iii .iipi'i! |iillil|iiiiliii!|i|ififiiiji!i|i!iliii ii iOillli! illipi IP liii^ P mmammmmm !'';.,jKpi::\^rn.}j!|i'}-n5f}[|n!j|imffUn;Ui{t!!;. liii:;;! iiiiijiiiifiiliiiiiilili 111 ipiliiim-" liiOliniilil isiiiiiiiiililli ^■''('i: 'Mm 1 'ill 1 mm III i lii!ll!Pi!B iiiiiiiiiiiii,,,, |!i|rlil|liiii! Pi pill iiipllljippiip ^^"-lilililiiiitlltlililiilililij lljlilljill Jiilliiiiiiiiiiii '*H ijililiipiliiyilljiljlll IfinHiinliilllilHllI I iliiliiliii i 'III ! lilji'iiiil I ||||jiii|{||jj|M piil!ii|ll!|l||ji[ilil[HHHi! .,-,.,„ m Is If' iiiili'iiiiiliii^ lli^ C8* LIB RARY OF THE UNIVERSITY OF CALIFORNIA. + x, means the sum of a, by and Xy whatever be the values of a, by and x. 1. Define Algebra. What symbols are used to denote numbers or quanti- ties ?— 2. What letters of the alphabet are employed to represent known or de- termined quantities ? — 3. How are unknown or undetermined quantities repre- sented ?— 4. What are coefficients V Wlien is the coefficient omitted ? — 5. What is the sign of equality ? — 6. What is the use of the sign + ? B 2 ALGEBRA. 7. The sign — (read minus) signifies that the quantity to which it is prefixed is to be subtracted. Thus 8 — 2 is the same thing as 1 ; a — ^ means the difference of a and ^, or b taken from a ; and a + 6 — a:, signifies that x is to be subtracted from the sum of a and b. 8. Quantities which have the sign + prefixed to them are called positive^ and those which have the sign — set before them are termed negative quantities. When there is no sign before a quantity + is understood ; thus a, stands for -\-'a. 9. The symbol X (read into) is the sign of multiplica- tion^ and signifies that the quantities between which it is placed are to be multiplied together. Thus, 6x2 means that 6 is to be multiplied by 2 ; and a X ^ X c, signifies that a, Z>, c, are to be muli plied together. In the place of this symbol a dot or full-point is often used. Thus, a .b .c^ means the same as a X ^ X c. The product of quantities represented by letters is usually expressed by placing the letters in close contact, one after another, ac- cording to the position in which they stand in the alphabet. Thus, the product of a into b is denoted by ab ; of a, b, and X, by abx, and of 3 «, a:, and y, by ^axy. 1 0. In algebraical computations the word therefore often occurs. To express this word the symbol .*. is generally made use of. Thus the sentence "therefore a + Z» is etjual to c -|- c/," is expressed by ** .*. a + Z> = c -|- ^•" EXAMPLES. Ex. I. In the algebraical expression, a-\-b — c,\Qta = 9, b = 7,c = S; then a+b—c= 9+7—3 = 16-3 = 13 Ex. 2. In the expression ax + ay — xy, let a = 5, a: =2, y = 7 ; then, to find its value, we have aar + ay — a;^ = 5X2+5 X7— 2X7 = 10+35 — 14 = 45 — 14 = 31 7. How is the symbol — read?— 8. What is meant by positive and what by negative quantities? — ^9. Write down the sign of multiplication ? Is any other mark u^ed to denote multiplication? When is no symbol used?— 10. What symbol is used to denote the word thertfore ? CH. I. DEFINITIONS. 3 Ex. 3. Ifa = 5, b = 4, c = 3, d=2,x= 1,^^ = 0, find the numerical values of the following expressions : (1.) a + b+c + x Ans. 13. (2.) a — b+c — x + t/ ... 3. (3.) ab + Sac — 6c + 4cx — xy ... ^b. (4:.) abc — abd -^bcd^ acx ... 29. (5.) 3a6c + 4aca7 — %bdx + ajt^ ... 176. 1 1 . The symbol -j- (read divided by) is the si^/i q/ fl?2t?2- szow, and signifies that the former of two quantities between which it is placed, is to be'~3ivided by the latter. Thus, 8 -7- 2 is equivalent to 4. But this division is^more simply expressed by making the former quantity the numerator, and the latter the denominator of a fraction ; thus —r means o a divided by b, and is usually, for the sake of brevity, read a by b. EXAMPLES. Ex. 1 . If a =2, 6 = 3 ; then, we find the values of ) '^«_- 3X2 __ 6 _ 2 bb 5X3 15 5 C2 ) ^^ + ^ -- ^X^+^ -, 4 + 3 _7__^ 8a--3Z> 8X2—3X3 16—9 7 Ex. 2. If a = 3, 6 = 2, c = 1 , find the numerical values of (1.) ^±^ Ans.-^ ^ 4.b + a 11 (2.) ^ + -^-^ ... 1 yf.y.ab-{-ac^bc 7 ^ 2ab — 2ac + be '" ~S 12. When a quantity is multiplied into itself any number of times, the product is called a power of the quantity. 11. By what symbo* is division denoted? What is its name? Is division ever expressed in au, other manner ? — 12. What is meant by the power of a quantity ? 4 ALGEBRA. 13. Powers are usually denoted by placing aboye the quantity to the right a small figure, which indicates how often the quantity is multiplied into itself. Thus a the^r^^ power of a is denoted by a (a') «X« the 2^ power or 5^Mare of a ... a' a X a X « the 3^ power or cube of a ... a' « X « X « X « ... the 4*^ power of a ... a* The small figures 2, *, *, &c., set over a, are respectively called the index or exponent of the corresponding power of a, 14. The roots of quantities are the quantities from which the powers are by successive multiplication produced. Thus, the root of the square number IG is 4, because 4X4 = 16, and the root of the cube number 27 is 3, since 3X3 X3 = 27. 15. To express the roots of quantities the symbol V^, (a corruption of r, the first letter in the word radix) with the proper index, is employed. Thus, ^a or A^a, expresses the square root of a, }/a *' the CM^e root of a. i^a " the /owr^^ root of a. &c. &c. EXAMPIiES. Ex. I. If 05 = 3, h=z1\ then a2==3X3 = 9, a» = 3 X3X3 = 27, ^>^ = 2X2X2X2=16. E x. 2. If g = 64; then >v/« = a/64 = 8, >ya = ^64 = ^4X4 X4 = 4, ^a=^64 = 2. Ex. 3. In the expression t 1 let a = 3, Z» = 5, c = 2, a; =6. What is the numerical value ? Here ax^ + b'' = S X ^ X ^ + o X ^ — ^OS +25 = 133, and 6:r — a^ — c = 5X6 — 3x3 — 2 = 30 — 9 — 2 = 19 • _^f!+i!_~_L21 = 7 "Z>x — a2 — c~ 19 13. How &re powers denoted?— 14. What are the roots of quantities V— 15. By whfit symbol are the roots of quantities expressed? What is the origin of this symbol ? CH :. I. DEFINITIONS. 5 Ex. 4. If a= 1, 5 = 3, c=5, ^ = 0, find the values of (I.) a^ + 2b'- c. Ans. 2. (2.) a2 + Sb^ - c2 ... 3. (3.) a2_j..2Z,2^3c2 + 4fl?2 ...94. (4.) 3« 2^, — 2^> 2c + 4c2 — 4a2fl? ...19 (5.) a' + ^»3 ...28. (6.) a' b^ , c« 3 "•" 3 "^ 3 ...51. Ex. 5. Let AT == 64, b = 8l, c=l : find the values of (1.) ^a + y/b Ans. 17. (2.) ^a + \^b + ^c ... 18 (3.) Vabc ... 72. 16. When several quantities are to be taken as one quan- tity/ they are enclosed in brackets, as ( )»{"})[ ]• Thus, (a -\-b — c) . (d — e) signifies that the quantity re- presented by « + /> — c, is to be multiplied by that repre- sented by rf — e; if then a = 3, 6=2, c= I, d = 5, e = 2, a-^- b — c = 4, d — e=S, and .*. (a-\'b — c) , (d — e) = 4X 3 = 12. Great care must be taken in observing how brackets are employed, and what effects arise from the use of them. Thus (a + b). (c -{- d), (a -\- b) c -{- d, a -{- be -^ d, are three very different expressions; for if a=3, 6=2, c=S, d=b, (1.) (a + 6).(c + c?) = (3-f2).(3-f5)=5x8 = 40. (2.) (a + 6)c-fc?=(3-f 2) 3-1-5 = 5X34-5=20. (3.) fl-t-6c + rf=3-f2X3-f 5 = 3+6-1-5 = 14. "17. Instead of brackets, a line called a vinculum is some- times used, and is drawn over quantities, which are taken collectively. Thus, a — 6 — c is the same as a — (6 — c). The line which separates the numerator and denominator of a fraction may be regarded as a sort of vinculum, cor- 16. When are brackets employed? — 17. What is a vinculum f May the line which separates the numerator and denominator of a fraction be regarded as a vinculum ? () ALGEBRA. responding, in fact, in Division to the bracket in Multipli- ^ a -\'b — c , „ , - ,- cation. Thus, — —. implies that the whole quantity a 4- 6 — c is to be divided by 5. 18. Like quantities are such as consist of the same letter, or the same combination of letters; thus, 5a, and 7a ; 4a6 and 9ab ; ^bx"^ and Gbx"^, &c. are called like quantities ; and unlike quantities are such as consist of different letters, or of different combinations of letters ; thus, 4a, 3^, lax, bbx"^, &c., are unlike quantities. 19. Algebraic (quantities have also diflPerent denomina- tions according to the number of terms (connected by the sign + or — ) of which they consist : thus, a, 2Z>, Qaxj &c., quantities consisting of one term, are called simple quantities. a-\-x, 3L quantity consisting of two terms, is called a binomial, bx -\-i/ -^ z,a, quantity consisting of three terms, is called a trinomial. CHAPTER II. ON THE ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION, OF ALGEBRAIC QUANTITIES. ADDITION. 20. Addition consists in collecting quantities that are like into one sum, and connecting by means of their proper signs those that are unlike. From the division of algebraic quantities into positive and negative, like and unlike, there arise three cases of Addition. Case I. To add like quantities with like signs. 21. In this case, the rule is *' To add the coefficients of the several quantities together, and to the result annex the 18. What are like and what are unlfke quantities?— 19. What is a simple quantity? What is a binomial and what a. trinomial?— 20. In what does ad* dltion of algebra consist ? Into how many cases is it divided ? CH. I. DEFINITIONS. common sign, and the common letter or letters;*' for it is evident from the common principles of Arithmetic, if -(-2*7, + 3a, and + 5a, be added together, their sum must be 4- 10a ; and if — Sb", — 46^, and — 8b^, be added together, their sum must be — 15^^. Ex.. I. Ex.2. Ex.3. 2x+ 3a- U 7x^+ 3x1/-- 5bc 4a»- 3a2+ 1 Sx+ 2a- ob 9x'^+ 2x1/— 7 be 2a3— a2+17 4x+ 8a— 7 b nx^+ bxy— 4bc 5a3— 2a2+ 4 9x-\- 4a- 6b x^-\- 4x1/— be 3a^- 7a2+ 3 5x+ la— 9b x^-\- 9xy— 2bc fl3_ «2_^io 23a:+24a-31* 29x^-\'23xy—l9bc 15a3— l4a2+35 Ex.4. Ex.5. Ex.6. 3:c'+4«2__ ^ 7a^-Sa^b+2ab^—Sb^ 2^2^—3^+ 2 2x^+ x^-Sx 4a3— a^b+ ab"— b^ 4x-y—2xr{- I 7x^+2x^-2x a'^2a^b+Sab^—5b^ 3a:2y-5ar+10 4X^+ X^" X 5a^-3a''b+4ab^-2b^ x-'y- a:+15 In these Examples it may be observed that some of the quantities have no eoefficient. In this case, unity or 1 is always understood. Thus, in adding up the first column of Ex. 2. we say, 1 + 1 + 1 1 + 9 + 7 =29 ; in the third, 2 + 1+4 + 7+5 = 19; and so of the rest. , Case II. To add like quantities with unlike signs. 22. Since the compound quantity a-\-b — c-\-d— e &c. is positive or negative, according as the sura of the positive terms is greater or less than the sura of the negative ones, the aggregate or sum of the quantities 2a — 4a + 7a — 3a will be + 2a, and of the quantities 7^2 — 5^2 Jj. 2^>2 — 8^>2 will be — 46^; for in the former case 22. State the rule in the 1st case. O ' ALGEBRA. the excess of the sum of the positive terms above the nega- tive ones is 205, and in the latter 4:b'^. Hence this general rule for the addition of like quantities with unlike signs, ** Collect the coefficients of the positive terms into one sum, and also of the negative ; subtract the lesser of these sums from the greater ; to this difference^ annex the sign of the greater together with the common letter or letters, «ind the result will be the sum required." If the aggregate of the positive terms be equal to that of the negative ones, then this difference is equal to ; and consequently the sum of the quantities will be equal to 0, as in the second column of Ex. 2. following. Ex. 1. —20:24- a?— 5 3a:2— 5a'+ 1 7a:2+2a;— 4 — j^2— 4j:-t-13 113-^—9^+ 9 Ex. 2. -7a6-t-36c— xy - ab-\'2bc-\-4:xy Sab — bc-\-2xy -2ab-\-4bc — Sxy 5ab^Sbc-{- xy —2ab J'l''^ Ex.3. — 5a:3-fl3ar2 — 2x»— Ax^ 7x^+ x^ 9j3— 14jc2 — 13j:3— 2^-2 — 4x^—6j^ Ex.4. 4j?3— 2x+Sy - i:»+ 4a:— y 7x^^ x-^-dy 9x^+2\x—2y Ex.5. 5a^—2ab+ ¥ — a'+ ab—2b'^ 4a'— 3a^»+ b"" 2a^+4ab—4b^ Ex. 6. 4x^y^+2xy—S — j:<2^2_ xy—l SxY+4xy—5 -9x2^2— 2xy-f9 Case III. 23. There now only remains the case where unlike quan- tities are to be added together, which must be done by col- lecting them together into one line, and annexing their proper signs; thus, the sum of 3a:, —2a, + 5^,-4^, is 3a: — 2a -I- 56 — 4jf ; except when like and unlike quantities 29. State the rule in the 2d and 3d cases. CH. II. SIMPLE EQUATIONS. hf are mixed together, as in the following examples, where the expressions may be simplified, by collecting together such quantities as will coalesce into one sum. Ex. 1. 3ab-\- X — y \ Collecting together /i^e quan- 4c — 22/+ X \ titles, and beginning with 3a6, bah — 3c + rf \^Q have 8G6 + 5a6=8«Z>; +a: ^y T^ ^1. , 1-2^=-^; 4c-3c = + c; 8«6 + 2a:— ^ -\- c -\- d -\- x"^ ] besides which there are the ===================== / two quantities + d and + a:^, which do not coalesce with any of the others ; the sum required therefore is 8a6 + 2a: — ^ + ^ + <^ + ^^• „ / Here \x'^—x'^=S^x'^ Ex. 2. ^ , I —2xy-\-xy=—xy ^y -{•Sx^^y-' +xy- X' ] _3^+4^+^== ^2^ 5 a:3-2a: +3/ -15+ y^ \ j^4.x'j^^x^^bx'^ + \2x^ Sx^—xy — 14+2^+12a:3— 2a: [ — ^2-fy^=0 - ~~~~ ~^^^^ \ — 2a:=— 2a:. SIMPLE EQUATIONS. 24. When two algebraic quantities are connected together by the sign of equality (=), the expression is called an equation. Equations, in their application to the solution of problems, consist of quantities, some of which are known and others unknown. Thus 2a: + 3 = a: + 7 is an equation in which x is an unknown quantity, and its value is such a number as will make 2a: + 3 and a: + 7 equal to each other. The number which here satisfies the equation, or is the value of a:, is manifestly 4, since 2 X 4 -f- 3 = II , and 4 + 7 = 11. The value of the unknown quantity, which has in this example been found by inspection, is usually obtained by a direct calculation, which is called the solution of the equation. 24. What is an equation ? What is meant bv the solution of an equation ? b2 10 ALGEBllA. 25. In effecting the solution the several steps of the pro- cess must be conducted by means of the following axioms, and in strict accordance with them : — (].) Things which are equal to the same thing are equal to one another. (2.) If equals be added to the same or to equals, the sums will be equal. (3.) If equals be subtracted from the same or from equals, the remainders will be equal. (4.) If equals be multiplied by the same or by equals, the products will be equal. (5.) If equals be divided by the same or by equals, the quotients will be equal. (6.) If equals be raised to the same power, the powers will be equal. (7.) If the same roots of equals be extracted, the roots will be equal. These axioms, exclusive of the first, may be generalized, and all included in one very important principle^ which should in every investigation in which equations are con- cerned be carefully borne in mind ; viz. that whatever is done to one side of an equation the same thing must be done to the other side, in order to keep up the equality. 26. If an equation contain no power of the unknown quantities higher than the first, or those quantities in their simplest form, it is called a Simple Equation. ON THE SOLUTION OF SIMPLE EQUATIONS CON- TAINING ONLY ONE UNKNOWN QUANTITY. 27. The rules which are absolutely necessary for the solution of simple equations, containing only one unknown quantity, may be reduced to four, each of which will in its proper place be formally enunciated and exemplified. 25. Wh;it are the axioms employed in the solution of equations, and stite the general principle which is based upon them? — ^26. What is a simple equation? SIMPLE EQUATIONS. ' 11 Rule I. '*If the unknown quantity has a coefficient, then its value may be found by dividing each side of the equation by that coefficient;" and the foundation of the Rule is, that *'if equals he divided by the same, the quotients arising will be equal." Ex. 1. Let 2a: =14; then dividing both sides of the 2a:_14 >, . 2a: _ i 14__7 equation by 2, we have-r ^? ^^^ -z ^> ^ "o — .-. a: = 7. Ex. 2. Let aa: = 6 + c ; then — = ; but — =a: ; b+c Ex.3. Leta: + 2a:+4a: + 6a: = 52 Collecting the terms, 13a: = 52 Dividing both sides of the equation by 13, a: = 4. Ex. 4. Let 6a: — 4a: + 3a: — ar = 36 The terms being added as in Case 2^ of Addition, 4a: = 36 Dividing each side of the equation by 4, a: = 9. Ex.5. 10a; =150 Ans. a:=15 Ex.6. 3a: + 4a: + 7a: = 84|>r- ... a:=6 Ex.7., 8a: — 5a: + 4a: — 2ar = 25 ... a: = 5 Ex.8. 12a: — 3a: — 4a: — a: = 24 ... a:= 6 28. Arithmetical questions may with great ease be ex- hibited under the form of an equation, and it will be seen by the subjoined examples in what relation arithmetical and algebraic operations stand to each other : as for instance, If 30s. be given for 5lbs. of tea ^ what is the price of lib. ? (1.) The j9r2ce q/'llb. is that which is to be found. 12 ALGEBBA. (2.) Then it is clear that the price q/* lit) X 5 must give the price of 5lt)s. (3.) But the cost of 5lbs. by the question =305. (4.) Therefore, the price of lib. in shillings X 5 = 30s. (5.) And therefore by dividing: by 5, we obtain the price of lib. = 6s. The several steps of this solution expressed algebraically would take the following more compendious form : (1.) Let a: = the price of lib. in shillings. (2.) Then 5a: = the price of lib. in shillings X 5. (3.) But the cost of 5lbs. is by the question = 30s. (4.) .-. 5a: = 30s. (5.) and .*. x = Qs. which is the price of lib., as was required. It will be seen by steps (2) and (3) of this example, that there are two distinct expressions for the same thing, and that in step (4) these expressions are made equal to each other. In framing equations from problems, this will in all cases take place. As a second example let this problem be taken : — A house and an orchard are let for £28. a year, but the rent of the house is 6 times that of the orchard. Find the rent of each. The rent of the house is equal to that of 6 orchards ; we may therefore change the house into 6 orchards, and we shall have Rent of the orchard -\- 6 times rent of the orchard = £2S. Taking the sum of the rents of the orchard, we get 7 times the rent of the orchard = £28 j and the 7*^ part of each side of the equation being taken, The rent of the orchard = £4 ; and .'. the rent of the house = 6 times the rent of the orchard = 6 times £4 = £24. Now to give to these operations an algebraical shape, Let X = the rent of the orchard in £ then 6a:= house CH. II. SIMPLE EQUATIONS. 13 But by the condition of the question, Rent of the orchard -\- 6 times rent of the orchard = £28. /. a; + 6x = £28. or, Tar = £28. and, dividing each side of the equation by 7, a; = £4, the rent of the orchard, and .-. 6x = 6 X £4 = £24, the rent of the house. Again, suppose the following arithmetical question was proposed for solution ; viz. ** To divide the number 35 into two such parts, that one part shall exceed the other part by 9." A person unacquainted with algebra might with no great difficulty solve this question in the following manner : — (I.) It appears, in the first place, that there must be a greater and a less part. (2.) The greater part must exceed the less by 9. (3.) But it is evident that the greater and less parts added together must be equal to the whole number 35. (4.) If then we substitute for the greater part its equiva- lent , viz. ^^ the less part increased by 9",'' it follows, that the less part increased by 9, with the addition of the said less part is equal to 35. (5.) Or, in other words, that twice the less part with the addition of 9, is equal to 35. (6.) Therefore, twice the less part must be equal to 35, with 9 subtracted from it. (7.) Hence, twice the less part is equal to 26. (8.) From which we conclude, that the less part is equal to 26 divided by2\ i. e. to 13. (9.) And consequently, as the greater part exceeds the less by 9, it must be equal to 22. But by adopting the method of algebraic notation^ the different steps of this solution may be much more briefly expressed as follows : (1.) Let the /e55 part =x. (2.) Then the ^rea^er part =x-{-9, (3.) But greater part + less part =35. 14 ALGEBRA. (4.) /. X +9+X =35. (5.) or2ar + 9 =3.5. (6.) ,\2x =35—9. (7.) or2a: =26. (8.) .'. X (less ipart) =?^=13. (9.) and a: -f 9 (^rea^er part) =13 + 9=22. 29. Having thus explained the manner in which the se- veral steps in the solution of an arithmetical question may be expressed in the language of Algebra, we now proceed to its exemplification. PROBLEMS. Prob. 1. A dessert basket contains 30 apples and pears, but 4 times as many pears as apples. How many are there of each sort ? Let X = the number of apples ; then, as there are 4 times as many pears as apples, 4:X = the number of pears. But by the question the apples and pears together =: 30, .-. X + 4:X=S0. Adding the terms containing ar, 5a: = 30. Dividing each side of this equation by 5, X = 6, the number of apple*, .-. the number of pears = 4ar = 4 X 6=24. Prob. 2. In a mixture of 16lt)s. of black and green tea, there was 3 times as much black as green. Find the quan- tity of each sort. Let a; = the number of tbs. of green tea, then Sx= black. But the black tea -\- the green tea =: 16lbs. .-. a: + 3a; = I6lt)s. Collecting the terms which contain x, 4:X= 16lbs. Dividing each side of this equation by 4, a: = 4tbs. of green tea, .-. the black tea = 3x = 3 X 4=12tbs. CH. ir. SIMPLE EQUATIONS. 15 Prob. 3. An equal mixture of black tea at 5 shillings a Tb. and of green at 7 shillings a tt>. costs 4 guineas. How many lbs. were there of each sort ? Let X = the number of lbs. of each sort ; then5a:=the cost of the black in shillings, and 7x= green But cost of black + the cost of green :^ 4 guineas = 845. .-. 5x^7x=S4 12^ = 84 .*. X = 7lbs. of each sort. Prob. 4. The area of the rectangular floor of a school- room is 180 square yards, and its breadth 9 yards. What is its length ? Let X = the length in yards ; then since the area is the length multiplied by the breadth, we have X X^ = the area of the floor, .•.9a: =180 and .*. X == 20 yards, the length required. Prob. 5. Divide a rod 15 feet long into two parts, so that the one part may be 4 times the length of the other. Let X = the less part, . 15ft. , then 4x = the greater part I ^' J^ "I Now these two parts make together 15 ft. > l :. X -\-4x=l5h. 5a: = 15 ft. ,\ x= 3 ft. the less part, and the greater part = 4 times 3 ft. =: 12 ft. Prob. 6. A horse and a saddle were bought for £40., but the horse cost 9 times as much as the saddle. What was the price of each ? Ans. £36. and £4. Prob. 7. Divide two dozen marbles between Richard and Andrew, so that Rictiard may have three times as many as Andrew. Ans. Richard's share = 18 ... Andrew's ... =6. Prob 8. A boy being asked how many marbles he had, said, KI had twice as many more, I should have 36. How many had he ? Ans. 12. Prob. 9. A bookseller sold 10 books at a certain price, ]6 ALGEBRA. and afterwards 15 more at the same rate, arid at the latter time received 355. more than at the former : what was the price per book ? Let X = the price of a book in shillings ; then 10a: = the price of the 1st lot in shillings, and I5x= 2d Now, if the price of the 1 st lot be taken from that of the 2d, there remains a difference in price of 35«. .-. 15a:— 10a: = 355. Subtracting the 10a: from the 15a:, we have 5a: = 355. .♦. X = Is. the price of a book. Pros. 10. Divide X300 amongst A, B, and C, so that A may receive twice as much as B, and C as much as A and B together. Let a: = B's share ih £ then 2a: := A's share ... and a: + 2a: or 3a: = C's share in <£. But amongst them they receive £300 ; .-. a: + 2a: + 3a: = £300, 6a: = £300; .'. a:::=£50 B's share ; .-. A's share = £100, and C's share = £150. Prob. 11. If to nine times a certain number, three times the number be added, and four times the same number be taken away, there will then be obtained the number 48. What is the number ? Let X = the number ; then 9x = nine times the number, 3a: = three times the number, and 4a: = four times the number ; .-. 9a: 4- 3a: — 4x = 48, 8a: = 48 ; .-. X = 6^ the number required. Prob. 12. The sum of £100 is to be divided among 2 men, 3 women, and 4 boys, so that each man shall have twice as much as each woman, and each woman three times as much as each boy. Find the share of each. Let each boy's share = x ; then each woman's share == 3a:, and each man's share = 2 times 3a: = 6x ; CH. II. SIMPLE EQUATIONS. 17 Hence we have, the share of the 4 boys = 4x, the share of the 3 women = 3 times 3 x= 9Xf and the share of the 2 men = 2 times 6x=\2x : But the sum of all these shares is to amount to £100 j /. 4a: + 9a: + 12a: "=£100, . 25a: = £100; .-.a: = £4; /. each boy's share =£4; each woman's = 3 times £4 = £12 ; and each man's =: 6 times £4 = £24. Pbob. 13. A gentleman meeting 4 poor persons gave five shillings amongst them ; to the second he gave twice, to the third thrice, and to the fourth four times as much as to the first. What did he give to each ? Ans. 6d.y 12c?., 18c?., 24d., respectively. Pros. 14. Divide a line 12 ft. long into three parts, such that the middle one shall be double the least, and the great- est triple the least part. Ans*. 2, 4, 6. Prob. 15. Divide 40 into three such^ parts, that the first shall be 5 times the second, and the third equal to the dif- ference between the first and second. Ans. 20, 4, 16. Prob. 16. A grocer mixed three kinds of tea, Bohea at 3s. per lb., Twankay at 5s., and Souchong at 75. per lb. The mixture contains the same quantity of each, and cost £6. How many lbs. are there of each kind ? Ans. 8 lbs. Prob. 17. A bill of £700 was paid in sovereigns, half- sovereigns, and crowns, and an equal number of each was used. Find the number. Ans. 400. Prob. 18. Two travellers set out at the same time from Guildford and London, a distance of 27 miles apart; the one walks 4 miles an hour, and the other 5 miles. In how many hours will they meet ? Ans. 3 hours. Prob. 19. A person bought a horse, chaise and harness, for £120 ; the price of the horse was twice the price of the harness, and the price of the chaise twice the price of both horse and harness ; what was the price of each ? C Price of harness = £13,6*8, Answer -J horse = 26*13, 4 ( chaise = 80 . . c 18 ALGEBRA. SUBTRACTION. 30. Subtraction is the finding the difference between two algebraic quantities, and the connecting them by proper signs, so as to form one expression : thus, if it were required to subtract 5—2 (i.e. 3) from 9, it is evident that the remain- der would be greater by 2 than if 5 only were subtracted. For the same reason, if b — c were subtracted from «, the remainder would be greater by c, than if b only were sub- tracted. Now, if b is subtracted from «, the remainder is a — b; and consequently, if ^ — c be subtracted from a, the remainder will he a — b -\- c. Hence this general Rule for the subtraction of algebraic quantities, " Change the signs of the quantities to be subtracted^ and then place them one after another, as in Addition.'* Ex. 1. From 5a-\'Sx — 2b take 2c — 4^. The quan- tity to be subtracted with its signs changed, is — 2c + 4^ ; therefore the remainder is 5a -|- 3a: — 2Z> — 2c + 4^. Ex. 2. From 7ar2— 2a; +5 take Sar^+Sj- 1 ; The remainder is 7a;2— 2a: +5 — 3a:2— 5a:4-l . or 7a:2—3a:2— 2a: —5a: -1-5 +\=:4x^—7x+6. But when like quantities are to be subtracted from each other, as in Ex. 2, the better way is to set one row under the other, and apply the following Rule ; * ' Conceive the signs of the quantities to be subtracted to be changed, and then proceed as in Addition.'* Ex. 3. Ex. 4. Ex. 5. From 7ar2— 2a:-t-5 I2a^—Sa+ b—\ 5i/^—4i/+Sa Subtract 3a:2+5a:— 1 6a^+ a—2b+3 Qy'^-^^y^ a Remainder 4a:2— 7a:-|-6 6a2_4a_|-3ft— 4 — ^2 * _j.4a Ex.6. Ex.7. Ex.8. From 7a:^+2a:— 3^ Ux+i/'-z— 5 13a:»-2a:2+7 Subtract 2a:^— x-{- y a:+^+z— 11 — a:'-|- a:^— 6 30. What is svbtr action ? State the rule for the subtraction of algebraic quantities, and explain the principle on which it rests. SIMPLE EQUATIONS. 19 ON THE SOLUTION OF SIMPLE EQUATIONS, CONTAIN- ING ONLY ONE UNKNOWN QUANTITY. Rule II. 31. '*Any quantity may be transferred from one side of the equation to the other, by changing its sign ;" and it is founded upon the axiom, that "if equals be added to or subtracted from equals, the sums or remainders will be equal." Ex. 1 . Let a: + 8 = 15 ; subtract 8 from each side of the equation, and it becomes a: + 8— 8 = 15—8 ; but 8—8=0, .-. ar=15 — 8 = 7. Ex. 2: Let :c— 7=20; addl to each side of the equa- tion, then ar— 7 +7=20 +7 ; but— 7 + 7=0; .-..1=20 + 7=27. Ex. 3. Let 3a:— 5=2x + 9; add 5 to each side of the equation, and it becomes 3.r— 5 +5 = 2:r + 9 +5, or ^x = 2a; +9 +5. Subtract 2x from each side of this latter equation, then 'dx —2x = 2jr— 2;r + 9+5; but 2 jr — 2x = 0, .-. 3^— 2a: = 9+5. Now 3x— 2x = ar, and 9 + 5 = 14; hence x = \A, On reviewing the steps of these examples, it appears (1.) That a: + 8= 15 is identical with ar=15— 8. (2.) a: — 7=20 with j' = 20 + 7. (3.) 3a'— 5 = 207 + 9 with 3a:— 2a: = 9+5. Or, that **the equality of the quantities on each side of the equation, is not affected by removing a quantity from one side of the equation to the other and changing its sign,"" From this rule also it appears, if the same quantity with the same sign be found on both sides of an equation, it may be left out of the equation; thus, ifa: + a = c + a, then x = c-\- a—a\ buta— a = 0, :.x = c. It further appears, that the signs of all the terms of an equation may be changed from + to—, or from — to +, without altering the value of the unknown quantity. For let x—b = c— a ; then, by the Rule, x = c-^a + b ; change 20 ALGEBRA. the signs oi all the terms, then b—x = a—c, in which case b — a -J- c = X, or a: = c — a + b, as before. Ex.4. 2x + 3 = ar+17 Ans. a:=14 Ex.5. 5x —4 = 4x+25 ... x = 29 Ex.6. 7^ — 9=6a:— 3 ... x = 6 Ex.7. 4x + 2a = 3x+9b ... af = 9Z» — 2a Ex.8. 15j:+4 = 34 ... x==2 Ex.9. 8a:+7=6a; + 27 ... a: = 10 Ex. 10. 9x — 3 = 4a; + 22 ... x = b Ex.11. 17ar — 4a: + 9 = 3a: + 39 ... a; = 3 b4-Sr Ex.12. ax^c = b + 2c ... a: = — ■ ' a Ex.13. 5a:- (4^ — 6) =12 The sign — before a bracket being the sign of the whole quantity enclosed, indicates that the quantity is to be sub- tracted; and therefore, according to the Rule, when the brackets are removed the sign of each term must be changed. Thus, the signs of 4ar and of 6 are respectively + and — , but when the brackets are removed they must be changed to — and + respectively. The equation then becomes 5a: — 4a: + 6 = 12. By transposition, 5ar — 4a: =: 12 — 6 ; .♦. x==6. Ex. 14. 6a: — (8 -f x) = 4a: — (a: — 10) By removing the brackets, and changing the signs of the terms which they enclose, the equation becomes 6a: — 8 — a: = 4a: —a: + 10. Transposing, 6a7 — a: — 4a: + a' = 10 + ^ ; .-. 2a: =18. Dividing both sides of the equation by 2, a: = 9. Ex.15. 4a: — (3a:+4)=8 Ans. a:=12 Ex.16. 8a: — (6a: — 8) =9 — (3 —a:) ... a:= — 2 Ex. 17. 4a:- (3a: — 6) - (4a: - 12) = 12 - (5a:- 10) "• Ans. a:= 2 Ex.18. 5a: — (3 + 3a:)=8 — ( — a: — 1) ... a: = 12 CH. II. SIMPLE EQUATIONS. 21 PROBLEMS. Prob. 1. There are two numbers whose difference is lo and their sum 59. What are the numbers ? As their difference is 15, it is evident that the greater number must exceed the less by 15. Let, therefore x = the less number ; then will ^ + 15 = the greater : But their sum = 59 ; .*. a: -j- a; + 15 =59, or 2x-\- 15 =59. And, transposing 15, 2:r = 59 — 15, or '2x = 44 ; .'. ar = 22 the less number, and ar + 15 = 22 + 15 = 37 the greater. Prob. 2. I gave to Richard and James 27 marbles, but to Richard 5 more than to James. How many did I give to each? Let X = the number I gave to James ; then X -\'5z= Richard: But together they receive 27 ; ... a: + ar+5=27, or 2ar + 5=27. Transposing, 2x = 27 — 5, or 2ar = 22 ; .-. x= 11, the No. James received, and :r-4-5=16 Richard received. Prob. 3. Four times a number is equal to double the number increased by 12. What is the number? Let X = the number ; then 4;r = 4 times the number, 2t = double the number, and 2ar + 12 = double the number increased by 12. Therefore, by the equality stated in the question, 4a: = 2:c+12. By transposition, 4ar — 2:r =: 12 '2x = 12 ; .-. ar = 6. Prob. 4. At an election 420 persons voted, and the sue- 22 AliGEBKA. cessful candidate had a majority of 46. How many voted for each candidate ? Ans. 187 and 233. Prob. 5. One of two rods is 8 feet longer than the other, but the longer rod is three times the length of the shorter. What are their lengths ? Ans. 4 ft. and 12 ft. Prob. 6. Five times a number diminished by 16, is equal to three times the number. What is the number ? Ans. 8. Prob. 7. A horse, a cow, and a sheep, were bought for £24 ; the cow cost <£4 more than the sheep, and the horse £10 more than the cow. What was the price of the sheep ? Let X = the price of the sheep in £-, thena:+4= cow and a; + 4 + 10= horse But these three prices taken together amount to £24 ; ...a: + (a; + 4 ) + (a: + 4 + 10) =24. «;., Adding together like terms, Jjk^. 3ar+ 18=24. jm y gy transposition, 3ar = 24 — 18, 3;r = 6; .•.ar = <£2, the price of the sheep. Prob. 8. A draper has three pieces of cloth, which to- gether measured 159 yards ; the second piece was 15 yards longer than the first, and the third 24 yards longer than the second. What is the length of each piece ? Ans. 35, 50, and 74 yds. Prob. 9. Divide £36 among three persons. A, B, and C, in such a manner that B shall have £4 more than A, and C £7 more than B. Ans. £7, £11, and £18. Prob. 10. A gentleman buys 4 horses; for the second he gives £12 more than for the first ; for the third £5 more than for the second ; and for the fourth £2 more than for the third. The sum paid for all the horses was £240. Find the price of each. Ans. £48, £60, £^5, and £67. Prob. 11. What number is that whose double is as much above 21 as it is itself less than 21 ? Let X = the number ; then 2x = double the number, 2:r — 21 =: what double the*number is above 21, and 21 — a; = what the number is less than 21 : CH. II. SIMPLE EQUATIONS. 23 But by the question these two values are equal to each other; /. 2x — 21 =21 —a;. By transposition, 2x -{- x =21 -{-21, 3a; = 42 ; .'.x = U. The answer may easily be proved to be correct, for 2x — 21=28 — 21 = 7, and 21 — ;r = 21 — 14 = 7 ; that •is, twice 14 is as much above 21, as 21 is above 14, namely 7. Prob. 12. In dividing a lot of oranges among a certain number of boys' I found that by giving 4 to each boy I had 6 to spare, and by giving .3 to each boy I had 12 remaining. How many boys were there ? Let X = the number of boys ; then, if I gave to each boy 4 oranges, I should give away 4 times X oranges ; /. 4:X = number of oranges distributed at first : But the total number of oranges is 6 more than this number ; .'. Total number of oranges = 4x -\- 6 : Again, if each boy received 3 oranges, there were 12 oranges left ; .'. Total number of oranges = 3.r + 12. These two values for the number of oranges expressed in terms of ar must necessarily be equal ; Axiom (1.) .-. 4a7 + 6 = 3ar + 12. By transposition, 4^-— 3a: = 12 — 6; .-. x = 6, the number of boys. Prob. 13. An express set out to travel 240 miles in 4 days, but in consequence of the badness of the roads, he found that he must go 5 miles the second day, 9 the third, and 14 the fourth day less than the first. How many miles must he travel each day ? Let x = the number of miles on the 1** day; thenar — 5= 2^ ... ar- 9= 3^ ... andar— 14= 4**^... Now the number of miles which he travels in 4 days = 240; .-.ar + a:- 5 + a; — 9+3: — 14 =3^0. _ Collecting the terms, 4a: — 28 = 240^-- , -v^. a : ^o /y^ OF '^^y' iftJHlV. 24 ALGEBBA. By transposition, 4x = 240 + 28, 4a: = 268; /• X = 67, the number of miles he goes on I«* day, ar — 5 = 62 2" ... a: — 9=58 3^ ... and a: — 14 = 53 4'*'... Prob. 14. It is required to divide the number 99 into five such parts that the first may exceed the second by 3, be less than the third by 10, greater than the fourth by 9, and less than the fifth by 16. Ans. The parts are 17, 14, 27, 8, 33. Prob. 15. Two merchants entered into a speculation, by which A gaiifed £54 more than B. The whole gain was £49 less than three times the gain of B. What were the gains? Ans. A*s gain = £157 ; B's = £l03. Prob. 16. In dividing a lot of apples among a certain number of boys I found that by giving 6 to each I should have too few by 8, but by giving 4 to each boy I should have 12 remaining. How many boys were there ? Ans. 10. MULTIPLICATION. 32. Multiplication is the finding the product of two or more algebraic quantities; and in performing the process, the four following rules must be observed. (1.) When quantities having like signs are multiplied to- gether, the sign of the product will be + ; and if their signs are unlike^ the sign of the product will be — .* * This rule for the multiplication of the Signs may be thus ex- plained : — I. If + a is to be multiplied by + 6, it means, that + a is to be ad(Ud to itself as often as there are units in 6, and consequently the pro- duct will be + ah. II. If — a is to be multiplied by +6, it means, that — o is to be added to itself as often as there are units in b, and therefore the product is — 06. 32. What is multiplication, and what are the Rules to be observed in mul- tiplication ? CH. II. MULTIPLICATION. 25 (2.) The coefficients of the factors must be multiplied together, to form the coefficient of the product. (3.) The letters of which they are composed must be set down, one after another, according to their order in the Alphabet. (4.) If the same letter is found in both factors, the in- dices of it must be added together, to form the index of it in t\\Q product. Thus, + a multiplied by + 6 is equal to + ab, and — a multiplied by — ^ is also equal to -\-ab; + 3a: X ■- ■5^= — ibxy; — 3«^> X +4cc?= — 12a6cc/; — 4a2^>2 x — 3a6c?2 = + 12a3 ^,3 fp . &c. &c. From the division of algebraic quantities into simple and compound^ there arise three cases of Multiplication. In performing the operation, the Rule is, "To multiply ^r^^ the sigiis, then the coefficients, and afterwards the letters." Case I. 33. When both factors are simple quantities ; for which the Rule has been already given. III. If +a is to be multiplied by — 6, it means, that + a is to be subtracted as often as there are units in 6, and consequently the pro- duct is — ab. IV. If — a is to be multiplied by — fe, it means, that — a is to be subtracted as often as there are units in b ; and, since to subtract a negative quantity is the same as to add a positive one, the product will be +ab. Or, these four Rules might be all comprehended in one ; thus. To multiply a — b by c — d, is to add a — b to itself as often as there are units in c — d; now this is done by adding it a times, and subtracting it d times ; But a — b, added c times . . . =iac — be, and a — b, subtracted d times = — ad + bd, /, a — bxc — d =:ac — be — ad + bd. i.e. +ax +c=:+ac' — b X + c=z — be t+a X — rf= — ad ^bx—d- + bd. 26 ALGEBRA. Ex. 1. 4ab Sa Ex.2. 2axi/ — 6axy^ Ex.6. 9xY -2^ Ex.3. — Sabc 5a^b Ex.4. - 5a^bc - 2b^x'- X^a'b — Iba'b^c + lOa^b^cx^ Ex.5. 4abc Sac Ex.7. — ^cdx 2c Ex.8. ^7ax^ — 2ac^x Case II. 34. When one factor is compound and the other simple, **Then each term of the compound factor must be multi- plied **by the simple factor as in the last Case, and the re- sult will be the product required." Ex. I. Ex. 2. Multiply Sab—2ac-\'d by 4a Sx^ -2a:2+4 — l^ax Product l2a'b—Sa''c+iad — 42aa:* + 28aa;'— bQax Ex.3. Multiply 7x^ —2x -|-4a by — Sa Ex.4. 12a3— 2a2-|-4a— 1 3a: Product — 21aar2+6aa:— 12a2 Ex.5. Multiply 9a'^x+Sa — ar-J- 1 by — a:2 Product Ex. 6. 4a:-j/-f3ar— 2y — Sxy Case III. 35. When both factors are compound quantities, each term of the multiplicand must be multiplied by each term of the CH. II. MULTIPLICATION. 27 multiplier; and then placing like quantities under each other ^ the sum of all the terras will be the product required. Ex. 1. Ex. 2. Ex. 3. Multiply a+ b a + * a^^ab-^-b'^ hy a -\' b a — b a — b 1st, by a... 0^+ ab a'^+ab a^^a'^b+ab'^ 2d, by^>... ab+b'' —ab—b'' —a:'b—ab''—¥ Product a'^+^ab+b'' a^ * —b^ a^ * * ^ Ex. 4. Ex. 5. Sx^+ 2x 3a:2— 2x +5 4:r 4" 7 6a; — 7 12ar*+ Sx"" 18a:3— 12a:2+30.r +2lx''+Ux — 21a;2+14:r— 35 l2x^+29x''+Ux ISx^—SSx-'+^ix—Sd Ex. 6. Uac— Sab + 2 ac — ab ■];- \ Ua^c^— 3a''bc+ 2ac •^Ua^'bc +Sa''b^—2ab + Uac — 3a&+2 UaV— 17a2^>c+16ac+3a252_5^^_|.2 Ex.7. +2x^— x+^ Ex. 8. Multiply a^+Sa^b+3ab^+b^ by ... a-\-b. Ans. a^+^a^b+ea^b^+'iab^+b*. 28 ALGEBEA. Ex.9. Multiply 4x2^+3:r^—l by 2^:'— or. Ans. &x*i/+2x^t/'-2x^^Sx^f/-\'X. Ex. 10 x^—x^+x—rj by 2a:2+;r+l. Ans. 2z'— :c*+2a:'— 10^2— 4a:— 5. Ex. 11 3a^+2ah--b'' by Sa^—2ab+b\ Ans. 9a*^4a^b^+4a¥—b*. Ex. 12 x^-^-x^tz+xi/^ +^' .-.by x—i/. Ans. x*—i/*. Ex. 13 x^^^x+l by x^-^^x. Ans. o:^— ^ar'+^a:^— Jr. ON THE SOLUTION OF SIMPLE EQUATIONS CON- TAINING ONLY ONE UNKNOWN QUANTITY. Ex. 1. 3a: + 4 (a: + 2)=36. The term 4 (a: -f- 2) means, that a: + 2 is to be multiplied by 4, and the product by Case 2** is 4a: + 8 ; .-. 3a: + 4a: +8 = 36. Adding together the terms containing a:, and trans- posing 8, 7x =36 — 8, 7a: = 28; .-. a: = 4. Ex. 2. 8 (a: + 5) + 4 (x + 1) = 80. Performing the multiplication, 8a: + 40 + 4a:+4 = 80. Collecting the terms, 12a: + 44 =80. Transposing, 12a: = 80 — 44, 12a: = 36; .-. a: = 3. Ex.3. 6 (a: + 3)+ 4a: = 58. Ex.4. 30 (a: -3) +6 =6 (a: +2). Ex. 5. 5 (a: +4) - 3 (a: - 5) = 49. Ex.6. 4 (3+2a:) — 2 (6 —2a:) = 60. Ex.7. 3 (a: -2) +4 =4 (3 -a:). Ex. 8. 6 (4 - a:) - 4 (6 - 2a:) - 12 =0. Ans a: = 4. a: = 4. ar :^ 7. x=5. X = 2. x = 6. CH. tl. SIMPLE EQUATIONS. 29 PROBLEMS. Prob. I . What two numbers are those whose difference is 9, and if 3 times the greater be added to 5 times the less, the sum shall be 35 ? Let X = the less number ; then ar + 9 = the greater. And 3 times the greater = 3|k(a: + 9) = 3ar + 27, 5 times the less =5a:.% But by the problem, 3 times the greater + 5 times the less = 35 ; .•.3a;+27+5ar = 35, 8a: + 27 =35. Transposing, 8a: = 35 — 27 = 8 ; .'. a: = 1 , the less number, and a: + 9 = 10, the greater. Prob. 2. A courier travels 7 miles an hour, and had been dispatched 5 hours, when a second is sent to overtake him, and in order to do this, he is obliged to travel 12 miles an hour. In how many hours does he overtake him ? Let X = the number of hours the 2"^ travels ; tliena:+5= P* .*. 12a:= the number of miles the 2** and7(a: + 5)= l'* But by the supposition the couriers both travel the same number of miles ; .-. 12a: = 7 (a: +5), 12a: = 7a: + 35. Transposing, 12a: — 7a: = 35, 5x = 35, a: =r 7, the number of hours the se- cond courier is in overtaking the first. Prob. 3. In a railway train 15 passengers paid <£3 125.; the fare of the first class being 6s. ^ and that of the second 4«. How many passengers were there of each class ? Let X = the number of passengers of the 1** class, then 15 — ar= 2^ .*. 6x = sum in shillings paid by P* class passengers, and4(15-a:)= 2^ 30 ALGEBRA. But these two sums amount to £3 125., or to 72«. ... 6a: + 4 (l5 — x) = 72, ex+60 — 4x = 72. By transposition, 6a: — 4a: = 72 — 60, 2a:=12; /. x = 6 No. of l** class passengers ; .*. the number of 2^ class passengers =15 — a: = 9. Pros. 4. What number is that to which if 6 be added twice the sum will be 24 ? . Ans. 6. Prob. 5. What two numbers are those whose difiference is 6, and if 12 be added to 4 times their sum, the whole will be 60 ? Ans. 3 and 9. Prob. 6. Tea at 6s. per Jb. is mixed with tea at 45 perltx, and 16 tbs. of the mixture is sold for £3 I85. How many lbs. were there of each sort ? Ans. 7 lbs. and 9 lbs. Pbob. 7. The speed of a railway train is 24 miles an hour, and 3 hours after its departure an express train is started to run 32 miles an hour. In how many hours does the ex- press overtake the train first started ? Ans. 9 hours. Prob. 8. A mercer having cut 19 yards from each of three equal pieces of silk, and 17 from another of the same length, found that the remnants taken together measured 142 yards. What was the length of each piece ? Let X = the length of each piece in yards ; .*. X — 19 = the length of each of the 3 remnants, and a: -- 17 = the length of the other remnant ; then 3 (a: — 19) + a: — 17 = 142, otSx — 57 +x— 17 = 142, 4a: -74 =142. Transposing, 4a: = 142 + 74, 4a: = 216; .-. a: = 54. Prob. 9. Divide the number 68 into two such parts, that the difference between the greater and 84 may equal 3 times the difference between the less and 40. Let x = the less part, * then 68 — x= the greater ; CH. II. SIMPLE EQUATIONS. 31 /. 84 — (68 — a:) = difference between 84 and the greater, and 3 . (40 — ar) = 3 times the difference between the less and 40. But by the question the differences are equal to each other; .-. 84 — (68 — a;) = 3 . (40 — x), or 84 — 68 + ar= 120 — Sx, By transposition, x -{-Sx = 120 + 68 — 84, 4x = 104 ; /. X = 2(i, the less part ^ and /. the greater = 42. Prob. 10. A man at a party at cards betted three shil- lings to two upon every deal. After twenty deals he won five shillings. How many deals did he win ? Let X = the number of deals he won ; ,'. 20 — j: = the number he lost ; /. 2x = the money won ; and 3 . (20 — x) = the money lost. But the difference between the money won and the money lost was 5s. :. 2x — S. (20 — ^) = 5, • 2^7 — 60 + 30:= 5, 5ar — 60= 5, 5x==65 ; :. ^=13. Prob. 11. A and B being at play cut packs of cards so as to take off more than they left. Now it happened that A cut off twice as many as B left, and B cut off seven times as many as A left. How were the cards cut by each ? Suppose A cut off 2a; cards ; then 52 — 2x = the number he left, and X = the number B left ; .*. 52 — a: = the number he cut off. But the number B cut off was equal to 7 times the number A left ; .-. 52 — a: = 7 . (52 — 2a;) 52 -a; = 364 — 14a;. Transposing, 14a; — a; = 364 — 52, 13a; = 312; .-. a; = 24; .-. A cut off 48, and B cut off 28 cards. 32 ALGEBRA. Prob. 12. Some persons agreed to give sixpence each to a waterman for carrying them from London to Greenwich ; but with this condition, that for every other person taken in by the way, threepence should be abated in their joint fare. Now the waterman took in three more than a fourth part of the number of the first passengers, in consideration of which he took of them but fivepence each. How many persons were there at first ? Let 4ar represent the number of passengers at first ; then 3 more than a fourth part of this number = a; + 3, and they paid 3 (x + 3) pence. .*. the original passengers paid 6 X 4ar — 3 (a: + 3) pence. But the original passengers paid 5 X 4a: pence ; .-. by equalizing these two values, we get 6 X 4ar - 3 (a: + 3) = 5 X 4a:, 24a: — 3a: — 9= 20a:. Transposing, 24a: — 3a: — 20a: = 9 ; .-. a: = 9 ; and .-. the No. of passengers were =4 X 9 = 36. Prob. 13. There are two numbers wliose difference is 14, and if 9 times the less be subtracted from 6 times the greater, the remainder will be 33. What are the numbers ? Ans. 17 and 31. Prob. 14. Two persons, A and B, lay out equal sums of money in trade ; A gains £120, and B loses £80 ; and now A's money is treble of B's. What sum had each at first ? Ans. £180. Prob. 15. A rectangle is 8 feet long, and if it were two feet broader, its area would be 48 feet. Find its breadth. Ans. 4 feet. Prob. 16. William has 4 times as many marbles as Tho- mas, but, if 12 be given to each, William will then have only twice as many as Thomas. How many has each ? Ans. 24 and 6. Prob. 17. Two rectangular slates are each 8 inches wide, but the length of one is 4 inches greater than that of the other. Find their lengths, the longer slate being twice the area of the other. Let X = the length in inches of the less ; thena: + 4= greater. CH. II. SIMPLE EQUATIONS. 33 Now the area of a rectangle is its length multiplied by its breadth ; .*. 8^ and 8 (a: + 4) are the areas of the slates. But the larger slate is twice the area of the less ; .-. 8arX 2 = 8 (ar + 4), . 16a; = 8a7 + 32; .'.&x = S2; :. a; = 4, the length of the less slate, and a: + 4= 8, greater slate. Prob. 18. Two rectangular boards are equal in area ; the ,breadth of the one is 18 inches, and that of the other 16 inches, and the difference of their lengths 4 inches. Find the length of each and the common area. __ Ans. 32, 36, and 576. Prob. 19. A straight lever (without weight) supports in equilibrium on a fulcrum 24 fbs. at the end of the shorter arm, and 8 tbs. at the end of the longer, but the length of the longer arm is 6 inches more than that of the shorter. Find the lengths of the arms. Let X = length in inches of the shorter arm ; then a; 4- 6= longer ... Now the lever will be in equililirium, when the weight at one end multiplied by the length of the corresponding arm is equal to the weight at the other end, multiplied by its corresponding arm ; .-. 24z = 8 (:r + 6), 24x = Sx+ 48, 16a: = 48; .♦. X = S inches, the length of the shorter arm ; anda:+6 = 9 longer ... Prob. 20. A weight of 6 lbs. balances a weight of 24 lbs. on a lever (supposed to be without weight), whose length is 20 inches ; if 3 lbs. be added to each weight, what ad- dition must be made to each arm of the lever, so that the fulcrum may preserve its original position, and equilibrium still be retained ? This problem resolves itself into two other problems : — (1.) To find the lengths of the arms in the original position : Let X = the lergth in inches of the shorter arm ; then20 — a:= longer ... D 34 ALGEBBA. Now, in order that there may be equilibrium, 24a: and 6 (20 — z) must be equal to each other ; /. 2ix= 120 — 6a:, 30:p=120; .'. x=: 4, the length of the shorter arm ; and 20 — x= 16, longer ... (2.) To find the addition to be made to each arm, so that there may again be equilibrium on the fulcrum in its original position, after 3 tbs. have been added to each weight : Let X = number of inches to be added to each arm ; then the lengths of the arms become 4 +a:, and 16 -f-.r inches respectively : and the weights at the arms have been respectively increased to 27 lbs. and 9 lbs. But by the principle of the equilibrium of the lever, 27 (4 + ^) and 9 (16 -|- ^) niust be equal to each other; ... 2f (4 4- a:) = 9 (16 + x). Divide each side of the equation by 9, and 3 (4+ a:) = 16+ a:, 12 4- 3a: =16+ a:, * 3a: — a: = 16 — 12 2a:= 4; .-. a: = 2. Prob. 21. The conditions being the same as in the last problem, how many inches must be added to the shorter arm in order that the lever may in its original position retain its equilibrium ? Ans. 1^ inch. Prob. 22. A garrison of 1000 men was victualled for 30 days; after 10 days it was reinforced, and then the pro- visions were exhausted in 5 days ; find the number of men in the reinforcement. Ans. 3000. Prob. 23. Two triangles have each a base of 20 feet, but the altitude of one of them is 6 feet less than that of the other, and the area of the greater triangle is twice that of the less. Find their altitudes. Ans. 6 and 12. N.B. The area of a triangle = J base X altitude. Prob. 24. A and B began to play with equal sums ; A won 125. ; then 6 times A's monej^ was equal to 9 times B's. What had each at first r Ans. £S. CH. II. SIMPLE EQUATIONS. 35 pROB. 25. A company settling their reckoning at a tavern, pay 4 shillings each, but observe that if there had been 5 more they would only have paid 3 shillings each ? How many were there. Ans. 15. Pbob. 26. Two persons, A and B, at the same time set out from two towns 40 miles apart, and meet each other in 5 hours, but B walks 2 miles an hour more than A. How many miles does A walk in an hour? Ans. 3 miles. DIVISION. 36. The Division of algebraic quantities is the finding their quotient, and in performing the operation the same circumstances are to be taken into consideration as in their multiplication, and consequently the four following Rules must be observed. (I.) That if the signs of the dividend and divisor be like, then the sign of the quotient will be + ; if unlike, then the sign of the quotient vrill be — .* (2.) That the coefficient of the dividend is to be divided by the coefficient of the divisor, to obtain the coefficient of the quotient. (3.) That all the letters common to both the dividend and the divisor must be rejected in the quotient. (4.) That if the same letter be found in both the divi- dend and divisor with different indices, then the index of * The Rule for the signs follows immediately from that in Multi- plication ; thus, If + ax +«>=+AthGn±^=+&,and-^=z+fl \ i.eJi/c, signs — uh , , — a6 (produce +, + a X -^b:=i—ab,. . ., — ; — t=— 0, and r-=: + a > + «o — b (and unlike -ax-b = + ab,..,. -±iLf!=_6,and-^!^=_a ] signs -. — a — b I 36. What is meant by the division of algebraic quantities ? State the Rules in division. 3<3 ALGEBRA. that letter in the divisor must be subtracted from its index in the dividend, to obtain its index in the quotient. Thus, (1.) +«*c divided by +ac or ^ =+^. Qabc (2.) + Qabc —2a or ^^-- z=^3bc. (3.) ^lOxyz +5i/. —2a — \Oxyz =— 2a:z. (4.) -20a^a:y -4ax^ ...or ^^^'^^ =^oaxy\ Of Division^ also, there are three Cases ; the same as in Multiplication. Case I. 37. When dividend and divisor are both simple terms. Ex. 1. Divide ISax"^ by 3aa:. 18a:r2 =6a:. 3aa: Ex. 3. Divide — 28a:y by —^xy. Ex. 2. Divide ISa^Z^^ by —5a + 15q^Z>- —5a Ex. 4. — 3a^»2. — 28^__ — 4a:^ f 7a:^2, Ex.5. Divide — Ua^^^c by 7 ac. — Ua^'^c _ +7ac Divide 25 a^c^ by —Sa^c. _+25a3c2__ — 5a2c "" Ex.6. Divide — 20x'^y^z^ hy—4yz. -^20xYz^ _ "•Ayz, ~ Case II. 38. When the dividend is a compound quantity, and the divisor a simple one ; then each term of the dividend must be divided separately, and the resulting quantities will be the quotient required. . 38. State the rule for Case 2d. CH. II. DIVISION. 37 Ex. 1. Divide 42a^+Sab+l<2a^ by 3a. Sa ^ ^ Ex.2. Divide 90a^x^—lSax^-[-4a^x—2ax hy2ax, 90ol^^_lSax^+^a^x-2ax^^^^^,__^^^^^_ ^ 2 ax Ex. 3. Divide 40:'— 2a:2+2a; by 2x 2x Ex. 4.. Divide —24:a^x^j/-'Saxi/-\-6x^i/^ by ^Sxi/, —24a'^x^i/ —^axy-\'Qx'^y''- __ — 3a;^ ~~ Ex.5. Divide Uah^J^la'^h''-'2\a''¥J^^5a^b by 7a^. Case III. 39. When dividend and divisor are both compound quan- tities. In this case, the Rule is, *' to arrange both dividend and divisor according to the powers of the same letter, be- ginning with the highest ; then find how often the first term of the divisor is contained in the first term of the dividend, and place the result in the quotient ; multiply each term of the divisor by this quantity, and place the product under the corresponding (i. e. like) terms in the dividend, and then subtract it from them ; to the remainder bring down as many terms of the dividend as will make its number of 39. When dividend and divisor are both compound quantities, what is the rule? 38 ALGEBKA. terms equal to that of the divisor ; and then proceed as be- fore, till all the terms of the dividend are brought down, as in common arithmetic." Ex. 1. Divide a'— 3a^b + Sab^ —b* hj a — b. a —b) a^—Sa^b + Sab^—b^ (a^— 2ab + b'' * — 2a2Z, + 3aZ>2 + 2a62 * ab^- ab^-- * ■b^ •b^ * In this Example, the dividend is arranged according to the powers of a, the first term of the divisor. Having done this, we proceed by the following steps : — (1.) a is contained in a', a^ times j put this in the quotient. (2.) Multiply a—b by a-, and it gives a?—d^b. (3.) Subtract a^^d^b from a^—Sa^b, and the remainder is — 2a2^. (4.) Bring down the next term + 3ab^. (5.) a is contained in ^2a^b, --2ab times; put this in the quotient. (6.) Multiply and subtract as before, and the remainder is a62. (7.) Bring down the last term —b^, (8.) a is contained in a¥, + b"^ times ; put this in the quotient. ^9.) Multiply and subtract as before, and nothing re- mams J the quotient therefore is a^ — 2ab -\- b"^. CH. II.. DIVISION. 39 Ex. 2. a2 + lax + x^ )a* + 5alr + lOa^t^ + lOaV + 5ai* + a^( a3 + 3flV + 3aa2+ 13 '^ ^a^x-+ la'x^+Sax^ * a'^x^+2ax^+x^ a^x^-\-2ax*+x5 Ex.3. -7x Ax'^'-lx ) 12a:5-13a:*~34a;'4.39ar2/3a:3+2ar«-5a:+-^^ M2a;5— 21a:* \ 4x2—7 + 8a:*— 34a:* + ^x^—Ux^ * -20a:'+39a:2 — 20a:'+35a:2 * + 4a;2 Ex.4. 3a:— 6\6a*— 96 /2a?»+4a:''+8a:+16 /6a:*— 12a:n * +120:3—96 4-12a:3— 24a:2 * 4-240:^-96 +24o:2— 48o: * +48o— 96 +48o:— 96 * When there is a remainder^ it must be made the numerator of a Fraction whose denominator is the divisor ; this Fraction must then be placed in the quotient (with its proper sign), the same as in com- mon arithmetic. 40 ALGEBRA. Ex.5. X6-l.X^-^X* X6-^X^-^X' — x^ — x*-\-x^ ■■ +x^—x^-\.2x -{-x^-^-x^ — x"^ * ^x^+'2x—l — x^ — x^-\-x * 4- x^+x-l _j_ x'^-^-x—l ♦ * ♦ Ex.6. ^^ix^x^-^^x^+^x^-ixf^x^-ix+l X* — hx^ * + x--ix + x'^—lx Ex. 7. Divide a*+4a^b+6a^b^+4ab^+b* by a+^. Ans. a34"3«'^+3«^'+*'- Ex. 8 a^—5a^x-\-l0a^x^—\0a^x^-\-5ax'-'X^ by a^^Sa^x+Sax'^-^x^. Ans. a2_2aa:+:r2. Ex. 9 25a:6— ar»— 2a:3— 8a:2 by Sx^— 4^;^ Ans. 5x^-{'4x^+3x-^2. Ex. 10 a*+8a'a:+24a2ar2+3$aa;'+16^* by a+'2x. Ans. a'+6a'ar+12aa:2+8x». CH. II. DIYISION. 41 Ex. 11. Divide a^^x^ by a— ar. Ans. a*-^a^x+a^x'^+ax^-\-x*. Ex. 12 ex^+dx^—^Ox by 3x^—Sx. 5x Ans. 2a;^+2^+5-g-^;33^ Ex. 13 9x^^46x^+95x^+\50x by x^—4x—5. Ans. Qx'—lOx^+bx^—SOx. Ex. 14 X*— ^a:3+a:2+^a:-2 by ^a:-2. Ans. Ja;^— Jx^+l. PROBLEMS PRODUCING SIMPLE EQUATIONS, CONTAIN- ING ONLY ONE UNKNOWN QUANTITY. Prob. 1. A fish was caught the tail of which weighed 9 lbs. ; his head weighed as much as his tail and half his body, and his body weighed as much as his head and tail. What did the fish weigh ? Let 2x = weight of the body in lt)s. ; .*. 9 + a: = weight of tail + | body = weight of head. But the body weighs as much as the head and tail ; .-. 2^ = (9 + a:) + 9, 2x = x+ 18; .-. ar=18, and .*. 2a: = 36, the weight of body in ttSs., 9 -l-ar = 27, the weight of head in tbs. and the weight of fish = 36 + 27 + 9 = 72 lbs. Prob. 2. A servant agreed to serve for £8 a year and a livery, but left his service at the end of 7 months, and re- ceived only £2 13s. 4d. and his livery ; what was its value ? Let 12a: = the value of the livery in d. But £8 = 1920c/., and £2 13^. 4d. = 6i0d, ; then, the wages for 12 months = 12a: + 1920 ; 12a: 4- 1920 .*. the wages for 1 month = r^ = a: -}- 160. and .*. the wages for 7 months = (a: -f. 160) 7. 42 ALGEBEA. But the wages actually received for 7 months = 12ar -|-640} /. 12a: + 640 = 7a: + 1120 ; .-. 5a: = 480, a: = 96 ; and .*. 12a:=1152c?. =:£4 165., value of livery. From the solutions of the two preceding problems, it will be seen, that by assuming\/br the unknown quantity, x with a proper coefficient, an equation free from fractions will be obtained. It is frequently not only convenient to make such an assumption, but a more elegant solution is generally thereby obtained. The coefficient of x must be a multiple of the denominator of all the fractions involved in the problem. Prob. 3. A cistern is filled in 20 minutes by 3 pipes, one of which conveys 10 gallons more, and another 5 gallons less, than the third per minute. The cistern holds 820 gal- lons. How much flows through each pipe in a minute ? Ans 22, 7, and 12 gallons, respectively. Prob. 4. A and B have the same income : A lays by ^ of his ; but B spending £60 a year more than A, at the end of 4 years finds himself £160 in debt. What did each an- nually receive ? Ans £100. Prob. 5. A met two beggars, B and C, and having a certain sum in his pocket, gave B ^ of it, and C |- of the remainder : A now had 20 + Z>2 Reduce to a mixed quantity. Here =za -\-b is the integral part, and — is \hQ fractional part ; b"^ . /. a -\-b -{• — is the mixed quantity required. 42. What is the rule for reducing an improper fraction to a mixed quantity P 46 ALGEBBA. Ex. 2. 15(22 -J_ 2a: 3c Reduce to a mixed quantity. 15^2 Here-r — =Sa is the integral part, , 2a: — 3c and — : is the fractional part ; . 2ar — 3c. , /. 3a -| r IS the mixed quantity required. ^7.2 _ Kq Ex. 3. Reduce to a mixed quantity. "^« Ans. 2a: — r-- 2x 12a« + 4a — 3c Ex.4 2 to a mixed quantity. - . 3c Ans. 3a -4-1 — :; — ' 4a 10x2y + 3a:«-2Z>2 Ex. 5 to a mixed quantity. 2Z»2 Ans. 10^-f-3a: -• 43. To reduce Fractions to a common Denominator. Rule. ** Multiply each numerator into every denomi- nator but its own for the new numerators, and all the de- nominators together for the common denominator." Ex. 1. _ , 2a: 5x , 4a Reduce -r>-T-> ana -r-» to a common denominator. o (> 5 2xXbX^=\0bx') f Hence the frac- 5a:X3X5=75a: > new numerators ; \ tions required are 4flX3X^=12a6 ) \ \obx 75x I2ab 3 XbX5=\5b common denominator ; ^ 15^' 1 5b* \bb ' 43. How are fractions reduced to a common denominator ? CH. III. ALGEBRAIC FRACTIONS. 47 Ex.2. Reduce — ?^ > and — - » to a common denominator. 5 4 ' Hence the frac- Here (2^+1) X 4= 8^+4 > „3^„^^erators; \ tions required 'Av V .'1= I /s.r V ^ are 3a:X5=15a: 5 I 8ar+4 , \bx 5X4=20 common denominator ; f — ^ , and- . Ex.3. Reduce ^^ , ^""^, and — , to a common denominator. a-\-x 3 2a: Here5xX 3X2^=30:.^ ) .'. the new frac- _30£^^ (a-:r) (a-|-x)x2:r=2a2a:-2a:3 f tions are Qax-^iSx' 1 X(«+^)X3 =3a +3a: r' 2agi?— 2:r^ , 3q+3a: (a+a;) X3 X2a: =6«a: +6a;2 jGo^'+e^' ^" Gax+Ga:^* 3^ ^bx 5x^ Ex. 4. Reduce — » — » and y to a common denominator. 5 3a a . Qa^ar 20abx .Ibax"^ AnS. r-;r^^ -TT— „-5 ^^d - 15a2 15a2 15^2 • Ex. 5. Reduce ^"^ , and ^"' ^ to a common denominator. 3 Ai.s.5^,and^£!±£ ^x 3a: • Ex.6. Reduce -^J—^, -^ 9 and — -, to a common denominator. 5 4 a 36 48aZ>^2 I 24a^>a: 456^:^ , 40aa: Ans ^ 5 -, and ——7. QOah 60ab 6i)ab Ex.7. Reduce 5 and ^2_, to a common denominator. 2x 2a2 . 14a2^2_2a2 , 8.r^— 2a:2+4a: Ans 5 and » ■' . ^a^x 4a^x 48 ALGEBEA. 44. To reduce a Fraction to its lowest terms. Rule. "Observe what quantity will divide all the terms both of the numerator and denominator without a remain- der ; Divide them by this quantity, and the fraction is reduced to its lowest terms." Ex. 1. Reduce hrz-T-' to its lowest terms. The coefficient of every term of the numerator and denomi- nator of this fraction is divisible by 7, and the letter x also enters into every term ; therefore Ix will divide both nume- rator and denominator without a remainder. 14V-|-7aa:-f21a:2 Now T^ =:2a:2+a+3ar, , 35a:2 and -;; — =5ar: Ix * , ^ . . . , . 2a:2_La-i-3a: Hence, the traction m its lowest terms is — — ■ . bx Ex.2. ^ , lOabc—ba^ + lOac . , Reduce r-^ to its lowest term. Here the quantity which divides both numerator and de- nominator without a remainder is ba ; the fraction therefore 4:bc—a-\-^c in its lowest terms is . ac Ex.3. a — b Reduce -r — Tc to its lowest terms. YIqyq a—b will divide both numerator and denominator, for by Ex. 2. Case HI. page 27. a^—b''- = {a-\-b) (a—b); hence r is the fraction in its lowest terms. a-\-b 10a:» . , Ex. 4. Reduce TFT to ^t lowest terms. Ans. — ♦ 3 44. Show how fractions are reduced to their lowest terms. CH. III. ALGEBRAIC FRACTIONS. 49 ^ „ , Sabx^ . , Ex. 5. Reduce — to its lowest terms. 6ax Ans. ^. 2 ^ 14a:V— 21ar'v2 Ex. 6 ^:r-i — to Its lowest terms. 7xy Ans. ^y-^^. X 51x^^17 x^+S4x . , Ex. 7 ,„ ' to its lowest terms. Ans. !f!=f±?. a:* ON THE ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION, or FRACTIONS. 45. To add Fractions together* Rule. ** Reduce the fractions to a common denominator, and then add their numerators together; bring the re- sulting fraction to its lowest terms, and it will be the sum required." Ex. 1. Sx 2ar x Add _, _, and-, together. 2xX^X^=S0xf eSx+S0x4-S5x 128ar . , ^ . xXdX7=S5x>''' Yo5 "^ToT ^^ fraction 5X7X3=105 3 required. a 2a J 5b Ex. 2. Add T5 ^5 and — , together. aXSbX'^a=:]2a'^b j 12a^6+8a^Z>+15^>^ _20q^^+156^ I2ab^ 12a^2 = (dividing by b) ^Q^'+^^^' is the I2ab sum required. aXSbX'^a=]2a^b \ 'laXbX4a= Sa^bf 5bXbXSb=l5b^ > /^X3^X4a=12a^>2 ; ~^ 45. State the rule for adding fractions. 50 ALGEBKA. Ex. 3. Add ^^ , ^"" 5 and y , together. (2a:4-3)x2a:X7=28a:2 +4:2x^ 28a:«+42a:+105ar~85+40 3r^ (3a:— 1)X5 X7=105ar— 35 f •*• 70^ 4xX5X2a:=40a:2 r _68a:2 + 147x-^35 5X2.X7=70. ) 705 -4^--' ' Ex. 4. Add — » —J and — , together. ^ ^ ^^ , 934a: ^°«- 693- ^ ^ 3a« 2a ,36 Ex.5 -^T-J -r-> and —, together. 105a84-28a«64-3062 70ab Ex. 6 ' > > and-, together. 169a:+77 ^^s. —105 Ex. 7 Q^ J and — ^ — , together. Ans Q7fl''+llj> 2ir— 5 or— 1 Ex. 8 g > and -^, together. Ans. 4^'-7^-3 6x Ex. 9 Tj and •— — -, together. a:— 3 x+3 ^ Ans. x^—O „ a+6 J a— 6 Ex. 10 —^9 and — -— , together. a— 6 a+o ° 2a24-26* CH. III. ALGEBRAIC FRACTIONS. 51 46. To Subtract Fractional Quantities. Rule. *' Reduce the fractions to a common denomi- nator; and then subtract the numerators from each other, and under the difference write the common denominator.*' Ex. 1. Sx n 14ar Subtract y from—. SxXl5=45x') 70x—45x 25x x, ^ ^.^ 14arX 5=70a ; > •*• 75 ="75" = 3 ^^ ^he difference 5 X 15=75 ) required. Ex.2. Subtract -^ from -^i-. «3 7 (2a:+l)x7=14a:+7^ 15^+6 -14ar --7 x—l . , (5ar+2)X3=15:r+6 ^ - 21 —~2r^ 3X7=21 ) fraction required. Ex.3. From . subtract 8 7 (10a:-9)x7=70a:— 63 j 70ar— 63— 24a:+40 463:~ 23 ( 3a:-5)x8=24x-40 ( - -qs -'~56~ 8 X 7=56 ) is the fraction required. Ex.4. From — -t_ subtract ;- ia+b) {a+b)=a^+2ab+b^) . a'+^ab+b^-^^+2ab^b^ ^ (^a-b) {n-b)=a''-2ab+b''y " '^'"■*' [a^b) {a+b)=a^~^b^ 1 a'^—b^ is the fraction required. 4«ir 9x X Ex.5. Subtract -r- from-— Ans. —• o 10 10 46. Give the rule for subtracting fractioxis. ^2 ALGEBBA. Ex.6. Subtract 5^ from ?i±±2. Ans. lHZf±lZ. Ex.7 ?£±iiromlf. ^4-1 5 Ex.8 ?£=:2from!£±2. 3^ 3 Ex. 9 —r-r from -. a-\-o a—o Ex. 10 — g— from y. 28 ATi-St- 4x2— 11a:— 5 5x+5 Ans, 4x^+3 ' 3x Ans. 2b Ans. llar+49 56 47. To Multiple/ Fractional Quantities. Rule. ** Multiply their numerators together for a new numerator, and their denominators together for a new de- nominator, and reduce the resulting fraction to its lowest terms." Ex. 1. 2xX'^3:=Bx' 7 X9 =63 Multiply y by y } :, the fraction required is ^ 8z^ Ex.2. Multiply — ^ by y . Here 2ix^+6x 2l = (dividing the nu- (4a;+l)X6a:_24ar2+6a; j ^^^.^^^^ ^nd denominator by 3) 3X7 =21 ' Sx''-{-2x is the fraction required. 47. State the rule for the multiplication of fractions. CH. III. ALGEBRAIC FRACTIONS. 53 Ex.3. Multiply ^3_ by _. By Ex. 2. Case III. page 27, (a^— 62)x3a«=(a+6) ia-V) X 3a- hence the product is 3«'X («+&) (a-6) ^ (dividingthe numerator and denominator bya+^>) ^r _ 3fl^— 3a^6 Ex. 4. Multiply -j^ by ^-__. / . 21a;r2_35a:p ,,. .,. Here " 28^3^42^ -= (dividing {^x'^''bx)yja=:Q\ax'^'-^bax\ the numerator and denomi- and ] , „ ,'^ax—ba, , ,^ , ^ . , , ^ I nator by 7a;) —r-^ — — is the (2a:3— 3ar)xl4=28ar»— 42a- f "^ ^ 4ar2-.6 ^ fraction required. Ex. 5. Multiply =- by -=-• Ans Ix—l. ^ ^ 3a:2-.a: , 10 3a:— 1 Ex. 6 by Ati Divide -^^— by -X_-. 5a^^^_5X(a+bH--'b)l .'. ^ ^^' ' ^ 4X (a+b) 2a 2a } . . -6r-= — 66 — ' f . . ?" . ^ ''" I the fraction required. Ex.4. ^. ., 4a: . 9x Divide y '^y y 20 Ans. -. Ex. 5. 4a+2 , 2x-hl 3 "^ 5a: lOar Ans. 3 . Ex.6. a:^-9 , r+3 5^4 * 4;r-l2 Ans. ^ Ex.7. 9a:2— 3a: ^"^ ♦ - — : by — • ^ 9x-3 Ana. • X CH. Til. SIMPLE EQITATIONS. 55 ON THE SOLUTION OF SIMPLE EQUATIONS, CONTAIN- ING ONLY ONE UNKNOWN QUANTITY. Rule III. 49. An equation may be cleared of fractions by multi- plying each side of the equation by the denominators of the fractions in succession. Or, an equation may be cleared of fractions by multi- plying each side of the equation by the least common mul- tiple of the denominators of the fractions. This Rule is derived from the axiom (4), that, if equal quantities be multiplied by the same quantity ( or by equal quantities), the products arising will be equal. Ex. 1. Let -=6. 3 Multiple/ each side of the equation by 3 ; then (since, the multiplication of the fraction — by 3 just takes away the de- 3 nominator and leaves x for the product) we have x = 6 X3 = 18. Ex.2. Let|+?=7. Multiply each side of the equation by 2, and we have ^+-=14. Again, multiply each side of this equation by 5, and it becomes bx-\-2x = 70, 7a: = 70; .-. a: =10. Ex.3. Let ^+^=13- 7. 2^3 4 2a: 1x Multiply each side by 2, then a: +—=26— -^« Qx 3, and 3a:4-2a:=78— — . 4, ... 12a; + 8a:=312— 6ar. 56 ALGEBRA. By transposition, l2x + &x + 6x=S\2y 26a: =312; /. ar= 12. This example might have been solved more simply, by multiplying each side of the equation by the least common multiple of the numbers 2, 3, 4, which is 12. Thus. 1+1= 13 -f. Multiply each side by 12, lH£+i|f ^^156-. 12i, or, 6x-\-4x=:l56 — Sx. By transposition, 6a: + 4a; + 3a; = 156, 13a; = 156; .-. a:= 12. Ex.4. Let |+f=22. Ans. a: = 24. Ex.5. 7x 5x 55 - 4 6 ""6 ' ... a:=10. Ex.6. 2^3 5 ... a; = 30. Ex.7. 5 6^2 * • ... a: = 60. 50. In the application of the Rules to the solution of simple equations in general containing only one unknown quantity, it will be proper to observe the following method. (1.) To clear the equation of fractions by Rule III. (2.) To collect the unknown quantities on one side of the equation, and the known on the other, by Rule II. (3.) To find the value of the unknown quantity by di- viding each side of the equation by its coefficient, as in Rule I. 50. Enumerate the three steps by which a simple equation containing only one unknown quantity may be solved. CH. III. SIMPLE EQUATIONS. 57 Ex. 1. Find the value of x in the equation -;= — h 1 = - -J- — . Multiply by 7, then 3^+ 7 =^ + ^. by 5, ... 15a: + 35= 7a; + 91. Collect the unknown quantities ") on one side, and the known > 15a: ■— 7a; = 91 — 35, on the other ; ) or 8a: = 56. 56 Divide by the coefficient of a:, a: = — = 7. Ex.2. a:-4-3 x Find the value of a: in the equation —^^ 1 = 2— -. Multiply by 5, then a:+3— 5 = 10 -. ......... by 7 ...7ar+ 21 —35=70 —5a:. Collect the unknown quantities! on one side, and the known > 7a: + 5a: =: 70 —21 + 35, on the other: j or 12a: = 84; ...a: =-=7. Ex. 3. Find the value of x in the equation col"oSr;' O'oir \ 40x-5.+. = 10.+ 4.-4+240. By transposition, 40a:— 5a:— 10a:— 4a:=240— 4— 5. or40a:— 19a;=231, i.e. 21a:=231; ••'^-21 ~4 As theirs/ step in this Example involves the ctse ''where the sign — stands before a fraction," when the unmerator 00 AXGEBRA. of that fraction is brought down into the same line with 40ar, the signs of both its terms must be changed, for the reasons assigned in Ex. 3, page 44 ; and we therefore make it —5a: + 5, and not dx-^d. Ex. 4. Find the value of x in the equation 2ar — - + 1 = 5ar — 2. Multiply by 2, then 4a:— a: + 2 = 10a:— 4. By transposition, 4 + 2 = 10a:— 4a: + a:, or 6 := 7a: ; 6 •••7=^- 6 OTX = J. Ex. 5. What is the value of x in the equation Sax+^bx = Sc+a ? Here Sax+2bx = (Sa-\--2b)Xx ; .-. (Sa+2b) Xx=: Sc+a. Divide each side of the equation by 3a+26, which is the 3c+a coefficient of x; then a:= iqa ' Ex. 6. Find the value of x in the equation Sbx-^-a = 2ax-\-4c. Bring the unknown quantities to one side of the equation, and the known to the other ; then, Sbx'-'2ax = 4c—a ; but Sbx—2ax = (3Z»— 2a) X^ i .-. (3^>— 2a)a: = 4c— a. 4c— a Divide by 8&— 2a, and a: = g^_,^^ » Ex. 7. Find the value of a: in the equation ^>a:-f a: = 2a:+8a. Transpose 2a:, then bx+x—2x = 3a, or bx— x = Sa; but ^a:— x = (b'-\)x; ,',(b'^l)x = Sa, 3a anda:= i__V"* CH. III. SIMPLE EQUATIONS. 59 Ex. 8. ar + ^+^=ll Ans. a? = 6. __ •V U/ *Xj w . Ex. 9. 5+4+3 = 2 + ^7 ... a: = 60. 3ar , 110 Ex. 10. 4a:-20=y + -y- ... a:=10. ^ X . X X \ 6 Ex.11. 2+3-4=2 ... ^ = 7' Ex.12. 3a; + - = -3 1 _ar+3 1 X' 9 3 ••• "'"S Ex. 13. -^^ 5 = 29-2ar ... a: = 14. ^x Ex. 14. 6a; — ;; 9 = 5ar ... a; = 36. 4 ^ a;4-3 , ^ 12a;4-26 Ex. 15. 2a; --^+15 = ■^— ... a- = 12. a;— 2 , X x^^ Ex. 16. -^ — '"3 — ^^ 2" - ^=1^- ^ ^ 2a;— 1 . , a;4-2 , Ex. 17. bx J-l=3a;+-^+7 ... a; =8. Ex. 18. 2aa; 4- & = 3ca; 4- 4a ... :g= ^^~ . ' ' 2a— 8c Ex. 19. 4 / iP— 1\ =5(^+-3-)'fi^^^- , 4 7a;-9 4/ . a;— 1 5 3 Mult, by 15, 45a;— 60— 28a;+36 = 72+4ar-4. 45a;— 28a;— 4a: = 72— 4+60— 36, I3a; = 92; .•.a; = 7J^. ^0 ALGEBRA. Ex. 20. ~9"+6^=a2 IT"' ^"^^• Mult, by 18, 8;r+6+l^=f?=8:r+19. 126£--522_ Mult, by 5a;— 12, 126a:— 522 = 65a:- 156, 126a:— 65a:=522 — 156, 61a:=366; .-. a:=6. Ex. 21. Mult, by 7 (a:-l), 7 - il^i=ii= 1. a:-f-7 Divide by 2, 3 = '-g=^. 3a;+21 = 73;— 7, 7ar-3a:=21+7, 4a: = 28; .-. x = l. Ex. 22. Mult, by 28, 16a:+10H — — -^==16a:+15+9. 196a: -84 , ^ = 14. 6a:+2 196a:-84 = 84a:+28, 112a:==>112; CH. III. SIMPLE EQUATIONS. 61 Sx-S 3ar-4 ^, 27+4ar Ex.23. —J 3~==^* 9 Ans. 9. 9a:+20 4ar— 12 , x Ex.24. -^=-^i:r+4- - ^• 20x4-S6 , 3a-+20 4a: , 86 . Ex.25. -^+^-^=- + 25. ... 4. 2a:+8^ 13^-~2 , ^ 7a:_ ar+16 Ex.26. — ^ I7a;-32'^3""12 36 ' "• *' Ex.27. 4(5ar-3)-64(3-a:)-3(12ar-4)=96. ... 6. Ex.28. 10(a:+J)-6a:(^-i)=23. ... 2. Ex.29. §2g£+?5±|£=i4+4-. ar+1 37+3 ar+l PROBLEMS. Prob. I. What number is that to which 10 being added, |.»h8 of the sum shall be 66 ? Let X = the number required ; then a: + 10 = the number, with 10 added to it. Now r of C^+IO) =2 (.+10)=^-^^= 2^. But, by the question, ^ths of (ar+10) = 66 ; Hence, — -f — = 66. 5 Multiply by 5, then 3a:4-30 = 330 ; 300 .-. 3a: = 330— 30 = 300; or a: =-5-= 100. o Prob. 2. What number is that which being multiplied by 6, the product increased by 18, and that sum divided by 9, the quotient shall be 20 ? Let X = the number required ; then 6a: = the number multiplied by 6 ; 6x -f- 18 = the product increased by 18, , .6a:+ 18 and — =that sum divided by 9. 62 ALGEBRA. XT , 1 . 6a:4-18 Hence, by the question, — f — =20. Multiply by 9, then 6a:+18= 180, 162 or 6a:= 180 — 18 = 162; ora: = -^ = 27. 6 Pbob. 3. A post is ^^ in the earth, ^^» in water, and 1:3 feet out of the water. What is the length of the post ? Let X = length of the post in ft. ; X then - = the part of it in the earth, Sx ^ „. . , — =the part of it in the water, 13 = the part of it out of the water. But part in earth + part in water + part out of water =: whole post; - (I) + m 13 = X. I5x Multiply by5, thenar+-y- + 65=5x ; by 7, ... 7x+\5x + 455=S5x, or 455=35a:— 7a:— 15ar=13a:. • 455 Hence x =——=35 length of post in ft. Prob. 4. After paying away ^*^ and -f*^ of my money, I had <£85 left in my purse. What money had I at first ? Let X = money in purse at first ; then - -|- - = money paid away. But money at first — money paid away = money remaining. Hence a: — {4 + 7/ = ®^» X X I.e. ^—7 — 7 = 85. Multiply by 4, then 4x — x — =-=340; by 7, ... 28a:— 7a:- 4a:=2380. .•.17a:=2380; » 2380 ora:=-=y = «£140. CH. Ill, SIMPLE EQUATIONS. 63 Pbob. 5. What number is that to which if I add 20, and from I"*' of this sum I subtract 12, the remainder shall be 10 ? Ans. 13. Pros. 6. What number is that, of which if I add ^**, ^*^, and f^^^ together, the sum shall be 73 ? Ans. 84. Prob. 7. What number is that whose ^<* part exceeds its ^th by 72 ? Ans. 540. Pbob. 8. There are two numbers whose sum is 37, and if 3 times the lesser be subtracted from 4 times the greater, and this difference divided by 6, the quotient will be 6. What are the numbers ? Ans. 21 and 16. Pbob. 9. There are two numbers whose sum is 49 ; and if 4-*^ of the lesser be subtracted from ^*^ of the greater, the remainder will be 5. What are the numbers ? Ans. 35 and 14. Pbob. 10. To divide the number 72 into three parts, so that J the^r5^ part shall be equal to the second, and S-^hs of the second part equal to the third. Ans. 40, 20, and 12. Prob. 11. A person after spending ^*^ of his income plus £10, had then remaining ^ of it plus £35. Required his income. Ans. X150. Prob. 12. A gamester at one sitting lost ^*^ of his mo- ney, and then won 10 shillings ; at a second he lost ^^ of the remainder, and then won 3 shillings ; after which he had 3 guineas left. What money had he at first ? Ans. £5. Pkob. 13. Divide the number 90 into four such parts, that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, may all be equal to the same quantity. Ans. 18, 22, 10, 40. Prob. 14. A merchant has two kinds of tea, one worth 9s. 6d. per It)., the other 13^. 6d. How many fbs. of each must he take to form a chest of 104 lbs. which shall be worth £56 ? Ans. 33 at 13s. 6d. 71 at 95. 6d. Prob. 15. Three persons, A, B, and C, can separately 64 ALGEBKA. reap a field of corn in 4, 8, and 12 days respectively. In how many days can they conjointly reap the field ? Leta:==No. of days required by them to reap the field; then if 1 represent the work, or the reaping of the field, ^ = the part reaped by A in 1 day, i= B tV = c •■•i + i+T'2 = ^"^'•e^ But the part reaped by all three in I day multiplied by the number of days they took to reap the field, is equal to the whole work, or 1 ; •••a + i + TV)^=l: Clearing of fractions by multiplying by 24, (6+3 + 2)a:=24, lla: = 24; .•. a: =2^ days. Prob. 16. a man and his wife usually drank a cask of beer in 10 days, but when the man was absent it lasted the wife 30 days; how long would the man alone take to drink it? Ans. 15 days. Prob. 17. A cistern has 3 pipes, two of which will fill it in 3 and 4 hours respectively, and the third will empty it in 6 hours ; in what time will the cistern be full, if they be all set a-running at once ? Ans. 2h. 24m. Prob. 18. A person bought oranges at 20d, per dozen ; if if he had bought 6 more for the same money, they would have cost 4:d, a dozen less. How many did he buy ? Let X = the number of oranges ; then 0:+ 6= at 4c?. less per dozen. Price of each orange in pt case = f § = ^d, and 2'* ... =|| = H .*. the cost of the oranges ='q"» But we have also the cost of the oranges =^^ (ar + 6). Two independent values have therefore been obtained for CH. III. SIMPLE EQUATIONS. 65 the cost of the oranges; these values must necessarily be equal to each other ; Multiplying each side of the equation by 3, 5x = 4(x + e), 5x = 4x + 24; /. a: =24, the No. of oranges. Prob. 19. a market-woman bought a certain number of apples at two a penny, and as many at three a penny, and sold them at the rate of five for two-pence ; after which she found that instead of making her money again as she expected, she lost fourpence by the whole business. How much money had she laid out ? Ans. 8s. 4d. Prob. 20. A person rows from Cambridge to Ely, a dis- tance of 20 miles, and back again, in 10 hours, the stream flowing uniformly in the same direction all the time ; and he finds that he can row 2 miles against the stream in the same time that he rows 3 with it. Find the time of his going and returning. Let Sx == No. of miles rowed per hour with the stream ; .*. 2x= against Now the distance divided by the rate per hour gives the time ; 20 .'. ^= the No. of hours in going down the river, 20 and 2^= coming up I But the whole time in going and returning is 10 hours ; . 20 , 20 ,^ ••3^ + 2^=^^- 2 1 Dividing by 10, — + J= 1. Multiplying each term of the equation by 3x, 2-f 3 = 3a:; .-. x = |= If. and .*. 3a: = 5, miles ^ hour down, 20 /. the time in going down the river = --- =4 hours, and con- sequently the time of returning = 10— 4^=6 hours. F 66 ALGEBRA. Prob. 21. A lady bought a hive of bees, and found that the price came to 25. more than ^^^ and y^ of the price. Find the price. Ans. £2. Prob. 22. A hare, 50 leaps before a greyhound, takes 4 leaps for the greyhound's 3 ; but two of the greyhound's leaps are equal to three of the hare's. How many leaps will the greyhound take to catch the hare ? Let ^ be the No. of leaps taken by the greyhound ; then — will be the corresponding number taken by the hare. o Let 1 represent the space covered by the hare in 1 leap ; g then- greyhound 4:X 4x •'• -^ X 1 or -g- will be the whole space passed over by the 3 3^ hare before she is taken ; and x X - ^^ — will be the space passed over in the corresponding time by the greyhound. Now, by the problem, the difference between the spaces respectively passed over by the greyhound and hare is 50 X 1, or 50 leaps ; Sx 4x ,.--- = 60, 9a: — 8^ = 300 ; /. X = 300 leaps. ON THE SOLUTION OF SIMPLE EQUATIONS, CONTAIN- ING TWO OR MORE UNKNOWN QUANTITIES. 51. For the solution of equations containing two or more unknown quantities, as many independent equations are required as there are unknown quantities. The two equa- tions necessary for the solution of the case, when two un- known quantities are concerned, may be expressed in the following manner : oo; + % = c a'x-Yb'y=:c'. Where a, b, c, a\ h\ c', represent known quantities, and CH. III. SIMPLE EQUATIONS. 67 or, y^ the unknown quantities whose values are to be found in terms of these known quantities. There are three different methods by which the value of one of the unknown quantities may be determined. FIRST METHOD. Find the value of one of the unknown quantities in terms of the other, and the known quantities by the rules already given. Find the value of the same unknown quantity from the second equation. Put these two values equal to each other ; and we shall then have a simple equation, containing only one unknown quantity, which may be solved as before. Ex.1. Given. + ^ = 8 ) K^ find . and ^. x—y = 4: (2)5 ^ From (1) y = S—x (a) (2) y = x-4. Putting these two values equal to each other, we get A'— 4 = 8 — a:, 2. = 12; .-. a: = 6. By W y = 8—. = 8—6=2. Ex.2. Leta: + 4j/=16 (I) 4.x -\. y = U (2) From equation (1), we have a: =: 16— 4y. (^) ^=2L-_.' Hence by the rule, — j— "^ 1^— 4^» 34—^ = 64 — 16?/, 15^ = 30; .-.3/ = 2. It has already been shewn that x=i 16—4^ = (since ^ =: 2 j and/. 41/ = 8) 16-8 = 8. 51. For the solution of equations containing two or more unknown quantities, how many independent equations are necessary ? State the fiist method of solution. 68 ALGEBRA. Ex. 3. 5x+S^ = 38 ) Ans J "^ = ' 4x- 2^=105 (!/ = ( r- Sx- Ex. 4. 2x—Si/=z—ll • 5 a: =4 c-2^ = 6 5 i^ = 3. SECOND METHOD. From either of the equations find the value of one of the unknown quantities in terms of the other and the known quantities, and for the same unknown quantity substitute this value in the other equation, and there will arise an equation which contains only one unknown quantity. This equation can be solved by the rules already laid down. Ex. 1. y—x=z2 (1) x+y=^8 (2) From (1) y = 2'\'X. («) This value of ^ being substituted in (2), gives a;+2+ar = 8, 2a: = 6; .-. a: = 3. And by («) ^ = 2+a:==2+3 = 5. Ex. 2. a:+2 , ^^ + I0a.==192 (2) Clearing equation (1) of fractions, ar+2+24^ = 93, or a:+24^ = 91 («). Clearing equation (2) of fractions, 2/+5+40a: = 768, or y+AQx = 763 (/3). From (a) a: = 9 1—24^. Substitute this value of x, according to the rule in equa- tion (iS); and 2^+40 (91 -24^) =763, or, y+3640—960z/ = 763 ; .-. 959^ = 3640-763=2877, and y = S. Uy rerernng to equation (a) we have j' = 91— 24^/ rK^^since2^ = 3; and .-. 24^ = 72>91--72 = 9. Enunciate the second method of solution. SIMPLE EQUATIONS. 69 Ex.3. f^+^^ = ^n Ans. ^^ = t THIRD METHOD. Multiply the first equation by the coefficient of x in the second equation, and then multiply the second equation by the coefficient of x in the first equation ; subtract the second of these resulting equations from the Jirst, and there will arise an equation which contains only ^ and known quan- tities, from which the value of y can be determined. It must be observed however, that if the terms, which in the resulting equations are the same, have unlike signs, the resulting equations must be added, instead of being sub- tracted, in order that x may be eliminated (i. e. expelled from the equations). Ex.1. Given 5^ -}- 4^ = 55 (1) 3ar-f2^ = 31 (2) To find the values of x and i/ Mult. (1) by 3, then 15a:-f 12^= 165 (2) by 5, ... 15ar+10^=155 .*. by subtraction, we have 2^ = 10 .-. 2/= 5. Now from equation (1) we have ^__55— 4£ 5 55—20 "■ 5 _35 ~" 5 = 7. Ex. 2. Let the proposed equations be ax+bi/ = c (1) ax-\-b'i/ = c' (2) Mult. (1) by a\ and aax-\-abi/ = a'c (2) by a, ... aax+ab'i/ = ac -, How are equations solved by the third method? 70 ALGEBRA, .*. by subtraction, (ab—ab')i/=:ac—a(f ac — ac ao—ao Mult. (1) by b\ and ab'x-\-bb'i/= b'c (2)by^, ... abx-\'bb'y=bc By subtraction, {ab'—dh)x = b'c^bc ; b'c — be .-. x = --, -• ab —ab Ex.3. Let3a:+4j/ = 29 (1) 17a:-3z/ = 36 (2) Mult. (1) by 3, then 9ar+12^=87 (2) by 4, ... 68a:— 12^=144. The signs of I2i/ in the two equations are unlike ; .*. to eliminate y from them, the two last equations must be added together; and then 772: = 231; .-. ar==3. From (1) we have 4^ = 29—3^, = 29—9, (since ar=3; and.-.3arr=9) = 20; :,y = 6. Ex.4. Let 6.+% = 33) ^^^ Ca: = 3 13a;— 4^=: 19 5 (y = 5, Ex.5. 4a;+3^ = 31> (^ = 4 ^ = 22^ ly = 5. Sx+2y 3x+2y 2x+Sy 5x^4:y 4x+2^ = 36 Ex.6. 3a:+2^ = 40> 5 a: = 10 ^ = 35 J ly = 5. Ex.7. 5a:— 4^=19^ ^x = 7 (^ = 4. Ex.8. 3x+7y=79l (a: = 10 2y-ix=^ 9S ^^ = 7. Ex.9. ^+1=6 3 ^ / ^a:=ll i=:^+3= 4) * (:.?^=4. CH. Iir. SIMPLE EQUATIONS. 71 Ex.10. -^-2^=2 Car =11 > Ans. •< Ex.11. -^-+2^ = 7 j ^^^8 67 Ex.12. -^=^^ .^^,3 8_£=^=6 j (y = 3. 52. When three unknown quantities are concerned, the most general form under which equations of this kind can be expressed, is ax-{- by -\^cz z=:d (1) a'x-\^b'y-\'C'z=id' (2) a"x-\-U'y'\-c"z = d" (3 ) , and the solution of these equations may be conducted as in the following example : Ex. 1. Let f +%+f = 29 (1) ^ ^^ ^^^ ^^^ ^^^^^^ 3x+2y+5z = 32 (2) \ o£x y z 4x+Sy+2z=25 (3) ) ^^ ^' ^' ^* I. Multiply (1) by 3, then ex+9y+l2z = 87 (4; (2) by 2, ... 6x+4y+i0z = 64 (5). Subtract (5) from (4) ... 5y+ 2z = 2S («). Multiply (2) by 4, then 12a:+8y+20z == 128 (3)by3, ... I2x+9y+ 6z=:75 Subtract —^+14^ = 53 (/3). II. Hence the given equations are reduced to, 5y+ 2z = 2S («) -^+14z = 53 (/3). 72 ALGEBRA. Again 5t/+ 2z = 23 Mult. (/3) by 5, then '-5i/+70z = 265 By addition 72z = 288, orz=^^®=4 From equation (/3) 2^= 14z— 53=56— 53 = 3. TTT -n .• /,x 29—Sy—4z 29—25 ^ III. From equation (1) ... z = ^ = — - — =2 Ex.2. 0:4-^+2: = 90 ^ Car = 35 2a:+40 = 3^4.20 > Ans. 'J^==30 2a:+40 = 4^4-10 ) (z = 25. Ex. 3. a:+ ^+ z= 53 ^ Car = 24 a:4-2^+3z=105 S- \y= 6 ar4-3y+4z = 134 ) ( z = 23. PROBLEMS. Prob. 1. There are two numbers^ such, that 3 times the greater added to ^^ the less is equal to 36 ; and if twice the greater be subtracted from 6 times the less, and the remainder divided by 8, the quotient will be 4. What are the numbers ? Let X = the greater number, y = the less number ; Then3a:+f=36 ) ^ I or, 9^+ ^=108 6j/— 2ar I '6^—2^= 32; — 8~= ^) Or, ^+9a:=108 (I) 63/— 2a:= 32 (2) Mult, equation (1) by 6, then 6^+o4a: = 648 Subtract (2) ... 6^/— 2a:= 32; then 56ar = 616; 616 .-.a; =^=11. From equation (1) ?/=108— 9^:== 108— 99=9. Prob. 2. There is a certain fraction, such, that if I add 3 to the numerator, its value will be ^^^ ; and if I subtract CH. III. SIMPLE EQUATIONS. 73 (me from the denominator, its value will be ^^, What is the fraction ? Let a: = its numerator, ^ ,^^^ ,^^ f^Minn ?« ^ y= denommator, Add 3 to the numerator, then ' > then the fraction is - » 3 y* or, 3^+9=^ Subtract one from denom'., and —^= * ^ ^' By transposition, i/—Sx=9 (1) ^— 5ar=l (2). Subtract equation (2) from (1), and we have 2j: = 8; 8 .-. a; =5= 4, the numerator. From equation (1) ^=9+3ar=9+12=21, the denominator. 4 Hence the fraction required is ~ • Prob. 3. A and B have certain sums of money ; says A to B, Give me £15 of your money, and I shall have 5 times as much as you will have left ; says B to A, Give me £5 of your money, and I shall have exactly as much as you will have left. What sum of money had each ? Let ar=A's "^o^^yKi ar-4-15 /what A would have after ^=B's / ' 1 receiving £15 from B. ^—15 = what B would have left. * . j^ ^ (what B would have after Again, 2^+ 5 — I receiving £5 from A. or— 5 == what A would have left. Hence, by the question, a:+15=5x (^—15) = 5^—75, > and ^4" 5 = X'—5, ) By transposition, 5y— a* = 90 0),\ and^— a: = —10 (2). ] Set down equation (1) 5z/— a: = 90. Multiply eq". (2) by 5, 5^— 5a: = — 50. Subtract (2) from (1) 4:r=140; 74 ALGEBRA. .-. x=- — j— = 35, A s money. From equation (1) 5y = 90+a: = 90+35 = 125 ; . .*. ^ = ~r"= 25, B s money. Prob. 4. What two numbers are those, to one- third the sum of which if I add 13, the result shall be 17 ; and if from half their difference I subtract owe, the remainder shall be two ? Ans. 9, and 3. Prob. 5. There is a certain fraction, such, that if I add one to its. numerator, it becomes | ; if 3 be added to the denominator, it becomes ^. What is the fraction ? Ans. /^. Prob. 6. A person was desirous of relieving a certain number of beggars by giving them 2*. 6c?. each, but found that he had not money enough in his pocket by 3 shillings; he then gave them 2 shillings each, and had four shillings to spare. What money had he in: his pocket ; and how many beggars did he relieve ? Let X = money in his pocket (in shillings); y = number of beggars. 5i/ TNo. of skills, which would have Ihen 2JX2^, o^ Y= j been given at 2^. 6d. each. and2X^jOr2^= at 25. each. Hence, by the question, -^=:^_j-3 (i) and 2^^ = a;— 4 (2). Sub*. (2) from (1), then |= 7, or 7/= 14, the No. of beggars. From eq'^. (2), a:= 2^+4=28+4=32 shillings in his pocket. Prob. 7. A person has two horses, /and a saddle worth £10); if the saddle be put on the Jirst horse, his value be- ' conies double that of the second; but if the saddle be put on the second horse, his value will not amount to that of the Jirst horse by £13. What is the value of each horse ? Ans. 56 and 33. Prob. 8. There is a certain nupiber, consisting of two digits. The sum of those digits is 5 5 and if 9 be added to CH. III. SIMPLE EQUATIONS. 75 the number itself, the digits will be inverted. What is the number ? Here it may be observed that every number consisting of two digits is equal to 10 times the left-hand digit plus the right-hand digit : thus, 34=10x3+4. Let X = left-hand digit. y = right-hand digit. Then \Ox-\-y = i\\Q number itself, and 10^4"^ = ^^® number with digits inverted. Hence, by the question, x-\-y = 5 (1), and 10a:+^+9=10^+:r, or9a:— %=— 9,ora:— ^=— 1 (2). Subtract (2) from (1), then 2y=Q, and ^=3, ^=5— z/ = 5— 3 = 2; .*. the number is {\Ox-\-y) =23. Add 9 to this number, and it becomes 32, which is the number with the digits inverted, Prob. 9. There are two numbers, such, that J the greater added to ^ the less is 13 ; and if ^ the less be taken from ^ the greater, the remainder is nothing. What are the numbers ? Ans. 18 and 12. Prob. 10. There is a certain number, to the sum of whose digits if you add 7, the result will be three times the left-hand digit; and if from the number itself you subtract 18, the digits will be inverted. What is the number ? Ans. 53. Prob. 1 1 . A merchant has two kinds of tea, one worth 95. Qd. per lb., the other 13s. Qd. How many pounds^ of each must he take to form a chest of 104 lbs. which shall be worth £bQ ? Ans. 33 at 13s. Qd. 71 at 9s. Qd. Prob. 12. A vessel containing 120 gallons is filled in iO minutes by two spouts running successively ; the one runs 14 gallons in a minute, the other 9 gallons in a minute. For what time has each spout run ? Ans. 14 gallon spout runs 6 minutes. 9 gallon spout ... 4 minutes. Prob. 13. To find three numbers, such, that the Jirst with J the sum of the second and third shall be 120 ; the second with ^^^ the difference of the third and first shall be 70 ; and ^ the sum of the three numbers shall be 95. Ans. 50, 65, 75. 76 ALGEBBA. CHAPTER IV. ON INVOLUTION AND EVOLUTION. ON THE INVOLUTION OF NUMBERS AND SIMPLE ALGEBRAIC QUANTITIES. 53. Involution, or ** the raising of a quantity to a given power,'* is performed by the continued multiplication of that quantity into itself till the number of factors amounts to the number of units in the index of that given power. Thus, the square of a=a X a^a^ ; the cube of. b=b X^X ^=^^ ; the fourth power of 2=2X2x2x2=16; the fifth power of3=3x3X3X3x3=«243; &c. &c. 54. The operation is performed in the same manner for simple algebraic quantities, except that in this case it must be observed, that the powers of negative quantities are alternately 4- ^.nd — ; the even powers being positive, and the odd powers negative. Thus the square of +2a is 4-2aX +2a or +4^2; the square of —2a is — 2aX--2a or +4a2 ; but the cube of —2a =— 2a X —2a X —2a = +40^ X — 2a= — Sa'. The several powers of a\ a a a a^ a a b^P &c.=&c. 4th power And the several powers of — — -> b b f}^ 2c 2c 2c Sc^' pourr- 2c^ 2^^ 2c^ 2c""'^16 &c.=&c. Cube=' 6c* ON THE INVOLUTION OF COMPOUND ALGEBRAIC QUANTITIES. 55, The powers of compound algebraic quantities are 53. What is involution? How is it performed ?— 64. In what manner is in- volution performed for simple algebraic quantities?— 55. How are the powers of compound quantities raised? CH. IT. INVOLUTION. 11 raised by the mere application of the Rule for Compound Multiplication (Art. 34). Thus, Ex. 1. What is the square Ex. 2. What is the cube of of a+26 ? d?-x ? a +2^> a'—x a +2b a'^x a'+lab a^—a^x +2a^>+462 — a'x+x'' Square = a^+4ab+4b^- Square = a*''2a^x+x^ — a^x+^a^x^—z^ Cube = a^—Sa^x+Sa'^x^—x^ Ex.3. What is the 5*^ power of a+b ? a+b a+b a2_|_ ab + ab+b^' a2+ 2a6+62=Square a+ b a^+2a''b+ ab^ + a'b+ 2ab''+b^ a^-^Sa^b-{- 3a62+^»-^=Cube a*'\'Sa^b+ 3a2^,2^_ ab^ + a^b+ Sa'^b^'^ Sab^+b^ a'+ia^b-\' 6a''b''+ 4a^+6*=4t^ Power a+ b a^+4a*b+ 6a%^'^ 4a''b^+ ab* + a'b^ 4a^b''+ Ga'^b^+Aab'+b^ a^+5a'b+l0a^b''+l0a''b^-{-5ab*+b' = 5^^ Power. 78 ALGEBRA. Ex. 4. The 4**^ power of «+3^ is a'^^l^a^b+b^a'^b'^'^lOSab^ Ex. 5. The square of ^x'^+2x-\-5 is 9a:*4-12a:3+34a:24-20a: +25. Ex. 6. The CM^>e of 3a— 5 is 27:^3— 135a;2+225ar— 125. Ex.7. The cwZ>e of a;^— 2a:+l h x^—Qx^ ~\.\bx^—2Qx^ + 15a:2-.6:r+l. Ex. 8. T\ie square oi a^b-\-ch a^+'2.ab-\-b'^-\-2ac-\-2bc'\-c'^. ON THE EVOLUTION OF ALGEBKAIC QUANTITIES. 56. Evolution, **or the rule for extracting the root of any quantity," is just the reverse of Involution; and to perform the operation, we must inquire what quantity mul- tiplied into itself, till the number of factors amount to the number of units in the index of the given root, will generate the quantity whose root is to be extracted. Thus, (1.) 49=7x7;.-.the5^. rooifof49(orbyDeri5,\/49)=7. (2.) -^b^=z—bX—bX—h',:. cube root oi —b^^ZZip^^—b 16a''_2a 2a 2a 2a , Yl6a*__2a ^^'^ SW~-3b'^Sb'^Sb^3br'' V 8lb''~ 8b' (4.) 32=2X2X2X2X2; .-.4/32=2 (5) a^=a''Xa^Xa''; :, ^a^=a^. Hence it may be inferred, that any root of a simple quantity/ can be extracted, by dividing its index, if pos- sible, by the index of the root. 57. If the quantity under the radical sign does not admit of resolution into the number of factors indicated by that sign, or, in other words, if it be not a coinplete power, then its exact root cannot be extracted, and the quantity itself, with the radical sign annexed, is called a Surd. Thus \/37, ^a"', ^b^, ^47, &c. &c. are Surd quantities. 66. What is Evolution f How is it performed? — 57. What is a Surd quantity ? CH. IV. EVOLXJTIOISI. 79 58. In the involution of negative quantities, it was ob- served that the even powers were all -|-, and the odd powers — ; there is consequently no quantity which, multiplied into itself in such manner that the number of factors shall be even, can generate a negative quantity. Hence quan- tities of the form //— a^, V— 10, ^—a^, V^^ V —a\ &c. &c. have no real root, and are therefore called impossible. 59. In extracting the roots of compound quantities, we must observe in what manner the terms of the root may be derived from those of the power. For instance (by Art. 55, Ex. 3), the square of a-\-b is a'^-\-2ab-{.b'^, where the terms are arranged according to the powers of a. On com- paring a-\-b with a2-f2a6-j-Z>-, we observe that the first term of the power (a^) is the square , of the first term of the root {a). Put ar-\-2ab+b''la+b a therefore for the first term of the or ^ root, square it, and subtract tnat square from the first term of the 2a+^ lab-^-b"^^ power. Bring down the other two •2a6-j-Z>2 terms 'lab-^b'^, and double the first ~^ ^ term of the root ; set down 2«, and =i having divided the first term of the remainder (2«Z>) by it, it gives b, the other term of the root ; and since ^ab-{-¥={'^a-\-b)b, if to 2a the term b is added, and this sum multiplied by b, the result is 2ab-\-¥ ; which being subtracted from the two terms brought down, nothing remains. 60. x\gain, the square of «+6-fc (Art. 55. Ex. 8.) is a^-\-1ab'\'b--\'2ac-\-2bc-\-c''- ; in this case the root may be derived from a''^2abJt'b''-\-'lac^2bc+c''(a+b+c the power, by aP \ continuing the _ . ,«^ , . ,„ process in the ^a+hfab+b^ last Article. \^ab-\-b- Thus, having 2a-|-26+cp«c+2^>c-t-c2 found the two |2qc4-2Z>g— c^ first terms (« +6) * * * of the root as — before, we bring down the remaining three terms 2ac-[-2^c 58. Explain the nature of an ?mpo5«6fe quantity. — 59. How are the roots of compound quantities extracted ? 8U ALGEBEA. +c2 of the power, and dividing 2ac by 2a, it gives c, the third term of the root. Next, let the last term (b) of the preceding divisor be doubled, and add c to the divisor thus increased, and it becomes 2a+2^+^; multiply this new divisor by c, and it gives 2ac-{-2bc-\-C', which being sub- tracted from the three terms last brought down, leaves no remainder. In this manner the following Examples are solved. Ex. 1. 4x* 89 4x^+-xj6x^+-^x^ 6x»+-'X^ 4x^+Sx+5 ] 20x^+l5x+25 I 20j:24-15:r4-25L 20j:2+l5ar+25L Ex.2. ar»+4ar*+2a:*+9:c2__4^^4(^3^2x2— a:+2 x^ 2:r'+2ar2)4a:^+2:r* 2^'+4a:2— a;)— 2a:*4.9x2— 4a: — 2a:*— 4a;3+ x"^ 2:t3_j_4^2_2a:+2) +4^:34-8x2— 4a:+4 +4^:3+8x2— 4a-+4 Ex. 3. The square root of 4x'^'\-4xy'\-y'^ is 2a:-|-^. Ex. 4 25a^+30a6+9^>2 is 5^+3^. Ex. 5. Find the square root of 9a:*+12a'3+22a:2-}-12a'+9. Ans. 3a^=+22:+3. CH. IV. ' EYOLUTION. 81 Ex. 6. Extract the square root of 4a:*— 16r'+24A-2~ 16a:4-4. Ans. 2a:2— 4x+2. 4 Ex. 7. Find the square root oi Sdx*— 36x^+17 x^—4x+ 9 2 Ans. 6x^—Sx+-, 3 32 16 Ex. 8. Extract the square root of x*+8x'^+24+ —-\ — -. Ans. x^+4+ — . ON THE INVESTIGATION OF THE RULE FOR THE EXTRACTION OF THE SQUARE ROOT OF NUMBERS. Before we proceed to the investigation of this Rule, it will be necessary to explain the nature of the common arithmetical notation, 61. It is very well known that the value of the figures in the common arithmetical scale increases in a tenfold pro- portion from the right to the left ; a number, therefore, may be expressed by the addition of the units^ tens, hundi^eds, &c. of which it consists. Thus the number 4371 may be expressed in the following manner, viz. 4000+300+70+1, or by 4X1000+3X100+7X10+Ii hence, if the digits* of a number be represented by a, b, c, d, e, &c. beginning from the left hand, then, A No. of 2 figures may be expressed by \Oa-\-b. 3 figures by lOOa+106+c. 4 figures by 1000a+100^>+10c+c4.cV100a4-10^^-c lOOOOa^ 2ma+l0b)2000ad+l00b' 2000ab+100b^ 200fl+20^>+c) 200ac+206c+c2 200ac+20bc-^c^ n. Letfl = 2) ^ ^ . . .. ^ . , b=zS> ^^^ ^^® operation is transformed into the ^ __ J ( following one ; 40000+12000+900+400+60+1 (2OO+3O+I 40000 400+30 J 12000+900+400 12000+900 400+60+1J400+60+1 400+60+1 in. But it is evident that this operation would not be aflfected by collecting the several numbers which stand in the same line into one sum, and . . . / leaving out the ciphers which are 53361 (231 to be subtracted in the several 4 ^ parts of the operation. Let this 43|T33 be done; and let two figures be J129 brought down at a time, after the .^ square of the first figure in the^ root has been subtracted; then" the operation may be exhibited in the manner annexed; from which it appears that the square root of 53,361 is 231. 461 461 CH. T. QUADRATIC EQUATIONS. 83 63. To explain the division of the given number into periods consisting of two figures each, by placing a dot over every second figure beginning v^rith the units (as exhibited in the foregoing operation), it must be observed, that, since the square root of 100 is 10; of 10,000 is 100; of 1,000,000 is . 1000, &c. &c. ; it follows that the square root of a number less than 100 must consist of one figure; of a number be- tween 100 and 10,000, o^ two figures ; of a number between 10,000 and 1,000,000, o^ three figures, &c. &c. ; and conse- quently the number of these dots will show the number of figures contained in the square root of the given number. Thus in the case of 53361 the square root is a number con- sisting of three fiorures. Ex. 1 . Find the square root of 105,625. Ans. 325. Ex. 2. Find the square root of 173,056. Ans. 416. Ex. 3. Find the square root of 5,934,096. Ans. 2436. CHAPTER V. ON QUADRATIC EQUATIONS. 64. Quadratic Equations are divided mio pure and adfected. Pure quadratic equations are those which contain only the square of the unknown quantity, such as x'^='36 ; x'^-\-5= 54; ax'^'—b = c'^ &c. Adfected quadratic equations are those which involve both the square and simple power of the unknown quantity, such as x'^-\-4:X=4i5 ; Qx"^ — 2x=21 ; aa;2+26a; = c+fl?; &c. &Q. 63. Explain the principle of the rule and the object of pointing off in ex- tracting the square root of numbers — 64. How are quadratic equations divided? What is an adjected quadratic equation ? 84 ALGEBEA. ON THE SOLUTION OE PURE QUADRATIC EQUATIONS. 65. The Rule for the solution of pure quadratic equations is this, "Transpose the terms of the equation in such a manner, that those which contain x^ may be on one side of the equation, and the known quantities on the other ; di- vide (if necessary) by the coefficient of x"^ ; then extract the square root of each side of the equation, and it will give the values of a:.*' Ex. 1. Let ar2+5 = 54. By transposition, 07^ = 54 — 5 = 49. Extract the square root J) of both sides of the > then a: = "I" V49 = + 7. equation, ) Ex. 2. Let 3a;2— 4 = 71. By transposition, Sa:^ =: 7 1 + 4 =: 75. 75 Divide by 3, :r2=y = 25. Extract the square root, j:' = + v'25 = + 5. Ex. 3. Let ax'^'—h-=iC\ then ax"^ = c -j- &, and j?2=-— — a HiX. , 4. bx'-X = 244 Ans. a: = + 7. Ex. 5. 9:^2+9 = 3^:^+63 :.= + 3. Ex. ,6. 4^:24.5 9 = 45 ••« a:=±10. Ex. 7. Z,a;2+c+3 = 26a:2+l ... x=± V" + 2 h ' 65. State the rule for solving pure quadratic equations. CH. V. QUADRATIC EQUATIONS. 85 ON THE SOLUTION OF ADFECTEI) QUADRATIC EQUATIONS. 66. The most general form under which an adfected quadratic equation can be exhibited is aar'^+^^zzrc ; where a, b, c may be any quantities whatever, positive or nega- tivCf integral or fractional. Divide each side of this equa- b c ^ b c , 1 . tion by a, then x^^-x=-- Let - =», -=g: then this equation is reduced to the form x^-\-px=:q^ where p and q may be any quantities whatever, positive or negative, integral or fractionaL 67. From the twofold form under which adfected quad- ratic equations may be expressed, there arise two Rules for their solution. Rule I. Let x'^-\'px=zq. Add ^ to each side ( , i , p'^ p"^ , p^+4q of the equation, then j Extract the square root i of each side of the "~2~ 2 equation, then ^ i'^P^ + ^^+i^ X ^ Hence it appears, that ** if to each side of the equation there be added the square of half the coefficient of x, there will arise, on the left-hand side of the equation, a quantity which is a complete square ; and by extracting the square root of each side of the resulting equation, we obtain a simple equation, from which the value of x may be de- termined." * Since the square of + a is + a*, and of — a is also + a^, the square root of + a^ may be either + a or — a ; hence the square root of ^+43' may be expressed by "T V y?^^ 4^. 66. What is the most general form of a quadratic equation ? Can it be re- duced to another form ?— 67. Enunciate the 1st Rule- OO ALGEBRA, 68. From the form in which the value of x is exhibited in each of these Rules, it is evident that it will have two values; one corresponding to the sign +, and the other to the sign — , of the radical quantity. Ex. 1. Let 072+8^= 65. Add the square of 4 {i.e. 16) to each side of the equa- tion, then . . . ar2+8.r4-16 = 65+16 =81. Extract the square root of each side of the equation, then a:+4=±\/81=+9, and a: =: 9 — 4 = 5 ; or A=— 9— 4=— 13. Ex. 2. Let a:*— 4a; = 45. Extract the square root, and a:— 2 = +>v/49=+7, and a: = 7+2 = 9; or, ar = 2— 7 = — 5. Ex. 3. x2+12a: = 108 ... Ans. xz=z 6 or —18. Ex.4. a:2— 14a:= 51 ar=17 or — 3. Ex. 5. a:2_ 8 /2\2 4 , , . ^ , 4 273 , side add \ y ^^ 25 '"^ " +5 "+ 25=^"+ 2 37 Extracting the square root, :z^+-=i — * 5 5 . ^-> + ~~-=7,or-7* Ex.14. 3a:2+2ar=161 Ans. a; = 7 or — 7§. SB ALGEBBA. Ex.15. 2x^—5z=U7 Ans.a:= 9or— i^. Ex.16. 3a:2— 2:r=280 a:=10or— 9^. Ex.17. 4a:2— 7a; = 492 ar=l2or — 10^. A quadratic equation seldom appears in a form so simple as those of the preceding examples ; it is therefore generally found necessary to employ in the solution of a quadratic the following reductions. (1.) Clear the equation effractions. (2.) Transpose the terms involving x^ and x to the left hand, and the numbers to the right-hand side of the equation. (3.) Divide all the terms of the equation by the coefficient ofx\ (4.) Complete the square. (5.) Extract the square root of both sides, and there will arise a simple equation, from which the value of x may be found. Multiply by 3, and 4a:2— 33 = x. By transposition, 4x'^—x==SS. Divide by 4, and x^^--x=-—» J ^ 4 4 Complete the ) ^ 1 . JL__25 . i.— ^4. ±= square, ] ^ "i 64"" 4 ■*"64 64 "^ 64 1 , 1 4-23 Extractmg the sq. root, a:— -=_E -^• . ^.= 1 + — =3or-2j. 9+l£±l=5.+5. 529 64 • CH. V, QUADRATIC EQUATIONS. 89 9X+4X+4: = 5x^+5x, 2_^ , 16__4 16__36 ^ 5^"+"25"~5"*"25— 25- 4 ,6 • ar— — = I - ' anda; = g±- = 2or--. / x^ J Ex.3. -gF— l=:r+ll Ans. a:=12 or — 6. T. . 2;r , 1 7 ^^•^- ■3-+i=3 ^=3ori. ^"•'- |--|=^ a: = 6or-^. Ex.6. rTT + -=3 a: = 2or— f x+l^ X Ex.7. ar2— 34 = ^ar. a:=6or — 5f. Ex.8. ? + - = 5i. ar = 25orl. o X 5' 24 Ex.9. a:+^^=3;2;— 4 x==5oT—2. X , x+l 13 Ex.10, — ttH — ^ = -^- a: = 2or-3. x-\-l X 6 3x x — l 11 Ex.11. — nr 7— =ar— 9. ... a:=10or— -=-. a:-j-2 D 7 Ex. 12. Given x''+8x = — 31 ; find x, x''+8x+\6 = 16—31 = — 15, a:+4=±VlIT5; .-. x = ^i+'^^-[5, &.r = -4-V:i45, both of which are impossible or imaginary values of x. 90 ALGEBRA. Ex.13. a:2-2a:=— 2 Ans. a:= l+VZTl. Ex.14. x^—l6x=:—\5. a:=15orl. Ex. 15. Let 13^:2+2^ = 60. ^. ., , , 2x 60 Divide by 13, ^'+ 13=73 • Add the ? 2,?f , J !^i JL=Z5^4-J 781 square ofJyr'^''+ 13+169"" 13"^ 169"~"l69"*' 169 ""169* Extract the ) J , >v/781 __i 27.94 square root ( ^+13 — — 13 ~— 13 +27.94-1 26.94 ^ ~ = -T;r-=2.07 or — 2.226. 13 13 Ex. 16. a:2-6ar+19 = 13. Ans. x = 4.732 or 1.268. Ex.17. 5x^+4x = 25. ... ar= 1.871. Any equation, in which the unknown quantity is found only in two terms, with the index of the higher power double that of the lower, may be solved as a quadratic by the preceding rules. Ex. 18. Let .r6-2a;3 = 48. Complete the square, a:®— 2ar'4-l ^49 ; Extract the square root, a:^— 1= + 7; .\x^ = 8 or — 6; and .*. X = 2 or V^6. Ex.19. 2a:— 7Va: = 99. 7 ^ 99 X — — va:= — ' 2^ 2 7 / Jiy 99 49 841 16 ' / 7 I 29 7 ,29 11 121 /. by squaring both sides, a: = 81 or — r-* CH. V. QUADKATIC EQUATIONS. 91 Ex. 20. x'+4x-' = 12. Ans. a: = + V2 or +V^. Ex.21. X6—8x^ = 5lS. ... a: = 3or^'^r9. RuiiE II. Let ax^ + bx = Cf Multiply each side of > , „ » i , the equation by 4«, \ then 4a^x^ + 4abx= 4: ac, ..^ Add &2 to each side, ? , ^ „_i_ ^ , . ,o a . 7„ we have \ 4a'x^±4abx+b^=4ac+b\ Extract the sq. root as before, 2ax + b = + V ^ac-^-b^ .-. 2az = + \^4ac+b^Zfb +y4ac+pT:^ and X = .-I— 2a From which we infer, that * * if each side of the equation be multiplied hy four times the coefficient ofx^, and to each side there be added the square of the coefficient ofx, the quantity on the left-hand side of the equation will be the square of 2aa:+&. Extract the square root of each side of the equation, and there arises a simple equation^ from which the value of ar may be determined.*'* If a =1, the equation is reduced to the form x'^-\'px=qi in this case, therefore, the Rule may be applied, by ** mul- tiplying each side of the equation by 4, and adding the square of the coefficient of a:." From the form in which the value of x is exhibited in this Rule, it is evident that it will have two values ; one corresponding to the sign -}-, and the other to the sign — , of the radical quantity. Ex. 1. Let3a;2+5a: = 42. Multiply each side of the equation by {4a) 12 ; > 36a:2+60a: = 504. then J * The principle of this Rule will be found in the Bya Ganita, a Hindoo Treatise on the Elements of Algebra. For a full account of that work, as translated by Mr. Strachey, see Dr. Hutton's Tracts, vol. ii. Tract 33. 92 ALGEBRA. Add (b'^) 25 to eacti side > of the equation, we have ^ 36^'+^30:r+25 =504+25=529. Extract the square root of each side of the equation, which gives 6x+5=±\/529 = + 23; .-. 6ar = + 23 — 5 = 18 or- 28; and. = ^=3, o ora: = r8__ 14__ Y ¥ ^^* Ex.2. Let:r2__i5^___54. Multiply by 4, then 4x'^—60x = — 216. Add (b^-) 225 ) ^^^ 4a:2-60a:+225 = 225-216 = 9. to each side ) ' Extract the square root 2t— 15 = + 'V/9= + 3; .-. 2a: = 15 + 3 = 18 or 12, A 18 12 o_^ and x = — or — = 9 or 6. ON THE SOLUTION OF PROBLEMS PRODUCING QUADRATIC EQUATIONS. 69. In the solution of problems which involve quadratic equations, sometimes both, and sometimes only one of the values of the unknown quantity, will answer the conditions required. This is a circumstance which may always be very readily determined by the nature of the problem itself. Problem 1. To divide the number 56 into two such parts, that their product shall be 640. Let X = one part, then 56— a; = the other part, and X (56 — x) = product of the two parts. Hence, by the question, x (56— a:) =640, or 56x— a:- = 640. CH. V. QUADRATIC EQUATIONS. 93 By transposition, a:^— 56ar= — 640. anda: = 28+12=40 or 16. In this case it appears that the two values of the un- known quantity are the two parts into which the given number was required to be divided. Prob. 2. There are two numbers whose difference is 7, and half their product plus 30 is equal to the square of the less number. What are the numbers ? Let X = the less number, then x-{-7 = the greater number, and ^ [-30 = half their product plus 30. Hence, by the question, ^ +30 = a:^ (square of less), or — ^+30 = 0:2. Multiply by 2 a;2+ 7^+60 = 2x\ By transposition ... x'^—7x = 60. Multiply by 4, and add ) 4^2^28^+49= 240+49 =289, 49 (Rule II.) 5 ' ' .-. 2a— 7 = >v/289 = 17 22: =17+ 7 =:24, or a;= 12 less number; hence x-{-7 = 12+7 = 19 greater number. Prob. 3. To divide the number 30 into two such parts, that their product may be equal to eight times their difference. Let X = the less part, then 30 — a: = the greater part, and QO—x—x or 30— 2:r = their difference. Hence, by the question, xy, (30—a:)=8x (30— 2a:), or 30a:— :r2_-240 — 16:r. 94 ALGEBRA. By transposition, x^—i6x= —240. .-. ar— 23= + V289=+17, and a: = 23 + 17 = 40 or 6 = less ipsirt ; 30— a: = 30 — 6 = 24 = greater part. In this case, the solution of the equation gives 40 and 6 for the less part. Now as 40 cannot possibly be a part of 30, we take 6 for the less part, which gives 24 for the greater part ; and the two numbers, 24 and '6, answer the conditions required. Prob. 4. A person bought cloth for £33 155., which he sold again at £2. Ss, per piece, and gained by the bargain as much as one piece cost him. Required the number of pieces. Let X = the number of pieces. . 675 Y[\Qw = the number oi shillings each piece cost, X and 48a: = the number o^ shillings he sold the whole for; .•.48a:— 675= what he gained by the bargain. 675 Hence, by the problem, 48a:— 675 = By transposition ) 2_225 225 and division, \^ ^6"^ \Q Complete the) 2_^ 50625 _225 50625 65025 sq. (Rule I.) ]^ "16 "^"^ 1 024 ~" le" "^ 1024 "" 1024 ' __225__ / 65025 _ 255 ••^ 32 "■ V 1024 ""'32 ' and x=: Ii-lll_= 15. 32 Prob. 5. A and B set off at the same time to a place at the distance of 1 50 miles. A travels 3 miles an hour^a^^er than B, and arrives at his journey's end 8 hours and 20 minutes before him. At what rate did each person travel per hour ? CH. V. QUADKATIC EQUATIONS. 95 Let X = rate per hour at which B travels. Thenj + 3= A 150 And = number of hours for which B travels. X 150 __ . j:+3"" But A is 8 hours 20 minutes (8^ hours) sooner at his journey's end than B ; „ 150 , Qi 150 Hence_^+8J = --, 150 , 25 150 or = • ^+3 ^3 X By reduction, ar2-|-3a:=54. 9 9 225 Complete the square, a;2+3a;+-=54+j=— -(RuleL): ,3 /225 15 and a:= =6 miles an hour for B, 2 ' ar+3=9 A. Prob. 6. Some bees had alighted upon a tree ; at one flight the square root of half of them went away ; at another 1*^" of them ; two bees then remained. How many alighted on the tree ? * Let 2a:2=the No. of bees. then :r + -^ +2 = 2:^2, or9a:+16a:24-18 = 18:r2 .-. 18a:2— 16a72-.9a:=18, or2A'2— 9:p= 18. (Rule II.) Multiply by 8, 16a;2— 72:r=144. * This question is taken from Mr. Strachey's translation of the Bija Ganita ; and the several steps of the operation will, upon comparison, be found to accord with the Hindoo method of solution, as it stands in that translation, p. 62. 5n) ALGEBEA. Add 81 ; then ICar'— 72a+8 1 = 225, or 4a:— 9= 15 ; .•.4a: =15+9 = 24, 24 and x = -r = 6 ; 4 ' .-. 2x^ = 72, No. of bees. Prob. 7. To divide the number 33 ihto two such parts that their product shall be 162. Ans. 27 and 6. Prob 8. What two numbers are those whose sum is 29, and product 100 ? Ans. 25 and 4. Prob. 9. The difference of two numbers is 5, and ^^^ part of their product is 26. What are the numbers ? Ans. 13 and 8. Prob. 10. The difference of two numbers is 6 ; and if 47 be added to twice the square of the less, it will be equal to the square of the greater. What are the numbers ? Ans. 17 and 11. Prob. 1 1 . There are two numbers whose sum is 30 ; and ^'^ of their product plus 18 is equal to the square of the less number. What are the numbers ? Ans. 21 and 9. Prob. 12. There are two numbers whose product is 120. If 2 be added to the less, and 3 subtracted from the greater, the product of the sum and remainder will also be 120. What are the numbers? Ans. 15 and 8. Prob. 13. A and B distribute £1200 each among a cer- tain number of persons : A relieves 40 persons more than B, and B gives £5 a-piece to each person more than A. How many persons were relieved by A and B respectively ? Ans. 120 by A, 80 by B. Prob. 14. A person bought a certain number of sheep for c£l20. If there had been 8 more, each sheep would liave cost him 10 shillings less. How many sheep were there ? Ans. 40. Prob. 15. A person bought a certain number of sheep for .£57. Having lost 8 of them, and sold the remainder at 8 shillings a-head profit, ho is no loser by the bargain. How many sheep did he buy ? Ans. 38. CH. V. QUADRATIC EQUATIONS. 97 Prob. 16. a and B set off at the same time to a place at the distance of 300 miles. A travels at the rate of one mile an hour faster than B, and arrives at his journey's end 10 hours before him. At what rate did each person travel per hour ? Ans. A travelled 6 miles per hour. B 5 Prob. 17. To divide the number 16 into two such parts, that their product shall be equal to 70. Let X = one part, then 16— a: = the other part. Hence x (16— a:) or IQx—x'^^lO. Transpose, and a:^— 16ar= —70. Complete the square, a:2-16a:+64=-70+64 = — 6, ... ar-8 = + ^^, orarzsS + 'V^^.* Prob. 18. To divide the number 20 into two such parts, that their product shall be 105 a:=10 + '^--3. Prob. 19. To resolve the number a into two such factors, that the sum of their nth powers shall be equal to b. Let a:=one factor, t then -= the other factor. X Hence x^A =&, ^ x^ or a:2»-|-a« zznbx"^ ; * It is very well known that the greatest product which can arise from the multiplication of the two parts into which. any given number may be divided, is when these two parts are equal; the greatest product therefore, which could arise from the division of the number 1 6 into two parts, is when each of them is 8 ; hence, in requiring *' to divide the number 16 into two such parts that their product should be 70," the solution of the question is impossible. f hy factors are here meant the two numbers which being mul- tiplied together shall generate the given number; if therefore X zz one factor, —must be the other factor, for a; x - = «. X X ■ 98 ALGEBRA. By Rule II. and 2a;«— ^»='V^Z>2«4^n^ or 2a;»=6+V^>2— 4a«, and ar«=— 2- — ; ^ 2 Prob. 20. To resolve the number 18 into two such fac- tors, that the sum of their cubes shall be 243. Ans. 6 and 3. ON THE SOLIJTION OF QUADKATIC EQUATIONS CON- TAININa TWO UNKNOWN QUANTITIES. The solution of equatfcns with two unknown quantities, in which one or both these quantities are found in a quad- ratic form, can only, in particular cases f* be effected by means of the preceding Rules. Of these cases, the two following are very well known. Case I. 70. " When one of the equations by which the values of the unknown quantities are to be determined, is a simple equation ;*' in which case, the Rule is, '* to find a value of one of the unknown quantities from that simple equation, and then substitute for it the value so found, in the other equation ; the resulting equation will be a quadratic, which may be solved by the ordinary Rules.'* * The most complete form under which quadratic equations con- taining two unknown quantities could be expressed, is this : ax'^ + by^-hcxy + d x-^ey=^m ax^ + b'y^ + cxy+d'x + ey=:m'; but the general solution of these equations can only be effected bv means of equa- tions of higher dimensions than quadratics. 70. There are two well-known cases, which admit of solution by the preceding rules; state them, and the rules employed for reducing the two equations to one quadratic of the usual form. CH. T. QUADKATIC EQUATIONS. 99 Ex. 1. Let a:+2^=7, ) and x'^'^Sxi/—y^=2S \ *° ^^^ ^^® values of x and i/. Prom pt equation, x=7—'2t/f :.x^= 49 —28^+41/^; Substitute these values for x and x^ in the 2^ equation, then 49— 28^+%2_|.2i^_6^2_^2-_23, or .V4-7y =49—23=26. By Rule IL 36^2^84^+49=3 12+49=36 1 , .-. 6j/+7=19 6^=19-7=12, or ^=2 ^=7-2^=7-4=3. Ex. 2. Let?^i^= 9 ? 3 > to find the values of x and ^. and 3^j/=210 3 From 1st equation, 2ar+^=27; .'. 2ar=27— ^. and x=z — ^^^ 2 27— V Hence, 8xi/=:SX-^Xy^2lO, or3X(27-^)X^ = 420 81^-3^/2=420 27^- 2/^= 140 ; or ^2_27^=— 140. By Rule IL, 4^2_io8^+729= 729-360=169; 27+13 .\ 2^-27=13, OTi/= ^ =20, , 27-20 ^, and a: = =3*. 2 ^ Ex. 3. There is a certain number consisting of two digits. The left-hand digit is equal to 3 times the right-hand digit ; and if 12 be subtracted from the number itself, the remainder 100 ALGEBRA. will be equal to the square of the left-hand digit What is the number ? Let X be the left-hand digit, ) then, by Art. 61, 10a:+^ and y the other ; \ is the number. Hence, a: = 3y ) , ., and 10^+^-12 =/ \ ^y *^^ ^^^s*^°" ' *\tftution } 30^+^-12=9z/S (for 10^=30^, and 0:^=9^^); 9^2-31^=-12; 31 12 ^ „ ^ 31 96J 961 12 961—432 529 By Rule L,y^-^y+ 324-324 - y=-32r-=324- 31 23 54 Hence,2/-jQ = -; ory=-=3, a: = 3^=9; . and consequently the number is 93. Ex. 4. Let 2a;— 3^= 1 \ 2a;24.ary— 5^2=20 \ *^ ^^^^ the values of a? and ^. Ans. x = 5, 2^=3. Ex. 5, There are two numbers, such, that if the less be taken from three times the greater, the remainder will be 35 ; and if four times the greater be divided by three times the less plus one, the quotient will be equal to the less number. What are the numbers ? Ans. 13 and 4. Ex. 6 What number is that, the sum of whose digits is 15, and if 31 be added to their product^ the digits will be inverted? Ans. 78. Case XL '1. When a;2, y"^^ or xy^ is found in every term of the two equations, they assume the form of ax"^^}) xy'\'Cy^=d, a'x'-\'b'xy-\-cy'^=d ; and their solution may be effected : — as in the following Examples : CH. V. QTTADBATIC EQUATIONS. 101 Ex. 1. Let 2x2+3^^+^2-- 20 20 Assume x=vy, then 2t;«^2^3v^2^^2^20, ory2= ^^3 _^^ j 41 and 5i;2^2^4^2=41,or^2==__- . 20 _. 41 ^ Hence 2j;2^3j;+i ■~"52;2+4' which reduced is, Gu^— 41i;= — 13; ^_41y 22 •'*" 6 "~ 6 * _ „ , 41u , 1681 1369 By Rule L, t?2- — +^^=-j^; _41 +37 _ 41+37 _13 1 '* ^ 12" 12 ' ^^^"~ 12 2 ^^3* T ^ 1 ^v, . 41 41 369 ^ Let .=-, then ^-^-^=_=_=9, or ^=3, ar=t?y=iX3=l. Ex. 2. What two numbers are those, whose sum multiplied by the greater is 77 ? and whose difference multiplied by the less is equal to 12? Let X = greater number, y = less. Theii (ar+2^) X ^ = ar^+a:^ = 77, and (x^y) Xy = xy—y^ = 12. Assume x-=vy\ ^_ ■ 77 Then v^y^+vy^ = 77, I ^^ ^ "^ v^+v '' andvw2_«2_,i2 C , 12 Hence, j^i = ,lj' 102 ALGEBEA. or 12v24-12v = 77?;— 77; which gives r^— To ^ ^ "" To . ^ 65 , 4225 529 and v^ vA = — . 12 ^ 576 576 » 65+23 88 or 42 H 7 /. v= — = — = = — or-. 24 24 3 4 Either value of v will answer the conditions of the question ; 7 , 12 12 48 48 _ buttaket;=j. ^^eT^2/^=:—[=zir[~7::ii'^Y^' 4 and i/=z 4, 7 x=v^=-X4 = 7. Hence, the numbers are 4 and 7. Ex. 3. Find two numbers, such, that the square of the greater minus the square of the less may be 56 ; and the square of the less plus J'"^ their product may be 40. Ans. 9 and 5. Ex. 4. There are two numbers^ such, that 3 times the square of the greater plus twice the square of the less is 110; and half their product plus the square of the less is 4. What are the numbers ?* Ans. 6 and 1. CHAPTER VI. ON ARITHMETICAL, GEOMETRICAL, AND HABMONICAL PROGRESSIONS. 72. If a series of quantities increase or decrease by the continual addition or subtraction of the same quantity, then those quantities are said to be in Arithmetical Progression. • For a great variety of questions relating to quadratic equa- tions which contain two unknown quantities, see Bland's Algebraical ' Froblenis, CH. VI. AKITHMETICAL PEOGBESSION. 103 Thus the numbers, 1, 2, 3, 4, 5, 6, &c. (which increase by the addition of 1 to each successive term), and the numbers 21, 19, 17, 15, 13, 11, &c. (which decrease by the subtrac- tion of 2 from each successive term), are in arithmetical progression. 73. In general, if a represents ^q first term of any arith- metical progression, and d the common difference, then may the series itself be expressed by a, a-\-d, a'\-2d, a +36?, a+4c?, &c., which will evidently be an increasing or a dc' creasing one, according as c? is positive or negative. In the foregoing series, the coefficient of d in the second term is one ; in the third term it is two ; in the fourth it is three, &c., i.e. the coefficient of d in any term is always less by unity than the number which denotes the place of that term in the series. Hence, if the number of terms in 7^ the series be denoted by (w), the nth. or last term in the ' progression will be a+ (^^—1) d: and, if the wth term be represented by / ; then Z=a+(n-l)d Ex. 1. Find the 50*^ term of4;he series, 1, 3, 5, 7, &c. Herea= O .-. /= 1+ (50— 1)2 d= 2V =1+49x2 w = 50) =99. Ex. 2. Find the 12'^ term of the series 50, 47, 44, &c. Here a = 50 V.-. / = 50 + (12 — 1) X— 3. =50-11X3 n= 12) =17. Ex. 3. Find the 25*^ term of the series, 5, 8, 11, &c. Ans. 77. Ex. 4 12*^ 15, 12, 9, &c. Ans. —18. Ex. 5. Find 6 arithmetic means (or intermediate terms) between 1 and 43. Here the number of terms is 8, namely, the 6 terms to be inserted, and the 2 given terms, and consequently 73. What is an arithmetic progression? Give an example of a series of quantiiies in arithmetical progression. 104 ALGEBRA. l=^^\ .-. l+7c/=43; w= 8) .*. c?=6. And the means required are 7, 13, 19, 25, 31, 37. Ex. 6. Find 7 arithmetic means between 3 and 59. Ans. 10, 17, 24, 31, 38, 45, 52. Ex. 7. Find 8 arithmetic means between 4 and 67. Ex. 8. Insert 9 arithmetic means between 9 and 109. 74. Let a be the first term of a series of quantities in arithmetic progression, d the common difference, n the number of terms, I the last term, and aS' the sum of the series : Then S=a+(a+d) + (a+2d)+ +1 and, writing this series in a reverse order, S = l + (l-d)+{l-2d)+ +a. These two equations being added together, there results 2 S== (a+l) + (a+l) + (a+l) + + (a+0 = (a+/) Xw, since there are n terms ; •••«=(«+0| 0). Hence it appears that the sum of the series is equal to the sum of the first and last terms multiplied by half the number of terms: And since /=a + (w— 1) d; .-. «=|2a+(n-l)rfj| (2). From this equation, any three of the four quantities a, d, n, s, being given, the fourth can be found. Ex. 1. Find the sum of the series 1, 3, 5, 7, 9, 11, &c. continued to 120 terms. S==\2a+(n-.l)d\xl. n=12o) =|2X1+(120-I)2|x^. =(2+119X2)X60=240x60=14400. CH. VI. ARITHMETICAL PROGRESSION. 105 Ex. 2. Find the sum of the series 15, 11, 7, 3, —1, ~5, &c. to 20 terms. = |2X15+(20~])X-4|x^ . ==(30~19X4)X10 = (30-76) X 10 = — 46 X 10 = — 460. Ex. 3. Find the sum of 25 terms of the series 2, 5, 8, 11, 14, &c. Ans. 950. Ex. 4. Find the sum of 36 terms of the series 40, 38, 36, 34, &c. Ans. 180. Ex. 5. Find the sum of 150 terms of the series ^, f , 1, I, h 2, h &c- Here a = i \ .■.S=^2a+ (n-1) d^^ d=i ( =j2Xi+ (150-1) xijH^ „=15o) =(| + l|£)75=I|lx75=3775. Ex. 6. Find the sum of 32 terms of the series, 1, H, 2, 2i, 3, &c. Ans. 280. PROBLEMS. Pros. 1. The sum of an arithmetic series is 1240, common difference —4, and 'dumber of terms 20. What is the Jirst term f Here *S= 1240 \ :, 6'=^2a+ (w-1) c?^ c^= - 4 I 1240=52a+ (20-1) X-4??2 w= 20/ =(2a-19x4)10 124 = 2a-76; .-. 2a =124+76 =200, and .-. a = 100. Hence the series is 100, 96, 92, &c. 106 ALGEBRA.. Prob. 2. The sum of an arithmetic series is 1455, the first term 5, and the number of terms 30. What is the com- mon difference? Here 8= 1455 \ \ 2a+ {n^\)d \^=^^ a= 5 \ 2X5+ (30-1) c/ ? y= 1455 n= 30/ (10+29^)15 = 1455. Dividing both sides by 15, 10+29c/= 97, 29^=87; .-. ^ = 3. Hence the series is 5, 8, 11, 14, &c. Prob. 3. The sum of an arithmetic series is 567, the first term 7, the common difference 2. Find the number of terms. \ { )n Here .9 = 567 1 .-.since ^ 2a+ (n—l) to find the number of terms. S= 198) Now ^2a+ (71-1)^^1 = 5', /. px30+(/i-l)x-2^|=198. (31-w)w=ic or, w2— 31/1 =—198, 169 4 ««-31„+(|)=^-1.8=: 31 ,13 31 ,13 .% n=—±—=z22 or 9. To explain the apparent difficulty arising from the two positive values of n, which gives us two differerit periods of the traveller's arrival at his journey's end, we must observe, that if the proposed series 30, 28, 26, &c., be carried to 22 terms, the 16^*^ term will be nothing, and the remaining six terms will be negative ; by which is indicated the rest of the traveller on the 16"^ day, and his return in the opposite direction during the six days following ; and this will bring him again, at the end of the 22*^ day, to the same point at which he was at the end of the 9*, viz. 198 miles from the place whence he set out. Prob. 9. How much ground does a person pass over in gathering up 200 stones placed in a straight line, at intervals of 2 feet from each other ; supposing that he brings each stone singly to a basket standing at the distance of 20 yards from the first stone, and that he starts from the spot where the basket stands ? It is evident that the space passed over by this person will be twice the sum of an arithmetic series, whose first term is 108 ALGEBRA. 20 yards (i.e. 60 feet), common difference 2 feet, and num^ her of terms 200. Here.^= 60 ) ., ^_ J 2a+(n-l)./| X |. n = 200 ) = (120+398) X 100. = 518X100 = 51800 feet. feet, miles, furlongs, feet Hence the distance required =103, 600= 19 . 4 . 640. Prob. 10. A person bought 47 sheep, and gave 1 shilling for the first sheep, 3 for the second, 5 for the thirds and so on. What did all the sheep cost him ? Ans. £110. 9^. Prob. 11. A gentleman began the year by giving away a farthing the first day, a halfpenny the second, three farthings the third, and so on. What money had he dis- posed of in charity at the end of the year ? Ans. £69. Us, 6J^. Prob. 12. A travels uniformly at the rate of 6 miles an hour, and sets off upon his journey 3 hours and 20 minutes before B ; B follows him at the rate of 5 miles the first hour, 6 the second, 7 the third, and so on. In how many hours will B overtake A ? Ans. In 8 hours. Prob, 13. There is a certain number of quantities in arithmetic progression, whose common difference is 2, and whose sum is equal to eight times their number ; moreover, if 13 be added to the second term, and this sum be divided by the number of terms, the quotient will be equal to the first term. What are the numbers ? hQ\.iheJirstterm—xX , . , .,, , and No. of terms = y;\ t^«" ^he second term will be a: + 2. In the expression 2a-\-(n — l)c?X -, substitute x for a, 2 for^, andy forn, and it becomes 2a:+(y— l)2x ^(=^^+2/'^— ^), for the sum of the series. By the problem, a:y"{-y'^—y = 8y, ox y=z9—x, a;4-2+13 and — - — ■ — = X, y CH. VI. GEOMETRIC PEOGEESSION. 109 Hence, ' ' = x, or x'^—'Qxz= — 13 ; .-. a:2-ar+16 = 16-15=1, anda:— 4 = +l; .'.a: = 5 or 3, ^=9 — a; =4 or6. From which it appears that there are two sets of numbers which will answer the conditions required; viz. 5, 7, 9, 11, or 3, 5, 7, 9, 11, 13. Prob. 14. There is a certain number of quantities in arithmetic progression, whose^rs^ term is 2, and whose sura is equal to 8 times their number ; if 7 be added to the third term, and that sum be divided by the number of terms, the quotient will be equal to the common difference. What are the numbers? Ans. 2, 5, 8, 11, 14. ON GEOMETRIC PROGRESSION. 75. If a series of quantities increase or decrease by the continual multiplication or division by the same quantity, then those quantities are said to be in Geometrical Progres- sion. Thus the numbers, 1, 2, 4, 8, 16, &c. (which increase by the continual multiplication by 2), and the numbers 1, i> h* ^> ^^- (which decrease by the continued division by 3, or multiplication by ^-), are in Geometrical Progression. 76. In general, if a represents the ^r^^ term of such a series, and r the common multiple or ratio, then may the series itself be represented by a, ar, ar^, ar^ ar'*, &c. which will evidently be an increasing or decreasing series, according as r is a whole number or a proper fraction. In the foregoing series, the index ofrm any term is less by unity than the number which denotes the place of that term in the seines. Hence, if the number of terms in the series be denoted by (n), the last term will be ttr"-^ 77. From the series given in the two preceding articles it is evident by mere inspection, that the common ratio can be found by dividing the second term by the first, or by dividing any term by that which precedes it. 75. Define a geometrical progression, and give an example. — 77. How is the common ratio of a series of numbers in geometrical progression found ? 1 10 ALGEBRA. Ex. 1. Find the common ratio of the geometrical pro- gression 1, 2, 4, 8, &c. 2 Here the common ratio =-=2, 2 4 8 Ex. 2. Find the common ratio of the series x » 7: > ^rr* &c. 3 9 27 4 2 2 In this example the common ratio =--t- -=-• 9 3 3 5 3 9 Ex. 3. Find the common ratio in the series - , 1 , - 5 ^;-- , &c. 3 o 25 , 3 Ans. -' 5 78. Let S be the sum of the series a, ar^, ar^^ &c., then a+ar-^ar^+ar^+ &c. . . . ar^-^+ar"-^ . . . = aS. Multiply the equation by r, and it becomes ar+ar2+ar'+ &c. . . . ar'"-^-\-ar^''^-{-ar"=rS. Subtract the upper equation from the lowers and we have, flr»— fl = r»S— 5^, or (r—l) S = ar^^a ; and therefore, S = ;-. If r is a proper fraction, then r and 2^s powers are less than 1. For the convenience of calculation, therefore, it is better . . o o — or^ m this case to transpose the equation into o = 9 by multiplying the numerator and denominator of the fraction ar'^—a rby-1. 79. If / be the last term of a series of this kind, then / z= ar»-^, .-. rl = ar"^ : hence S = ( — ) = . From \ r— I / r— 1 this equation, therefore, if any three of the four quantities S, fl, r, /, be given, the fourth may be found. 78. What is the expression for the sum of » terms of a series of numbers in geometrical progression ? CH. VI. GEOMETRIC PROGRESSION. Ill Ex. 1. Find the sum of the series 1, 3, 9, 27, &c. to 12 terms. Here^= 1 ) .^__ arn^a _ lXS''-l k ,o i 8P-1 ~" 2 _ 531441 — 1 531440 _ ""2 "" 2 ' Ex. 2. =265720. 2 4 8 Find the sum of ten terms of the series 1+- -< 1 j &c. ^3^9^27 a=z 1 2 f /2\io 3 I w=IO , /2\io /2\io l-r Nowg)^' l-r ^_2 3-2 —1 y X^' 3 /2\io 2^0 1024 \3/ 3^o"~39049» and 5 = 10 1024 58025 ^ ^""59049 ""39049' 3X58025 ___ 174075 39049 39049 Ex. 3. Find the sum of 7 terms of the series, 1, 3, 9, 27, 81, &c. Ans. 1093. Ex. 4. Find the sum of 1, 2, 4, 8, 16, &c. to 14 terms. Ans. 16383. Ex. 3. Find the sum of 1, -> -^ r;=' &c. to 8 terms. 3280 Ans. 2l87- Ex. 6. Find three geometric means between 2 and 32. Here a= .2) And ar"-^ = / /=32> .-. 2r* =32, nz= 5) r* =16, .*. r = 2. And the means required are 4, 8, 16. 112 ALGEBBA. Ex. 7. Find two geometric means between 4 and 256. Ans. 16 and 64. Ex. 8. Find three geometric means between J and 9. Ans. ^,1,3. Ex. 9. Find a geometric mean between a and /. Let X be the geometric mean required ; Then a, x, /, are three terms in geometric progression, A ^ ^ and - = - a X orar2 = a/ :,xz=\/'ai, Ex. 10. What is the geometric mean between 16 and 64 ? Ans. 32. Ex. 11. Insert four geometric means between ^ and 81. Ans. 1, 3, 9, 27. PROBLEMS. ' Prob. 1. Find three numbers in geometric progression, such that their sum shall be equal to 7 ; and the sum of their squares to 21. Let ->a:, xy, be the numbers. Then by the problem, ^•\-x+xy =7 ... (1) i'+^2+^2^. = 21 ... (2) From equation ( I ) , ^' (" + ^ +^ j = "^ .-.by squaring, x''\—^ + -+^+'ly+y~)=^9 From (2) x'^[j, +1 +^^)=21 .-. by subtraction, x\- +2+2^) = 28, or 14a: = 28; .-. a:= 2. CH. VI. GEOMETRIC PBOGKESSION. 113 This value of :c being inserted in (1), 1 7 5 + 3 o^, 1 .*. yz=z^.= — = 2 or - •^4 2 Hence, the numbers are 1, 2, 4; or 4, 2, 1. Prob. 2. There are three numbers in geometric pro- gression whose product is 64, and sum 14. "What are the numbers? Ans. 2, 4, 8; or 8, 4, 2. Prob. 3. There are three numbers in geometric pro- gression whose sum is 21, and the sum of their squares 189. What are the numbers? Ans. 3, 6, 12. Prob. 4. There are three numbers in geometric progres- sion ; the sum of the Jirst and last is 52, and the square of the mean is 100. What are the numbers ? Ans. 2, 10, 50. Prob. 5. There are three numbers in geometric progres- sion, whose sum is 31, and the sum of the first and last is 26. What are the numbers ? Ans. 1,5,25. ON THE summation OF AN INFINITE SERIES OF FRACTIONS IN GEOMETRIC PROGRESSION; AND ON THE METHOD OF FINDING THE VALUE OF CIRCULATING DECIMALS. 79. The general expression for the sum of a geometric series whose common ratio (r) is a fraction, is (Art. 78) S = . Suppose now n to be indefinitely great, then r" (r b^ng a proper fraction) will be indefinitely small, * ?o * When r is a proper fraction, it is evident that r» decreases as n increases; let r=^ for instance, then r2=-^» ^^ = 7^ ' ^'* = J5555' &c., and when n is indefinitely great, the denominator of the frac- tion becomes so large with respect to the numerator, that the value of the fraction itself becomes less than any assignable quantity. I 114 ALGEBBA. that ar» may be considered as nothing with respect to a in the numerator a^ar"" of the fraction expressing the value of S ; the limit, therefore, to which this value of 5 ap- proaches, when the number of terms is infinite, is -^— . 1— ?• Ex. 1. Find the sum of the series l+o + 7+^' &c. ad infinitum. Here a = 1 ^ a 1 2 Ex.2. Find the value of ^ + 25 + J^ + &c- ««? infinitum. Herert = r J i ^ . o_ 5 _ 1 1 1 ( " "", 1 5-l""i' Ex. 3. Find the value of 1 + -+ - +^-^+&c. ad infinitum- Ans. -. 3 9 27 Ex.4 l+-+Tg + gT+ &c. afl? 2>i/fm7wm. Ans. 4. 2 4 8 Ex. 5 5"*" 25 ■^125'^ ^^* ^^^^'fi^^^'^' 2 ■ ^ 3* 80. These operations furnish us with an expeditious me- thod of finding the value of circulating decimals, the num- bers composing which are geometric progressions, whose HO. What is the expression for the sum of a geometric series, when the number of terms is infinite ? CH. VI. GEOMETRIC PROGRESSION. 115 common ratios are j^> -j^j looo' ^^' ^^^^^^^"^ *^ ^^^ number of factors contained in the repeating decimal. Ex. 1. Find the value of the circulating decimal .33333, &c. This decimal is represented by the geometric series _-j 1 4- &c. whose /?r5/ dermis 77:' and common 1 ratio -j-7v g \ g Hence «=y^' . ^^ ^ ^ 10 3 3__1 1 (* * ^ 1-^ , 1 10—1 ~^9 3* "•^To' ; ^"lo Ex. 2. Find the value of .32323232, &c. ad infinitum. Here«-j()Qi ^^ ^ ^ iqq ^ 3^ 3^ L I * ^""^ 1 L 100-1~"99' ''""'loo; 100 Ex. 3. Find the value of .713333, &c. ad infinitum. The series of fractions representing the value of this 71 S S decimal are Y00+ (geometric series) Yooo"^ 10000 "*" ^^' 71 , -. ^ 1^, X- 4. 1l— 4. . 3L^« 4tTr *^100"^ '^ '^*^ ''^^ „ 3_\ _3_ Here fl— jQ()Q i 1000 _ 3 _ 8 1 l_ i *' ' ~" J^"~1000-100"~90b""3'00* ''—10 ; 10 Hence the value of the decimal =(-^+. 9)^ + 3^5= i^-- 107 'l50* i(M> \><«' <^0^ '-5"^ 116 ALGEBRA. "Ex. 4. Find the value of .81343434, &c. ad infinitum. ;34 a 10000 34 _ 34 ''^-'l—r r~"10000— 100 9900* ''"lOO ^ ^""loo And value of the decimal=— 4.5— -^ 4- Ji-=??^. 100 ^ 100 ^ 9900 9900 100 ; ^^ Ex. 5. Find the value of .11111, &c. ad infinitum. a.rfc><- y 7 7 . 7 ci^ 7 V ^■'■/^^ ^fCr-c-c'^fc-trt^'^^^' Ans. 3- "5 Ex.6 : .2S2S23, kc. ad infinitum. ^ .23 ^ Ans. ^. V Ex.7 83333, kc. ad infinitum. i ^ ^ t Ans. z • , Ex. 8 7141414, &c. «c? infinitum, % * 707 ^ Ans. ^. "^ Ex.9. 956666, kc. ad infinitum. ^°^^- 300- t^' are the arithmetic means to be inserted '6 9' 18 11 9 18 between g and „' and therefore their reciprocals 6, 9'~r' are the three harmonic means required. Ex. 4. Find a harmonic mean between 12 and 6. Ans. 8. Ex. 5. The numbers 4 and 6 are two terms of a har- monic progression ; find a third term. Ans. 12. Ex. 6. Find two harmonic means between 84 and 56. Ans. 72 and 63. Ex. 7. Insert three harmonic means between 15 and 3. 2 . ^"^^ f ' ^' T. IcM^ 82. Let a, 6, c, d, e, &c. be a series of quantities in ^^ttZ^ul^ ^111,1 . ^ Aarwiomc progression J then -» r> -' i' -> &c. are in ariYA- CH. VI. HARMONIC PROGHESSION. 119 metic progression, and according to the definition of an arithmetic progression (Art. 72), we have b a c b ^ ' \-hH « d c-e d ^^^ &c. = &c. _ a—h b—c From(l) ^=-^ a_ a'—b ^^^ C, c 6— c * or, converting this equation into a pro})ortion, a : c::a—b : b--c :d::b — c:c — d /. j / / .««p fUi.c '"*"* 120 ALGEBRA. But by geometric progression ac = h"^ and .-. 2 = - + - ' X z 2/ 3: z or- -1 = 1-^ y X z y' CHAPTER VII. ON PERMUTATIONS AND COMBINATIONS. 83. By Permutations are meant the number of changes which any quantities «, b, c, rf, c, &c., can undergo with re- spect to their order, when taken two and two togetherTT^ree anS^three^ &c. &c. Thus ab^ ac^ ad, ba, be, bd^ ca, cb, cd, da, dby dc, are the different permutations of the fou?- quan- tities fl, b, c, d, when taken two and two together ; abc, acb, bac, bca, cab, cba, of the three quantities a, b, c, when taken three and three together, &c. &c. 84. Let there be n quantities, a, b, c, d, e, &c. : then, by Art. 83, it appears that there will be («— 1) permutations in which a stands first ; for the same reason there will be (n— 1) permutations in which b stands first; and so of c, d, e, &c. Hence there will be n times (n—l) permuta- tions of the form ab, ac, ad, ae, &c. ; ba, be, bd, be, &c. ; ca, cb, cd, ce, &c. ; i.e. " the number of permutations ofn things taken two and two is n {n — 1)." 85. If these n quantities be taken three and three to- gether, then there will be n (n—l)'i(n—2) permutations. Forif («— 1) be substituted for n in the last article, then the number of permutations of ?/ — 1 things taken tuw and two together will be (w — 1)"^^— 2) ; hence the number of permutations of b, c, d, e, &c. taken two and two together, CH. VII. PERMUTATIONS. 121 are (^—1 ))((«— 2), and consequently there are (w— l)x (w— 2) permutations of the quantities «, ft, c, d, e, &c. taken three and three together, in which a may stand first ; for the same reason there are (/i--l)(w— 2) permutations in which b may stand first ; and so of "c, c?, e, &c. The num- bers of the permutations of this kind will therefore amount to n (w~l) (w— 2). 86. To find the number of permutations ofn things taken r together. By Arts. 85 and 86— The No. taken ttvo together = n (n — l) , three = w (w — 1 )X(w — 2) Similarly four =n (n — l)y(J^n^2y:^n-^S) If the law, which is observed in these particular cases, be supposed to hold generally, that is, if the number of per- mutations ofn things «, ft, c, d, &c. taken r—l together, be n(n-^l) («— 2) (w— r4-2) Then, by omitting «, it is equally true that the number of permutations of n—l things ft, r, dy &c. taken r—l together, will be by putting w— 1 for w in this last expression (« — 1) (72-2) (w— r+l) Now, if a be placed before each of these permutations, there will be (n—l) (n— 2) (n— r+1) permutations of things taken r together, in which a stands first. It is clear that there will in like manner be the same number of permutations of things taken r together, in which each of the other things ft, c, d, &c. stand respectively first ; and as there are n things the entire number of permuta- tions of n things taken r together, will be the sum of the permutations taken r together in which the n things a, ft, c, d, &c. respectively stand first, that is, n times ('?— i) («-2) {n^rJ^l). or « (w — 1) (7«— 2) (w— r-fl) It has thus been proved, that if the law by which the ex- pression for the number of permutations of n things taken r — 1 together is found, be true, it is also true for the next superior number, or when n things are taken r together ; but the law of the expression has been found to hold for 122 ALGEBRA.. the number of permutations of n thinsrs taken two togrether, and for the number of permutations of n things taken three together: it is therefore true by the theorem just demon- strated when the // things are taken fow together, and if true when taken four together, it is true also when taken Jive together, and so on for any number not greater than 71, which may be taken together. This proof aflPords an excellent example of demonstrative induction^ a method of reasoning of great importance in the mathematical sciences. 87. If r=w, i.e. if the permutations respect all the quan- tities at once, then (since n—r = 0) the number of them will be /2(« — 1)(/2— 2) &c 2.1. Thus, the number of permutations which might be formed from the letters com- posing the word ''virtue'' are 6X5X4X3X2X1 =720. 88. But if the same letter should occur any number of times, then it is evident that we must divide the whole num- ber of permutations by the number of permutations which would have arisen if different letters had occurred instead of the repetition of the same letter. Thus if the same letter should occur twice, then we must divide by 2X1; if it should occur thrice, we must divide by 3x2X 1 ; if^^ times, by 1.2.3... p\ and so for any other letter which may occur more than once. Hence the general expression for the number of permutations of n things, of which there are p of one kind, r of another, q of another, &c., &c, is n (« — 1) (;?— 2) (??— 3)....2.1 ^, 1.2.3..pXl.2.3..rxl.2.3..9- ^^^' *^^ permutations which may be formed by the letters composing the word ''easiness'' (since s occurs thrice, e twice) are 8.7.6.5.4.3.2.1 1.2.3.X1.2 =^^"- Ex. 1. What is the number of different arrangements which can be made of 6 persons at a dinner table ? The number = 1x2X3X4X5X^ = 720. Ex. 2. Required the number of changes which can be rung upon 8 bells. The No. of changes = 1X2X3X4X5X6X7X8 = 40320. CH. Yll. COMBINxVTIONS. 123 Ex. 3. With 5 flags of different colours, how many sig- nals can be made ? The number of signals, when the flags are taken, Singly, are = 5 Twjo together ...=5.4 = 20 Three =3.4.3 = 60 Four =5.4.3.2 = 120 Five =5.4.3.2.1... = 120 .*. the total number of signals = 325 Ex. 4. How many permutations can be formed out of 1 letters, taken 5 at a time ? Ans. 30240. Ex. 5. How many permutations can be formed out of the words Algebra and Missippi respectively, all the letters being taken at once ? Ans. 2520 and 1680. ON COMBINATIONS. 89. By Combinations are meant the number of collections which can be formed out of the quantities, «, Z>, c, d, c, &c., taken two and two together, three and three together, &c. &c., wi thou t having regard to the order in which the quantities are arranged in each collection. Thus ab, ac, ad, be, bd, cd, are the combinations which can be formed out of the four quantities a, b, c, d, taken two and two to- gether ; abc, abd, acd, bed, the combinations which may be formed out of the same quantities, when taken three and three together ; &c. &c. 90. From the expression (in Art. 86) for finding the number of permutations of n things taken r and r together, we immediately deduce the theorem for finding the number of combinations of n things taken in the same manner. For the permutations of n things taken two and two together being n (7^ — 1), and as each combination admits of as many permutations as may be made by two things (which is 2X 1), the number of combinations must be equal to the number of permutations divided by 2 ; i. e. the number of combinations of n things taken two and two together is — ^^-r . For 124 ALGEBBA. the same reason, the combinations of n things, taken three and three together, must be equal to —^^ — , — ^ and in general, the combinations of n things taken r and r to- n{n — \)(n^2).„.(n—r+\) gether must be equal to 1 ^ » ^^ ; — • Ex. 1. Find the number of combinations which can be formed out of 8 letters, when taken 5 at a time. _, . 8X7X6X5X4 The number = j^axaxl^ = ^^- Ex. 2. What is the total number of combinations which can be formed out of 6 colours taken in every possible way ? No. of combinations when the colours are taken — 1 at a time = 6 2 =T— ^ ...i = 15 = 20 6 .3 .2 .5 . 1 6. 4 1. 6 .2. .5. 3 .4, .3 1.2.3.4 6.5.4.3.2 "I. 2. 3. 4. 5" 6.5.4.3.2. 1 '1.2.3.4.5.6 = 15 = 6 = 1 Hence the total number =^3 Ex. 3. Find how many different combinations of 8 let- ters, taken in every possible way, can be made. Ans. 255. *J* Several other useful and interesting subjects of an elementary character yet remain to be treated of. The Editor is pre- paring for publication a Second Part, which will embrace these subjects. THE END. Printed by A. Swebtino, Bartlett'^-^ijHdiDgs, i^>pclon. ^, 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. OCJ ?1 iqiR ilH! Illyillil! liiilililHliilliilJH^ iiiliiiiiii III I iitiSliiwHpi if liliil lljilillililiir , J j ill II 1 1 1 nil 111 lij: lliii lili! 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