and obtain for the second term of the quotient + 4 a^ 6 ; multi- plying the divisor by 4 a^b^ and subtracting the product, -J- 16 a^ 6* — 28 a3 ^3 -[- 12 a2 ^4^ from the second dividend, in the same manner as before, we obtain for a remainder — 36 a^b^-\-63a^b^ — 27 a 6^, which forms the third dividend. We now divide — 36 a^ b^ by 4 a^ 6 and obtain — 9ab^ for the third term' of the quotient ; multiplying the divisor by — 9 a 62, and subtracting the product, — 36 a^ 63 -[- 63 a^ 6^ — 27 a b^y from the third dividend, we have no remainder. The entire quotient, therefore, is 5 a3 -|- 4 a^ 6 — 9 a 62. As another example, let it be proposed to divide a^ — 6^ by a — 6. Operation. a5 — a^b i a^-\-a^b + a^b^ + ab^ + 6- + aH — b^ + a^b — a^b^ + a3 62_55 4-a3 62_a2 53 + «2 63-65 -|-a2 63_a64 + a64 — 65 + a 64 — 65 0. Ill the last example, several terms are produced in the course of the operation, which are not found in the dividend ; these terms disappear, by reduction, when the quotient and divisor are multiplied together. 1. Divide a3 -|- 3 a^ m + 3 a »»' + 1»3 by a + m. ZCITI. DIVISION OF POLYNOMIALS. 55 2. Dividea3 + 5a4x+10a3 22_[_i0a2x3 + 5ax4-f.x5 by 3. Divide m"^ — 4 m^ z -|- 6 w^ x^ — itnx^ -^-x^ by m — x. 4. Divide (3 x'* — 96 by 3z + 6. 5. Divide264 — 9a2i2^6a4 + 4a36 — 3a63 by 2a^ + a Divide20a5_4ia4 5_|_50a3 62__45fl2 53_|.25a64 — 665 by 5a^ — Aa^b + 5ab^ — Sb^ 7. Divide4x4 — 9a22;2+6a3x — a4 by 2x2 + 3izx — a^ 8. Divide a^ -{- 2 a^ z^ -^- z^ by a^ — a 2 + 22. 9. Divide a^ __ I6 a3 2;3 _|_ 64 x^ by a2 _ 4 ^ x + 4 x2. 10. Divide x^ — x^ + x3 — x2 + 2 x — I by x2 -(- x — 1. 11. Divide 10a4 _ 48 a3 6 _^ 51 ^2 62 + 4 a 63 — 15 6^ by 8a6 — 2a2_565J. 12. Divide ni^ — x"^ by we — x. 13. Divide 5a5 63c5__22a4 63 c6_^5a363c7^12a263c8_ '7«262c8_|_28a62c9 by a26c2_4a6c3. Art. 45. In the preceding examples, the division could be exactly performed. Let it now be proposed to divide a by Operation. l—x a / 1 — X a — ax ( a-\-a + ax ax — ax^ X + « t;^ + a x3 -(- 1 — X + ax2 a x2 — a x3 + ax3 a x3 — ax* + a X*.. i « the exanjple above, it is manifest that the operation would never terminate. This quotient is called an infinite series, and we see that any term, except the first, may be formed by multi' plying the preceding term by x, remembering to place the divi' 56 DIVISION OF POLYNOMIALS. XllJ sor under the last remainder, to indicate the continuation of th nexing the remainder placed over the divisor. 8. Divide a by \-\-xy^ finding 6 terms of the series, and an nexing the remainder placed over the divisor. Art. -47. It is manifest, that when a product is represented in its factors, dividing one of the factors, divides the whole pro- duct; also, that we may, in any case, divide the dividend first by one factor of the divisor, then divide the resulting quotient b> another, and so on. XIV MULTIPLICATION OF FRACTIONS. 69 Thus, 7 8 . 3 = 168 ; dividing the factor 8 by 4, we have 7 . 2 . 3 zn 42, which is ^ of 168. Also, in dividing 6 . 12 or 72 by 2 . 4 or 8, we may first di vide by 2, which gives 3 . 12, and then this quotient by 4, whicb gives 3.3 = 9 = ^. In like manner, (m — n) (a^ — 6^) divided by a — 6, gives {m — n) {a -\- 6), to obtain which, we divide the factor a^ — b^ by the divisor. Also, (x2 _ y2) (a2 _j_ 2 a 6 _|_ ^2) divided by (x + y ) (a + 6), gives (x — y) (a-^-b); to obtain which, we divide the factor a;2 — y2 by 2; _j_ y ^ and the factor a^ _|_ 3 « 5 _|_ 52 ^y a + 6. 1. Divide a^C^J + y) ^Y ^ -{-!/• 2. Divide (m2 — l)(rt + 6) by m — 1. ^. Divide 14(a;3_l) (a + ni) by 7(a; — 1). 4. Divide («2_62) (a:3_|.y3) by (a — 6) (x + y). ^. Divide 27 (m^ — n^) (x^ + yS) by 3 (m^ + n2) (i + y ). 6. Divide 30 m2 (x^ — yS) (;;j4__„4) by 10w(x3 + y3) (m2 — w2). 7 Divide25(a + 6) (m4— 1) (x5+l) by5(a + 6)(m2+l)(x+l). SECTION XIV. MULTIPLICATION OF FRACTIONS BY INTEGRAL QUANTITIES. Art. 4:8. Fractions have the same signification in algebra, that they have in arithmetic. Thus, - signifies that one unit is divided into b equal parts, and that a of those parts are useil ; or, it represents division, and signifies that a is divided into i equal parts. How much IS 5 times -^j ? Ans. -f ^. How much is 3 times y « Ans. — -. 6 6 -80 MULTIPLICATION OP FRACTIONS. XIV How much is c times -? Ans. -r-. o What is I of 4? | of 4 is ^, and f of 4 is ^. Ans. What is ^ of a? ^ of a is -, and f of a is — -. Ans, TTTi ' 1 ^ /.»l/..c . a „ . ac What is the -r part of c f - of c is — , and - of c is -^. o 6 o In the first three questions, the object was to multiply a frao tion by an integral quantity, and in the last three, to find a frac- tional part of a quantity, that is, to multiply an integral quantity by a fraction ; and we perceive that both objects were accom- plished by multiplying the numerator and the integral quantity together. •* Hence, to multiply a fraction hy an integral quantity ^ or an integral quantity hy a fraction ; multiply the numerator hy thi integral quantity, and write the product over the denominator, 1. Multiply — by 6 c, 2. Multiply by mx, 3. Multiply • by m-\-n, 4. Multiply -^^ by 4c + 3a;. 5. Multiply by wi-j-n. 6. Multip.y r—-^ by 12a3 + 25a2 6. ^ h^-\-^ac 7. What is the ^-i^ pa^ of a^--h^1 h -\- c a What is the o^To"^^^'' o part of 2 a c -|-2 ft cl 9 Multiply 25 x2 + 13 z y by '^^^~^-. X.IV MULTIPLICATION OF FRACTIONS. 61 10. Multiply ^ by 5. 15 The fraction — signifies that a is divided into 15 equal parts . but if the denominator be divided by 5, a will then be divided into 3 parts, or ^ as many parts as it was before ; consequently, each of the parts will be 5 times as great as before ; that is, the a . ^ . a fraction ^\ao times --. o 15 1 1. Multiply r- by h. a c Here a is divided into 6 c equal parts ; but if the denominator be divided by 6, a will then be divided into t as many parts as it was before ; the parts, therefore, will be h times as great as they were before: that is, the fraction - is 6 times t-. c be HevCf to multiply a fraction and an integral quantity togetJir er, divide the denominator by the integral quantity ^ if possible. Art 49. Combining this rule with the preceding, we have a GEN-J AL RJIiE TO MULTIPLY A FRACTION AND AN INTE6Rii.l- QUANTITY TOGETHER. Divide the denominator by the integral quantity , if it can be dona ; if noty multiply the numerator by the integral quantity. The following examples may be performed by dividing the denominators. t Multiply — 5^ by m. 3 a 5 4- 4 a V 2. Multiply z^ y by ^g ,3 y. 1 19 ^4 y» - 3. Multiply !!!!=^±^* by a + 6. 4 Multiply Xi^ ^ + 25 6° + 43 6 MULTIPLICATION OF FRACTIONS. XIV 6- Multiply ^j-p^^^i^l+J^^j^^,-^, by a3 + 3«M 7. Multiply ^^l^j by .= + ,=». 9- Multiply ^^^,"^^^3^ by m + y. 10- Multiply ^^-^^.i^^ ^1- Multiply ,^ (,!+)\:r,,.) by 7(a-6) (c^d), ^^' Multiply (^.!.'J)|jj^3^5) by (m^n) (x + y). 13. Multiply T by 6. Dividing the denominator by 6, the fraction becomes - or a 14. Multiply ^ by x^ y. Dividing the denominator by x^ y, the fraction become^' !!^±A£,thatis,m2 + 6c. From the last two examples we see, that. If a fraction is multiplied by a quantity equal to its denomh fMtor, the product will be its numerator, 15. Multiply ^^^ by c. 16. Multiply — ^ _A by a — 6. 17. Multiply 3^^^^ + ^=^y by m^ + Sme. ^^ m^-\-3mc ^ ' XIV MULTIPLICATION OF FRACTIONS. f>3 18. Multiply — ^ by 127 a. (Z c Art 50. Multiply — g— by mi/. First multiply by m ; the product is ; multiply this pro- mx duct by y, and the result is — - (Art. 40). The result would have been the same, if we had divided the integral quantity and the denominator of the fraction both by m, and then proceeded according to the first rule in Art. 4:8. Therefore, When an integral quantity and a fraction are to he multiplied together, if the integral quantity and denominator of the fraction have common factors, those factors may he omitted in both before multiplying. 1. Multiply ^2^2 hymxy. 2. Multiply j2 ,3 rf hy ah^c\ 3. Multiply ac^xy . :; t: — • 1 -«. 231 a^ m2 x ' 33 a3 — 66 a m^' In the preceding examples the greatest common divisors of the numerators and denominators are monomials ; in those which follow, the greatest common divisors are polynomials, but the quantities may be easily separated into factors, so as to ^exhibit the common divisors (Art. 55). 4a2_462 . 4(«2-_62) 13. — — -— -. This IS the same as —-. ~- 3a — 36 3 (a — 6) i ja — b) (a + b) _ 4 (a + 6) __ ^ a + Ab ■ 3(a_5) — 3 — 3 j^ Sx^y + Sxy^ ^^ i^ — x^ ^x^+Gxy + Sy^ a^-^'Zax + x^ a2_2a6 + 62* 16(23_|_y3)- 5a3-f.l0a26_|.5q^2 45 a^ 63 (a; -f y ) 8«34_8a26 • 1^- 25a6(x4 — y4y Art. 64. The actual division of one monomial by another is impossible, when the coefficient of the former-is not divisible by 'hat of the latter, when the divisor contains any letter not found ax MULTIPLICATION OF FRACTIONS BY FRACTIONS. 71 in the dividend, or when the exponent of any letter in the divisoi exceeds the exponent of the same letter in the dividend. A pol- ynomial cannot be divided by a monomial, unless the latter will divide every term of the former ; and the complete division of one entire polynomial by another is impossible, whenever the first term of the original or of any partial dividend, cannot be divided by the first term of the divisor, the quantities being arranged according to the powers of the same letter. In such cases the division is expressed in the form of a frac- tion, the divisor being placed under the dividend. The result should then be reduced to its lowest terms. With- regard to polynomials, however, we sometimes partially execute the division, placing the last remainder over the divisor^ and annexing it to the entire part of the quotient. Express the division in the following examples and reduce the results. o ^ 1. Divide Sa^b hy lab, Ans. — -. 2. Divide 13 2; y by 26 a 6 c. 3. Divide 33 x y^ by 66 x^ y^. 4. Divide 45 a 6 c by 180 a 6 x\ 5. Divide abc hy m-\-n, 6. Divide 46 x2 by 4 x^ y _[. 6 a: y^. 7. Divide 3 z + 3 y by 6 (z^ _ y'^y 8. Divide 12 1^ + 12 y'^ by 36 x^ + 36 yK 9. Divide 13 {x — y) by 39 (xS — y^). SECTION XX MULTIPLICATION OF FRACTIONS BY FRACTIONS. Art.65. Whatisfof^? ^ of |-=: g^, and § of J = |> ft /• I re Wha is the - part of - 1 The r- part of - = 7-, , and the o a b ^ d bd 7* 78 MULTIPLICATION OF FRACTIONS BY FRACTIONS. XX a . c ac . . a c ac rii T part of -5 = r— ,; that is, v • j = i—r ^n like manner a oa o a oa a c m acm b ' d ' n b d n Hence we deduce the RULE FOR MULTIPLYING FRACTIONS BY FRACTIONS. Multiply all the numerators together for a new numerator ^ ana all the denominators together for a new denominator. Remark. As the results should be reduced to the lowest terms, It is convenient to represent the operation and omit the common factors, previous to the actual performance of the multiplication Thus -5^ ^+^ - ^^^("^ + ^) - ± %\so 3a 14 86 3.14.8«6 1.2.8 16 7 • 56»i * 9a3— 7.5.9a36m"" 1.5. ^a^m" l^a^m* Another mode of proceeding is the following, viz : when frac- tions are to be multiplied together, if the numerator of one and the denominator of another have common factors, omit those factors before multiplying. 36 a2 ft 25 2 y3 Thus, if — —r — is to be multiplied by — ; — ^, divide the 5x2y *^ 16ao3c numerator of the first and the denominator of the second by 4 a 6 J also, the numerator of the second and the denominator ot the first by 5 X V. The fractions then become — and ttk~9 ^Le ^ ,. ^ . 45ay2 product of which is . -^ . 4 o^ c X . ,T 1 . , 26c ^ 6a63 1 Multiply— ^ ^^"tW 2. Multiply ^ by -^. 3. Multiply —1^- by ^^-p^. 4. Multiply —-i— by—. XXI. ADDiriON AND SUBTRACTION OF FRACTIONS 5. Multipy -^^ by^-p-^. 3x2 — 4a; ^ - 7^ 6.. Multiply -jj^ by2^3_3^. ^- •^"'"P'y 5j=To ''^ -2r- 8. Find the product of -^ , ^^ and ^^. ^ TV 1 , :. ^ 44-2; 3w2 ^ a4-6 9, Find the product of — ^-^-, — ^-7 and -r— J — . '^ y3'a_j_5 44-x .« T. .. 1. J /. 39x2 a2__2;2 49«6c 10. find the product of -zr-x-i -^tk — and ; . ^ 7a2* 13x a-j-x 11. Find the product of , — ' ^ „ and -5 5 ^ 9m3 16 a^ a2 — ^9 12. Find the product of li^^+i^), ^-!z^' and -^ 2m3 X— y ' »i2 S{x+y) SECTION XXI, ADDITION AND SUBTRACTION OF FRACTIONS. COMMON DENOM- INATOR. Art. 66. To represent the addition and subtraction of frac- tions, we merely write them after each other, with the signs -(- and — between them, being careful to place these signs even with the line separating the numerator and denominator. Thus, tt C £ - + -7 — -p. But if the denominators are alike, the addition oaf and subtraction may be performed upon the numerators. 2 4 2 + 4 6 Add together — ■ and — -. Ans. ,' = -— . ^ 11 11 11 11 4 ^ 7 ^ 7 — 4 3 Subtract ^ from -. Ans. "^ = jg- 80 ADDITION AND SUBTRACTION OP FRACTIONS. XXI All 1 « , ^ * «4- 6 Add together - and -. Ans. — - — . ^ c c c r^ I ^ *. c . c — 6 Subtract — from — . Ans. . m m m Add together — and -. Here tlie denominators are differ* a ent ; but if the numerator and denominator of the first fraction be multiplied by d, and the numerator and denominator of the second be multiplied by 6, the denominators will be made alike, without changing the value of the fractions. The first fraction becomes 7—, and the second 7-7; then adding, we have — ^—-^ bd hd' ^' bd Ans. Add together -r , - and -^. If we multiply the numerator and denominator of each fraction by the denominators of both the 1 1/. • , adfbcf.bde others, the fractions become 7-77., ^-f^and 7-7>, the sum of o a J oaj a J -. , . adf-\-bcf-\-bde . „. j • which IS — - — ' /.. ' , Ans. H^nce we derive a bdf RULE FOR THE ADDITION AND SUBTRACTIOUT OF FRACTIONS. Reduce them to a common denominator ^ and then add or sub- tract the numerators. Art. GT. The preceding examples give also the following rule for reducing fractions to a common denominator. Multiply the denominators together for a common denominator , and multiply each numerator by all the denominators except its own. For this is equivalent to multiplying the numerator and denominator of each fraction by the denominators of all the oth- ers, which does not alter the value of the fractions (Art. OS). This rule for reducing fractions to a common denominator, will uniforndy give correct, but not always the simplest results. Suppose it required to reduce -z — -„ - — and - — — to a com< '^'^ * 4 m^ 6 m m^ x XX J. COMMON DENOMINATOR. 81 men deaoininator. The product of the denominators will, in this case, give a common denominator much greater than is ne- cessary. In order to obtain the least, we must, as in arithmetic find the least common multiple of all the denominators (Art. 61). The least common multiple of 4m3, 6 m and 3m^x is 2^ ^m^x ; this therefore is the least common denominator sought. To produce this denominator, the first denominator must be multiplied by 3 z, the second by 2 m^ z, and the third by 4m; these, therefore, are the quantities by which the nu- merators are respectively to be multiplied. The fractions then 3ax 'Hbnt^x . 4cm become —r — -— , -— — r— and 12wi3x' I2m^x 12w»3x Art. 68. Hence we deduce a RULE TO REDUCK FRACTIONS TO THE LEAST COMMON DENOMINATOR. JFHnd the least common multiple of all the given denominators, and this will be the least common denominator sought ; then mul- tiply each numerator by the quantity, by which it would be neceS' sary to multiply its denominator, in order to produce the least common denominator. Remark. The quantity by which any numerator must be multiplied, may be found by dividing the common denominator by the denominator of the given fraction. It is to be observed also, that fractions must be reduced to their lowest terms, before we apply the rule for reducing them to the least common denom- inator. 1. AddP-,™ andf. q n h ^ ... 3a 2a . x 2. Add -— , — - and -. 7 5 y o Ajj 4 3x , lly 3. Add -— — — and -— ^. 7ab 14 a^ 21 b^ A * jj 9x2y 3a6 ^ Ixy^ ^•^^^■4^'67?""^42P,^ i9' 82 ADDITION AND SUBTRACTICN OP FRACTIONS XX' 6. Add "+4, "_ and ^lif-. fi AAA ^ — 6a-j-* i« ^•"^^^ 7^'5^'^0T3^^2577»• 9. Add— -^ and-. 10. Add 1±^^ and Izif!. 1 — x2 l + a;2- 11. Add -i— and ~. l-f-z I X 12. Add -5^, and ^^. 13. Add 1^ and ^"^ 4 62 4:ab + 8a 14. Subtract |^ from — . 15. Subtract ?!=i%romf±^. 49 28 16. Subtract -^ from r^^. 17. Subtract?^ from ii+B. 18. Subtract ifi^* from ^ 19. Subtract ?|l^^romi±. 20.. Subtract =-1^ from 1^. 7a — 14x 21 21. Subtract!^ from 1±^. 1 + l2 1 — x« XXII. DIVISION BY FRACTIONS. BU 22 Add — ; — and mp. In this example, the integral quantity must be reducea to a fraction having x-\-y for its denominator. 23. Add a + 6 and ^^. 4 24. Add m2 + w2 and ^j^^ + ^» . ^ , ' Im — 2» ^K AAA 1 A 3a2 — 362 - ^ ^^^ ^ ^^ a2 + 62 ■ 26. Subtract ^^""^ from 3w»« 27. Subtract -^— - from 5 A3 „3 x+l 7 a 6 4- 4 c2 28. Subtract 2 from T » 29. Subtract a+6 from ^^ + ^ 65^ 30. Subtract »t— 1 from ^^ . SECTION XXII. DIVISIOIV OF INTEGRAL AND FRACTIONAL QUANTITIES BY FRAC- TIONS. Art. 69. How many times is f contained in 8 ? ^ is con- tained in 8, 40 times, and f is contained A^ times. How many times is ^ contained in a? ^ is contained in a, 9 a tmes, and f is contamed in a, - - times. How many times is - contained in c ? t- *" contained in c, o 64 DIVISION BY FRACTIONS. XXII b c times, and - is contained in c, — times. This result is th« o a same as the product of c by — How many times is f contained in f ? Reduce the fractions to a common denominator ; f = f ^ and f =:: f | ; f ^ is contained in f f as many times as 21 is contained in 32, which is ff times. c ct How many times is - contained in - ? Reduce the fractions a , c b c . a ad b c . to a common denommator ; — = -r~ii a^" r ^^ 7— 7 j 7— 7 ^s a bd b b d bd : , . ad . , . . , . contamed m — , as many times as be is contained m /za, which is ^ — . This result is the same as the product of - be '^ b by-. From the preceding questions, we deduce the following RULE FOR DIVIDING AN INTEGRAL OR FRAOTIONAL QUANTITY BY A FRAOTJOJf. Invert the divisor and then proceed as in multiplieation. Remark. When it is required to find what part one quantity is of another, make the quantity which is called the part, the dividend, and the other the divisor : also, when it is required to find the ratio of one quantity to anotner, make the quantity which is expressed first, the dividend, and the other the divisor. Perform the following questions, taking care after invei^ng the divisor to omit common factors as in Art. 65. 1. Divide m by §. 2. Divide a^bc by *-y~2' 3. Divide x^ + y^ hy j-. XXII. DIVISION BY FRACTIONS. !!i5 4. Divide 3(a2 — 62) by 51?;Zl5) 5 Divide — * by 4- a X 6. Divide ^ by -g-r-. 7. Divide — — f by .- _ . 1 ah ■' 49 a2 ^ ,-.._, 11x2^0 ^ 992;m2n3 8. Divide ^ . ,/^ by —. — -rf-, 9 a^ 63 -^ 4 a 6* r. T^• -J x^-^-^xy + if . 3(x+y) 10. Divide 141? by ^. 3 •' 5x 11. Divide — - — by — 7—. 5 4 .c Tx. .. 9x2_3x ^ z2 12. Divide = by — . o o 13. -J is what part of -77; — r ? 4/» 16 wi^ .. 5x4 «4 35z«5 14. r^=— ^ IS what part of ^. / ^ ? 17 a w3 '^ 34 a3 w^ ,^ 3(x + y). ^ r9(x2 — y2). 15. ^^ "T^^ IS what part of \^ ^ ' f 7 x2 *^ 14 /» ,^ „,, . \ . ^76c + 2162c 1462c2 + 426«, 16. What IS the ratio of ?- to k-^ ? 4 X y 9 x"^ y* »5 , „„ . , . ^155a2 62 65 a 63 17. What IS the ratio of -— r to ^„ » , _ A ,0 «ru . t- • rlO(m+p) 55(m2_«2) 18. What IS the ratio of ^.^y to —^-—-tU j Ibcx"^ 56 62 c2 X (9. What IS the ratio of —z — to r— — ? »i2 — 1 w -|- 1 20. What IS the ratio of ^ ^"' +i ^/ + ^'^ 736 c 1 2(x3 + 3x2y + 3xy2 + y3) 36562^3 B6 LITERAL EQUATIONS. XX m SECTION XXIII. LITERAL EQUATIONS. Aft. 70. Let the learner find the value of z in the following equations. he — z 3x4-4m ,,,.,. a 2 (2) 6 z -h 15 y = 102. $ X^XIV. WITH TWO. UNKNOWN QUANTITIES. 89 The coefficients of y in the two equations will be alike, if the 1st be multiplied by 5 and the 2d by 2; or the coefficients of x will be alike, if the 1st be multiplied by 3 and the 2d by 2. But the best way, in this question, is to divide the 1st by 2 and the 2d by 3, which gives (3) 22; + 3y = 30;) (4) 2 z + 5 3^ = 34. f Subtract the 3d from the 4th, 2y = 4, and y = 2 cents, price of a peach. Substitute 2 for y in the 3d, 2 X -j- 6 := 30, from which xr= 12 cents, price of a melon. 4. Says A to B, ^ of my money and $10 is equal to ^ of yours , yes, says B, but \ of my money and $10 is equal to | of yours. H«>w much money has each ? Let X •=. A's money, and v ^ B's. Then, Remove the denominators, (3) 2x4- 60 = 3y;) _ (4) 33^ + 120 =: 8 X. ) By transposition, (5) 2x — 3y = — 60; > (6) 3 y — 8 X = — 120. S Add the 5th and 6th, — 6 X = — 180 ; change the signs, 6 X = 180 ; hence, x = $30, A's money. Substitute 30 for x in the 3d, 60 + 60 r= 3y ; from which y = $40, B's money. We see, that, in the preceding problems, the conditions, in each case, give rise to two distinct equations, which may be called the original equations ; the others which follow, are de- duced from these, or are mere modifications of them. From the two original equations containing two unknown quantities, we obtained one with only one unknown quantity. This is called eliminating the unknown quantity, which this new equation does not contain. Thus, in the solution of the first 8* 00 EQUATIONS OF THE FIRST DEGREE XXIV question in this article, we eliminated y, and thus obtained an equation with x only and known numbers. We perceive, moreover, that when the quantity to be elimina- ted, is in the corresponding members of the two equations, that is^ o'.*;h^T in the first or second members of both, and is found in only one term of each, the following rule will, enable us to effect ihe elimination. FIRST METHOD OF ELIMINATION". Art. 72. Multiply or divide the equations y if necessary , so as to make (he coefficients of the quantity to be eliminated the sanijs in the two equations ; then subtract one of the resulting equations from the other ^ if the signs of the terms containing this quantity are alike in both equations, or add them together, if the signs ai't different. In applying this rule, the equations should first be freed from fractions, if they .contain any, and it is advisable to transpose all the unknown terms into the first members ; moreover, if the un- known quantity to be eliminated, is found in several terms in one or both of the equations, these terms, in each, must be re- duced to one. The coefficients of any letter in the two equations will be made alike, if, after the equations are prepared as prescribed above, each equation be multiplied by the coefficient of that let- ter in the other equation ; or, if each equation be multiplied by the number, by which the coefficient of that letter in this equa- tion must be multiplied, in order to produce the least common multiple of the two coefficients of the letter to be eliminated. For example, in the 3d question, the least common multiple of 4 and 6, the coefficients of x, is 12, which may be produced by multiplying 4 by 3, or 6 by 2. If, therefore, the 1st equation be multiplied by 3 and the 2d by 2, the coefficients of x will be ftlike. 1. A shoemaker sold 3 pairs of shoes and 4 pairs of boots for $26 ; and, at the same rate, 5 pairs of shoes and 3 pairs of boots ♦or $25. What was the price of the shoes and the boots a pair \ XXIV WITH TWO UNKNOWN QUANTITIES. ^ 91 Let X = the price ofa pair of shoes, and y = the price of a pair of boots. Then^ (1) 32; + 4y=:26;) (2) 5 X + 3y =: 25. ) Transpose 4y in 1st and divide y 3 26 4 y (3) z = — -, Transpose 3 y in 2d and divide by 5, (4) x = ?^^. Now, since the second members of equations 3d and 4th aie each equal to a;, they are equal to each other. Hence, 25 — 3y 26 — 4y t», , . . ,. ^ j« — - = 5 — -. Multiply by 5 and 3, or 15, o o 75 — 9y = 130 — 20 y; transpose, reduce and divide, y = $5, price of a pair of boots. Substitute 5 for y in the 3d, 26—20 ^^ . ^ ... 1^ X = — = f 2, price of a pair of shoes. o 2 What fraction is that to the numerator of which if 4 be added, the value of the fraction will be ^; but if 7 be added to the denominator, the value will be ^ ? Let X z=z the numerator, and y = the denominator. X The required fraction then will be expressed by - Hence X ft (2) — p-- = i- \ Multiply the 1st by y, (3) x-}-4:=.^; transpose the 4, (4) a;=|— 4. Multiply the 2d by y + 7, y-L7 (5) X = — ^ — Put the two values of x equal, |-4 = ^±I; multiply by 10, 92 EQUATIONS OF THE FIRST DEGREE XXIV 5y — 40=z2y-\-i4; transpose, reduce and divide, y = 18, the denominator. Substitute 18 for y in the 4th, X = J/ — 4 1= 5, the numerator. The fraction sought then is -^^. In the solution of the last two questions, we found the valufl of X from each of the original equations, as if y were known ; that is, we found from each an expression for z, consisting of y g and known numbers ; then, by equalizing these two values of x, we obtained an equation without x. We might have eliminated y in a similar manner, and found an equation without that.lettei Hence, we have a SECOND METHOD OF ELIMINATION. Arl. 73. Mnd the value of one of the unknown quantities, from each of the equations y as if the other unknown quantity were determined ; then form a new equation hy putting these two val- ues equal to each other. Observe, however, that the unknown quantity itself must not be contained in any expression for its value. 1. Says A to B, give me one dollar of your money, and I shall have twice as much as you will have left ; yes, says B, but give me one dollar of your money, and I shall have three times as much as- you will have left. How much money has each? Let X =. A's money, and y = B's. Then after B has given A $1, A will have a;+ 1, and B will have y — 1 dollars. But if A gives B $1, A will have z — 1, andB,y + l. Hence, from the conditions of the question, (1) z+l = 2y-2;) (2) 3 z — 3 = y-\-l. ( The first equation gives (3) x = 2y — 3. Now this value of z may be put instead of z in the 2d eoua ton ; but is z in the 2d is multiplied by 3, we must mult ■ li XXIV WITH TWO UNKNOWN QUANTITIES. 95 this value by 3, and the result, 6y — 9, may be substituted 'of 3 z, in the 2d, which gives 6y — 9 — 3 = y-(~l> ^''o™ which we deduce y = $2|, B's money. Substitute 2f for y in the 3d, X = 5| — 3 = $2|, A*s money. 2. The mast of a vessel consists of two parts ; \ of the lower part, added to ^ of the upper part, makes 28 feet ; and 5 times the lower part, diminished by 6 times the upper part, is equal to 12 feet. Required the length of each part. Let X =z the lower, and y z= the upper part. Then, (1) 3- + f = 28;> 2) Si— 63/= 12. ) The 1st gives (3) y = 168 — 2x. Multiply this value of y by 6, and substitute the result for- 6 y m the 2d ; but, as 6y has the sign — , the value of 6y must be subtracted, that is, its signs must be changed when the substitu- tion is made. Hence, we have 5x — 1008 + 12 z = 12, which gives a: = 60 feet, the lower part. Substitute 60 for x in the 3d, y = 168 — 120 = 48 feet, the upper part. From the solution of the foregoing questions, we deduce a THIRD METHOD OF ELIMINATIOIf . Art. 74L. Findy from one of the equations^ the value of thf, quantity to he eliminated, as if the other unknown quantity were determined, and substitute this value in the other equation, in- stead of the unknown quantity itself 1. There are two numbers, such that if 15 times the 2d be added to the 1st, the sum will be 53; and if 3 times the 1st be added to the 2d, the sum will be 27. What are these num- bers? 94 EQUATIONS OP THE FIRST DEGREE XXIV 2. Two men talking of their money, the first says to the sec ond, ^ of mine and ^ of yours make $6 ; but | of mine and ^ of yours make only $5f . How much money has each ? 3. A farmer sells to one man 5 cows and 7 oxen for $370 and to another, at the same rate, 10 cows and 3 oxen for $355 Required the price of a cow and that of an ox. 4. Says A to B, give me $4 of your money, and I shall have as much as you ; yes, says B, but give me $4 of your money, and I shall have three times as much as you. How much money has each? 5. I can buy in the market 3 bushels of potatoes and 4 bush- els of corn for $5, and, at the same rate, 6 bushels of potatoes and 7 bushels of corn for $9. What is the price of a bushel of each? 6. A man bought some wheat at 8s. per bushel, and some rye at 5s. per bushel, to the amount of $20 ; he afterwards sold, at the same rate, ^ of his wheat and f of his rye for $9^. How many bushels of each did he buy, and how many of each did he sell? 7. A laborer wrought 8 days, having his son with him b days, and received for both $10; he afterwards wrought 10 days, hav- ing his son with him 9 days, and received $13. What were the daily wages of himself and son ? 8. What fraction is that, whose numerator being dovliled, and the denominator increased by 8, the value becomes § ; but the denominator being doubled, and the numerator increv^d by 2 the value becomes ^ ? 9. What fraction is that, whose numerator being diminished by 3, the value becomes f ; but the denominator bei.*g dimin- ished by 3, the value becomes ^? 10. A man bought coffee at 12 cents and tea at 'to cents a pound, and paid for the whole $249; the next day he disposed of I of his coffee and f of his tea for $180, which was $10«iO more •ban it cost him. How many pounds of each article did he buy and how much of each did he sell ? XXIV WITH TWO UNKNOWN QUANTITIES. 95 1 1. A market-woman bought eggs, some at two for a cent; and some at five for three cents, and gave for the whole f I '60 ; she afterwards sold them all for $3-10, and thereby gained ^ a cent on each egg. How many of each kind did she buy? 12. A grocer having two casks of wine, drew out 25 ga!?ong from the smaller and 30 from the larger, and found the number of gallons remaining in the former to the number remaining in the latter as 5 to 7; he then put 10 gallons of water into each cask, and found the number of gallons of the mixture in the smaller to the number of gallons in the larger as 3 to 4. How many gallons lid each cask at first contain ? 13. A man would sell 4 bushels of wheat and 9 bushels of oats for 63 shillings ; or he would exchange 4 bushels of wheat for 8 bushels of oats and 12 shillings in money. At what price did he estimate the wheat and oats per bushel ? 14. A grocer has two casks of wine, the larger at 12s. and the smaller at 10s. per gallon, and the whole is worth ,£136 ; from the larger he draws 60 gallons, and from the smaller 20 ; he then mixes the remainders together, adds 40 gallons of water to the mixture, and finds it worth 9s. per gallon. How many gallons were there at first in each cask ? 15. A sportsman has a fishing rod consisting- of two parts; twice the upper part exceeds the lower by 5 feet ; moreover, 3 times the lower part added to 4 times the upper part, exceeds twice the whole length of the rod by 55 feet. What is the length of each part 1 16. A owes $600, and B $800 ; but neither has sufficient to pay his debts. Says A to B, lend me ^ of your money, and I shall be enabled to discharge my debts ; yes, says B, but lend me i of your money, and I can discharge mine. How much money has each in possession ? 96 'EUUATIONS OP THE FIRST DEGREE XXV SECTION XXV. CQUATIOirs I F THE FIRST DEGREE WITH SEVERAL UM&KOWN Q.VAN TITIES. Art. 75. 1. A drover bought at one time an ox, a cow and a calf for $65 ; at another, and at the same rate, two oxen, three cows and a calf for $145; and, at a third time, three oxen, two cows and a calf for $165. Required the price of an ox, a cow, ind a calf. Let X =. the price of an ox, y = the price of a cow, and z = the price of a calf. Then, (1) x+ tj + x= 650 (2) 2x + 3y + z=145;> (3) Sx + 2y-\-z=l65. ) Here we have three distinct equations, containing three dif- ferent unknown quantities; and the first step in the solution, is, to deduce from them two equations containing only two different unknown quantities. Let us eliminate z, that is, obtain two equations without z. - First method. The coefficients of z being alike in the three equations, by subtracting the 1st from the 2d, also the 2d from the 3d, we have (4) x + 2y = 80;) (5) X— y = 20. i Since equations 4th and 5th do not contain z, we may obtain om them the values of x and y. Subtract the 5th from the 4th, 3y = 60 ; hence, y = $20, price of a cow. Substitute 20 for y in the 5th, X — 20 =: 20 ; hence,' x :=l $40, price of an ox. Substitute the valu3s of a? and y in the 1st, 40 -f 20 + 2 = 65 ; hence, z = $5, price of a calf. XXV WITH SEVERAL UNKNOWN QUANTITIES 97 ■ o = 165. ) I Second method. Resume the original equations, (1) ^+ y + 2^ = (2) 2x + 3y + z: (3) Sx + 2y + z Deduce the value of z from each equation, as if x and y were known, (4) z=z 65 — X — yO (5) 2=145— 22 — 3y;> (6) zz=165 — 3x — 2y. ) Put equal to each other the values of z in the 4th and 5th i also, the values of z in the 5th and 6th, (7) 65 — X — yi=145— .2x — 3y; (8) 145— 2x — 3y = 165 — 3x — 2y. Transpose and reduce in the 7th and 8th, (9) x + 2y = 80;) (10) X — y = 20. / Find the value of x in the 9th, also in the 10th, (11) x = 80-25^;) (12) x = 20+ 7/. } Put the values of x in the 11th and 12th equal, 20 + 3/ = 80 — 2 y ; hence, y = $20, price of a cow. Substitute 20 for t/ in the 12th, X = 20 + 20 = $40, price of an ox. Substitute 40 for x and 20 for y in the 4th, 2 = 65 — 40 — 20 = $5, price of a calf. Third method. Take the original equations, (1) ^+ I/ + ^= 65 (2) 2x (3) 3x Deduce the value of 2 from the 1st, as if x and y were known (4) 2 = 65 — X — y. Substitute this value of z in the 2d and 3d, (5) 2x-i-3y-f 65 — X — y=145; (6) 3x + 2y + 65 — X — y=165. 9 + y + 2= 650 + 3y + 2=]45;^ + 2y + 2=165. ) 98 EQUATIONS OF THE FIRST DEGREE XXV Transpose and reduce in the 5th and 6th, (7) x + 2y=: 80;) (8) 2x + i/=100. } Deduce the value of x from the 7th, (9) x=:80 — 2y. Double this value of z, and substitute the result in the 8th, 160 — 4y+y=100; which gives y = $20, price of a cow Substitute 20 for y in the 9th, and we have X = $40, price of an ox. Substitute 40 for x and 20 for y in the 4th, and we have z =z $5, price of a calf. 2. There are three men. A, B and C, whose ages are such, that iff of A's, ^ of B's and f of C's be added, the sum will be 70 years ; if twice A's be added to B's, the sum will be 5 times C's ; and if ^ of A's be subtracted from ^ of B's, the remainder will be ^ of C's. Required their ages. Let ar, y and z reptesent the respective ages of A, B and C. Then, (1) ^^ + i!/ + i^ = '70;) (2) 2x+ y = 5z; > (3) i!/-ix = iz. ) Remove the denominators in the 1st and 3d, and bring down the 2d, (4) 16x + 21y + 18«=1680 (2) ^x + y = 5z; (5) — 2x + 3yr=3«. JFHrst method. Transpose all the unknown quantities into the first members, and bring down the 4th, (4) 16x + 21y + 18z=1680; (6) 2a: + y — 52 = 0; (7) ~-2z + 3y-^32=:0. To eliminate x, add the 6th and 7th ; also, add the 4th to 8 imies the 7th, (8) 4y — 8z = 0; ) (9) 45y — 6z=1680. ) 1 XXV. WITH SEVERAL UNKNOWN QUANTITIES. 99 To eliminate z from these two last equations, divide the 8th by 4 and the 9th by 3, (to) y-2z = 0; ) (11) 15y — 2z = 560. i Subtract the 10th from the 11th, 14 y = 560 ; hence, y = 40 years, B's age. Substitute 40 for y in the 10th, 40 — 2 z z= ; hence, z = 20 years, C's age. Substitute 40 for y and 20 for z in the 2d, 2 2 -|- 40 = 100 ; from which, x = 30 years, A's age. Second method. Take the original equations eleared of frac- tions, that is, the 4th, 2d and 5th, inverting the order of the members in the 2d and 5th, (4) 16a: + 21y+l8z=1680;^ (2) 52 = 2a: + y; \ (5) 32 = — 22: + 3y. ) Deduce the value of z froin each of these, (7) . = '-^; (8) z = ^I^^^. Put equal to each other the values of z in the 6ih and 8th • also, the values of z in the 7th and 8th, — 2a; + 3y _ 1680— IGz — 21y ^^^ 3 ■- 18 ' (10) 2xj^^zz?i_tl3i. Clear the 9th and 10th of fractions, transpose, lednee, and find »he value of x from each, (U) ,^lC80-39y. (,2) x = «J! I*K) EQUATIONS OF THE FIRST DEGREE XXV Put the values of x in the 11th and 12th equal, 3y 1680 — 39 V , =1 ^ J hence, y = 40 years, B s age. Substitute 40 for y in the 12th, a; := 30 years, A's age. Substitute 30 for x and 40 for y in the 7th, 60 + 40 100 ^^ z = ^ = — - = 20 years, C's age. Third method. Resume the 4th, 2d and 5th" equations, (4) 16 a; + 21 y+ 18 2 = 1680 ; ^ (2) 5z = 2x + y; [ (5) 32 = — 22: + 3y. ) Deduce the value of z from the 5th, (6) . = ^^1^. ■ , Substitute this value in the 4th and .2d, (7) 16x + 21y— 12 2+18^ = 1680,0 (8) -'«- + '^^ = 2.+y. ■ \ Clear the 8th of fractions, and reduce the 7th and 8th, (9) 42 + 39y = 1680; ) (10) 42 = 3y. S Deduce the value of x from the 10th, ^ (U) . = % Substitute this value in the 9th, 3 y + 39 y = 1680 ; hence, y zr: 40 years, B*6 a^je. Substitute 40 fory in the 11th, 2 = f of 40 = 30 years, A's age. Substitute 30 for x and 40 for y in the 6th, — 60 + 120 „„ ^^ z = ^ ='A|^ = 20 years, C s age. XXV. WITH SEVERAL UNKNOWN QlTANTITIES. lOl 3. A boy bought of one man 3 apples, 2 peaches, 4 pears and 2 oranges for 22 cents ; of a second, at the same rate, 2 apples, 3 peaches, 2 pears and 4 oranges for 24 cents ; of a third, 5 ap- ples, 1 peach, 6 pears and 10 oranges for 36 cents; and of a fourth, 4 apples, 3 peaches,. 2 pears and 8 oranges for 32 cents. What did he give apiece for each 1 Let X represent the price of an apple, t/ that of a peachy z thit of a pear, and u that of an orange. Then, (1) Sx + 2y + Az+ 2m = 22 (2) 2x + Si/ + 2z+ 4w = 24 (3) 5x+ !/ + Gz+10u=:S6: (4) 42; + 3y + 22+ 8m = 32. Let us eliminate y, that is, obtain three equations without y. The three equations below may be found as follows. To obtain the 5th, multiply the 3d by 2, and subtract the 1st from the re- sult, To obtain the 6th, subtract the 2d from the 4th. To ob- tain the 7th, multiply the 3d by 3, and subtract the 4th from the result. (5) 7x + 8z+l8u = 50: (6) 2x + 4 (7) llx+16z + : Now let us eliminate z from the last three equations. But since the 6th does not contain z, we may divide it by 2 and place the result below. To obtain the 9th, multiply the 5th by 2 and subtract the 7th from the result. (8) x + 2u = ^; ) (9) Sx + 14u = 2i.i To eliminate x from the last two equations, multiply the fiAh l>y 3 and subtract the result from the 9th. . 8 z< = 12 ; hence, u=^l^ cent, price of an orange. Substitute the known value of w in the 8th, and we have z 1= 1 cent, price of an apple. Substi ,ute the values of x and u in the 5th, and we have z zz: 2 cents, price of a pear. 9* z+18w=:50; ^ u = 8; [ 16z-f-22w = 76. ) 102 EQUATIONS OF THE FIRST DEGREE XXV Substitute the values of x, z and m, in the 3d, and we have y = 4 cents, price of a peach. Let the learner solve this question by the second and third methods of elimination. From the solution of the foregoing questions, it is easy to see^ that the three modes of elimination given in the last section, may be extended to any number of equations. To apply the Jirst method for the purpose of eliminating a par- ticular quantity from several equations, it is only necessary to operate upon these equations taken two and two. In applying the second method to several equations, we must find, from each equation that contains it, the value of the un- known quantity to be eliminated, and then put any two of these expressions for its value equal to each other. To extend the third method, we must, after having found from one of the equations the value of the unknown quantity to be eliminated, substitute this value in every other equation that con- tains this unknown quantity. If a question involves five unknown quantities, and gives rise to five different equations, we should first deduce from them four equations with only four different unknown quantities ; secondly, from these we should deduce three equations with only three un- known quantities ; thirdly, from these three, two with only two unknown quantities ; and, finally, from these two, one equation with only one unknown quantity, from which the value of this unknown quantity might be determined. Or if six equations containing six unknown quantities, were given, we should first obtain from them five containing only five unknown quantities, and then proceed as before ; and so on, if a fitill greater number of equations were given. If either of the equations does not contain the unknown quan- tity to be eliminated, that equation may be put aside to be placed with the next set of equations, viz : those which contain one lesi unknowi quantity. XXV. WITH SEVERAL UNKNOWN QUANTITIES. 103 Either of the methods of elimination may be used, but the first will generally be found most convenient, because it does not give rise to fractions. It will, however, be useful for the learner to perform every example by each method, in order to familiar- ize him with the process, and enable him to judge which will be best in any particular case. It is not necessary that the same mode of elimination should be pursued throughout the solution of a question, but either of them may be resorted to whenever it shall seem the most convenient. 4. A merchant bought at one time 4 barrels of flour, 3 barrels of rice, and 2 boxes of sugar for $72 ; at another, 2 barrels of flour, 5 barrels of rice, and 3 boxes of sugar for $84 ; and at a third time, 5 barrels of flour, 9 barrels of rice, and 8 boxes of Bugar for $187. What were the flour and rice per barrel, and what was the sugar per box 1 5. There are three numbers, such that if 3 times the 2d be subtracted from 4 times the 1st, and twice the 3d be added to the remainder, the result will be 9 ; if twice the 1st and 5 times the 2d be added, and from the sum 3 times the 3d be subtracted, the remainder will be 4 ; and if 5 times the 1st and 6 times the 2d be added, and from the sum twice the 3d be subtracted, the remainder will be 18. What are these numbers? 6. Three boys. A, B and C, counting their money, it was found that twice A's added to B's and C's, would make $525 ; that if A's and twice B's were added, and from the sum C's were subtracted, the result would be $3'00 ; and the three together had $325. How much money had each 1 7. Three men owed together a debt of $1000, but neither of them had sufficient money to pay the whole alone. The first could pay the whole, if the second and third would give him ^^ of what they had ; the second could pay it, if the first and third would give him ^^ of what they had ; and the third could pay it, if the first and second w^ould give him ^^ of what they had Hovv much money had each 1 104 EQUATIONS OF THE FIRST DEGREE. XXV 8. The ages of three men, A, B and C, are such, th^t ^ of A's, I of B's and ^ of C's make SO years ; ^ of A's, ^ of B's and -| of G's make 78 years ; and ^ of A's, ^ of B's and ^ of C's make 35 years. Required the age of each. 9. Four men could earn together in one day 26 shillings. If the 1st wrought 6 days, the 2d 8, the 3d 9, and the 4th 12, they would all earn 237s. ; if the 1st wrought 2 days, the 2d 5, the 3d 7, and the 4th 9, they would earn 161s. ; and if the 1st wrought 4 days, the 2d 3, the 3d 2, and the 4th 1, they would earn 60s What were the daily wages of each ? 10. A merchant had four kinds of tea, marked A, B, C and D. If he mixed 7 pounds of A, 10 of B, 12 of C and 18 of D, the whole mixture would be worth $22*90 ; if he mixed 4 pounds of A, 5 of B, 8 of C and 11 of D, the whole would be worth ^1380; if he mixed 4 pounds of A and 9 of C, the mixture would be worth $5*70 ; and if he mixed 18 pounds of A, 12 of B and 36 of D, the mixture would be worth $31 80. What was each kind worth a pound ? 11. I find that I can buy in the market 1 bushel of wheat, 2 bushels of rye, 3 of barley, 4 of oats and 6 of potatoes for $12 ; 3 bushels of wheat, 4 of rye, 8 of barley, 3 of oats, and 4 of po- tatoes for $24^ ; 5 bushels of wheat, 2 of rye, 10 of barley, 6 of oats and 8 of potatoes for $32 ; and 8 bushels of wheat, 7 of rye, 6 of barley, 5 of oats and 4 of potatoes for $35^ ; moreover, that a bushel of wheat and a bushel of oats cost as much as a bushel of rye and a bushel of barley. Required the price of a bushel of each. KXVI. SUBSTITUTION OF ALGEBRAIC QUANTITISS. 105 SECTION XXVI. NUMERICAL SUBSTITUTION OF ALGEBRAIC QUANTITIES Art. 76. Find the numerical value of the following quanti- 163, when a=z5, 6 = 6, c = 7, and d=. 10. 18. a + bcd. 19. 2ab + Sd. 20. a^-^-b + c + d. 21. ab^+cd\ 22. a + b + d^. 23. a-^b — c. 24. a + b — c — d. 25. a^—b — c + (^. 26. (a-^b + c)d. 27. lb + c + d)a. 28. ab(c-{-d). 29. (a + *)(c + ^)- 30. {a + b)(d-'c), 31. (a + 6)(c — £/). 32. (a — fe)(c — /). 33. (62_a2)(c + j). 34. (a + 6)(c2 + rf2). 35. (62_a2)(rf2__c2) 36. («2_52)(c2__^^. 37. (a+bf. 38. (a-|-6)2c(/. 17. abc + d. Find the value of the following expressions, when a = ^ t := 2, w = 4, and w = 6 1. ab, Ans. 6. 5: = 30. 2 a2 6c. Ans.52 .6 .7 = 1050. 3. abed. 4. a^bcd. 5. ab^cd. 6. a^b\ 7. acd^. 8. ab 9. abc d • 10. e^ a U. a^b^ c^d' 12. a bd' 13. a + b-\-c. 14. «6 + c. 15. a + bc. _ 16. ab-\-cd. 39. a-\-b — n + wi 42. 6a- -5abn. 40. 3a + 26 + n — »L 43. 3m- -1 abn. 41. ab-\-mn. 106 SUBSTITUTION OF ALGEBRAIC QUANTITIES. XXVI 44. a + b m — n 45. -6 + a n — m AR bm 47. 48. a — 4« m^ — 4 « 6 » a_27 63 49. {a + b){m — n) ^ ~^' G-jr-ma ' (fii — a)2 Substitute numbers in the following equivalent expressions, showing their identity whatever numbers are put instead of the letters, observing however to give the same value to the same letter in the two members of an equation. 50. {a-\-b) c = ac-^b c. Suppose a=z2, 6 = 3, and c = 6. Then, (a + 6) c = (2 + 3) 6 = 5 . 6 = 30. Also, ac + 6 c = 2. 6 + 3. 6 = 12+18 = 30. 51. (a+6)(a — 6) = a2_62. 52. (« + *)^ = «^ ^2 aft + 62. 53. (a — m)2 = aa_2am + m2. 54. a + - = ! — . c c 55. m — 1 ' • 56. r— - = m3 — m^-\-m — 1. m + 1 ' 57. J: =a^^ab + b^ 58. ^^f"f =a2 — 2« + 4. a + 2 ' 59. (« + 6)(a + c) = a2 + a(6 + c)+6c. 60. (a + 6) (a + c) (a + rf) = a3 _(. ^2 (J ^ c + (^ 4 9(bc-\-bd-]-cd)-{-bcd. 61 6 , « — 6_ «2 "^- a + 6"»~ 6 — a6 + 62- • fl + 6 fl-6_ 2(«2 + 62) '^' a — 6'T"a + 6- a2__52 • XXVII GENERALIZATION. 107 ^--2a+l __ a-~_I 64 « + 6 « — & 4fl6 _ a_6 a + 6 a2 — 62~" • SECTION XXVII. ^kAxralizatiox. Art. 77. In the problems of the first six, as also in those of the 24th and 25th sections, letters have been used to represent unknown quantities only, and the results, expressed in definite numbers, correspond to the particular questions only, from which they were derived. But in pure algebra, letters may also represent known quan tities, or they may be used indefinitely, and afterwards any num- bers may be substituted in their place. Also the results of pure algebra, which are called ybrmw/te, show by what operations they were obtained, and furnish rules for the solution of all questions of the same kind. 1. Two men, A and B, are to share $420, of which B is to have 3 times as much as A. Required the share of each. Here the object is to separate $420 into two parts, such that one shall be 3 times as great as the other. Let X =z A's share ; then, 3 X =z B's share. Hence, a; -|- 3 x = 420 ; x= $105, A's share; 3 X = $315, B's share. Instead of 420, in this question, put the letter a ; then the problem will be, to separate the number a into two parts, one of which shal be 3 times the other. 108 GENERALIZATION. XX V^IJ Representing the shares as before, we have x-\-Sxzz^a. Hence, z = 2, A's share, J o / General ^ormulae. Sx=z—, B's share. J If we now put $420 instead of a in these formulae, we have 420 x = — = $105, A's share, ^ „ .jj„ • > Particular answers. 3x = -^- — = $315, B's share. ) We perceive, from the general formulae, that one part is a fourth, and the other three-fourths, of the number to be divided, without regard to the particular value of that number. Let the learner put other numbers instead of a in the formulae, and find the two parts. Any number divisible by 4 will give whole numbers for these parts. 2. A father left by his will $4500 to be divided between his son and daughter, with the condition that the son w«s to receive $500 more than the daughter. What was the share of each ? In this problem it is required to separate $4500 into two parts, such that one shall exceed the other by $500. Instead of $4500, let us suppose that the number to be separated into parts is indefinite, and that it is represented by a ; also, that b repre- sents the excess of the greater part above the less. Then the problem is, to separate the number a into two parts, such that the greater shall exceed the less by 6. Let X = the less part ; then, x-\-bz=. the greater part. Hence, x-^x-\-b=:a. Reduce and transpose 6, 2x=za — b; divide by 2, a b a — b , . xz=^ — -= -^, the less part XXVII. GENERALIZATION. 1(W Tc obtain the greater we add h to the less, and we have CL h z-\'h=i- — - 4- 6. Change h to halves, x + h=z- — --{- — ', reduce, • t a , h a-\-h _ 1 + 6=- + -== — g— , the greater part. If we examine the formulae for the two parts, and recollect that a and b may stand for any numbers, provided that b is less than fl, we see that they furnish the following rule for separating a quantity into two parts. The less part is found by subtracting half the excess of the greater above the less from half the number to be separated ; or, by subtracting the excess of the greater above the less from the number to be separated^ and dividing the remainder by 2. The greater part is found by adding half the excess of the greater above the less to half the number to be separated; or, by adding the excess of the greater above the less to the number to be separated, and dividing the sum by 2. Let the learner separate each of the following numbers into I wo parts by means of the formulae, or by following the rule. Numbers to be separated. Excess of one part over the other 3. 150 30. 4. 230 . 60. 5. 1200 120. 6. 27 ...'..... 5. 7. 35 ........ . 3. 8. 70 3. 9 47^ 13. 10. 99 33^. 11. Separate a number a into three parts, such that the mean shall exceed the least by b, and the greatest shall exceed the mean by c. 10 110 GENERALIZATION. XXVIL Let X = the least part ; then x-{-b=. the mean part ; and x-\-b -\-c=: the greatest part. Hence, x-\-x-\-b-\-x-\-b-\-c = a. Reducing, ^x-\-2b-\-c=^a; transposing, Sx==.a — 2 6 — c; dividing, a 2b c a — 26 — c , , ajr=- ^— _ = , the least part. Adding b to the least, we have ,, a 2b c , ^ a 26 c,3i a . h c a-\-b — c . = ^ + ^ — -^ = — '— , the mean part. Aiding c to the mean, we have ,,, a . b c, a, 6 c,3c x + 6 + c=- + ---+.= -+-_-+^ a , 6 , 2c a + 6 + 2c ^ = 3 + 3-+ 3- = — '-3-' , the greatest part. Let the learner translate these formulae into rules, recollecting that a represents the number to be separated, 6 the excess of the mean above the least, and c the excess of the greatest above the mean. 12. A man bought sugar at a cents, flour at 6 cents, and co^ fee at c cents per pound, and the whole amounted to d cents. How many pounds of each did he buy, if he bought the sam« quantity of each 1 Let X =z the number of pounds each. Then, ax-\-bx-\-cx = d. Separating the 1st member into factors, one of which is x, (a -{- b -\- c) X =z d. Dividing by the coefficient of x, x = — , - . — , the number of pounds of each. a-\-b-\-c This formula may be translated into the following rule, viz : Divide the price of the whole by the sum of the prices of a pouna gf each sort ; the quotient will be the number of pounds of each XXVII. GENERALIZATION. 11 If in the formula we substitute the numbers 7, 6, 10 and 92 for a, 6, c, and d respectively, we have 92 X z=z ^ , ^. , ,^ rr: f 4 = 4 Ibs., particular answer. 7 + G+lO ^^ *^ 13. A farmer found he had a times as many cows as oxen, and b times as many sheep as cows, and that h.s whole stock amounted to the number c. Required the number of each. Let X ==. the number of oxen ; then ax=i the number of cows ; and abx=z the number of sheep. Hence, x-\-ax-\-abx=ic. Here x is taken 1 -f- a -f" ^ ^ times ; therefore, {l-{'a-\-ab)xz=c. Dividing by the coefficient of x, X = -—. • r, number of oxen. 1 ^a-f-ao /• ax= —-. , r X a, number of cows. 1 + a + ao abx=z -— ; X ciby number of sheep l + « + «o If 3 be put for a, 4 for 6, and 128 for c, in these formulae, we have 128 the number of oxen = = ^^ =. 8 ; 1 — j— o~[~ 1«* the number of cows = 3 . 8 = 24 ; the number of sheep = 4 . 3 . 8 = 96. 14. What will be the particular answers in the preceding ques lion, ifa=:5, 6 = 7, andc = 369? 15. Two men had engaged to perform a certain piece of work ; the first could do it alone in a days, and the second in b days. How long would it take both working together to do it ? Let X =z the number of days in which both would do it. Then, as the first could do the whole in a days, in 1 day he H'ould lo - of it ; and, as the second could do the whole in i a 112 GENERALIZATION XXVD days, in 1 day he would do - of it; so that both would per 11 X z form — I- - of it in 1 day, and in x days, - 4- t- But in i a ^ b -" ^ ' a * b days, we have supposed that they would perform the vvhol« Hence, 7 7 — |- T = 1> piece of work. Multiplying by a ; and 6, bx-{-ax=zab; or. ab T rrr . A n dividing by b + a, . b + a -—' From this formula we derive the following rule for any similar case, in which two workmen are employed. Divide the product of the numbers expressing the times in which each would perform it^ by the sum of those numbers. Let the learner find the answers to the following questions, by means of the preceding formula. 16. If A could perform a piece of work in 6 days, and B could perform the same in 5 days, how long would it take both together to perform it ? 17. By a pipe A, a certain cistern will be filled with water in 5^ hours, and by another pipe B, it will be filled in 8^ hours; in what time will it be filled, if water flow through both pipes at the same time 1 18. Let it be proposed to find a rule for dividing the gam or loss in partnership, or, as it is commonly called, the rule of feU Icwship. First, take a particular case. Three men tr.ided in company and put in stock in the follow- ing proportions, viz ; A put in $5 as often as B put in $3 and as ofteL as C put in $2. The company gained $650. Required (►ach man's share of the gain. XXVII. GENERALIZATION. 113 Let X =z A's share. Then, since B furnished f as much stock as A, he must have f as much gain ; therefore, — - = B's share. In J ike manner, o 2 X -— = C's share. Hence, o x-\----\-~=z 650. Multiplying by 5, DO 52_[-3a;-|-2x = 3250, or 10x = 3250; X = $325, A's share. -? = $195, B's share. o ^ = $130, C's share. 5 'i*o generalize this question, suppose that A put in m as often as B put in n and as often as C put in p dollars, and that the? pained a dollars. Then B puts in — , and C — as much as A. Let X =r A's gain ; then, — = B's gain, and ^— = C's gain. Hence, X-] f- — = a. Multiplying by m, mx-\-nx-^px=:7na; or, {m-\-n -\-p) x=ima; dividing by the coefficient of 2, ma « .. , -, or ni X — i J — , A's share. m-\-n -\-p m-\-n-\- p B's share is - of A's ; — of wi X — 7 ; — is — : ; — m m m-\-n-\-p m-f-n -}- p n ind — of it is n times as much ; therefore, m nx a T», 1 — = 71 X — f i — , B s share. m m-\-n-\-p , 10* 1 14 GENERALIZATION. XXVIl In like manner, C's share is — of A's : or m - — z=zp X -^—, ; — , C s share. m m-f-n-\-p By examining these formulae, we perceive that the whole gair a is divided by m-\-n -{-p, the sum of the proportions of th€ stock furnished by all the partners, and" that this quotient is multiplied by m, A's proportion, by n, B's proportion, and by ^, C's proportion of the stock, to obtain their respective shares of the gain. Hence, observing that a may represent the loss as well as the gain, to find each partner's share of gain or loss, we have the following rule. Divide the whole gain or loss by the sum of the proportions of the stocky and multiply the quotient by each partner's proportion. This rule is applicable, whatever be the number of partners. 19. Suppose A put in $400, B $300, and C $200, and that they gained $450. By the preceding formulae, or by the rule, what was each partner's share of the gain ? Remark. When the sums actually put in are given, the sim- plest proportions of the stocks will be found by dividing these Bums by the greatest number that will divide them all without any remainder. Thus, 400, 300, and 200 are all divisible by 100 ; and the quotients, 4, 3, and 2, express the proportions of the stock. 20. What would be each man's loss, if A furnished $300, B $150, and C $100, the entire loss being $99? 21. What would be each man's share of $500 gained, if foui partners furnished respectively $800, $600, $400, and $200? 22. A put in $200 for 6 months, B $150 for 5 months, and C $300 for 2 months. They gained $27^ ; what was each man's share of this gain ? Remark. It is evident, that, if the stocks are employed une- qual times, each partner's stock, or his proportion of the stock, must be multiplied by the number expressing the time during whish it is in trade, and that then the proportions of these pro Impels tnusit be used. XXVII. GENERALIZATION. 115 In general, known quantities are represented by the first, and unknown quantities by the last letters of the alphabet. But, in some cases, it is more convenient to use the initial letters of the names of quantities, whether known or unknown. In the following questions relating to simple interest, let p rep- resent the principal, r the rate, t the time, i the interest, a the amount, and d the discount. In these questions, r is supposed to be a fraction, as '06, 05, &c., according as the rate is 6 per cent., 5 per cent., &/C., and the time is supposed to be expressed in years and fractions of a year. 23. What is the simple interest of p dollars, for t years, at r per cent. ? The principal multiplied by the rate gives the interest for one vear; hence, rp^=. the interest for 1 year ; and trpz=. the interest for t years. Therefore, i=ztrp. This formula gives the following rule. To Jind the interest when the principal^ ratCy and time art known, multiply together the principal, time, and rate. 24. The principal being $256*25, the time 4^ years, and the rate 6 per cent., what will be the interest 1 In the equation trp=zi, provided any three of the quanti- ties are given, the other may be found. Let the learner find the formulae and make rules for the following general questions, and solve by the rules the particular examples subjoined. 25 The interest, time, and rate being given, to find the prin- cipal. 26. If the interest, for 7 years at 5 per cent., is $26*25, what is the principal ? 27. The interest, time, and principal given, to find the rate. 28. The interest being $74*4711, time 6 years, and the prin- cipal $225-67, what is the rate 1 29. The interest, rate, and principal given, to find the time. 30. If the interest is $102, rate 4^ per cent., and the principal $320, what is the time ? 116 GENERALIZATION XXVII 31. What is the amount of p dollars, for t years, at the rate r, simple interest \ The amount being the sum of the principal and interest, we have a z=p -\-trp; or, a =p (1 + < r). This formula gives the following rule. To Jind the ainount^ toJien the principal^ timcy and rate art J'tiowTiy multiply the time and rate together ^ add 1 to the product, and multiply this sum by the principal. 32. The principal being $650, rate 4^ per cent., and the time 7 years and 3 months, what is the amount by the preceding rule? The equation, p-^-f rp=::a, contains four different quanti- ties, any three of which being known, the other may be deter- mined. Let the learner find formulae and rules for the following gen- eral questions, and solve the particular examples by the rules. 33. The amount, time, and rate being given, to find the prin- cipal ; that is, to find what sum of money put at interest, at a given rate, and in a specified time, will amount to a given sum. N. B. The principal, in this case, is sometimes called the present worth of the amount. 34. What is the present worth of $300, due in 3 years and 4 months, the rate being 6 per cent. ? 35. The amount, principal, and time given, to find the rate. 36. The amount being $405 09, principal $321 50, the time 4 years, what is the rate? 37. The amount, principal, and rate given, to find the time. 38. Amount $352, principal $275, and rate 8 per cent., re- quired the time. 39. The amount, time, and rate given, to find the discount. Remark. The formula for the discount may be found by sub* trading the formula for the present worth from the amcmt a^ and simplifying the result XXVII GENERALIZATION. 117 40 Required the discount on <£100, due in 3 months, tl e rata being 5 per cent. 41. At a given rate, in what time will a sum be doubled 1 In what time will it we tripled ? Remark, Take the formula for the amount, put 2p and 3^ successively instead of a, and then find the value of t, 42. In what time will a sum be doubled at 6 per cent. ? In what time will it be tripled ? 43. In what time will a sum be doubled at 5 per cent. ? In what time will it be tripled? 44. Separate the number a into two parts, one of which shall be n times the other. 45. Separate a into two parts, so that the second may be the - part of the first. n '^ 46. Separate a into three parts, such that the second shall be m times, and the third n times the first. 47. Separate a into two parts, so that if one of them be di- vided hy 6, and the other by c, the sum of the quotients shall be d. 48. Separate a into two parts, so that the mth part of one shall exceed the nth part of the other by b, 49. What number is that whose mth part exceeds its nth part hy pi 50. After paying away — and - of my money, I had a guin eaa left. How many guineas had I at first? 51. After paying away the — and the - parts of my money, I had a dollars left. How much money had I at first? 52. A and B together could do a piece of work in a days ; B could do it alone in b days ; in how many days could A do it alone ? 53 A company at a tavern paid a shillings each ; but if there had been b persons less, each would have had to pay c shillings. How many persons in the company ? 1 18 GENERALIZATION. XXV II 54 A gentleman has 6 sons, each of whom is a years older than his next younger brother, and the eldest is b times as old aa ihe youngest. Required their ages. 55. A person borrowed as much money as he had in his purse, and then spent a shillings ; again he borrowed as much as he had left in his purse, after which he spent a shillings ; he bor- rowed and spent, in the same manner, a third and fourth time ; after the fourth expenditure he had nothing left. How much money had he at first? 56. A man agreed to work n days, with this condition, that he should receive a shillings for every day he worked, but should forfeit h shillings for every day he was idle. At the end of the time agreed on, he received a balance of c shillings. How many days did he work, and how many was he idle ? 57. A gentleman gave some beggars a cents apiece and had h cents left ; but if he had given them c cents apiece, he would have been obliged to borrow d cents for that purpose. How many beggars were there? The following questions may be solved by means of tfwo un- known quantities. 58. Said A to B, the sum of our ages is a years, and their dif- ference is b years. Required their ages, A being the elder. 59. One pair of boots and a pairs of shoes cost b dollars; and c pairs of boots and one pair of shoes cost d dollars. Required the price of the boots and shoes a pair. 60. There are two numbers, such that if y part of the second be added to the first, the sum will be a ; and if - part of the firs^ c be added to the second, the sum will also be a. Required the wo numbers. 61. What fraction is that, to the numerator of which if a be added, the value of the fraction will become — ; but if a be ad n n ded to the denominator, the value of the fraction will be — T 9 XXVIII. NEGATIVE QUANTITIES, ETC. ^ 119 62. What fraction is that, from the numerator of whith if a be Bubtracted, the value of the fraction will become — ; but if a be subtracted from the denominator, the value of the fraction will P n oecorae - s 9 63. What will be the particular answer to the 61st, if a=r 2 — z= f , and ~ = -^; and what will be the particular answer lo the62d,ifa = 3, -=T^, and ^ = Wl SECTION XXVlll. NEGATIVE QUANTITIES AND THE INTERPRETATION OF NEOATIVB RESULTS. Art. T8. It may happen, in consequence of some absurdity or inconsistency in the conditions of a problem, that we obtain, for a result or answer to the question, a quantity affected with the sign — . Such a result is called a negative solution. Negative results not only indicate some absurdity or incon^is ten,cy in the conditions of a question, but also teach us how to modify the question, so as to free it from all inconsistency. As such negative quantities frequently occur, we shall proceed to show, that, when isolated or standing alone, they are subject to the same rules as when connected with other quantities. We remark, in the first place, that negative quantities are de- rived from attempting to subtract a greater quantity from a les? The greatest quantity that can be taken from another, is that quantity itself. Thus, 7 is the greatest number that can be sub- tracted from 7, and a is the greatest number that can be sul)- tractei' from a. In such a case the remainder is zero; thu3^ 7 — 7 = 0, anda — = 0. i20 NEGATIVE QUANTtTlES AND THE XXVlll If it were required to subtract 9 from 7, we represent it thus, 7 — 9, or 1 — 7 — 2; this being reduced becomes — 2. Tha sign — before the 2, shows that there were 2 out of the 9 units, which could not be actually subtracted. If 7 be subtracted from 9, the remainder is the same, except that it has the sign -}-• In like manner, if we subtract b from a, b being the greater, the remainder, a — 6, will be negative; but if we subtract a from 6, the remainder, b — a, will be the same as before, except thai it will be positive. Art. 79. Suppose it were required to add b — c to a. It is evident, that we are to add to a the quantity 6, and subtract from the sum the quantity c, and the result is a-\-b — c. Now, as the reasoning does not depend at all upon the value of 6, the method of proceeding must be the same when 6 = 0, which reduces the expression b — c to — c or — c, and a-\-b — c becomes a-f-O — c or a — c; that is, — c added to a gives a — c, which accords with the rule already given for ad- ding polynomials. Hence, Adding a negative quantity — c, is equivalent to subtracting an equal positive quantity -\- c. Art. 80. Since b — 6 = 0, a-\-b — b is of the same value as a, and may be regarded as the quantity a under a different form. Now, in order to subtract + b from a, it is sufficient to strike it out from the eixpression a-\-b — 6, and we have a — b\ or if we would subtract — 6, strike that out, and we have a-\-b; that is, -\-b subtracted from a gives a — by and — b subtracted from a gives a-\-b, which accords with the rule already given for subtracting polynomials. Hence, Subtracting a negative quantity — 6, 25 equivalent to adding an equal positive quantity -\- b. Art. 81. With regard "to multiplication, in Art. 30, we have alieady seen that the product of a — A by c — d is ac — be — ad-^-bd. XX VIII. INTERPRETATION OF NEGATIVE RESULTS. 12i Now it is manifest that the sign of any term in this product IS entirely iiidependent of the absolute value of the letters, a, 6^ c, and d. Let us suppose then, in the first place, that a and d are each equal to zero. Upon this supposition, the quantities to be mul- tiplied together become — b and c — 0, or — h and c; and the product becomes O.c — he — 0.0-|-6.0, or — be. Hence, *- h multiplied by -}- <^> produces — he. Secondly, suppose 6 and c each equal to zero. Then the quan tities to be multiplied together become a and — d\ and the pro- duct, ac — be — ad-\-bd^ is reduced to — ad. Hence, -{- a multiplied by — d, produces — ad. Lastly, let the value of each of the letters a and c be zero> We then have to multiply — b by — d\ and the product, ac — he — ad-\-hd, is reduced to '■\- h d. Hence, — 6 multiplied by — d, produces -\-hd. From these several results we deduce the same rule for the signs in multiplication, as that given in Art. 31. Art. SS. Since in division, the product of the divisor and quotient must reproduce the dividend, it follows from the prece- ding demonstration, that the rule for the signs in the division of isolated quantities, is the same as that given for polynomials. We conclude then, in general, that the four fundamental op- erations are performed upon algebraic quantities when isolated, according to the same rules, in respect to the signs, as when they constitute terms of polynomials. Art. 83. It is manifest from what precedes, that addition in algebra does not always imply the idea of augmentation; for, if we add — 6 to a, the result a — 6 is less than a by the quan- t'ty h. Nor does subtraction in algebra always imply the idea of dimi- nution ; for, if we subtract — h from a, the result a-]- his greater ihan a by the quantity b. To distinguish these results from those of addition and sub- traction in arithmetic, we use the terms algtbraie sum and alge* 11 122 NEGATIVE QUANTITIES AND THE XXVIIl braic difference. Thus, a — 6 is the algebraic sum of a and — b ; and a + 6 is the algebraic difference between a and — b Art. 84:. 1. If a rectangular field is 10 rods long and 7 rodji wide, how much must be added to the length, in order that the field may contain 49 square rods ? Suppose X rods added to the length ; then 10 -j- 2; = the length after x rods are added ; hence, 7(10 + x)r=49, or 70 + 7x = 49. This equation gives x = — 3 rods. Here the value of x being negative, indicates some absurdity in the question. If we return to the equation 70 -|- 7 x = 49, we perceive that the absurdity consists in supposing, that something must be arithmetically added to 70 to make it equal to 49. The result, 2; zi: — 3, shows that — 3 rods must be algebraic- ally added to the length, that is, 3 rods must be subtracted from it. Let us then put — x instead of -|-x in the original equation, and it becomes 7(10 — x) = 49, or 70 — 7x=:49. This gives x r= 3 rods. The question therefore should have been as follows : If a rectangular field is 10 rods long and 7 rods wide, how much must be subtracted from the length, that the field may con- tain 49 square rods ? We are conducted to this modification in the question, merely by changing the sign of x in the original equation. We see, moreover, that both equations give the same result, except with regard to the sign. 2. If a field is 9 rods long and 5 rods wide, how much must be subtracted from the length, so that the area of the field may be 65 square rods t Kx = the number of rods to be subtracted, we have 5(9 — x)=65. or 45— 5x=:65. XXVIII. INTERPRETATION OF NEGATIVE RESULTS. 123 This equation gives x =r — 4 ; hence, — 4 rods *are to be sub- tracted from the length, that is, 4 rods are to be added to i^ Indeed the equation 45 — 5x=i65 is evidently absurd, since it supposes that something must be taken from 45 to make it equal to 65. Let us put + ^ instead of — x in the original equation ; thiw equation then becomes 5 (9 -|- a;) = 65, which gives 2=4 rods. We see therefore that the inquiry should have been, how much must be added to the length. 3. A father whose age is 68 years, has a son aged 20; iK how many years will the son be one fourth as old as his father t Suppose X = the number of years ; then 68 4-x 20 -f- X := — J—. This equation gives x = — 4. Changing the sign of x in the original equation, we have 20 — x = — J — , which gives x = 4 years. The question therefore should have been; how man/ yeara ago was the son one fourth as old as his father ? 4. A laborer wrought for a gentleman 7 days, having his son with him 4 days, and received 27 shillings ; at another time he wrought 9 days, having his son with him 6 days, and received 33 shillings. What were the daily wages of the laborer and his sof lespectively ? Let X = the daily wages of the man, and y = the daily wages of the boy. Hence, 7x-j-4y=:27, and 9x-t-6y = 33. These equations give x =r 5, and i/=. — 2. Changing the sign of y in the original equations, we have 7x — 4y = 27, and 9x — 6y = 33. These equations give x = 5 shillings and y = 2 shilhngs. 124 NEGATIVE QUANTITIES AND THE XXVlll It Appears then that the son was an expense to his father, and that the inquiry should have been : how much did the laborer receive per day for himself, and how much did he pay per day for his son ? 5. What fraction is that, to the numerator of which if 1 be added, the value of the fraction will be f j but if 1 be added to the denominator, the value will be f ? Let - be the fraction. y Then ^±J = I, y and —^T—i = I- These equations give %=. — 5, and y = — 9. Here the values of x and y are both negative. Changing the signs of X and y in the original equations, we have — 2:4- 1 ^ , — a; ^ -r* =. f, and -— - = #. But we may — y — y+A change the signs of the numerators and denominators of the first members without altering the value of the fractions; we then have = I, and = f . y y— 1 The question should, therefore, have been as follows : What fraction is that, from the numerator of which if 1 be subtracted, the value will be | ; but if 1 be subtracted from the denominator, the value will be |? The preceding problems render it manifest, that a negative result indicates some absurdity in the conditions of the question, and show us, that the conditions are modified so as to remove the absurdity, by rendering subtractivCf quantities which had been previously considered as additive, or by rendering additive, quan- tities which had previously been considered as subtractive. We see moreover, that, in order to ascertain what the condi- tions should have been, we have only to change, in the original equations, the signs of those (juantities for which we have ob- tained negative values, ana modify the question accordingly. X^XVIII. INTERPRETATION OF NEGATIVE RESULTS. lZ5 Negative quantities are sometimes said to be less than zero^ and, in an algebraical sense^ they may be so considered. But strictly speaking, no quantit} can be less than zero. When we say, for example, of a bankrupt, that he is worth $5000 less than nothing, we mean simply, that he owes $5000 more than he can pay- Negative quantities do not, in reality, differ from positive quan tities, and are merely positive quantities taken in a sense differ ent from that first supposed. Let the learner solve the following questions, and show how the negative results are to be interpreted. 6. What number is that, /^ of which exceeds ^ of it by 5 ? 7. A man, when he was married, w^s 30 years old, and his wife 20. How many years before their marriage was his age to hers as 7 to 6 ? 8. What fraction is that, whose value, if its denominator be diminished by 2, will be ^, but whose value, if its numerator be dnninished by 2, will be ^^? 9. Find two numbers, such that their difference shall be 20, and the difference between 6 times the greater and 3* times the less shall be 96. 10. A cistern is provided with two stopcocks, A and B, through which water flows. After the stopcock A had been open 5 min- utes, and B 3 minutes, there were found 24 gallons in the cis- tern ; but if A had been open 7 minutes and B 5, there would have been 32 gallons in the cistern. How much water flows into the cistern through each stopcock in a minute? il. Three times A's money, twice B's, and four times C*a make $13000; four times A's, three times B's, and twice C'a make $25000 ; and six times A's, four times B's, and once C's mane $40000. Required the estate of each. 11* *26 DISCUSSION OF PROBLEMS. XXIX SECTION XXIX. DISCUSSIOX OF PROBLEMS. Art 85. "When a question has been resolved generally, that IS, by using letters to represent the known quantities, we some- times inquire what values the unknown quantities will assume, in consequence of particular suppositions with regard to the known quantities. The determination of these values, and the interpretation of the remarkable results which we may obtain, constitute what is called the discussion of the problem. The discussion of the following problem, which is originally due to Clairaut, presents many remarkable circumstances. 1st case. Two couriers set out, at the same time, from two points, A and B, which are a miles asunder, and travel towards each other. The courier from A goes b miles per hour, and the courier from B, c miles per hour. At what distance from A and B will they meet? A R B Let R be the point of meeting. Suppose* z = the distance from A to R, and y = the distance from B to R. Then we have (1) z + y = a. As the courier from A goes b miles per hour, he will be -j- hours in going x miles ; in like manner, the courier from B will be - hours in going y miles ; and since they are equal times on the road, we have (2) r = - • Multiply the 2d equation by 6, 2 = — ; substitute this value of x in the let, c ^XIX DISCUSSION OF PROBLEMS. 127 — — -\-i/ =za'f multiply by c, by-\-ci/=:ac, or (b-\-c)i/z=ac; hence, Qi C y = = distance of the point of meeting from B. h y Substituting the value of y in the equation z = -^, or 2 = c - X y, we have c b ac abc ab ' x=z~ . =—. — = —77—, — ^ = r- 1 — = distance of the point c o-j-c c(o-j-c) b-\-c ^ of meeting from A. As the sign — does not occur in the values of x and y, these values will always be positive, whatever numbers are put instead of a, b and c. Indeed it is evident, that since the couriers travel towards each other, they must necessarily meet between A and B. 2rf case. Suppose now that the couriers, setting out from the points A and B, as in the diagram below, proceed both in the same direction, and travel towards the point C, at the same rates as before. What distance will each travel before one overtakes the other ? A B R C Suppose R the point where they come together. Let z = A R, and z^ =: B R. Then, (1) z — yz=. a, and (2) -=^. ^ ' be These equations being solved, give a ab ; ac -, and y = b — c' ^—b—c Here the values of x and y will not be positive, unless b ia greater than c ; that is, unless the courier from A travels faster Ihan the courier from B. ^ -r5=:?^— "i^- 128 DISCUSSION OF PROBLEMS. KXIX Suppose b =. 10, and c = 8 ; then we have 10 a _ 10a_ 10 — 8 2 , 8a 8a ^ But if we suppose that b is less than c, and that 5 =: 8 and := 10, we have 8a 8a . , ^=8^ro==¥ = -^"'^^ 10a lOa , . -5 a. ^ 8—10 —2 Here the values of x and y are both negative, and show that there is some inconsistency in the question ; and indeed it is ab- surd to suppose, that the courier from A can overtake the courier from B, both travelling towards C, unless the former travel faster than the latter. In order to see how the question is to be modified, let us change the signs of x and i/ in the original equations. We then have y — X zr: a, and -7— =. — -. The last equation, by a change of the signs, becomes b ~ c It is evident that the 2d equation will remain the same as be- fore, because it merely expresses the equality of the times Tue equation y — xz=a shows that y is greater than x, or thit the point where they come together, is further from B than It is from A ; and since this point cannot be between A and B, it must be on the other side of A with respect to B, as at R' in Ihe subjoined diagram. C R' A B R C When, therefore, 6 is less than c, the question should be aa follows : XXIX DISCUSSION OF PROBLEMS. 129 Two couriers set out from the points A and B, a miles distanl from each other, and travel towards C ; the courier from A goes b miles per hour, and the courier from B, c miles per hour ; how far will each travel before the courier from B overtakes the one from A ? [n this case, the equations, y — 2 = «, and - = -, will give ah . ac X = -, and ij = . c — 6 c — 6 If 6 = 8 and c = 10, we have 8a ■ ^ 10 a We see, in this question, that a change of sign indicates a change in direction. Numerous instances of similar indications occur in the application of algebra to geometry. ^d case. Resuming the formulae, ab . ac . . . X =: , and i/ = , let us suppose b =z c ; then b — c being 0, we have ab - ac In order to interpret these results, let us go back to the orig- inal equations, x — y z=. a^ and - = -. By putting b instead of e in the second equation, it becomes - in -, which gives i:=.y\ and substituting x for y in the first equation, we have X — xz=.a, or :=z a. This result is manifestly absurd, since we have a known quan- tity equal to zero ; and it is evident, that since the couriers travel equally fast, it is impossible one should overtake the other TTl We regard therefore — , or any similar expression, as a syrri' bol of impossibility ;^a.nd when a question gives Ozna (a being 130 DISCUSSION OF PROBLEMS. XXIX any known |uantity different from zero), or when the unknown quantity is found = -, the question is to be considered as im- possible. There is, however, another signification of such an expression m ..... . as — , which It IS important to notice. Let us take the expressions for x and y, viz : 2 = , and o ' c 10 a 10 a ,.^ Making 6 = 10 and 1 ""10 — 99 '1 c =z 9-9, we have \ 99 a 99 a ^ _ 10 a _ lOf — innn Making 6 = 10 and 1 ""—lO — 9 99~ 01 —'""""' c = 9-99, we have \ 999 a 999 a ,^^ '»" =15,f=10000«; Making 6 = 10 and 1 — 10 — 9-999 — -001 c =z 9-999, we have i 9999 a 9999 a „-. ( ^-ro'=9^999--^oor=^^^ We here perceive that the value of the fraction increases in pro-' portion as the denominator is diminished ; if then the denomina- tor be less than any assignable quantity, that is 0, the value of the fraction will be greater than any assignable quantity, or infi»« nitely great. Hence mathematicians consider a fraction, whose lumerator is a definite quantity, and whose denominator is zero, as a symbol of infinity. Thus, Tit Tit 7" , are Symbols of in* finity. Remark. In the problems of geometry, solved by the aid of algebra, there are many instances, in which an in/inite quantity Mislead of denoting an absurdity, is the true Result sought. XX^X DISCUSSION OF PROBLEMS. 131 If a definite quantity be divided by an infinite or impossible quantity, the quotient will be zero. Thus a divided by - gi'ei m m 4tth case. Suppose now, that, in the formulae for 2 and y, 6 = c and a = 0. The distance between the points A and B being nothing, thesa points must be coincident, as in the following diagram, A B and the formulae for x and y become 0.6 ^ O.c n A u . x=r — = -, and y = — = -; or x . = 0, that is, = 0, and y . = 0, that is, = 0. Now, as the couriers set out from the same point, and travel equally fast and in the same direction, they cannot be said to come together at any particular point, since they are constantly together throughout the whole route. But in order to see the general import of the expression ^, let us return to the original equations, x — y^=.a^ and - = -., Putting instead of a, and h instead of c, we have x — y = 0, , « y and- = -. The first equation gives x = y, and this value of % being sub- y V Btituted in the second, gives -r=z^. This last equation, in which the two members are precisely alike, is called an identical equation, and is verified by putting any quantity whatever instead of y. The value cf t therefore cannot be determined from this equation. Moreover the equation - = | , gives x = y, an which is the same as 6 (2 a + 6). Dividing 62 by 6, twice the tens, we have for a quotient 10 ; but as the unit figure cannot exceed 9, we put 9 in the root at the right of the 3, and we have 39 for the entire root. In order to determine whether 9 is the proper unit figure of the root, we observe that the divisor 6 (tens) corresponds to 2 a, and 9 is the figure which we have found for h ; hence, 60 -f- 9 or 69 corresponds \.o2a-\-h\ therefore we place 9 at the right of the divisor and multiply 69 by 9; the product 621 answers to 6 (2 a -|- 6) ; this subtracted from the dividend leaves nothing. Therefore the true root is 39. Let the learner extract the roots of the following numbers bf the process last explained. ' 1. 784. 4. 841. 2. 2809. 5. 1296 3. 6084 6. 8649 7 What the second root of 127449? 136 EXTRACTION OF THE XXX The second {owers of 10, 100, 1000 are respective.y 100 10000, 1000000; hence the second povve of any whole numbef between 10 and 100, that is, consisting of two figures, will be between 100 and 10000, that is, it will contain three or four fig- ures; also, the second power of a number consisting of three figures, will contain five or six figures. We canj therefore, as- certain the number of figures in the root of any proposed num- ber, by beginning at the right, and separating 'it into parts or periods of two figures each. The left hand period may consist of one or two figures. There will be as many figures in the root, as there are periods in the power. Separating 127449 into periods, we see that the root must contain three figures, or hundreds, tens, and units. Let a represent the hundreds of the root, b the tens, and c the units ; then a-\-h-\-c^ regard being paid to the rank of the fig- ures, will represent the root. The second power ofa-|-6-f~^ 's a^-\-^ak-\-h'^ -\-^cic-\- 2hc-\-c^, or a2 + 2a6 + 62-|-2{« + 6)c + c2. Hence, the second power contains the second power of the hundreds, plus twice the product of the hundreds by the tens, plus the second power of the tens, plus twice the sum of the hundreds and tens multiplied by the units, plus the second power of the units. We proceed now to extract the root. Operation. 1274^49(357. Root. _9 = «2. 37'4(65 i=::2« + 6. 325 =:(2« + 6)6. 494'9(707 = 2(a + 6) + c. 4949 • =:[2(« + 6) + c]c. 0. We first seek the second power of the hundreds of the root, which must be foui?d in the 12, (120000) ; the greatest second power in this part i 9, (90000), the root of which is 3, (300). XXX SECOND ROOTS OF NUMBERS. 137 Putting 3 as the first figure of ^he root, and subtrac in^ its second power from, 12, we bring down the next period at the. right of the remainder. Wp now consider 374 as a dividend. This dividend contains 2ab-^ b^, or twice the product of the hundreds by the tens, plus the second power of the tens, togethei with the hundreds arising from multiplying twice the sum of the hundreds and tens by the units. It is now our object to find 6, or the tens of the root ; and for this purpose, we divide by 2 a or twice the hundreds. But as the product of twice the hundreds by the tens, can have no sig- nificant figure below the fourth place, in dividing we reject the right hand figure of the dividend, separating it by an accent. We double the hundreds, and obtain 6 for a divisor, which is contained in 37 six times. But if we put 6 at the right of the divisor and multiply 66 by 6, we obtain a product greatef than 374. We next try 5, which we place in the root and also at the right of the divisor, and we have 65 corresponding to 2 a -\-b; this multiplied by 5 gives 325, corresponding to (2 a -f- b) b. We now subtract 325 from the dividend, to 4;he remainder annex the figures of the last period, and obtain for a new divi- dend 4949. This dividend contains 2 (« + 6) c + c^ — [2 (a + 6) + c] c, or twice the sum of the hundreds and tens multiplied by the units, plus the second power of the units. To obtain the units, therefore, we must divide by twice the hundreds and tens already found. But as hundreds and tens multiplied by units, can have no significant figure below tens, we reject the right hand figure of the dividend, separating it by an accent. Double the hundreds and tens makes 70, (700), = 2 («-|- 6), which is contained in 494, (4940), seven times. We then put 7, which corresponds to c, in the root, and also at the right of the divisor, and we have 707 = 2 («-|- 6) -{- c; this multiplied by 7 gives 4949 zzi [2 {a -\- b) -\- c] c, which sub- 12* 138 EXTRACTION OF THE XXX tracted from the dividend, heaves no remainder. Therefore 357 IS the root sought. If the root contains more than three figures, representing the figure of the highest rank by a, the next by h, &c., we have for the second power a^ -\- '^ a h -\- h'^ -\- '^ [a ~\- h) c -\- c^ -\- 2(a + 6 + c)^+rf2_|_2(a+i + c + (/)e + c2, &c., ona2^ (2a + 6)6+ [2(a + fe) + c] c+ [2(a + 6 + c) + ^] rf + \jl {a -\- b -\- c -\- d) -\- e] e &lc. ; the first form of which shows, that, after the first figure has been found, each of the successive figures is obtained by dividing by twice the whole root already found; and the second form shows, that, in each case, the quo- tient is to be placed at the right of the divisor, and that the divi- sor thus increased, is to be multiplied by the quotient. Moreover, from a consideration of the rank of the figures, it is plain, that twice the root already found, multiplied by the next lower figure, can produce no significant figure below the second from the right in each dividend. a What is the second root of 1024832169? Operation. 1024832169( 32013 . Root. 12'4(62 124 • 8321(6401 6401 192069( 64023 192069 0. ^neraimg in this question as in the preceding ones, we find that the second divisor 64, is not contained in the dividend 83,. the right hand figure being rejected, which shows that there are no hi»ndreds in the root sought ; in this case, we place a zero in the root, also at the right of the divisor, and bring down the nexf two fip'Tres to form a dividf id. XXX SECOND ROOTS OF NUMBERS. l39 We may observe, that, if the last figure of the preceding aivi- sor be doubled, the root, so far as it is ascertained, will be doubled ; for that divisor contains twice this root, with the ex- ception of the figure last found. Art. 87. From the preceding analysis we derive the follow- ing RULE FOR EXTRACTING THE SECOND ROOTS OF NUMBERS. 1 . Begin at the right, and, by means of accents, separate the number into periods of two figures each. The left hand period may contain one or two figures. 2. Find the greatest second power in the left hajid period, place its root at the right of the proposed number, separating it by a line, and subtract the second power from the left hand period. 3. To the right of the remainder bring down the next period to form a dividend. Double the root already found for a divisor. Seek how many times the divisor is contained in the dividend, re- jecting the right hand figure. Place the quotient in the root, at the right of the figure previously found, and also at the right of the divisor. Multiply the divisor thus increased by the last figure of the root, and subtract the product from the whole divi- dend. 4. Bring down to the right of the remainder the next period, to form a new dividend. Double the root already found for a divisor, and proceed as before to find the third figure of the root. Repeat this process until all the periods have been brought down. Remark. If the dividend will not contain the divisor, the right hand figure of the former being rejected, place a zero in the root, also at the right of the divisor, and bring down the next tteriod. Extract tfte roots of the following numbers. 140 EXTRACTION OF THE SECOND ROOT& OF NUMBERS. XXX 1. 1369. , 7. 36100 2. 2401. 8. 1100401 3 361. 9. 1432809. 4. 123201 10 151905625 5. 502681. 11. 901260441. 6. 11881. 12. 2530995481. Art. 88. There are comparatively but few numbers which are exact second powers ; and the roots of such as are not per* feet powers, cannot be obtained exactly either in whole numbers or fractions. For example, the root of 42 is between 6 and 7 ; but no number can be found, which, multiplied by itself, will produce exactly 42. We shall however see hereafter, that the root of any positive number may be approximated to any degree of exactness. Since the roots of numbers, which are not perfect powers, can- not be obtained exactly, either in whole or fractional numbers, they are said to be irrational^ or incommensurable ; that is, these roots and unity have no common divisor. Roots of other de- grees, besides the second, are also called irrational ^ when they cannot be exactly obtained. The second root of a quantity, whether that root can be found exactly or not, is indicated either by the exponent ^, or by this character ^, called the radical sign. Thus, (25)^^ or i/25 =r 5 ; and (3)^ or ^3 means the second root of 3. But the second root of a negative quantity cannot be obtained, even by approximation, since there is no number, which, multi- plied by itself, can give a negative quantity. The second roots, therefore^ of negative quantities are called imaginary, in opposi- tion to those of positive quantities, which are rcaZ, although they cannot be exactly obtained. Thus, ( — 16)^ or ^ — 16 is im- aginary. These imaginary quantities, except in some of the higher branches of analysis, indicate absolute absurdity in the questions from wHch they arise. J fi)' XXXI. SECOND ROOTS OF FRACTIONS, ETC. 141 SECTION XXXI. • BCOND ROOTS OF FRACTIONS — AND THE EXTRACTION OP SECOND ROOTS BY APPROXIMATION. Art. 89. The second power of a fraction is found by raising both numerator and denominator to the second power ; for this ia equivalent to multiplying the fraction by itself. Thus, ^ a a a^ h' b~W Hence, the second root of a fraction is found by extracting the root of the numerator and that of the denominator. Thus, the root of ^f is ^, and that of -^ is -. Let the roots of the following fractions be found. 1- a- 4. iff. 2. «. 5. /^V- 3. T%V- 6. T^V- Art. 90. If, however, either the numerator or denominator is not a perfect second power, the root of the fraction can be ob- tained by approximation only. Thus, the root of ^f is between ^ and ^. It is nearer to f . The denominator of a fraction, however, may always be ren- dered a perfect second power, by multiplying both numerator and denominator by the denominator, which does not alter the value of the fraction. For example, f = f ^, the approximate root of which is f — . By this mode, the root has the same de- nominator as the given fraction. Re7nark. The sign -|-, placed after an approximate root, sig- nifies that it is less, and the sign — , that it is greater than the true one. When a greater degree of exactness is requisite, we may, aftei having multiplied both terms of the fraction by its denominate f, multiply both terms of the result by any second power. 142 APPROXIMATE SECOND ROOTS OF NUMBERS. XXXI 3 21 144.21 3024 , ^hus, ^ = 49 = 1^14749 or ^^, the approximate -oot of which IS — — . o4 Let the learner find the roots of the following fractions, in the denomination of their respective denominators. 1. J. 4. ^. 2. ^, 5. f^. 3. A. 6. U. Art Ol. The root of a whole number may be approximated in the same way, by converting it into a fraction, having a sec- ond power for its denominator. If, for example, we would find the root of 5, exact to 12ths, we change 5 to the fraction ^^, the approximate root of which is f|- — . But it is most convenient to change the number into a frac- tion, having the second power of 10, 100, or 1000, &,c., for a de- nominator ; that is, convert the number into lOOths, lOOOOths, or lOOOOOOths, &c., and the root will be found in decimals. ' Thus, 5 = U^ = mU = mUU ; that is, 5 = 5-00 = 5-0000 z=: 500000Q ; the approximate root of the first is f f + = 2-2 +, of the second f f ^ + = 223 +, and of the third fHf + = 2-236+. In the example just given, we perceive that twice as many /eios are annexed to the number, as we wish to have decimals in the root. Indeed, it is plain, that there must be half as many decimals in the root as there are in the power, because the sec- ond power of lOths produces lOOths, the second power of lOOtha produces lOOOOths, &c. Moreover, we neeE6R£i£ Art. 93. A pure eqitation of the second degree, or a pure quadratic equation, is one which contains the second power, but no other power, of the unknown quantity. 1. A's age is to B's as 7 to 9, and the sum of the second pow- ers of their ages is 1170 years. Required their ages. 2. Two couriers set out, at the same time, from two places 220 miles asunder, and traveled towards each other till they met ; when it was found that the first had traveled only f as fast as the second, and that the number of hours they had been on the road, was equal to the number of miles the first traveled per hour. Required the rate per hour and the distance each traveled in the whole. 3. A gentleman has two square rooms, the sides of which are as 5 to 6; and he finds that it takes 11 square yards more of carpeting to cover the floor of the larger, than it does to cover ifiBt of the smaller. Required the length of one side of each loom. 4. A farmer had an orchard, in which the number of trees in each row was to the number of rows as 6 to 5; and the number of bushels of apples, gathered from each tree, was to the number of rows as 4 to 5; moreover, the number of bushels in the whole was equal to 80 times the number of trees in one row iiXXII. PURE EQUATIONS OF THE SECOND DEGREE. 145 How many rows were there, how many trees iu each row, and how many bushels of apples were gathered ? 5 A gentleman has u rectangular piece of land 50 rods lona and 18 wide, which he wishes to exchange for another of the same area and in a squure form. What must be the length of one side of the square? 6. A man wishes to make a cistern containing 800 gallons, the bottom of which shall be a square, and the height 6 feet. Required the length of one side of the bottom. Note. A gallon wine measure is 231 cubic inches. 7. An acre contains 160 square rods. What is the length of one side of a square containing an acre of land ? 8. What would be the length of one side of a square contain- ing 12 acres? 9. What number is that, to which if 10 be added, and from which il 10 be subtracted, the product of the sum and difference will be 150? 10. The product of two numbers is 900, and the quotient of the greater divided by the less is 4, What are those numbers 1 11. There is a house, whose breadth "is to its length as 5 to 6, and whose height, exclusive of the roof, is to its breadth as 4 to 5. Required the dimensions of the house, supposing that it takes 2200 square feet of boards to cover the four sides. 12. A merchant bought two pieces of cloth of equal length ; the one cost 5 shillings a yard more, and the other 5 shillings a yard less, than the number of yards in each piece. The price of the whole being ^136 18s., how many yards were there in each piece, and what was the price of each per yard? 13. A company at a tavern found that their whole bill was $45, and that each had to pay 5 times as many cents as there were parsons in the company. How many persons were there, and how much had each to pay ? 14. There are two numbers, the sum of whose second powers is 5274, and the difference of these powers is 1224. Required the numbers. 13 l46 U ri:CTED EQUATIONS OF THE XXXIII 15. A mail lent a certain sum of money at 6 per cent, a year, and found that if lie multiplied the principal by the number rep resenting the interest for 8 months, the product would be $900 Required the principal. SECTION XXXIll. AFFECTED EQUATIONS OF THE SECOND DEGREE. Art. 94:. The equations of the second degree, ivhich we have hitherto considered, involved the second power only of the unknown quantity. But, in its most general sense, an equation of the second degree, with one unknown quantity, is composed of three kinds of terms, viz : one kind containing the second power of the unknown quantity ; another containing the Jirst power of the unknown quantity ; and a third composed wholly of known quantities. Such are called affected et^uations of the second degree, or af- fected quadratic equations. 1. There is a rectangular field whose length exceeds its breadth by 8 rods, and whose area is 180 square rods. Required the length and breadth. Let X 1= the breadth in rods ; then X -|- 8 := the length. Hence,. »2 4_8x=: 180. If we compare the first member of this equation with the sec- ond power of z -|- a, which is z^ -j- 2 a x -|- a^, we see that it con- tains two terms which correspond respectively to the first two terms of this second power, viz : x2 = x2, 2 a X = 8 X. Hence, 2a = 8, a=z 4, XXXIII. SECOND DEGREE. 147 Now since 16 corresponds to a^, if we add 16 to bMh mem- bers of the equation, x^ -\-Sxzzz 180, the first member becomes a perfect second power corresponding to x^ ~{- 2 a x -\- a^, and we have x2 + 82;+ 16= 180+16= 196. We now take the root of each member. The root of the first member is x -|- 4, for (z -(- 4) (x -(- 4) = a;^ -j- 8 z -|- 16 ; and tha^ of the second member is 14. We have therefore, X + 4=:±14. Every positive quantity has two second roots, one positive and the other negative; for the second power of — a, as well as that of -|- «, is -|- a^. Therefore, since in an equation such as x -j- 4 = db 14, the value of x is not determined until the known quan- tity is transposed, and it may happen that the negative as well as the positive root will answer the conditions of the question^ we place the double sign dn before the second member. This sign is read plus or minus. In the above equation, transposing 4, we have x=. — 4 dc 14. Calling 14 plus, xz=z 10 rods, the breadth ; and z-|-8=: 18 rods, the length. Calling 14 minus, x=:— 18, andx + 8=: — 10. The firsv value only of x answers the conditions of the que& lion. The second value will however satisfy the equation ; for. (_ 18)2 + 8 (— 18) = 324 — 144 = 180. In order to interpret the negative valae, we substitute — x for -j-zinthe original equation, and we have x^ — 8xr=180, .or x^x — 8)= 180. This shows that x now represents the longer Bi4e instead of the shorter. The solution of the equation, jc2 — 8 X r= 180, will be similar to that of the following question. 2. Twenty times a certain number exceeds its second powet liy 75 What is that number ? Let X = the number. Then, x^ + 75 = 20 x. Transposing, x2 — 20x.= — 75. i48 AFFECTED EQUATIONS OF THE XXA.Q1 In this equation the term containing the first power of x being negative, in order to render the first member a perfect second power, we compare it with the second power of x — a, which ia x^ — 2ax-\- a^^ and we have z2 = x\ • — 2ax = — 20 a;. Hence, — 2a =—20, — a = — 10, a2 = 100. Adding 100 to each member, we have a;2_20x-f 100 = — 75+100=i25. We now take the root of each member. The root of the first isz — 10; for, (x — 10) (x— 10) = 2;^— 202 + 100, and that of the second is zt 5. Therefore, X — 1 = ± 5. Transposing, X = 10 rt 5 ; hence, X = 15, or X == 5 Both values of x are positive, and, therefore, both answer the conditions of the question. Indeed, 15. 15 + 75 = 20. 15; also, 5 . 5 + 75 = 20 . 5. Hence we see the propriety of giving the double sign to the root of the second member. Art. 95. Any affected equation of the second degree may be reduced to the form of x^-\-px=z q, p and q being any known quantities, positive or negative. ll is manifest that an equation may be reduced to this form in the following manner. 1. Clear the equation of fractions if necessary ; transpose all the terms containing x^ and x into tht first member y and the knoion terms into the second ; reduce the tttms which contain x^ into one term^ and those which contain % into another ; also, reduce the known quantities in the secona mejuher to one term. 2. If the term containing x^ is not positive, make it so by changing all the signs. 3. If tlie coefficient of r^ vs not 1 , divide all the terms by that coefficient. A.XXIII. SECOND DEGRL 1. 149 1. A draper bought a quantity of cloth for c£27 If he haj bought 3 yards less for the same sum, it would have cost him 15 shillings a yard more. How many yards did he buy ? Let X =. the number of yards. 27 Then — = the price per yard in pounds, and 27 = the price per yard, if he had bought 3 yards X— 3 27 27 3 less. Hence, r-= \- -. Clear the equation of fractions, X — o X A^ 108x= 1082 — 324 + 3x2 — 9xj tzanspose, reduce, and change the signs, 3 x2 _ 9 x = 324 ; divide by 3, x2 — 3x = 108. Here p = — 3, and q z=. 108. Comparing the first member with x^ — 2ax-|-a2, we have x2 = x2, — 2ax = — 3x, — 2a = — 3, — a =-t, Adfting f to each member of x^ — 3x = 108, we have x2 — 3x + f = l08 + f =^1^ + 1 = ^J We now take the root of each member. The root of the first is X — 1^, because (x — ^) (x — ^)=ix^ — 3 x -[- } : and that of the second is db ^. Hence, x — | = ± V"* Transposing — f , x = f±^=:^=12; orx = — J^ = — 9. The fir jt value only of x answers the conditions of the ques- iion In order to interpret the second value of x, viz : xz=z — 9, we Bubstitute — X for -{-x in the original equation, which then be- 27 27 . 3 27 27 , 3 . comes r= h T. or r-^ = + -j . smc(j — x — 3 _x~4' x + 3 X ^4' 13* /.'>0 AFFECTED EQUATIONS OF THE XXXIII when the denominator is negative, it is the same as if the de- nominator were positive and the whole fraction were preceded by the sign — . This. last equation becomes by transposition, 27 27 3 — =z -\- - , which answers to the following question. A draper bought a quantity of cloth for ^7. If he had bought 3 yards more for the same sum, it would have cost him 15 shil- lings per yard less. Let the learner solve the question as now stated. 2. Since p may always represent the coefficient of the first f ower of the unknown quantity, and q the known term, let us solve the general equation x^-\-px=:q, by comparing the first member with x^-\-2 ax-\-a^. We hare X2 = xs. 2az = px, 2a — p. a p -2' flS "~4* Adding ^ to each member, we have ^^+P^ + ^' = 2 + j- We now take the root of each member. That of the first member is a; + |, since ^z + 1 W x + 1 j = x^ +p x + 1- The root of the second member can only be indicated, until defi- nite valued are assigned top and q. We have then. + | = ±( Third powers. 1, 8, 27, 64, 125, 216, 343, 512, 729. 5 The numbers in the 2d line are the 3d powers of those imme- diately over them, and the numbers in the first line are the third roots of those immediately beneath them. The third power of 10 =z 1000 ; that of 100 — 1000000, and that of 1000 = 1000000000. Hence, the third power of an in- tegral number between 10 and 100, that is, of a number consist- •ng of two figures, must be between 1000 and 1000000 ; that is, ^^ z= a^ The second power of ab is a 6 . fl 6 z= fli X 2 &! X 2 -, ^2 ^2 . the third power of 2 6 c is 26c .26c .26c = 21X3 61X3^1X3 — 23 63^3 =^8 63 c3; and the fourth power of 4 6^ c3 d^ is = 4^ 6^x4 ^3x4 ^4x4 _ 256 6^ c^^ d^^. In these examples, adding the exponent of any quantity to itself, is the same as multiplying this exponent. Hence we have the following RULE FOR RAISING A MONOMIAL TO ANY POWER. Raise the numerical coefficient to the required poioer^ and mul- tiply the exponent of each letter by the number which marks the degree of that power. It is moreover manifest that any power of a product ^ is the pro- duct of that power of each of its factors. Thus, the fourth power of 4 62 c3 d^, which is 256 6^ c^^ ^16^ ig the product of the fourth powers of 4, 6^, c3, and d^ Remark. With regard to the signs, when the index of the power is even, the power will always have the sign + ; but when I he index is odd, the power will have the same sign as the root This manifestly follows from the rules for multiplication. 1. Find the 2d power of 7 a m^. 2. Find the 2d power of 8 6^ c z*. 3. Find the 2d power of 15 a'* m^ p^, 4. Find the 7th power of 2 x'^ y^. XXXVIII. POWERS OF POLYNOMIALS. 16^ 5. Find the 13th power of h^ c^ cP. 6. Find the 10th power o^^b^c^d^. 7. Find the mth power of ^j'^g-^x. Ans. /»'*'" gr' ^x*". 8. Find the mth power of p" g-*. 9. Find the 7«th power of 2x2 y3. Ans. 2'"x2'"y3« Remark. In the pre.'.eding example, since m is indefinite, ih« powei of 2 must be represented merely. 10. Find the wth power of 3^^ q^. 11. Find the 4th power of —^p^qK 12. Find the 5th power of — 2 x^ y^, 13. Find the 3d power of — 7 a^ 62 ^ d. 14. Find the 6th power of — 2 a m^ n^p^ (p x y. 2 a 4 flSs 15. Find the 2d power of ^. Ans. ^-t^ 16. Find the 2d .36c power of ^^^ . 17. Find the 3d ^5m2n 18. Find the 4th P"^^^ "^21^,2x3- SECTION XXXVIII. POWERS OF POLYNOMIALS. Art. 106. Any power of a polynomial may be indicated by enclosing it in a parenthesis, and placing the index of the power over it at the right. Thus, (2 6 4" ^)^ represents the second power of 2 6-|-c. The same thing may be indicated by a vin- 2 culum and the exponent, thus, 2 6 + c . Powers thus indicated may be raised to other powers in the same manner as simple quantities, that is, by multiplying the exponents. For example, the fourth power of {a-\- 6)3 ig 15 170 POWERS O* POLYNOMIALS. XXX VII J (a~^ b)^^'* z=i (a-\-by^. Moreover, when several compounc quantities are represented as multiplied together, the whole is laised to any power by raising each factor to the power required. Thus, the second power of (2 c + + 4 ^) 3/3, &,c. By substituting now for (1 4-^)*", its value, given in equatioa [1], and putting 1 instead of Aj equation [7] becomes m{l+Bi/ + Cy^ + Di/^ + Ei/,6LC.) = B + (B + 2C)y + (2 C+ SD)y^ + (3D + iE) yS, &c., or, [8] m-{-mBi/ + mC2/^ + mDi/^+mEi/'^,&,c. =B + (B + 2C)y + (2C+3D)3/2 4.(3 2>4.4£)y3,&c. But, as was proved at the commencement of this section, the terms not involving y are equal, as are also the coefficients of the same powers of y. Hence, B=im; B + 2C=mB; hence, C=^^) = ^^^li^) ; iC+3Z>..mC; hence, Z>=g(^).^^ -y(^^^^ 'iD + iE = mD; hence, E= :5i^=l) -- m (m— 1) (m— 2) (ot — 3) 1.2.3.4 These results are sufficient to enable us to continue the forma ti m of the coefficients as far as we please. The next coefficient ij -J 1 1 ntlm — l)(m — 2) (m — S)(m — 4) , , would evidently be — ^ —. — _ J -r— e~^- -, and the succeeding one "^C^.-!) (m-2) (^-3M«»-4) (^-5)^ Substitute these values of ^, B, C, «fcc. in equation 1], and A becomes, {m-l){m-^){m-S) 1.2.3.4 ^ ' *-°- Restoiing the "alue of y, viz : y = -, we have, 176 BINOMIAL THEOREM. XXXIA (■ '■?•)"= . , m a m(m — 1) a^ m(m — l){m — 2) q3 m(m — \)(m — 2)(m — S) a^ 1.2.3 * x3~r 1.2.3.4 ' 1^' Multiplying both members by z'", ^i(m— l)(m — 2) x^a^ m(m--l)(m — 2)(m — 3) '^ 1.2.3 a;3^" 1.2.3.4 ^ Reducing the fractions to lower terms, (x + a)- = 2» + m X— 1 « 4 ^?i^-=ll) z^-s a2 _^ 1 . Z m(m— ])(m — 2) , ^^ (;;,— 1) (;;^_2) (m — 3) 1.2.3 "^ 1.2.3.4 ^ x"'-'*a\&,c. Art. lis. Such is the formula for any power of a binomial ; from which we readily deduce the law both of the letters and the coefficients. First, with regard to the letters, we see that, in the first term, J, which is the first term of the binomial, is raised to the power to which the binomial was to be raised, and that the powers of 7 in the successive terms go on decreasing by unity. Secondlv>_r y. the second term of the binomial, is found in the second term of the power with 1 for its exponent, and, in the successive terms, the powers of a go on increasing by unity. Moreover, the sum of the exponents of x and a in the same term, is always equal to m, the exponent of the power to which the binomial is raised. With regard to the coefficients ; we perceive, that the coeffi- cient of the first term is 1 ; that of the second term is equal ia •»», the exponent of the power to which the binomial is raised. XXXIX. BINOMIAL IHEOREM. L71 To obtain the coefficient of the third term, we multiply t^at:' m — 1 of the second, which is m by — - — ; that is, we multiply by m — 1 and divide by 2. To obtain the coefficient of the fourth term, we multiply mat m 2 of the third by — ^— ; that is, multiply by wi — 2 and divide by o 3, and so on Art. 113. Hence, having one term of any power of a bino- m al, the succeeding term may be found by the following RULE. Multiply the given term by the exponent of x. in. that term ^ that is, by the exponent of the first or leading quantity of the bino- mial, and divide the product by the number which marks the place of the given term from the first inclusive ; diminish the exponent of X by 1, and increase that of a by 1. The coefficient of the first term always being 1, and that of the second being the same as the index of the power required, we can, by the preceding rule, write any power of a binomial. Let it be required, for example, to find the 9th power of The first term is x^ ; the second is 9x^ a ; the third is found Dy multiplying 9, the coefficient of the second, by 8, the expo- nent of x in the same, dividing the product by 2, which marks the place of the second term, diminishing the exponent of x and increasing that of a each by unity. The third term then is 9.8 2 36.7 c? a2 =3 9 . 4 z7 a2 = 36 x'^ n^. The fourth terra is x^ a^ = 12 .7 x^ a^ z=. 84 x^ a^. Finding, in a similar manner, the succeeding terms, we have (x + a)9 =:i-9 + 9 x8 « -f 36 a:7 a^ + 84 x^ a^ + 126 x^ a^ -f 126 x4 aP + 84 2:3 ^6 _|_ 36 ^2 ^7 _|_ 9 j; ^8 _|_ ^9, Since any quantity with zero for an exponent is 1, we may vuppose a^ to enter into the first term, and x^ into the last If 178 BINOMIAL THEOREM. XXXIX we should attempt to find another term succeeding a^ or x^ a®, we should obtain for its coefficient — ^ = — = 0. No addi- tional terms therefore can be obtained. Applying the rule and remembering that odd powers of nega- tive quantities are negative, we have also (a — . 6)10 = «io _ 10 «9 6 _j_ 45 ^8 ^,2 _ 120 «7 ^3 _|. 210 a6 54 _ 252 a^s + 210 «4 £6 _ 120 a3 ^7 _^ 45 ^2 68 _ 10 a 69 -j_ ftio. From the preceding examples, as well as from the table of 'powers given in the Art. 108, we infer, 1. That the number of terms in each power of a binomial ex- ceeds by 1 the index of that power. Thus, in the fifth power, there are six terms ; in the ninth power, there are ten terms. 2. When the number of terms is odd, there is. one coefficient, in the middle of the series, greater than any of the others ; but, when the number of terms is even, there are two coefficients in the middle, of equal value and greater than any of the others. Moreover, those which precede and those which succeed the greatest or greatest two, are the same, only arranged in an in- verse order. Therefore, when half, or one more than half of the coefficients have been found, the others may be written down without the trouble of calculation. 1. Find the seventh power of a -|- 6. 2. Find the tenth power of x -\- y. 3. Find the fifth power of m — n. 4. Find the eleventh power of 6 -|- c. 5. Find the thirteenth power of 2; — y 6. Find the sixth power of 3 a -|- 6. In this last example, the numerical coefficient of a must be raised to the requisite powers by multiplication. First write the power, merely indicating the operations with egard to 3 a, and we have (3«)6 + 6 (3 af 6 + 15 {^af 62_|_29 (3 afb'^+l^ (3a;,2 6< + 6 (3 a) 65 -j- 66. X-XXIX. BINOMIAL THEOREM. 170 Raising 3 a to the several powers indicated, and substituting the results, 729 a6 _|_ 6 . 243 a5 6 _(_ 15 . 81 a* b^ + 20 . 27 a^ h^ + 15 . 9 a2 64 _|. 6 . 3 a 65 _^ 66. Performing the multiplication, we have for the final result, 729 a6 + 1458 a5 6 + 1215 o^ 62 + 540 a? P + 135 a^ b* + 8a65_|_66. 7. Find the fifth power of x + 2 y. • 8. Find the third power of 6 a + 5 x. ' 9. Find the fourth power of a-{-b — 2 c. When a quantity containing several terms, as a-j-b — ^ ^ Iz to be raised to a power, it is convenient to sfLj!titrte cthe:- '.ot- ters, so as to render the quantity a binbmial, *aioO this binonlaJ. to the required power, and then restore the value of the Ict*'^f& substituted. Thus, in the present example, let 6 — 2 c := m ; then a -|- 6 - 2 c = a + wi. Now (a + m)* == a^ + 4 a^ m + 6 a^ rr? -j- 4 a w^ 4" ^^' ^^^» mz=:b — 2 c; m2 = ( 6 — 2 c)2 = 62 — 4 6 c + 4 c2 ; m3 z= (6 — 2 c)3 = 63 — 3 62 (2 c) + 3 6 (2 c)2 — (2 c)3, or, ;;j3 — ^3 _ 6 62 c + 12 6 c2 — 8 c3 ; m4 == (6 — 2 c)4 = 64 — 4 63 (2 c) -f 6 62 (2 c)2 — 4 6 (2 c)^ + (2c)4,or, i»4 — 64 — 8 63 c +24 62c2 — 32 6c3+ ICc^. Putting these values instead of m, ff|2, &c., and performing the miiltiplication by 4 a^, 6 a^, &c., we have (0 + 6 — 2c)4z=za4_j.4a3 6 — 8a3^. + 6«2 62__24a3 6c-f 24a2c2+4a63— 24a62c+48a6c2_32ac3+64— 863c + 24 62 c2 — 32 6 c3 + 16 c^. 10. Find the fifth power of 2 a + 3 x. 11. Find the third power of 4 x — 3 y -)- a. 12. Find the third power ofa-j-64-c-|-d 180 NUMERICAL ROOTS TO AN IT DEGREE. XL Let a -\- b z= m, and c -\- d =: n; then a -f- ^ -f c -f- «? =s m -\-n. 13. Find the sixth power of a -|- 2 6 — c. 1 i. Find the fifth power ofa + 6 — 2c — 38,,35- The preceding examples, as well as what was said in Art 10*>, relative to the powers of products, show, that any root of a product will be the product of the roots, to the same degree, oj ruch of the factors of this product. Thus, the second root of XLl. ROOTS OF MONOMIALS. 185 «^ b^ c* IS ah c2, which is the product of the second rvK).3 v f a^, 6^ and c**, the factors of a^ b^ c'*. In like manner, if any numerical quantit} is divided into fac- tors which are exact powers of the required degree, (and this may always be done, when the number itself is an exact power of that degree,) we may extract separately the roots of these fac- tors, and then multiply these roots together. For example, 1764 = 36 . 49, the second root of which is 6 . 7 = 42. Art. 118. From the preceding mode of finding the roots of literal quantities, it follows, that, if the exponent of any factor is not divisible by the number which expresses the degree of the root, the division can be expressed only, and gives rise to frac- tional exponents. Thus, the second root of a is a^; the third root of a is a* ; the fourth root of a^ is a*. The expression a* indicates either the fourth root of a^^or the third power of a* ; for the third power of a^ is a^ = a*. li\ like manner, a* denotes either the fourth power of q^ or the fifth root of a^. The radical sign may be used to indicate a root of any degree, if we place over it a figure denoting the degree of that root. — 3 — Thus, ^ denotes the second root; i/ , the third root; Y , the fourth root, and so on. Hence, v/«" = «^. ^a3 = a^, - ^«6 ,1 Observe, that ^a is the same as j^a, the 2 over the sign be- ing generally understood. We see, therefore, that in the prece- ding equivalent expressions, the nuTiber over the radical sign is the same as the denominator, and the exponent under the sign is tie s^me as the numerator, of the fractional exponent 16* 186 ROOIS OF POLYNOMIALS. XLil Art. 110. By means of exponents eith'rT entire or fractional, any quantity may be expressed in a greU variety of forms Thus, c?:=.c^.az=:a.a.a-=:.cu^.c^.(j^z=:a,cfi.c?.(j^,c^ -zz a^ . a^ . a^ . a^ . a^ . a^", &.c. Also, ^a^ b^ z=z a^ b^ =. a^ b^ ^=1 a . a^ . b^ . b^ z=z a^ . a^ . a^ . b^ .6^, &c. Hence, any quantity mav be separated into an indefinite num* ber of factors ; the only restricuon is, that the sura of the expo- nents of those factors, which are alike except with regard to their exponents, shall be the same as in the given quantity. In the first example given above, the sum of the exponents must be uniformly 3; in the second, the sum of the exponents of a must be f , and that of the exponents of b must be f . 1. Separate a^ into three factors. 2. Separate a* into seven factors. *S. Separate a? into six factors. 4. Separate a into three factors. 2 5. Separate d^ b into four factors. 6. Separate 3 a^ into six factors. 7. Separate 35 into three factors. 8. Separate 10 into seven factors. SECTION XLII. ROOTS OF POLTNOMIAXiS. Art. ISO. Let it be required to find the second root ol 4 m* |-12»inf9ii2. Operation. 4ywS+12mri + 9nS (2m + 3n. Root. 4n^ 12mn + 9n3(4m + 3n. I2mn-i-9n^ 0. tLII ROOTS OF POLyNOMIALli. IfcO By recurrii.^ to the second power of a -{- 6, which is g^ f-SaA •}- 62, we see that 4 m^ corresponds to a\ We therefore take the second root of 4 m^, which is 2»i, and place it at the right, as the first terra of the root sought, and subtract its second power from the given quantity. The remainder, 12 wt w -|- 9 w^, answers to 1 ab-\-h^y or {^ a -\-h)h. Dividing the first term of this remain- der by 4 m, corresponding to 2 a, we have 3 n for the second term of the root, which we annex to the 2 m in the root, and also to the divisor. The divisor thus increased, becomes 4 m + 3n,.=: Za-\-h. We then multiply 4 m + 3 n by 3 w, = 6, and we have 12 m w 4~ 9 '^^j which subtracted from the dividend, leaves no re- mainder. Hence, the second root of 4 m^ -j- 12 m » + 9 w^ is 2 m + 3 w, or — 2 m — 3 w ; or rather, i 2 m zh 3 w. The double sign may be omitted, until the operation is com- pleted, and then all the signs of the root may be changed, if both roots are required. When there are more than three terms in the power, the sec- ond root will contain more than two terms. But the mode of proceeding will be almost the same as that for finding the second roGls of numbers. We form a second dividend, in the same man- ner as the first was formed, and for a divisor double the whole root found. The division will give the third term of the root. The process is manifest from the formula, \a-\-b-\-c-\-d,6i,c.)^ = a2-|-2a6 + 62^2 (a + 6)c + c2 + 2(a + 6 + c)c?+d2, &c., in which a, 6, c, &,c. represent the teinis of the root. The following example will serve to illustrate the process. Required the second root of a'* + 6 a^ z +1 1 a^ a;^ -|- 6 a x^ -|- a^ Operation. a4 Qa^T+nd^x^ + Qax^ + x ^'ia^+^ax . 6q3 z+ 9q^^^ 2a2 2;2_|_6a2;3_|_2;4 (2a2-f6 gx + x^ 2a2a:S-|-6az3-t-2;^ a The root required is a^^Sax-^xK 188 ROOTS OF POLYNOMIALS. XLIl From the preceding analysis we derive the following RULE FOR EXTRACTING THE SECOND ROOT OF A. POLYNOMIAL. 1. Arrange the quantity according to the powers of somt letter. 2. Find the root of the first term^ and place it as the first term of the root sought, subtract the second power of this term from the given polynomial, and call the remainder the first dividend. 3. lyouhle the term of the root found, for a divisor, by which divide the first term of the dividend, and place the quotient, with its proper sign, as the second term of the root, also at the right of the divisor. Multiply the divisor, with the term annexed, by the second term of the root, and subtract the product from the dividend. 4. The remainder will form a new dividend, which is to be divided by twice the whole root found, and the quotient is to be placed as the next term of the root, also at the right of the divi' sor. Multiply the divisor, with the term last annexed, by the last term of the root, and subtract the product from the last dividend. 5. The remainder will form a new dividend, with which pro- ceed as before ; and thus continue, until all the terms of the root are found Remark 1. Each of the remainders must be arranged in the same order as the given polynomial was first arranged. If the given quantity contains no fractions, and a dividend oc* curs, the first term of which does not contain all the letters o\ the first term of the divisor, or which contains any one of then» with a less exponent than it has in that term of the divisor, wp may be assured that the given quantity is not an exact second power, and, therefore, does not admit of an exact root. Remark 2. In dividing we merely divide the first term of the dividend by the first term of the divisor ; and, since double the first, the first two, the first three, &c. terms of the root, will have the first terms alike, it is manifest that the successive divisors will have their first terms the same. Find the second roots of the fVlowing quantities KlJl. ROOTS OF POLYNOMIALS. 189 1. 3iz^+9x* + 20x-{-\2x^-{-25. 2. a4_|_54«252_|_i2a36+108«63 + 8164. 3. ]03;4_l0z3_i2x5 + 5a;2-|-92;6_2x+l, 4. 9a4 — 20a63_12a36 + 34a262 + 2564. 5. 4x4 + 8az3 + 4a2x2+1662x2_|_i6a622;^1064. 6. x^ + 4x5-{-2x^+9x^ — 4x-{-i. 7. 4z4-f 6 a:3_j_ 5^x2.^15 2; _|. 25. Art. 121. The rule for extracting the third roots of numbers, might, with slight modifications, be applied to the extraction of the third roots of algebraic polynomials. But it is generally the most convenient to use the rule, derived from the binomial theo- rem, for the extraction of roots to any degree. This rule, ap- plied to literal quantities, will, as is evident from the formula for the wth power of x -|- «, be as follows. BULE FOR EXTRACTING ANY ROOT OF A POLYNOMIAL. 1. Arrange the quantity according to the powers of some letter. 2. Find the mth root of the first term^ place it as the first term of the root sought ^ and subtract the mth power of it from the poly" nomial. 3. The remainder will form a dividend^ which is to he divided by m times the (m — \)th power of the term of the root found, and the quotient is to be placed as the second term of the root. 4. Raise the whole root to the mth poioer, and subtract the rcr suit from the given polynomial. 5. The remainder will form a new dividend^ which is to be di- vided by m times the (m — \)th power of the whole root already founds and the quotient placed as the third term of the root. 6. Raise the whole root to the mth power, subtract the result from the given polynomial, and with the remainder proceed as before; and thus continue until all the terms of the root are found. Remark. It is manifest that the first term of each successive divisor will be the same ; and, since we always divide the first term of the dividend by the first of the divisor, it is sufficient to find the first term of the first divisor and use that throughout ; 190 ROOTS OF POLYNOMIALS. XLU and^ in subtracting, only one term of the remainder reeds to be brought down, viz • that which contains the highest power of the letter according to which the given quantity was arranged. As an example, let it be required to extract the thir^ root of 8 x3 + 60 x2 y -|- 150 xi/^-{- 125 y3. Ojperation. 82;3-f 60 z2y_^ 150 xy2_j_ 125^3 (2^_j_5y, Root. 8x3 60xQy (12a:2. Divisor. The index m of the general formula, when applied to this question, is 3 ; and, after having arranged the quantity according to the powers of x, we find the third root of Sx^, which is 2x; subtracting the third power of 2 x, we have, for the first term of the remainder, 60x2y, which we divide by 12x2, ■= three times the second power of 2 x. The quotient is 5 y, which we put as the second term of the root, and raise 2 x -|- 5 y to the third power * The result is the same as the given quantity, and, when subtracted, leaves no remainder. Therefore, 2x-|-5y is the root sought. As a second example, we shall trace the operations for extract- ing a root consisting of three terms. Let it be required to extract the fifth root of x^^ — 10 x^ a -|- 45 xS «2 _ 120 x7 a3 _}_ 210 x^ a^ — 252 x^ aS _[_ 210 x^ a^ — ' 120 x3 a? -f 45 x2 a8 _ 1 X a9 _^ a^o. The quantity being arranged according to the powers of x, we find the 5th root of x^^, which is x^, and subtract the 5th power of this root from the given quantity. The first term of the re- mainder is — 10 x9 a. This term we divide by five times the 4th power of x^, which is 5 x^. .The quotient, — 2 ax, we place as the second term of the root, and raise r^ — 2 a x to the 5th power. The 5th power of x^ — 2 a x is x^*^ — 10 x^ a -|- 40 x® a^ — 80 x^ fl3 _|_ 80 x6 a'* — 32x5a5^ which subtracted from the given * Let the learner use the binomial theorem for finding the powers of ai y quantity conrfsing of more than one term. X.LIIL SIMPLIFICATION OF IRRATIONAL QUANTITIES. 191 quantity, gives a remainder the first term of which is 5x^a^ This tp«m being divided by 5 x^, the first term of five times the 4th power of x'^ — 2 ax, gives for a quotient a^, which we place as the third term of the root. We then raise x^ — 2ax-\-a^ to the 5th power, and it produces the whole of the given quantity Hence, x^ — 'Hax-j-a^ is the root sought. 1. Findthe3drootof27a3-|-81az2_|.81a2z + 27 23. 2. Find the 4th root of 16 x^a + 1000 7^ cfi -\- 600 x^^ a^ -\^ 160 29a2 + 625a«. 3. Find the 4th root of 625 c^ — 1 000 c^ y z + 600 c* y^ z^ — 160 c2y3 23_|_ 16^4^4. 4. Find the 5th root of 32 a^^ — 80 a^ h^ + 80 a^ je _ 40 a4 59 -J. 10 a2 612 — 615. 5. Find the 6th root of 729 x^ -f- 2916 x^ y + 4860 x^y^-\- 4320 x3 y3 _|_ 2160 x2 y^ + 576 x yS _|_ 54 ^6. SECTION XLIU. SIMPLIFICATION OF IRRATIONAL OR RADICAL QUANTITIES. Art. IS^. When a quantity is not an exact power of the de- gree required, its root cannot be found exactly. In such a case, the root is commonly expressed either by a radical sign or by a fractional exponent. Expressions indicating roots which cannot be accurately obtained, are called, as has already been stated^ ir- rational or incommensurable quantities. They are also sometimes called surds or simply radical quantities. Thus, ^2 or 2* and 3—1 /4 or 4^ are irrational quantities. In like manner, we are obliged to express the second root of o by a sign, thus y/« or a^ ; although, perhaps, when a has been replaced by its numerical value, the root may be exactly found. Algebraically considered, however, such expressions are in th* condition of irrational quantities. 192 SIMPLIFICATION OF X LTIl Expressions of this kind may, in many cases, be simplified. The root of a product, as was shown in Art. 117, is formed b) multiplying together the roots of all the factors of that product Hence, we may take the roots of such factors as are exact pow- ers, and indicate the roots of the other factors^ leaving these roots to hi approximated afterwards if necessary. Let it be required, for example, to find the second root of 1 92 0.2 6-3 t. The root is indicated* thus, ^192 a^b^c. But 192«263c =r 64 . 3a262^c or 64a^b^.Sbc. Now the first three factors, 64, a^ and 6^, are second powers^ we may, there- fore, take the roots of these and place their product as a coeffi- cient to the expression indicating the root of 3 6 c. We have then 1/192 a^b^ c =z8ab i/3 b c. It only remains now to ap- proximate the root of 3 6 c, the value of the letters supposed to be known, and multiply the result by 8 a 6. In separating an irrational quantity into factors for the pur- pose of simplifying, the learner has merely to find the greatest numerical factor that is an exact power, and the greatest expo- nent of each literal factor, not exceeding its given exponent, that is divisible by the number which marks the degree of the root. 1. Simplify ^125a3 65. 2. Simplify (80«6c4)*. 3 3. Simplify |/108a9 66 c2. 4. Simplify ^45 « 62 c2. 3 5. Simplify ^320 a3 6 — 64 a^ bK The greatest numerical factor in this quantity that is a third power is 64, and the greatest literal factor that is a third power is a3. HenQe, 320 a^b — Qi a^ 6^ = 64 a^ (5 6 — a^ 63). Tak- ing the root of 64 a^y and indicating that of 5 6 — a® 53^ ^q h^ve ^320«3 6_64a5 63 _ 4 ^ ^5b — a^b'^, or 4 a (56— a2 63)* G. Simplify ^24«4__8a36. 7. Simplify (2 a^ b^ + a^b c)K JtLIII IRRATIONAL QUANTITIES. 193 8. Simplify ^a^ -\- d^ b^. 9. Simplify y/TeSa^ — 256aX 10. Simplify y/3456 a^b—\ 728. If the quantity which is under the radical sign, or which is en. closed in a parenthesis with a fractional exponent, is a fraction, the expression may be simplified in the following manner, viz : Multiply both terms of the fraction bp such a quantity y as will render the denominator an exact power of the requisite degree^ then take the roots of the denominator and of such factors of the numirator as are exact powers. Remark. This preparation of the fraction is rarely advisable, except when the denominator is a monomial. Thus, 1 /^ = t /6«^* = . /'^'_ eab = ' V 8b Y 1662 \/ 1663°"" h.^^'- _ la like manner, I Vl^^ = . V 12^^ = K 94 V- 27 63 V^. 2^L..12a«63=i-^^12„«6.. 11 Simplify y |f 12. S.mpl.f, {-^^f. U. Simplify ^g. 14. Simplify (ll)*. 15. Simplify y/^~. 16. Simplify I yK^'-l?4^. 17 194 IRRATIONAL QUANTITIES XLIV 17 Simplify (- 500^4_25 0^7r) 18. Simplify ^ 27^5^^ 320 a3 6 4- 640 a^ 54 c3 m^ ' Art. 1S3. As we can extract the root of any factor and place it as a factor before the radical sign ;.so, if we would put under the sign any factor standing before it, we must raise that facto* ♦o a power of the same degree as the radical. Thus a b ^c = \J(j^ b^ c ; and | ya m =. \/ — wr- . Reduce the following quantities entirely to a radical form. 1. 3azy/6r ' 5. 2a (36)^. 2. §v/^^- 6. S{xy)^, 3. 4. (« + 6)^/3^. 8. ||(m^)i SECTION XLIV. OPERATIONS ON IRRATIONAL QUANTITIES WITH FRACTIONALi EX- PONENTS. Art. 1^4:. In general, operations are performed on quantitiei with fractional exponents, in the same manner as if the exponents rere entire. Add 4a^ and3a^. The sum is 4 a^ + 3 a^ = 7 a^. Add a b^ and c b^. The sum is ab^ + c 6^ = (a-f-c)6^. XLIV WITH FRACTIONAL EXPONENTS. I9& From 9 z* subtract 4 z*. The difference is 9 z^ — 4 z^ = 5 z^. From 3 a z^ y^ subtract 2 6 z- y^. The difference is 3az^y* — 26z^y*=(3a— 26)z^^^ Add 3 . 8^ and 5 . 18^. The sum indicated is 3 . 8^ + 5 . 18^. But these terms may be simplified and the expression reduced. For, 3.8^ = '). 4^. 2^ =3. 2. 2^ = 6.2^; and 5.18^ = 5.9^.2^ = 5 . 3 . 2* = 15 . 2^ Hence, 3 . 8^'+ 5 . 18^ = 6 . 2^ + 15 . 2^ = 21(2)^. In a similar manner, ( 192 63) * + (24 c^)^ = 4 6 . 3* + 2 c . 3^ =:(46 + 2c)3t Add (^\)^ and ii)^. The sum expressed is (^)* + (i)^. But (^)^ = (f f)^ = (^^.6)^ = 1. 6*; and(# = (^e)^ = (^.6)^= J.6^ Hence (^)^ + a)^ = f.6^ + i.6*=:A.6^ + A.6^ = /f (^) • 'T^® result therefore in its simplest form is -^^ (6)* We deduce therefore the following RULE FOR THE ADDITION AND SUBTRACTION OF IRRATIONAi^ Q.UANTITIKS. Express the addition or subtraction as usual by signs ^ sim plify the terms if possible ^ and reduce similar terms. Remark. Irrational quantities, indicated by means of frac- tional exponents, are similar, when the factors having fractional exponents are alike in all, and have severally the same expo- nents. Multiply a^ bv a^. This is performed by adding the exponents. Thus, a^ a» 196 IRRATIONAL QUANTITIES XLIV Multiply 3a^6^ by 5a^6^. The product is 15 J^^ b^^^ = lo J 6* Multiply 2 a^ by 3 a^. Reducing the exponents to a common denominator, wt ha\ e 2a^ = 2 a^^, and 3 a^ = 3 a^^ ; therefore, 2 a^ . 3 a* = 2a^ .3a^^=:6«^t Divide a^ by a^. ' This 'is performed by subtracting the exponent of the latter J from that of the former. Thus, — = a^ ^ = a' Divide 15 a^ b^ by 3a^ 6^ 1 he quotient is ——-z=ba^ •*. 3«^6* Divide 3a^6^ by 5a^ fti Reducing the exponents of the similar factors to a common 3a* 6^ sJb^ sJb^ denominator, we have — - — - = — - — — = . 5«*6* 5a^b^^ ^ 2 Find the second power of 3 a^. This is performed by raising the coefficient to the required power, and multiplying the exponent by the index of the power. Thus, (3 a^) ^ = 9 a^. In like manner, (3 a^ b^) * = 81 J b^. Conversely, the root of an irrational quantity is found, by taking or expressing the root of the numerical coefficient, and dividing the exponents of the other factors by the number which marks ♦he degree of the root. T ms, the second root of of is a^, and the fourth root of a" XLIV vvnn fractional exponent**. 197 is a^. Also the third root of 21 a^ b^ is 3a* b^^ ; and the fifth root of 7 J 6* is 7^ a"^ 6^^. From what precedes, we see that the following operations, viz : multiplication, division, finding powers, ahd extracting roots ^ are performed upon quantities with fractional exponents, in the same manner as if the exponents were whole numbers. Art. Id^. In the multiplication of irrational quantities, we assumed that their fractional exponents might be reduced to equivalent ones having a common denominator, without chang- ing the value of the quantities. This, however, may be.easily proved. For, multiplying the numerator of the exponent c f any quan- tity, raises that quantity to a power, and multiplying the denom- inator, divides the exponent, and therefore extracts the root. Consequently, when both terms of a fractional exponent are mul- tiplied by the same number, which is done in reducing to a com- mon denominator, the quantity to which the exponent belongs, is raised to a power of a certain degree, and then the root of the result is extracted to the same degree ; or the reverse. The value of the quantity, therefore, remains unchanged. Accordingly, the exponents of all the factors in any product, may be reduced to fi actions having a common denominator. Thus, 2a^ b^ — 2^ a^ b^ = 64* «* b^ z= (64 a^ b^)^. Moreover, it is manifest that the fractional exponents may be reduced to decimals, and that the value of the result will be either exactly or approximately the same as that of the given quantity. For example, a* =z a°'^^, and a® = aO-875^ If \i ^q^q required then to multiply a* by a^, we should have a* . a^ == a^ 25 . afwi _- ^0-25 + 0-875 — - al*125^ 1. Add {27 a* x)^ and (Sa^x)^. 2 Add (128)^ and (72)^. 3. Add (135)^ and (40)i 17* 19S IRRATIONAL QUANTITIES XLl*/ 4. Add (12)^, 2 (27)^ and 3 (75)^. 5. Add2(8)*— 7(l8)*and5(72)^ — (5()^ 6. Add- 7 (54)^, 3 (.16)^ and (2)^ — 5(128)^ 7. Add (4 a2 6)^, 3 a 6^, (27 a^ 6)^ and ( 125 aS ft) 8. Add 3 (^V)^ and 4 (f)^. 10 .Add {9abf, (c^ab)^, (^^Y and (xy)* * 11. From (18)^ subtract 8^ 12. From ( 108 a x'^)^ subtract (48 ax^)^. 13. From (432 a^ b)^ subtract (16 a? b)^. 14. From ( 192 a^ ftsji subtract (24 a^ b^)t 15. From f (f )^ subtract f (^)^. 16. From 5 (20)^ subtract 3 (45)^. 17. From (16 a 6)^ — (343 m)^ subtract (9 a 6)* — (1000c3m)t ^^- ^^^™ (l92 j -(l35)^"^*'^^K-27-) -(W 19. Multiply 7 a* 6^ by 3 a^ 6^. 20. Multiply 2 a 6 c by 5 a^ 6^ c^. 21 Multiply m a^ c* by 3 m a^ A 22. Multiply 25 x^ y by 3 x^ y*. 23. Multiply 10 (108)* by 5 (4)^. The product is 50 (108)* (4)* = 50 (432)* = 60 (216 2)^ = 50 6 . 2* = 300 (2)^. 24. Multiply 5 . 5* by .3 . 8^, and simplify XLIV. WITH FRACTIONAL EXPONENTS. 199 25 Multiply 2 . sHy 3 4^ The product is 6. 3*, 4^ = 6 3^.4^ = 6.27* 16* = 6(432)* 26. What is the product of 4, 2 (3)* and 72*? 27. What is the product of 5 (3)*, 7 (f )^ and 2*t i. 1 28. Multiply (T by o". m p 29. Multiply 3 a" by 5 a^ . 30. Multiply together 2^, 3* and 5* 31. Multiply (a + 6)* by (a + 6)i 32. Multiply 3 (c — d^ by 4 a (c — rf)*. 33. Multiply 4 aS 6 (x—y)* by 3(0 + 6)* (x— y)*. * 34. Multiply5(m + n)*(c — rf)*by7(i» + »:*(c — rf)^. a5. Multiply 3 + 5* by 3 — 5*. 36. Multiply 7 + 2 (6)^ by 9 — 5 (6)* 37. Multiply 9 + 2(10)* by 9 — 2(10)*. 38. Divide a^ by a*. 39. Divide a6*c by a^ 6^ c*. 40. Divide 6 a^^ 6^ c^ by 3 a* 6 c^, 41. Divide 3 a^ fe^ c^ by 4 a* 6^ c* 42. Divide 10(108)* by 5 (4)^. 43. Divide 10 (27)* by 2 (3)*. 44. Divide 8 (512)^ by 4(2)* 45. Divide a"* by a". 46 Divide 3 a" 6" by 4 0^6'. 47 Divide ^(§)* by ^ ( J)* 200 IRRATIONAL QUANTITIES XIJV 48. Divide (a — 6)^ by (a — 6)i. 49. Divide 12(x — y)^by 4a(x — y)i 50. Divide 13 {a + b)^ (c — d)^ by 39 m (a -f- b)^ (c — rf)* 51. Find the 2d power of 2 a^ b^. 52. Find the 3d power o{5a^b^ A 2a^6^ 53. Find the 3d power of Sx^f/ 54. Find the 4th power of ^ . 3^. 2 . 3^ fli 6 c2 55. Find the 5th power of — '- — , 50. Find the. 3d power of {a-\-b)^. 57. Find the 3d power of 3 (x — y)^. 58. Find the 5th power of 2 (x2 — y9)i (b — r)*. 59. Find the »»th power of a^ b^ c. 60. Find the 7«th power of a* 6* c «?. 61. Find the 3d power of ^-^^^-i^. 62. Find the 2d power of ?ii±y)j£Zlf(l^. 5(m + n)*(x— ^)t 63. Extract the 2d root of a^ b^. 64. Extract the 3d root of 27 a^ b^. 65. Extract the 3d root of 2 a^ b^, 66. Extract the 4th root of 3 a^ b^ c, 67. Extract the wth root of 10 a* x y^. 68' Extract the with root of 6 a' 6*. XLV WITH THE RADICAL SIGN. 20 1 69 Extract the 2d root of {a-\-b)^. 70 Extract the 2d root of 16 (a — 6)^c — rf)^. 71. Extract the 3d root of Sa^{m — n)^ (c — d)^. 72. Extract the 3d root of ^ (-' - y)U-' + d)\ 5b{m + n}^ SECTION XLV. OPERAriONS UPON IRRATIONAL QUANTITIES WITH RADICAL :>IGNt Art. 130. Although radical signs may be wholly dispensed with, and fractional exponents used instead of them, yet, as these signs occur in almost all mathematical treatises, and are some- times very convenient, we shall show how to perform the various operations on quantities affected with them. Irrational quantities affected with radical signs, are commonly called radical quantities, the mode of simplifying which has already been shown. The addition and subtraction of radical quantities are, it is manifest, performed in the same manner as when fractional ex- ponents are used. We observe, however, that radical quantities are said to be similar, when they are of the same degree and nave the quantities under the sign in all respects similar. 3 3 — Let it be required to add together ^192 and y/24. The sum expressed is y^l92 + ^24. But ^i92 z=z ^64.3 = 4 ^S, and ^24 = ^873 = 2^3; hence, ^192 +^24 = 4^3 -f- 2^3r=6y^3. In like manner, the sum of m \^a^ h^ c and 6^ „ */q3 ^ ^ ^ ah^ m -\- ah^ n) \^a c, or ah^{m-\- n) y^a c. 202 IRRATIONAL QUANTITIES X L V 8 3 — Subtract ^108 from 9^4. The difference expiessed ia 3 — 8 8 — 3 — 9^4 — ^108, which simplified becomes 9^4 — 3i/4 = 6^4. In like manner, i/o^ 6 subtracted from 3 m wc^ b, leaves' 4 - 4 — 4 - dmc \/h — a \^h =. (3 m c — a) ^b. Art 137. Rules for other operations on radicals, may be easily deduced from the modes given in the preceding section, for performing corresponding operations on irrational quantities with fractional exponents. 2%c exponents of the quantities under the radical sign and the index over that sign, may both be multiplied or divided by the same number without affecting the value of the expression. 2 8 15 10 For example, ^a^ or i/a^ z= a^ =. dP^ = ^a^^, which might have been obtained by multiplying the 2 and 3 of the expression 2 — ^a3 both by 5. In like manner, i/a^ b = i/a^ b^. 10 15 3 2 Again, ^a^^ = d^^ =z a^ z=z \Jd^ or ^a^, which is obtained 10 by dividing the 10 and 15 of the expression ^a^^ both by 5. In a similar manner, i/a^ 6^ ■=. wa^b. Art. 128. Hence, two or more radical expressions may be — 3 made to have the same index over the sign. Thus, ^a^ and ^b^ 6 — 6 — 4 6 are respectively the same as wa^ and i/6^ ; also i/a^ b and Wx y^ 12 12 are respectively the same as i/a^i^ and ^x^y^. This process is evidently the same as reducing fractional expo- Qents to a common denominator, the indices over the sign being considered as denominators, and the exponents of the quantities under the sign, as numerators. The common index will therefore be the product of all tho indices over the sign, or their least coramor multiple. a^ b^ XLV WITH THE RADICAL SIGN. 203 Art. 139. Multiply ^a by ^6. The product is Wa b ; for ^a =. a^, and ^b = b^ ; there* fore, ^a .y/b = a^ 6* = (a 6)^ = ^^. Multiply 2 y/'^ by 3 y^P! We first render the indices over the sign alike ; we then have 2 j/^ . 3 v/P = 2^^ . 3 y/'fts = 6a^^ 6 * = 6 (a^o 69)^^ ~ 6 ya^K Divide Q^ab by 3^/a. The quotient expressed is V^ 3^a Rut 6y/^ =6a^ 6^, and 3^a = 3a^; therefore ^V^"_^ == =r26^=:2y/6: 3a^ — 3 — Divide 3i/a by 5i/6 Making the indices alike and then dividing, we have -J- — z=i — 5- ■=. - . — =z:L(_A^ — Hence, we have the following RULE FOR THE MULTIPLICATION AND DinsiOW OF RADICALS. Make the indices over the radical sign alikcy if they are not Stj ; then multiply or divide one coefficient by the other ; also take the product or quotient of the quantities under the radical sign, placing the latter result under the common sign, which is to be preceded by the product or quotient of the coefficients. Art. 130. Find the fifth power of ^^^T. Since 2 ^^^ =2d^ b^, we have (2)/^Tf = (2 a^ 6*)* = 25 . J^^ b^^^ = 22a^ b^ = 32 (a^^b^)^ = 32^aiop 204 IRRATIONAL QUANTITIES XLV This result might have been obtained from 2 i/a^ b by rais« hig the coefficient 2 to the fifth power, and multiplying the expo* nents o/*the quantities under the radical sign by 5. 9 — Raise 3 wa^ to the third power. Since S^^ = sJ, we have (S^a^f = (3 J)^ i= 32» X J^ = 27 a^ = 27 ^^. This result is obtained from S^^, by raising the coefficient 3 to the third power, and dividing the index 9 by 3. Hence we have the following RULE FOR RAISIIVG A RADICAL TO AKTY POTVER. Raise the coefficient to the power required^ and either raise the quantity under the radical sign to the same power , or divide the index over it by the number expressing the degree of the power. Art. 131. Since extracting a root is the reverse of finding a power, we have the following RULE FOR EXTRACTING ANY ROOT OF A RADICAl,. Extract the root of the coefficient^ and either extract the root of the quantity under the radical sign, or multiply the index over it by the number expressing the degree of the root. Thus, the third root of 8 ya^ is 2 ya^ ; and the fourth root of 81 ^flS" is zfc 3 p'os. The fifth root of 4 /a2 is ^4 . ]^a^=\^'^ . ]^a^ :i= Art. 13S. The division of irrational quantities often gives rise to fractions, whose numerators and denominators are both irrational. In such a case, it is often desirable to convert the fraction into another equivalent to it, but of a simpler form. This may be accomplished by multiplying both terms of the fraction, by any quantity, which will rtnder oni of them ra" ion a] X.LV. WITH THE RADICAL SIGN. 205 Thus, if both terms of — _- or I / — be multiplied by v/«, we have — =, or if both be multiplied by ^x, we have V^^ \/ax X . V^ In like manner, multiplying both terms of the fraction — b} \/{a + xy\ gives 3 " = ^"p^ = l/¥ . \/(a-\-xY __ ^b^(a-\-xY ^ _^ ^~^ _ i ^= i » ^so, — 1= — 5— J = a + x a + x J, -i —^ - n- 2" X" X n— 1 X Since the product of the sum and difference of two quantities is the difference of their second powers, Art. 33, if we would render the denominator of — 1 _ rational, we multiply both 3-^2 terms of the fraction by 3+i/2. We have then — ?- — - =: V/2(3 + v/2) _ 3v/2"H-2 _ 3^/2+2 (3~\/2) (3 + V/2) ~ 32_(^2")2 9^=2~ ~. 3i/2+2 , , . . a — \/'h -1 . Also, multiplymg both terms of the fraction - 7 _ a + t/j by a — 1/6, we have 1 . To render the denominator of. ^ _ rational ^10—^2 — ^3 first multiply numerator and denominator by ^10-|-t/2 +l/3 18 206 IRRATIONAL QUANTITIES XLV V/30 + »/6 + 3 ivhich gives 1 L__ , then multiply both terms of thia last bj 5-j-2i/6 , which gives 5—2/6 5/30 + 2/180+11/6 + 2'; 25 — 24 = 5/30+12/5 + 11/6+27. Simplifications of this kind may be made in fractions nvolvin^ Radicals of other degrees than the second ; but, except when the quantity to be rendered rational is a monomial, the process be- comes so complicated as to be inconsistent virith the design oi this treatise. Remark. In the following questions, let the learner simplify his results, when it can be done. 1. Add /8 and /50. 2. Add /16^ and /46. a Add /36 a2 y and /25y. 4, Add /500 and /108. 5 Add 4/147 and 3/75. 6 Add 3/1 and 2/^. 7 Add 9/243" and 10/363 8 Add 12 ^i" and 3^^. 9. Add ^/a26"and ^/46x4 10. Add /12 + 2/27 and 3/75 — 9/48. 11 Add 7^54 + 3^/^16 and ^2 ^5^128. 12L Add ^ST — 2^24 and /28 +2/63. 13. Add /18a5 63 and /50 a^ ¥. 14. Add /45c3— /8073 and /So^ 15. Add y/ -p- and-y/ S^ —\/ T^- XLV WITH THE RADICAL SIGH 2(0 16. From y/50 subtract ^8. 17. From ^448" subtract ^112.' 18 From ^192 subtract ^24. 19. From 5 ^20 subtract 3 y/45. 20. From ^320 subtract ^40. 21. From ^f subtract \^^, 22. Fromy/8 subtract 2y/|". 23. From ^72 subtract 3 y^^ 24. Fromy/SOa^a; subtract y/20 aS x3. 25. From S^o^y subtract 2^a6T. 26. From ^256 subtract y^32. 27. From ^^ subtract \/\ 28. From 7^54 + 3 ^Te subtract 5 ^128 — ^%. 29. From ^| —\/l subtract y/l —^\, 30. Multiply 3 ^2 by 2 ^2. 31. Multiply ^2 by ^8. 32. Multiply ^2 by ^4. 33. Multiply ^o" by \/b, 34. Multiply 2 ^a6 by 3^ac. 35. Multiply 5 ^a^ c^ by a ^ac, 36. Multiply 5^5 by 3y/a 37. Multiply 2 ^3 by 3^4! 38. Multiply 2 a + 3^6 by 2 « — 3^6, 39 Multiply 7 + 3^12 by 3 + 4^2. 40. Multiply 3 + ^5" by 2 — ^5^ tub IRRATIONAL dUANTlTlEo. XLV 41. Multiply y/2 H-^3 by 2^2" —y/f . 3 & — 42. Multiply fci/a^ by cy/fl. 43. Multiply ^f by Yi". 44. Multiply ly/f by 3^/1: 45 Multiply together ^2, ^6 and ^12. 46. Multiply together 2^3^ 3^4 and 6y/8. 47. Multiply together 3 y/rt 6, 4^a6 and 2^m _ 3 _ 7 — 48 Multiply together ^^^, ^^7 and ^6. 49 Divide 6 ^a~ by 3 ^oT 50 Divide 8 ^oT by 2 ^a. 51 Divide S^x^ by 5^^^. 52. Diride 8^108 by 2^6. 53. Divide ^^5" by ^^2. 54. Divide ^7" by ^tT 55. Divide 3^^ by 2^^ 56. Divide a i/x y by b C/4: c. 57. Divide a + y/6 by a — ^h. Remark. In this and the two following examples, first repre- lent the division, and then simplify by rendering the denomina- ors rational. 58. Divide 4 + ^2" by 3 + ^27 59. Divide ^3 by 3 + ^3. 3 _ GO Find the 2d power of ya. 61. Find the 2d pow its absolute value is greater. Thus, ]> — 2, and — 3 ^ — 7 814 INEQUALITIES. XLVIl With a few exceptions, the principles established relative to equations, are also applicable to inequalities. We shall proceed to notice these principles and the exceptions. The quantities separated by the sign ^, are called memhers oftlic inequality. An inequality is said to continue in the same sense, when that member which was the greater previous to a particular operation, continues so afterwards; and two inequali- ties are said to exist in the same sense with regard to each other ^ when the corresponding members are the greater members. Thus, a'^b and c^d exist in the same sense, because the first member of each is greater than its second. X. The same quantity or equal quantities may he added to both mchtbers, or subtracted from both members of an inequality , and the inequality will continue in the same sense as before. Thus, if 5^3, by adding 4, we have 5 + 4 ]> 3 -{- 4, or 9^7; also, if a > 6, we have a -\- c^ b -\- c Again, if — 3> — 7, by adding 8, we have 8— 3>8 — 7, or 5>1; also, if — a ^ — 6, we have c — a^ c — b. Moreover, the inequalities 10]>7, and a^b, give, by sub- traction, 10 — 5^7 — 5, or 5 ]> 2, and a — c^b — c. Hence, we may transpose from one member to the other any term of an inequality, taking care to change its sign ; because that is equivalent to subtracting the same quantity from both members, or adding the same quantity to both members. Thus, ?f 3 X -|- 20 ^ 40 — x, we have, by transposition, 3 x -|- 2; ]> 40 — 20, or 4 x> 20. 2. The corresponding members of two or more inequalities, existing in the same sense with respect to each other, may he added, and the resulting inequality will exist in the same sense as the given inequalities. Thufl, by adding the two inequalities 5^3 and 15^7, w<^ have 5 -f 15>3 + 7, or20>10. Also, ifa>6, c>f/, and c ^fy ^ve have a -\- c -\- e'^ h -\- d -\-f. 3, But if two inequalities existing in the same sense, be sub tr acted, member from member, the resulting inequality will not always' exist in the same sense as the given inequalities. XL VII. INEQUALITIES. 215 Indeed the resul may, according to circumstances, be an ine- quality in the same sense as those given, or one in a different Bense, or it may be an equation. Thus, 13 > 4 and 20 > 7 give, by subtraction, 20 — 13 > 7 — 4, or 7^3, which is an inequality in the same sense as the two proposed. Again, 15 > 12 and 10 > 3 give, by subtraction, 15 — 10 <^ 12 — 3, or 5 <^ 9, an inequality in the opposite sense to the pro- Finally, 20> 17 and 12 > 9 give 20 — 12 = 17 — 9, or 8 =: 8, an equation. In general, let a ^ 6 and c'^ d; then, according to the par ticular values of a, 6, c and rf, we may have a — c^ b — d, a — c<^h — rf, or a — c =:: b — d. 4. The two members of an inequality may be multiplied or di vided by the same positive quantity ^ or by equal positive quanti- ties, and the result will be an inequality in the same sense as the proposed. For example, multiplying both members of 11^7 by 8, we have 88 ^ 56. Also, if a^b, ac^b c. Again, dividing both members of 35 ^ 21 by 7, we have 5^3. Also, if am^cm, we have a^ c; and if m ^ w, m n p p' 5. Butf if both members of an inequality be multiplied by the same negative quantity, or by equal negative quantities, the re- lult will be an inequality in a sense opposite to that of the pro posed. Thus, if 7^5, and we multiply by — 3, we have — 21 < — 15 Also, if a ^ 6, multiplying by — m, we have — am << — b m. In these examples, the sens«^ is inverted, because a negative quantity is less in proportion as its absolute value is greater. 6. Htnce it follows, that the sense of an inequality will be in- 216 INEQUALITIES. XLVIl verted, if all the signs of both members be changed ; because thii is the same as multiplying both members by — 1. 7. Both members of an inequality, if they are positive quanti^ ties, can be raised to the same power, and the result will be an in^ equality in the same sense as the proposed. Thus, from 7>2 we have 72>22, or49>4; and fJGto a^ b, a'"^b'^. 8. But if both members of an inequality are not positive, and both be raised to the same power denoted by a whole number, the *'esulting inequality will not always exist in the same sense as the proposed. Thus, 3 > — 2 gives 3^ > (— 2)2, or 9 > 4, in the same sense as the proposed. But — 3 > — ^5 gives ( — 3)2 <;^ ( — 5)®, or 9 <^ 25, in the reverse sense of the proposed. 9. Roots to the same degree, of the two members of an inequal- ity, may be extracted, and the resulting inequality will be in the same sense as the proposed. Thus, 27 > 8 gives ^27 > ^8^ or, 3 > 2, and, in gen n , n eral, a^b gives i/a ^ r/6. Tf the root be of an even degree, it is necessary that both members of the given inequality be positive ; otherwise one or both of the roots would be imaginary, and they could not be com- pared. Art. 136. There are some problems, the solution of which involves the principles of inequalities. The following are of this kind. 1. Three times a certain number added to 16, exceeds twice that number added to 24, and two fifths of the number added to 5 is less than 11. Required the number. Let X represent the number ; then Sx -\- 16^2x-|- 24, and 2 X \- ^ <^ 11. The first inequality, by transposition and re- 5 duction, gives z ^ 8. The second, multiplied by 5, becomes XLVIII. EQUIDIFFERENCE. 217 2z -\-25<^^o, which, by transposition, reduction, and division, gives x<^ 15. Any number, therefore, entire or fractional, which is greater than 8«and less than 15, will fulfil the conditions of the question. 2. Says A to B, I have an exact number of dollars in my |j!jrs«; if I had twice as many and $10, 1 should have more than $19; but if I had three times as many, my number would be less than the number I now have increased by $41. Required the number of dollars in his purse. 3. A certain number divided by 17 gives an entire quotient, which quotient increased by 2, exceeds 4 ; but if the number be multiplied by 2, and the product be increased by 4, the result will be less than the number itself increased by 56. What la the number? SECTION XLVIII. EQUIDIFFERENCE. Art. 137. The difference between two quantities is some times called their arithmetical ratio, or ratio oy subtraction Thus, the arithmetical ratio of 9 to 7 is 9 — 7 or 2, and that of a to 6 is a — 6. Four quantities, such that the difference between the first and second, is the same as that between the third and fourth, consti- tute what is called an equidifference, sometimes called also an arithmetical proportion. Thus, 9, 7, 5 and 3 form an equidiffer- ence; for 9 — 7 = 5 — 3. This is sometimes expressed thu 9.7:5.3, in which one point denotes difference, and two points denote equality. But this notation is objectionable, because these characters are sometimes used, the one to represent multi plication, and the other division. In like manner, if the quantities, a, 6, c and c?, are such thai a — bzrzc — rf, or — a-\-bz=: — c -}- d, these four quaniiiiei constitute an equidifference. 19 218 EQVIDIFFERENCE. XLVIU The quantities, «, b, c, d, are called terms of the equidiffe^ euce. Also, a and d, the first and last terms, are called the ex^ tremeSf because they occupy the extremities ; b and c, the second and third terms, are called the means, because they occupy the middle in the equidifference. Remark. In the definition of equidifference, it is supposed, that, if the second term is greater than the first, the fourth is greater than the third ; but, if the second is less than the first the fourth is less than the third. From any equidifference, a - — b =: c — d, or, —ra-{-b=z — c -\-d, we deduce, by transposition, a-\- d =i b -\- c; also, « =r 6 -f- c — d, and d=ib -\-c — a, b=: a-\-d — c, and c =r. a-\-d — b. Hence, In any equidifference, the sum of the means is equal to the sum of the extremes. Moreover, either mean is equal to the sum of the extremes, diminished by the other mean ; and either extreme is equal to the sum of the means, diminished by the other extreme. Suppose we have a-\- dz=b-\-c; by transposition we obtain a — bz=: c — d. Therefore, if the sum of two quantities is equal to the sum of two other quantities, the first two may be made the means, and the last two the extremes, or the reverse, of an equidifference. When three quantities, a, b, c, either increasing or decreasing, are such that the difference between the first and second s equal to that between the second and third, that is, a — bz=.b — c, they constitute what is called a continued equidifference, and the quantity b is called the arithmetical mean between a and c. Thus, 3, 5, and 7, or 12, 8, and 4 form a continued equidifference. Take, for example, a — 6 = 6 — c. From this we deduce (I I c i = — ^— ; alsOj az=z^b — c, and cz=z^b — a. Hence, In any continued equidifference, the mean is half the sum of the extremes, and either extreme is found by subtracting the other cav. treviefrom twice the mean XLIX PROPORTIONS. 21P 1 The means of an equidifference are 10 and 12, and he known extreme is 6. Required the other extreme. 2. The extremes of an equidifference are 7 and 4^, and one of the means is 6. What is the other mean ? 3. The means of an equidifference are 8 and 12, and t\e last term exceeds twice the first by 5. Required the extremes. 4. In a continued equidifference, the extremes are 10 and 15^. Required the mean. 5. In a continued equidifference, the mean is 7 and onf ex- tieme is 8. Required the other extreme. 6. The mean of a continued equidifference is 14, and the third term exceeds the first by 8. Required the extremes. SECTION XLIX. RATIO AND PROPORTION. Art. 138. The quotient arising from the division of one quan- tity by another, whether the division can be exactly performed or can only be expressed, is called the ratio of these quantities. It is sometimes called ratio hy division y or geometrical ratio. But when the word ratio simply is used, it signifies ratio result- ing from division. A ratio is most appropriately expressed in the form of a frac- tion. ThuiS, f is the ratio of 3 to 5, and j- is that of a to b. An equation formed by two equal ratios, is called 2i proportion. Sometimes the term geometrical proportion is used, to distin- guish it from arithmetical proportion or equidifference. Thus, 3 9 J a c - zr: --, and 7 = - are proportions. 7 21 a For the sake of convenience in writing and printing, most au« thors express division by the sign : , placed between the quanti- ties, and, instead of the sign =, use the sign : : Thus, a:b: . 220 PROPORTIONS XLIX c:d 18 read, a is to i as c is to d^ and is the same as y =^ ~i o a The signification in both cases, is, that a divided by b, is equal to c divided by d. In this treatise points will sometimes be usea to denote division, but the sign =: will always be preferred rather than : : . In any proportion, a : b z=z c : d, the quantities a, b, c, and to its consequent. a c 16. J{a : b =: c : d, and e :f= g : h, that is, - = — , and -^ = ^, by multiplying these two equations together, we obtain lJ^Jf,^^^<^^''^f=^S''dh. The same result would have been obtained by multiplying to- gether the corresponding terms of the two given proportions. This is called multiplying the proportions in order. Hence, If two or more proportions are multiplied in order , the result will form a proportion. From this it follows, that. If proportions are divided in order, the result will form a prO' portion. a c 17. Given, a : b = c : d^ or - = —. a Raising both members to any oower denoted by rw, we have or c"* Therefore, Similar powers of proportional quantities form a proportion. From this it follows, that. Similar roots of proportional quantities will form a preptn tiom. PROGRESSION BY DIFFEREI^CE. 2^9 SECTION L. PRyGRJCSSION BT DIFFBREHTCB. Art. 140. A series of quantities, such that eaci is greater ihan that which immediately precedes it, or such that each ex- ceeds that which immediately follows it, by the same quantity, "ionstitutes what is called a progression hy differencCf or aritltr nrtical progression. Thus, the natural numbers, 1, 2, 3, 4, &c., form such a pro- gression, since the difference between any two contiguous num- )e.rs is unity. Progression by difference may be either increasing or decreas- m+\' m+l ' m+1 ' '* As an example, let it be required to insert six mean differen tials between 4 and 25. Here d =. — r— ; . becomes d =r — - — = 3 ; and he pro m+ 1 7 ^ ;?ression is 4, 7, 10, 13, 16, 19, 22, 25. It is manifest from what precedes, that, 20* 234 EXAMPLES IN PROGRESSION LI If between the terms of a progressitm by difference taken two and twOf the same number of mean differentials be inserted, the result will be in progression. For example, let it be required to insert between every two adjacent terms of the progression, 3, 9, 15, 21, two mean differ- entials. In this case, d = — r— - becomes d z=z — — — =. 2 ; and the wi-j- 1 3 orogression is 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. SECTION LI. EXAMPLES INVOLVING PROGRESSION BY DIFFERENCK. Art. 14:^. 1. How many strokes does a clock strike in 12 houf s ? 2. Find the 10th term and the sum of the first 10 terms of the series, 20, 25, 30, &c. 3. Find the 16th term and the sum of the first 16 terms in the series, 100, 98, 96, &c. 4. Find the last term and the sum of the series, 12, 13^, 14|, &c. the number of terms being 30. 5. The number of terms being 28, find the last term and the sum of the series, 3, 3|-, 4f , &,c. 6. Insert six mean differentials between 20 and 55. 7. Insert five mean differentials between 6 and 10. 8. Insert five mean differentials between every two adjacent terms of the progression, 5, 17, 29, 41. 9. Suppose t"hat, as in Venice, a clock denoted, by the number of strokes, the hours from 1 to 24, how many strokes would such a clock strike in 24 hours ? 10. A farmer wished to set out, upon a triangular piece of ^and, 25 rows of apple trees, he first row containing 2 trees, the second 5, the third 8, and so on. How many trees did he re^ quire for his purpose? LI BT DIFFERENCE. 235 11 A gardener has 100 plants and a reservoir of watei all in a straight line, the plants being 3 feet asunder, and the reservoir 10 feet from the first plant. How far must he walk in order to water these plants, if he commence at the reservoir, and return lo it for a new supply of water for each plant, finally coming to the reservoir after having watered the last one? 12. A falling body descends, in vacuo, 16^ feet the first sec- ond, and in each succeeding second, 32^ feet more than in the preceding. How far will a body descend in 10 seconds? 13. We observe that, in the preceding question, the difference is just double the first term. Let the learner generalize that question, by substituting, in the second formula for jS^, 2 a instead ofd 14. Two travelers, A and B, 188 miles asunder, set out at the same time with the intention of meeting. A goes regularly 10 miles per day ; but B goes 3 miles the first day, 6 the second, 9 the third, and so on. In how many days will they meet? 15. Two men, 135 miles asunder, set out at different times and travel towards each other. One starts 5 days before the other, and goes 1 mile the 1st day, 2 miles the 2d, 3 miles the 3d, and so on. The other travels 20 miles the 1st day, 18 the 2d, 16 the 3d, and so on. How many days and what distance will each have traveled when they meet ? 16. Divide 51 into three parts, which shall form a progression by difference, the common difference being 5. 17. Find three numbers in arithmetical progression, such tha their sum shall be 18 and their continued product 192. Remark. Let y =: the common difference, and x z= the mea^ term. Then x — y, a:, and x-\-i/ will represent the numbers 18. Divide 50 into five parts, which shaJl form a progression by difference, of which the first term shall be to the last as 7 to 3. 19. There is a number consisting of threp digits, which form a decreasing arithmetical progression. The sum of the digits is 9, and if 396 be subtracted from the number, the digits will be inverted. Required the number. SK6 PROGRESSION BY QUOTIENT LIl SECTION LII. PROGRESSION BY QUOTIENT. Art. 14G. A progression by quotient , called also geometricai progression, is a series of quantities such, that, if any one of them be divided by the next preceding, the quotient will be the same in whatever part of the series the two successive terms are taken. Progression by quotient may be either increasing or decreas ing. Thus, 2, 4, 8, 16, 32 form an increasing, and 60, 20, ^^, ^, a decreasing progression by quotient. The quotient arising from the division of any term by that which precedes it, is called the common ratio. The ratio in the first of the two progressions given above, is 2, and that in the second is ^. In general, let a, b, c, d, &.c. be the successive terms of a pro- gression. Let q represent the constant ratio ; then, since each term is q times the preceding, we have b=:aq, c=iaq^, dzzzaq^, ez=zaq^, &lc. Now representing the last term by Z, and supplying by points the place of the indefinite number of terms omitted, the terms of the progression will be, 1st 2d 3d 4th Sth 0th 7th a, aq, aq^, aq^, aq^, aq^, aq^, , /. We readily perceive, that the exponent of q in any term is ess by unity, than the number which marks the place of that term. Thus, the 4th term is aq^, the Sth aq^. Consequently, if n rep- resent the number of terms, the nth or last term will be aq''~^. Therefore, the formula for the last term is Hence. Any term of a progression by quotient , may be found, by mul- tiplying the first term by that power of the ratio, denoted by a number 1 less than that which marks the place of the term. LII. PROGRESSION BY QUOTIENT. 2«i^ Ex. What is the sixth term of the series, 3, 6, 12, itc? Here a = 3, ^r = 2, and nz=zQ\ therefore, 1=: aq"'^ be- comes / = 3 . 25 = 3 32 = 96. Ex. 2. Required the 7th term of the sesies 3645, 1215, 405, &c. In this question, a = 3645, gr = ^, n = 7. Then / = 3645 X 'i)6 = 3645.^^=5. Art 147. To find the sum of any number of terms, denote chis sum by ^S^ ; then 8z=.a-\-aq-\-aq^^-^aq^-{-aq^-\-aq^-\- -^ a 3"~2 -|~^ 3'*~^' Multiplying this equation by g, we have 1 Sz=zaq-{'aq^-\'aq^-\' aq* -\- aq^ -\- aq^ -{- -\-aq'^^-\-aq^ Subtracting the former of these equations from the latter, ob- ierving that the terms of the second members all cancel except a and ay", we obtain ^8 — Sznaq^ — a; or {q — \) 8 z=i a q^ — a; consequently, _ qg" — fl _ «(g"— 1) _, ag" — a aq^^^q — a ... ,. , n- But — • = —5 ^- ; substitutmg / mstead of lUi equal, ag"~^, we have Sz=: — — . We have then two formulae for the sum of a geometrical pro- gression, viz : a(«" — 1) , ^ ql — a __ 5 = -^ — ^, and S = ± -. Hence, g-— 1 q — l To find the sum of a progression hy quotient^ subtract unity from that power of the ratio ^ denoted hy the number of terms, multiply the remainder by the first term^ and divide the product hy the ratio minus unity ; or^ multiply the last term hy the ratio subtract the first term from the product ^ and divide the remainda hy the ratio minus unity. 238 PROGRESSION BY QUOTIENT. LIl Ex. Required the sum of the series, 1, 2, 4, 8, &c., the number of terms being 10 ? In this question, a = 1, j = 2, and n = 10. Therefore, S=. 1J21"-- 1) _ 1 (1 024-1) _ 2_f-- -2-1 -1*^^- Or we may first find the last term, and then use the formula S=: il^. We have then /= 1 . 29 = 512, and S- 2. 612-1 _ 1024-1 _ 2-1 - 1 -^^^' Ex. 2. Required the sum of the first six terms of the series, lO, 5, f , &c. ? Here a = 10, j = ^, and n = 6. Using the first formula for S. we have ^^ lofW"-!] ^ IQ(A-l) ^ 10 (-ff) ^ Art. 14:8. If y is a proper fraction, q — 1 will be negative, q^ — 1 will also be negative, since any power of a proper frac- tion, if the index is greater than unity, is less than the fraction itself. Changing the signs of numerator and denominator, in 1/. 1/.^ , « «(l — Q^) ^ ^ — <2 7" the formula for S, we have Sz=z —\ i-^, or <»=z — ^. l_-.gr 1— gr Now, as the powers of a fraction less than unity become less and less, the greater the index of the power, it follows that if w, the number of terms, is infinitely great, 5" must be infinte y small, that is, zero. In this case, Sz=z — — will become *^=x. 1 — q a — a .0 a 1-gr l-q Since q is supposed to be a fraction, let it be represented b} ~ , so that q= —. We shall then have o = z= %* ^ n 1 — ^ n — fit 111. PROGRESSION BY QUOTIENT. 239 The formala, therefore, for the sum of a decreasing progres sion by quotient, continued to infinity, is n — m Hence, Tojind the sum of an infinite decreasing series in progression by quotient y multiply the first term hy the denominator y and di' vide the product by the difference between the denominator ana numerator of the ratio. If 9 is a fraction whose numerator is 1, the formula for he sum of an infinite decreasing series, becomes Sz:z -. n — 1 Ex. What is the sum of the series, 5, -^, ^, &c., continued lo infinity. 5 3 In this example, a = 5, and J = f ; therefore, S z= = 3 — 2 15. Art. 14:9. From the formula for the sum of an infinite de- creasing progression by quotient, may be deduced the rule given in arithmetic, for reducing periodical fractions, sometimes called repeating and circulating decimals^ to vulgar fractions. Let us take the decimal "333 continued infinitely. This is a Drogression by quotient, in which the first term a is -^^, the sec- ond y^^, &/C., the common ratio q being -j^. Hence, by sub- a n ^ 10 stitution, the formula, S =. , becomes S =r ^-^ == n — 1 10 — I Again, in the fraction, '0404 &c.^ a =. y^^y, and q z=: ^^^ ; lience, S=: -^ becomes^^ = ^qV— i = ^• In like manner, the sujr )f -290296 &c., = i^inny ♦ 1000 1000—1 240 PROGRESSION BV QUOTUNT. LL Let us take the fraction •428571428571 &lc. Here a — NutMt^* and as the second period is -000000428571, q =r. TuuifUTjTi- Therefore iS=: ioooooo_i ~ UUih which reduced, is f. Consequently we see, as in arithmetic, that the repeating or circulating figures are to he made the numerator of a fraction^ whose denominator is as many 95 as there are repeating figures^ and thtn the resulting fraction is to be reduced to its lowest terms. ■ If the repeating figures do not commence immediately after the decimal point, we have only to find the value of the repeat- ing part, and add it to the decimal which precedes, reducing them both to the same denominator. For example, -5333 &c. = -^^ -f- y^^y + ^^n &lc., in which the first term of the progression is y^^, and the ratio ^ j in tbis ^ . 10 case, the sum of the whole is <^jy -\- ^^^ ' = ^ -|- f^tj - Art. l^O. Suppose that a and / were given, and it were re quired to insert between them any number of terms, such tha^ Ihe whole should form a progression by quotient. The term? thus introduced are called mean proportionals, or geometric means * From the equati9n l = aq^—^j in Art. 140, we have Now, if it be required to insert m terms between a and /, since there would be m -f- 2 terms in the whole, we put »i + 2 instead of n, in the value of g, just fouiid. LII. PROGRESSION BY QUOTIENT. 24' We have then m+l _ q z= y/ - . Hence, When any number of mean proportionals is to be inserted he» ttoeen two quantities ^ tojind the common ratio j divide the greater quantity by the less, and extract the root of the quotient to the de- gree denoted by the number of terms to' be inserted plus unity. Knowing the common ratio, it is easy to write the progression, which is expressed in general terms as follows, viz : m-\-l m-j-l OT+l «,«V/ i.«V/ («)• "via) '■ Ex. Insert five mean proportionals between 2 and 128. OT+l In this example the formula, q =z\X - , becomes 5=. ^JL|I — ^64 = 2 ; and the progression is 2, 4, 8, 16, 32, 64 128. It is manifest, that the same number of mean proportionals may be inserted between the terms of a progression by quotient taken two and two, and the result will be in progression. Ex. Between every two adjacent terms of 3, 24, 192, insert two mean proportionals. 3 — 3 - In this case, q =: y^ = y^S = 2 ; and the resulting pro- gression is 3, 6, 12, 24, 48, 96, 192. Art. 151. In the formulae already given, a, q and n were ■supposed to be known, and it was required to find / and S. Bui if any three of the five quantities, a, q, n, / and S, are knowa. thf» remaining two may be found. There may be, therefore, ten different problems ; but the stu- dent, at this stage of his progress, is capable of solving only four .>f them. Four of the remaining six will be solved under thfl 21 242 « EXAMPIES IN LIIl head of logarithms ; but the two others give rise to equations too difficuh of solution to be admitted into an elementary trea- tise. Let the pupil solve the following problems. 1 . Given a, q and n ; find / and 8. Note, This question has already been solved, and the results may be assumed in solving those which follow 2. Given a, n and /; find q and 8. 3. Given q, n and /; find a and 8. 4. Given y, n and t tions, i/^ -j- 5 — ^ := rb I ; ransposing and reducing, i/i -[- 5 = 3, or i/a; + 5 = — 2; squaring both equa- x-(-5=:9, orx-|-5 = 4i transposing and reducing, =1 4, or X = — 1. Ans. Another method. Resume the equation, x-\-^ — x/x -\-^:=zQ. Substitute some letter, as y, instead of ^x+^5 ^^^^ we have ^x -|- 5 = y, and consequently x -)- 5 = v'^ ; hence, the eq'iation becomes y^ — y = 6. This gives, by the formula, y = ^± v/6 + 1 = 3, or y = — 2 ; therefore, y2 = 9; ory^ — A. But y'^z=.x-\-^\ hence, x4-5 = 9; orx-|-5=:4; transposing and reducmg, x = 4t; orxz=: — 1. Ans. The latter method of solution is preferable, as it saves the ne« cessity of repeatedly writing a polynomial. 33. Given2x2 + 3x — 5^2x24-3x4-9 = — 3; to find z Adding 9 to each member. 2x2 4-3x4-9 — 5y/2x2 4-3x4- 9 = 6. Lety = ^2x2 4-3x + 9; then y2 — 2x2 4-3x4-9; hence y2 — 5y = 6; then, by formula 2d, Art. 96, y = ^=tv/6 + V==6,oryr= — I; hence, y^ = 36, or y2 _ 1, Taking the 1st -alue of y^ 2x94-3x4-9rr 36; x2 4-|x = V| z = 3 ; or X = — |. 250 EXERCISES IN EQUATIONS L\ Taking the other value of y^^ viz : 1, — 3d=v/— 55 -= — t — • Solve the following equations. 34. 2 + 16 — 7^z + 16=:10 — 4^x + 16 35. x + ^i+r6=:2 + 3y/x + 6. 36. x2 — 2z + 6^x2— 2ar + 5 = ll. 37. (lO + a;)^— (10 + a:)^ = 2. 38. (a;— 5)3— 3(2— 5)^ = 40. v/a:24.^_|,6 18- (|v/x2 + 2 + 6-^^2) 3 - ^2:2 + 2 + 6 40. fl;2_a;_^5^2x2— 52 + 6 = ^"'^^ . SECTION LV. CXERCISKS IN EQI7ATION8 OF THX SECOND DEGREE WITH TWO tJlT KNOWN QUANTITIES. Art 154. 1. Given ^ ""J^ 7-^1^^^ \ ' *^ ^^^ * ^""^ ^ From the 1st equation, X =:2y ; substitute this value of y in the 2d equaticn, 4y2_y2 = i2; 3y2=,i2; y2 = 4; y = ±2. LV OP THE SECOND DEGREE. 251 Hence, x =: 2 y = ± 4. In the preceding question, the value of one unknown quantity was found in one equation, and substituted in the other ; and in this way the solution can be effected, when one of the given equations is of the first degree, and the other of the second. Find X and y in the following equations. 42^ + 23^ 2. 5. 2y + 4z=:ll. + 33^^ When both equations are above the first degree, different ex pedients are to be adopted, which will be best learned from examples. ((1) zy = 50. Adding twice the 1st to the 2d, (3) x2 -f- 2 z y 4" J^^ = 225 ; taking the square root (4) x + y = ±:15. Su )tr acting twice the 1st from the 2d, (5) x2 — 2 X y + y2 ::- 25 . taking the square root, (6) X — y = ±:5. Adding tho 4th and 6th, and dividing by 2, x = ±10. Subtracting the 6th from the 4th, and divid ng by 2, 3^==h5. 252 EXERCISES N EQUATIONS L\ By taking all the possible combinations of the signs in the 2d members of the 4th and 6th, we have X = 10, and y = 5 ; or, x=z — 10, and y =: — 5 ; c r, z = 5, and yzz: 10; or, X = — 5, andy = — 10. 5(1) z^+xy = li. h2) xy + y^ = 2i. Aoding the 1st and 2d, (3) x2 _j_ o 3. y _^ y2 _. 36 . taking the root. (4) x + y = ±6. But, x^-\-xi/= 12 is the same as {x-\-7/)x=i 12; substituting in this dt 6 instead of =b6x=12; x = db2. Substituting this value of x in the 4th, db2+yr=d;:6; i/ = ri=6iF2,or, i/ = rt 4. In the last equation but one, the upper signs correspond, as also do the lower. ((1) x + y = s. Squaring the 1st, (3) x^-\-^xy-\-y^z=:s^\ subtracting from this 4 rimes the 2d, (4) X* — 2xy-|-y^ = ^'^ — 'ia^'f taking the square root. (5) X — y — dcv/^'^ — 4a2. Adding the 1st and 5th, and dividing by 2, s-j=i\/s'^ — 4a^ V — !1 LV. • O/ THE SECOND DEtJREE. 25o Subtracting the 5th from the 1st, and d.viding by 2, 2 l('Z) 2:3 + 2/3 —189. Adding 3 times the 1st to the 2d, (3) 23 -^ 3 x2 y + 3 X y2 _|_ y3 _ 709 ^ taking the 3d «xa 7 32768 15 8388608 LVI * lOGARITllMS. 25*J Suppose now it were required to multiply 1^28 by 1024. Looking in the table, we find against 128 the exponent 7, and against 1024, the exponent 10; these exponents being added give 17. We now find 17 in the column of exponents, and against it, in the column of powers, we find 131072, which is the pro- duct of 128 by 1024 ; that is, 128 . 1024 = 2^ . 2^^ =z 2^^ =z 131072. Divide 2097152 by 64. Looking iff the table, we find 21 for the exponent correspond- ing to the dividend, and 6 for that corresponding to the divi- sor ; subtracting the latter from the former, we have 15 for the exponent corresponding to the quotient; we now find 15 among the exponents, amd against it stands 32768, which is the quotient 2097152 221 required. That is, J^ ='^=2'^ = 32768. Find the third power of 64 The exponent against 64 is 6, which multiplied by 3 gives 18 ; against the exponent 18 we find 262144, which is the power re- quired. That is, (64)3 _ (26)3 _ 218 ~ 262144. Find the fifth root of 32768. Against 32768 we find the exponent 15, which divided by 5, gives 3 ; against the exponent 3 stands 8, which is the root re- quired. That is, (32768)^ = (2^5)i = 2^ zn 2^ = 8. Let the learner find the answers to the following questions by means of the table. 1. Multiply 16 by 128. 2. Multiply 1024 by 64. 3. Multiply 512 by 2048. 4. Multiply 256 by 16384 5. Multiply 256 by 512. 6. Multiply 32768 by 128. 7. Divide 2097152 by 65536 8. Divide 32768 by 1024. 9. Divide 262144 by 16384. SWM) LOGARITHMS . LVl 10 Divide 524288 by 512. 11. Divide 4096 by 256. 12. Divide 8:388608 by 131072. 13 Find the 3d power of 32. 14. Find the 2d power of 128. 15. Find the 4th power of 16. 16. Find the 2d powder of 1024. 17. Find the 4th power of 32. 18. Find the 5th power of 16. 19. Find the 2d root of 1024. 20. Find the 3d root of 512. 21. Find the 6th root of 262144. 22. Find the 4th root of 65536. 23. Find the 5th root of 32768. 24. Find the 7th root of 2097152. Art 1^6. The number 2, which is raised to the several pow- ers in the preceding table, is called the base of the table ; and the exponents of the several powers, are called logarithms of the numbers to which those powers are equal. Thus, if 2 is the base of the table, the logarithm of 256 would be 8, and that of 16384 would be 14. A table might be made, having for its base 3, 5, or any num- ber except 1. Unity would not answer for a base, because all the powers as well as all the roots of 1 are 1. Tables of logarithms in common use, are constructed upon the number 10 as a base. Hence, The common logarithm of a number, is the exponent of tht power to which 10 must be raised, in order to produce that num* her. Thus, 3 is the logarithm of 1000, because 10^ = 1000 , and 5 is the logarithm of 3162277, because lO^s = ^10 = 5 162277, nearly. Remark. We shall hereafter use the expression log for the words "logarithm of." LVI. LOGARITHMS. 261 Now 100 -- 1^ 10' = 10, 102 _ 100, 103 — 1000, 10* =l lOOOO, &LC. Therefore, log. 1 = 0, log. 10 = 1, log. 100 = 2 log. 1000 = 3, log. 10000 = 4, &c. Again, 10-i = tV = -1, 10-2==^^^ = -01, 10-3 = .^^^^^ = •001, 10-4 = ^^^= -0001, Art. 133. Therefore, log. -1 = — 1, log. -01 = — 2, log. -001 = — 3, log. 0001= — 4. Hence, the logarithm of a number between 1 and 10 must be a fraction, that of a number between 10 and 100, 1 + a frac- tion, that of a number between 100 and 1000, 2 -}- a fraction, and so on It also appears, that the logarithm- of a fraction less than unity, must be negative, and that the logarithms of intermediate numbers between '1 and 01, 01 and '001, -001 and 0001, will consist of whole numbers and fractions. Art. 1^7. To form a conception of the construction of loga- rithms, let us place some of the powers of 10 in one line, and their exponents or logarithms in another beneath. Thus, 1 or 100, 10, 100, 1000, 10000, 100000. 0, 1, 2, 3, 4, 5. If we examine these two series, we shall perceive that the for- mer is a progression by quotient, and the latter a progression by difference. . Now if we insert between the terms of the former series taken two and two, any number of mean proportionals. Art. l«iO, and between the terms of the latter taken two and two, the same number of mean differentials. Art. 14:4r, the terms of the latter result will be the logarithms of the corresponding terms of the former result. Thus, the mean proportional between I and 10 =z i/l . 10 = 8 162277 ; and the mean differential between and 1 ^z -L^ zzr j — -5. Then, log.' 3' 162277 = 5. If however we were to insert a very great number of mean proportionals between 1 and 10, we should find among them terms which differ very little from 2, 3, 4, 5, 6, 7, 8 and 9, and ivhich therefore might be considered equa to these numbers i^'-l LV.GARITHMS. LVl Indeed this difference would be less, in proportion as the numbei of means inserted was greater, so that the approximation might be carried to any degree of accuracy. If we insert now between and 1 a number of mean differen- tials, equal to the number of mean proportionals inserted be- tween 1 and 10, the terms of the result would be the logarithms of the terms of the series previously found, and those correspond- ing to 2, 3, 4, &c. would be the logarithms of these numbers. This process which we have given, is designed to show the learner the possibility of constructing logarithms, rather than aa a mode which can conveniently be reduced to practice. The methods by which logarithms are actually calculated, are in general very different from that given above, and are too com- plicated to be introduced into an elementary treatise. Suppose then that we have a table containing the logarithms of all the natural numbers, 1, 2, 3, &/C., to any definite extent, [n such a table the logarithm of 2, for example, is '30103 ; that 30 103^. is, IQT^^^^^ = 2. This signifies, that, if 10 were raised to the 30103d power, and then the 100000th root were extracted, the result would be very nearly 2. Art. 158. Since logarithms are exponents, they are subject to the rules previously given for exponents. Hence, 1. To multiply numbers together ^ add their logarithms; the sum. will he the logarithm of the product. 2. To divide one number by another ^ subtract the logarithm ^f the divisor from that of the. dividend; the difference will be the logarithm of the quotient. 3. To raise a number to any power, multiply its logarithm by the number expressing the degree of the power ; the product will he the logarithm of the power required. 4. To extract any roof of a number , divide its logarithm by the number expressing the degree of the root, or, what amounts to the same, multiply its logarithm by the fractional exponent which indicates the root ; the result will be the logarithm of the root rt \uired. '/\'\. LOGARITHMS. 2(^3 <>. Since a fraction expresses division^thc logarithm of afrac* lion is foundy by subtracting the logarithm of the denominator from that of the numerator. 6. The logarithm of either extreme of a proportion is found bp adding the logarithms of the means, and from the sum subtract- ing the logarithm of the other extreme ; also the logarithm of either mean is found, by subtracting that of the other mean from thi sum of the logarithms of the extremes. Art. loO. In constructing a table of logarithms, it is only necessary, in the first place, to calculate those of the prime num- bers ; from these the logarithms of all compound numbers may o*» deduced by addition and multiplication. Thus, the logarithms of 2 and 3 being found, by adding them we have that of 6. In fact, the log. 2 == 0-3010300, and log. 3 = 0-4771213; hence log. 6 = 03010300 + 0-4771213 = 0-77H1513. Again, 2 X log. 2 — 0-6020600 == log. 4, and 3 X log. 2 = 0-9030900 = log. 8, &c. Hence, from the logarithms of 2 and 3, we easily obtain those ot all the powers oi these numbers, as well as those of all the combinations of these powers. From the mode of performing multiplication by means of log- arithms, it follows that the logarithms of those numbers which are 10, 100, 1000, «fec. times the one of the other, must have their decimal parts the same, and can differ only with regard to their integral parts. Thus, the logarithm of 2 being 03010300, the logarithm of 10 . 2 or 20 =r log. 10 + log. 2=1 + 0-3010300 — 1 '3010300; log. 200 = log. 100 -f log. 2 — 2-3010300; log 2000 — log. 1000 + log. 2 z=z 3-3010300. In like manner, the og. 3 being 0-4771213, we have log. 30 = 1 4771213, log. 30(: =: 2-4771213, log. 3000 i= 3-4771213, &c. Again the logarithm of 371250 being 5-5696665, we have log. 37125 — log. (3X|2^) = 4-5696665, log. 3712-5 = log. («-^\p) = 3-5696665, LOGARITHMS 371-25 = log. (3-V^) = 2-5696665, 37125 ^log. (3VV2^) = 1-5696665, 3-7125 = log. (^7^^25) • :=, 0-5696665, -37125 =:\og.{^\h'') = — 1-5696665, •037125 = log. (HF-^) = — 2-5696665, •0037125 = log.(-^A^^) = — 3-5696665, 264 LOGARITHMS LVi log log log. log. log. log. log. 00037125 = log. ( aa3 7j^) _ _ 45696665. n dividing by 10, in each instance, we have subtracted the . )garithm of 10, which is 1, from the logarithm of the dividend. In the last four examples, the subtraction is represented only, the decimal part being positive. Art. i60. We have before shown, that the logarithms ol fractions less than unity, are negative; as represented above, however, the integral part alone is negative. But the negative part being greater than the positive, the expression as a whole is still negative. Since negative logarithms do not occur in the tables, we use the logarithms of decimals in the form given; and, to distinguish them from logarithms wholly negative, we place the minus sign over the integral part. Thus, log. •0037125 = 3-5696665. The integral part of a logarithm is called its characteristic, because it always determines the order of units, expressed by the first significant figure of the corresponding number. From wh?»t precedes we see, that. The characteristic if positive^ is always one less than the nunt' bet of integral figures in the corresponding number ; but, if the characteristic is negative, it is always equal to the number of places by which thefrst sigvif cant figure is removed to the right of the decimal point. ■ Thus, if 2 be the characteristic, there would be three figures, preceding the decimal point in the corresponding number ; but if the characteristic be 1, the first figure of the number would be tenths, if it be 2, the first figure would be hundredths. In logarithmic tables the characteristic is usually omitted. Bint© we can never be at a loss to determine it, and since the l^Vll UBE OF LOGARITHMIC TABLES. 2C^•^ same c.ecimal part of a logarithm, correspor. ds to several difTerent numbers, composed of the same figures, but in which the figures express different orders of units. SECTFON LVIl. C8E OF THE TABLES IN FINDING THE LOGARITHMS OF GIVEN NUMBERS, AND THE REVERSE. Art. IGl. Logarithmic tables are usually accompanied with directions for using them, which are somewhat different m aifTor- ent works, according to the arrangement and extent of the tables The principle, however, is in all cases the same. In some tables, logarithms are carried only to five, in others to six, and in others to seven decimal places. The student is supposed to be provided with a table of loga- rithms carried to seven decimals, extending to the number 10000. If, however, his tables are carried only to five or six decimals, he may disregard the last two, or the last decimal, in the logarithms which occur below. Art. lOS. To ^nd from the tables the logarithm of a given number. If the number consists of less than four figures, whether it be integral, mixed or decimal, find the figures in the left hand col- umn marked N. or Number, and, on the same horizontal line, in the next column to the right, will be found the decimal part of its logarithm, to which prefix the proper characteristic. In this, and all other cases, zeros to the right or left of the number will have no effect on the decimal part of the logarithm. Thus, log. 37 z= ] -5682017; log. 3700 =z 35682017; log. 385 — 2-5854607 ; log. 385000 = 55854607 ; log. 257 — 0409933 1 ; og. 0573 = 2 7581546. To find the logarithm of a number consisting of four figures ook for the first three figures in the left hand colamn, md the 23 J!00 • USE OF LOGARlTUimC TABLES. LVll fourth at the top of the page ; then, on the same horizontal line with the first three, and beneath tiie fourth, that is, in the same vertical line with it, will be found the decimal of the logarithm, to which prefix its proper characteristic. Thus, log. 4790 z= a 0808792; log. -03745 = 2-5734518*. When the number contains more than four figures, find tli* decimal part of the logarithm of the first four, as already direct- ed ; then consider the remaining figures of the number as a frac- tion, placing a decimal point before them ; multiply the difference between the logarithm already found and the next greater by thia fraction ; and, rejecting as many figures on the right as there are dtscimals in the multiplier, add the product to the decimal of ihe logarithm corresponding to the first four figures, remembering tc place the right hand figures of the decimals to be added undei each other ; prefix the appropriate characteristic, and the result will be the logarithm sought. For example, in finding the logarithm of 3745I2G, we take the decimal part of the logarithm corresponding to 3745, and add to it -120 of the difference between that logarithm and the neyt greater, and to the result prefix as a characteristic. The reason of this method of proceeding will be seen as fel- lows. The decimal logarithm of 3745000 is -5734518; t^ie next greater decimal logarithm,* corresponding to 3740000, is 5735678. The difference between these numbers is 1000, rnd the difference between their logarithms is '0001160; moreover, 3745126 exceeds 3745000 by 126. Wherefore, if, when the number increases 1000, the logarithm increases -0001160, r hen the number increases 126, the logarithm should increase . /^g, - -126, of -0001160, which is -0000146160; retaining onl' sev HX decimals, we have '0000146, which added to '5734518 gives In the mor<» e«cended tables, as those of Callet, four figures of tl nura- cer are found in the left hand column, and the fifth at the top. JVl iover fropoftional parts of the differences, are placed on the right hand sic «^ the LVll. USE OF LOGARITHMIC TABLES. 267 •5734G04; to this sum prefix the characteristic G, and we hare .eg. 3745126 = 6 5734664. The result would evidently have been the same, if we had dis- regarded the rank of the decimals in the difference of the loga- rithms, nmltiplied this difference by '126, rejected the three right hand figures of the product, and added the reserved part of the pioduct to 65734518, placing the right hand figure updei the 8. In like manner, we shall find log. 327983 =:5'5158514 ; also, log. 0379426 = 2 5791271. Remark. This mode of finding the logarithms of large num- bers, as well as that to be given for finding numbers correspond- ing to given logarithms, supposes that logarithms increase in the same ratio as the numbers themselves, which, though not strictly true, is nearer the truth, the greater the numbers and the less their difference and gives results sufficiently accurate for most practical purposes. Let the 'earner find from his tables the logarithms of the fol- lowing numbers 1. 1273. 6. 12710-63 2. 57293. 7. 274967. 3. •01273. 8. 333 333 4. •00279. 9. 435-501. 5. 327496. 10. 111-3734 Art. 1G3. To find the number corresponding to any given logarithm. Look for the decimal part of the logarithm in the tab e, and if it be found exactly, the first three figures of the number will be found in the left hand column, marked N., in the same hori- zontal line with the logarithm, and the fourth at the top, directly above the logarithm, the rank of the figures being shown by the characteristic. Thus, 3-5860244 being the given logarithm, the correspond mg number is 3855. Had the characteristic been 1, the nurn' 268 USE OF LOGARITHMIC TABLES. lA II ber would have been 38'55 ; had it been 6, the number vvouIH have been 3855000 and had it been 2, the number would have been 03^55. If the decimal part of the logarithm is not found exactly in the table, take the difference between the given logarithm and th<» next less tabular logarithm, for a numerator, and the difference between the next less and the next greater tabular logarithms, for a denominator. Reduce the fraction thus formed to a decimal, and, rejecting the decimal point, place the result at the right of the four figures corresponding to the less tabular logarithm ; lastly, place a decimal point, if necessary, according to the char- acteristic of the given logarithm. For example, let 2-4716423 be the logarithm, the correspond- ing number to which we wish to find. The next less decimal in the table is '4715851, the difference between which and the given logarithm, the characteristic being neglected, is 572 of the lowest order of decimals in the logarithms ; the difference between the next less and next greater tabular logarithms, is 1466 of the lowest order of decimals in the logarithms ; reducing i^?¥F ^^ ^ decimal, we have the figures 39, which placed at the right of 2962, the figures corresponding to •4715851, gives 296239 ; but as the characteristic of the given logarithm is 2, we point off three figures for integers, and obtain 296239 for the required number. The reason of this method is obvious. For, if, when the loga- rithm increases 1466, the number increases I unit of any order, when the logarithm increases 572, the number ought to increase i^-^wE ^^ ^ ""'^ ^^ ^^® same order. The order of the unit of which we find a fractional part, is always determined by the characteristic. In the example just given we found the fractional part of 01, viz. "039 ; but had the characteristic been 3, the frac- tion would have been a part of 1. Let the learner find the numbers corresponding to the follow- ing logarithms, carrying the numbers to six significant figures when the decimals are not found exactly in the tables. LVII. USE OF LOGARITHMIC TABLES. 269 Common tables will generally give numbers with sufficient accuracy to six or seven figures. 1. 1-4771213. 6 02134445. 2. 3-3010300. 7. i-4840150. 3. 1-4991217. 8. 27734667. 4. 3-1171167. 9 32276677. 5. 5-3458726. 10. 43334475. An. 164:. Since the logarithm of a vulgar fraction is found by subtracting the logarithm of the denominator from that of the numerator, it follows that the logarithm of any proper fraction, like that of a decimal, must be negative. But we ordinarily make the characteristic only negative. Thus, log. 2^ = log. 2 — log. 257 = 0-3010300 — 2-4099331 r= 389 10969. In order that we may be able to subtract the lat- ter decimal from the former, we may suppose the characteristic of the logarithm 0-3010300 to be changed into — 1 + 1, so that the decimal -4099331 can be taken from the positive part of 1 + 1-3010300; then subtracting the 2 from — 1, we have 3 for a characteristic. Or, as is more commonly done in subtrac- tion, after having borrowed 1 in subtracting the left hand deci- mal, we may carry I to the 2, and then subtract the 3, which giv? 3, the same as before. Art. 100. But there is another form for the logarithm of any proper fraction, by which the negative characteristic is avoided. This form is obtained by increasing the true characteristic by 10. For example, the logarithm of -3 is 1-4771213; adding 10 to the characteristic and reducing gives log. -3=3 9-4771213. In like manner, log. -03 = 84771213 ; and log. 003 = 7-4771213. ilence, in this way, if the first figure of the decimal is tenths, the characteristic of its logarithm is 9 ; if the first figure is hun« dredths, the characteristic is 8 ; if the first figure is thousandths, the characteristic is 7, and so on. That is : 23* *'"U USE OF LOGARITHMIC TABLES. LVH Tht characteristic of the logarithm of a decimal fraction, is IC di7niftished by as many units^ as are equal to the number of places by which the first significant figure of the fraction is removea frmn the decimal point. « Likewise, in finding the logarithm of a vulgar fraction, we may increase the logarithm of the numerator by 10, and then subtract the logarithm of the denominator. Thus, the logarithm of fIt would, in this way, be 7-8910969. But we must recollect, that every such logarithm is, in fact, 10 too great, and that the result of any operation in which it is used, would be affected accordingly. Art. 160. In division, we have seen that the logarithm of the divisor is to be subtracted from that of the dividend ; but, in- stead of this, we may add the arithmetical complement of the log- arithm of the divisor to the logarithm of the dividend, dropping 10 afterwards in the result. The arithmetical complement of the logarithm of a number, is what remains, after the logarithm of that number has been sub' traded from 10 Thus, the arithmetical complement of the logarithm of 17, is 10 — log. 17=10— 12304489 = 8-7695511. The arithmetical complement of a logarithm may be found, by biibtracting the first right hand significant figure of the logarithm from 10, and all the others from 9; so that we may, if we please commence the subtraction at the left hand. Wc must bear in mind that each arithmetical complement added, makes* the result 10 too great, and allow for this in any operation, in which arithmetical complements of logarithms are used. The fact that adding the arithmetical complement of a loga- rithm and afterwards subtracting 10, is equivalent to subtracting the logarithm itself, may be easily proved. For, let / represent any logarithm, and /' a logarithm which is to be subtracted from it; the result would be / — /'. Now the ai-thmetical complement of /' is 10 — /'; adding this to /, LVII. USE OF LOGARlTHMlJU TABLES. 271 we have ^+10 — I' \ subtracting 10, we have /-|- 10 — /' — 10 , which reduced becomes / — /', the same result as when I' was subtracted immediately from /. If however we add the arithmetical complement of the loga- rithm of a fraction with its characteristic 10 too great, the result, without dropping 10, will be the same as if the logarithm of the fraction had been subtracted. To prove this, let /' be the true logarithm of any fraction ; then lO-f-Z' would be its logarithm with a characteristic 10 too great ; the arithmetical complement of this is 10 — 10 — /', which added to any logarithm /, gives /+ 10 — 10 — /' or / — /', which is precisely the same as if /' were directly subtracted from /. Art. 1 67. Let the learner find the logarithms of the follow- ing numbers. When the fractions are less than unity, the loga- rithms may be given in both forms, viz : with the negative char- acteristic, and with the characteristic 10 too great. 1. •7234. 6. ¥/• o •00576. 7. 3A- 3. •00087926. 8. 456+f. 4 h 9. 145J. 5. Iff- 10. •Nt^- Art. 168. Find the numbers correspor iding to logarithms, the characteristics being each 10 too great. Six sig- nificant figures may be found in each case, when the decimal j^it is not found exactly in the tables. L. 94371213. 5. 5-4771213 2. 87294179. 6. 9-8878879. 3. 6-3010300. 7. 8-830037^ 4. 26734217. 8. 7-4378678. B. It will be observed that fmding the logarithm of a \ ul- N. gar fraction, aid then obtainiiig t;.5 number corresponding to th^t logarithm, converts the fract 3n m*o a -^o^ ina' 2'"2 APPLICATION OF LOiJARITHMS LVlll SECTION LVlll.. APPLICATION OF LOGARITHMS TO ARITHMETICAL OPKRATIOMS Art. 109. 1. Multiply 456 by 723. Log. 456 = 2 6589648 Log. 723 — 2 8591383 Prod. z=z 329688 ---5-518I03L Adding the logarithms of 456 and 723, we find the sum to be 5 5181031 ; we then find from the tables the number correspond- ing to this logarithm, viz : 329688, which is the required pro- duct. 2. Multiply 2678 by 03745. Log. 2678 =3-4278106 Log. 03745 = 25 734518 Prod. = 100 29----- 20012624. In adding, there is 1 to carry when we arrive at the character- istics; this 1 is positive, and, being added along with the 3 and 8, gives for a characteristic 4 — 2 or 2. The same without the negative characteristic. Log. 2678 == 3-4278106 Log. 03745 = 8-5734518 12 0012624 Subtract--- 10 Prod. = 100-29 2 0012624. It would have saved labor to drop the 10 at the time of adding. 3. Multiply -0374 by 277. Log. -0374 = 2-5728716 Log. -277 = 1-4424798 Prod. = 0103598 ---2-0153514. The figures answering to the decimal of the resulting loga^ nthm are 103598; but since the characteristic is 2, the first LVIII. TO ARITHMETICAL OPERATIONS. 273 figure of the number must be hundredths, therefore a zero pre- ceded by the decimal point, must be placed before the figures found. The same without negative characteristics. Log. •0374 = 8-5728716 Log. -277 = 9-4424798 Prod. = •0103598---8-0153514. In the suiu of the logarithms, the characteristic becomes, in fact, two tens too great ; but we drop only one of them, and then the characteristic 8 shows that the first figure of the number must be hundredths. . 4. Divide 48945 by 65. 5 By Log. 48945 = 4 6897083 Log. 65 z= 18129134 S By subtraction, auot =753 2-8767949. The same with the arithmetical complement of the logarithm of the divisor. The contracted expression, comp. log.^ will some- times be used to signify arithmetical complement of the loga- rithm. Log. 48945 = 4-6897083 ^ Comp. log. 65 = 8-18708 66 5 Add. auot. z= 753 2-8767949 5. Divide 775 by -025. Log. 775 = 2-8893017 Log. 025 = ^-39794 00 auot. = 31000 - - - 4 4913617. The sign of 2, in subtracting is changed to -|-, and then the two characteristics are added. The same without the negative characteristic, and with the comp. log. of the divisor. Log. 775 = 2-8893017 Log. -025 = 8-3979400, comp. log. = 1-6020600 auot. = 31000 4 4913617. 274 APPLICATION OF LOGARITHMS LVIll In this example, the resulting characteristic is not too great because the logarithm of the fraction was taken without the neg ative characteristic. 6. Divide 005127 by 0559. Log. •005127 = 3-7098633 Log. 0559 ^z=z 2-7474118 Quot. = -091717 2 9624515. In subtracting, there is 1 to carry to the 2, which makes it I, and this subtracted from 3, gives 3 + 1 or 2 for a characteristic Or, if the learner is accustomed, when he borrows 1, to dimin- ish by 1 the next figure in the minuend, he will take 1 from 3, which gives 4; from this he will then subtract 2, and obtain 4 + 2, or 2. The same with positive characteristics and comp. log. of the divisor. Log. -005127 = - - - 7-7098633 Log. -0559 = 8-7474118, comp. log. = 1-2525882 auot. = -091717 8-9624515 In this case the resulting logarithm is 10 too great, because the logarithm of the dividend was taken with a positive chaiao teristic. 7. Required the third power of 27. Log. 27= I 4313638 3 Mult, by 3. Power = 19683 - - 4-2940914. 8. Required the third ppwer of -271. Log. -271 = 1-4329693 3 Power = -0199025 1 1 - - - - 2-2989079. LVIIl TO ARITHMETICAL OPERATIONS. 275 The same with positive characteristics. Log. -271 = 9-4329693 Power = -019902511 - -82989079. By thi« last method, the characteristic after multiplication, becomes 28, which is three lOs or 30 too grtat ; dropping two 10s or 20, we have the characteristic 8, which shows that the fiist figure of the number is hui dredths. 9 Required the fifth root of 15. Log. 15 = 1 1760913 (5 Divide by 5 Rpot =: 1-71877 - - - 2352183. 10. Find the third root of -000729. Log. -000729 = 4-8627275 = 6 + 2-8627275 (3 Root =09 2-9542425. In this question a difficulty occurs in dividing the logarithm 48627275, since the integral and fractional parts have different signs, and the negative characteristic is not divisible by 3. To obviate this difficulty, add 2-)- 2, which is zero, to the charac- teristic; the logarithm then becomes 6 + 28627275. Dividing now the negative and positive parts separately, we have the re- sult as above. In all cases of finding the root of a fraction, if its logarithm ia taken with a negative exponent, and that exponent is not divisi- ble by the number expressing the degree of the root, we must make it so, by guding to the logarithm 1 + 1, 2 +2, 3 -f- 3, or *ome equivalent expression. T'lifi same with positive characteristics. Log. 000729 z=z 6-8627275 2 - - - - Add. 26-8 627275 (3 Root = Oi -8 9542425. 276 APPLICATION OF LOGARITHMS I Vlll By the second method, the logarithm when first found, is too great by 10; we then add two more 10s, which makes it three lOs too great ; this divided by 3 gives a result 10 too great as re- quired. Whenever we use the positive characteristic in finding the root of a fraction, before dividing the logarithm, it is necessary to make the characteristic as many 10s too great as there are units in the number which marks the degree of the root. The divi- 81 jn will then leave the result one 10 too great. 11. Find the value of x in the expression, x=z (^)^. Log. 2 = 0-3010300 • Comp. log. 7 r= - - 9- 1549020 Log. f - - char. 1 too great, :^ 9-4559320 3 Log. (f )3 - - char, three 10s too great, = 283677960 20 Add. 48-3677"960(5 * X = -471584 - 9 6735592. 12. Find the value of x in the expression, x = /45 . 13 . (•75)\f Vl9. 117. 11 / • Log. 45 = 1-6532125 Log. 13 = 1-1139434 Log. -75 = 9-8750613 Comp. log. 19 = 8 7212464 Comp. log 117 = 7-9318141 Comp. log. 1 1 = 8;9586073 8^2538850 3 24-7616550 10 34-7616550(4 % z= -0490245 - - - 6^904137. LVIIL TO ARITHMETICAl. JPERATIONS. 277 In this question we have used the logarithm of one fract'cn, with the increased characteristic, and three corap. logS. of whole numbers ; the sum of the six logarithms added, will therefore be 40 too great. Dropping 30, multiplying by 3, adding 10 to the product, and dividing this sum by 4, will leave the final 'oga- rithm 10 too great. 13 Find the value of x in the expression. (12)=.(|)*.V'i Log. 2 = 0-30t0300 Comp. log. 3 = 9-5228787 9-8239087 10- 19-8239087 (2 Log.v/S — Q'Ql lQ'^/1*^ . . . . . UQ1IU^^ i~4oo"7 Log. 569-333 = 27553664 Comp. log. 400 = 7-3 979400 01^53306 4(9 14-r = 104- - - - 0-0170340." 1 r= 04. 6. ilnv many years must $1000 remain at compound intere^ , tie rate ueing 6 per cent., in order to amount to $1191'016T ■««-(f) ^ Substitutmg ^^ -he formula, t =: . -r-^, — .-, we obtain ^ — ' log. (1-06) • Log. 1 191016 — 30759176 Comp. log. 1000 = 7-0000000 -^t;) 00759176 Log. lOi'. — 00253059. Hence, t = :§MHf = ^ years. Or, performing this layt division by logarithms we have Log. 0759176 = 8-8803424 • Comp. log. -0253059 =: 1-596778^ / = 3 0-4771207. ,o this question the operation would have been shorter, if we Had divided 1191 016 by 1000, before applying logarithms i86 COMPOUND INTEREST. LIX 6. Find the amount, at compound interest^ of $357'50, for 8 years at 6 per cent. 7. Find the amount of $1573 for 4 years, at 5^ per cent, com pound interest. 8. Required the compound interest on $1000, for 7 years and 4 months at 4 per cent. 9. What sum of money will, in 6 years, at " per cent, com pound interest, amount to $*2745-90? 10. What sum of money will amount, in 10 years, to $417 712, O'^mpound interest being reckoned at 6 per cent. ? 11. In how many years will $75 amount to $149*495, atSfer cent, compound interest? 12. A principal of $108'50 amounted, in 12 years, at com- pound interest, to $22045; what was the rate per cent.? 13. In what time would any sum be doubled at compound in- terest, the rate being 6 per cent. ? In this question the amount is to become double the princi- pal; therefore, in the formula for ty we substitute 2p instead of Ay which ffives t = z —-^. — r, or, by reduction, t = ^ log. (l+r/ ^ log.S log. ( i f- rV 14. In how many years will any sum, at compound interest be tripled, the rate being 6 per cent. ? 15 I'l how many years will any sum be doubled, at 5 pei cent, compound interest? 16. What would $357 amount to in 10 years at compound in- terest, the interest b6ing reckoned semi-annually, at the rate of 6 per cent, a year ? 17. The population of Boston in 1830 was 61*^92; what was it in 1840, supposing the annual rate of inert J.»e ja >.e 3^J>r ner rent. ? 18. The population of P\.ladeipnia m 1830 was lfeS797, and »> 1840 it was 258832: what was the annual rate of increase ? LIX. COMPOUND INTEREST. 287 19. In 1830 New York contained 202589, and in 184C t «onf ained 312234 inhabitants; if the population continue to increase at the same rate as it did from 1830 to 1840, in how many years from the latter date will it amount to lOOOOOO r Art. 179. 1. A man saves annually $300 which, at the end of each year, he deposites in a bank, and is allowed 5 per cent, compound interest. How much would be due him froin the bank, at the end of 12 years from the time of the first deposit? To generalize this question, let a be the sum annually depos- ited, t the time, and r the rate. Then the amount of the sum first deposited would, according to the principles already given, be a (1 -(-'•)*• The second deposit remaining in the bank one year less, would amount to a(l-|-r)*~i. The amount of the hird deposit would be a(l-j-r)*~2, and so on. The last de- posit but one, remaining in the bank two years, would amount to a (1 -(- r)'2 ; and the last deposit would amount to a(l +r). B °nce, if A represent the gross amount, we have A=«(l+r) + «(l+r)2+ +«(l+ry-2^ ^(l+r)'-i + a(l+ry. The second member of this equation is a progression by quo- lent, in which the first term is a(l + r), the ratio 1 + r, and ie last terra a (1 + r)*. In the formula, >S^= — — r-, substi- tuting A instead of >S^, a(\ + r)* instead of /, a{\-\-r) instea« of a, and \-\-r instead of 9, we have ^ 1 + r-. • °' _4^ «(l+'-)[(l-t-'-)'-l] . r Substitutmg in this formula the numbers given in the question Droposed, we have _ 300, (105) [(l'05)ia_i] 05 288 ANNUITIES. LX In applying logarithms, it is best to commence with the quau • ity between the brackets. Log. 1()5 = 0021 1893 12 ( 1 05)^2 _. 1 -795856 - - 0- 25427 16 1 Log. -795856 = 99008345 Log. 1-05 = 00211893 Log. 300 ==: 2-4771213 Comp. log. 05 = 1-3010300 A =1 $5013-893 .... 3*7001751. 2. If a young man, by omitting some useless expense, saves 25 •?ents every day, and, at the end of each year, deposits'his savings m an institution which allows 6 per cent, compound interest, how »iiuch would be due him from the institution, at the end of 20 years from the time of the first deposit, a year being considered 365 days] SECTION LX. ANNUITIES. Art. 173. An annuity is a certain sum of money payable an- lually, or at other regular periods, for a stated number of years, or during a person's life, or forever. The following question is one of annuities. 1. A man wishes to put at compound interest such a sum of money, as will afford him annually $500 for 20 years, at the end of which time the principal and interest shall be exhausted What sum must he put at interest, the rate being 6 per cent, v It is manifest that the amount of all he rpr;eives, must ue thf same %8 the amount of the sum put at interest. LX. ANNUITIES. 289 To generalize this question, let a be the sum received annu' ally, r the rate of interest, and t the time. As the first sum is drawn out at the end of the first year, the drawer must be considered as having received, at the expiration of the whole time, the amount of that sum at compound interest for t — I years, which according to Art. 171, is a(l +'*)'""• In like manner, that drawn out at the end of the second year, amounts to a(l -\-ry-~^; that at the end of the third year, to a (1 -\-ry~^y and so on; the sum drawn at the end of the last jear is simply a. The gi OSS amount of the whole drawn out, is, therefore, «(l + ry-i_|-a(|-|-r)'-2 + a(l + r)*-3+ . . . .+ a{\-^rY ^ a(i-\-r) -\- a; or, by a change in the order of arrangement, a + a{\ + r) + a{\+Yf+ . . , . +«(l + ry-3 + aa+r)«-2 + a(l+ry-i. This is a progression by quotient, in which the first term is a, the ratio 1 -\-r, and the last term a(l+r)*-^ Substituting these in the formula, dition that he who loses, shall forfeit to all the rest as much money as they already have. First A loses, then B, then C, then D, and finally E. Yet, at the end of the fifth game, each has left; $32. How much has each at first? 68. A and B sold 100 eggs, and each received the same sum. If A had sold as many as B, he would have received 18 pence for them; and if B had sold as many as A, he would have received only 8 pence for them. How many did each sell ? 69. Separate 24 into two parts, whose product shall be 35 times their difference. 70. What two numbers are those, whose product is 4 times their difference, and whose product multiplied by their difference is 16? 71. The sum of three numbers is 21 ; if the first be subtracted fifom the second, and the second from the third, the latter re- mainder will exceed the former by 3 ; moreover, the sum of the squares of the first and third is 137. Required the numbers. 72. There are two rectangular vessels, which together hold 180 cubic feet ; the bottom of each is a square whose side is equal to the height of the other vessel. If each vessel were a cube whose side was equal to one side of its bottom, the two ves- sels would contain 189 cubic feet. Required the dimensions of each. 73. There are two numbers, such that the square of the greater, multiplied by the lesf?, is 30 more than the square of the STlO MISCELLANEOUS QUESTl dN8. less, multiplied by the greater ; moreover, the 3d power of the greater exceeds that of the less by 98. What are the numbers 1 74. What number is that whose fourth power exceeds ten limes its second power by 936 ? 75. Find a number, such that if its square root be increased by 4, the cube root of the sum shall be 2. 76. The first year a man was in trade he doubled his money , the second year he gained $5 more than the square root of the number of dollars he had at the commencement of that year, when he received a legacy of as many dollars as were equal to the square of the number he then had, and found that his whole fortune amounted to $ 13340. With how much money did he commence business? 77. If the sum of two numbers be increased by 2, and the sec- ond power of this result be added to the sum of the numbers, the amount will be 154. Moreover, the difference between the sec- ond powers of the two numbers is 40. What are the numbers f 78. The sum of three numbers in progression by difference is 15 ; and the sum of the squares of the extremes is 58. Required the numbers.! 79. Four numbers are in progression by difference ; the sum of the squares of the first two terms is 10; and the sum of the squares of the last two terms is 74. What are the numbers? 80. Find three numbers in progression by quotient, whose sum is 26, and the sun\ of whose second powers is 364. 81. Four numbers are in progression by quotient; the sum of the first two is 30, that of the last two is 120. Required the numbers 82. Required the compound irrterest on ^120, for 10 years at 6 per cent, annually. 83. What will $300 amount to in 10 years, at compound ii lerest semi-annually, the yearly rate being 5 per cent. ? •gSlVBB.SIT7| THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. OCT 1 ft i«| •* *^9m OCT 2* jojr * » ^♦JD F£Q Q , ^^ 8 J936 MAY 11 1938 HFH 2<^ ^Vt{ LD21-100m-7,'33