is unlimited, but if the equation be of the form ax + by = c, the number of solutions is always limited, and in some cases no solution can be obtained. Let a x-by= c be the equation in which a is less than b; then we have where my and n are the nearest quotients, whether in excess or defect, arising from dividing by and c by a; hence d and e are each less than Let the last fraction, which must obviously be a whole number, be put=v, then we have dyte = a v, and as before a. fvg where ƒ and g are each less than d. Put the last fractional expression, which must also be a whole number, v1, and repeat the process until we arrive at a fractional expression in which the coefficient of the subsidiary unknown quantity v, is unity. Let this fraction = p, then we have an expression of the form, v„ =hp±k; and by reversing the steps, we get in succession the values of the subsidiary unknown quantities which have been employed, and finally those of x and y. = EXAMPLES. 1. Given 7 x + 12y = 50, to find the values of x and y in whole numbers. Here the smaller coefficient is 7, and therefore we have =p, then v = 2p+1, and hence y = 3 v + p = 7p+ 3 ; x=7-2y + v = 2 · 12p. Consequently we see that if x is to be a whole positive number, the value of p is limited to zero; hence when p=0, we have x= 2 and y = 3, which are the only integer values of the two unknown quantities. 2. A person purchased between 50 and 60 horses and oxen; he paid 31 dollars for each horse, and 20 dollars for each ox, and he found that. the oxen cost him seven dollars more than the horses; how many of each did he buy? Let x denote the number of horses, and y the number of oxen; then In these expressions for x and y, we see that p may have all possible values from 0 to any extent; hence making p we get = 0, p = 1, p = 2, etc., x = 3, 23, 43, 63, 83, 103, etc.; = 36, therefore by the limitation in the question, we have x = 23 and y the respective numbers of horses and oxen that were purchased. The total number of solutions is unlimited. 3. How many ounces of gold, of 17 and 22 carats fine, must be mixed with 5 ounces of 18 carats fine, so that the composition may be 20 carats fine? Let x = the number of ounces of 17 carats fine, and y = the number of 22 carats fine, then by the condition of the question we get If p = 1, 2, 3, 4, etc., then we get x= 2, x = 4, x = 6, x 8, etc. 134. If there be only one equation to determine three unknown quantities, as a x + by + c z=d; then by transposition, ax + by =dcz, and by giving to z all its different integer values, we shall obtain in the usual manner all the corresponding values of x and y. The values of x and y cannot be less than unity; therefore the highest value of z cannot exceed the value derived from the equation 3. Given 11x+5y= 254, to find all the possible values of x and y. Ans. x 19, 14, 9, 4; y = 9, 20, 31, 42. 4. In how many ways can 207. be paid without using any other coin than half-guineas and half-crowns. Ans. In 7 different ways. 2 = 7, to find the values x = 5, 6, 7, 8 Ans. y = 3, 6, 9, 12, ... 6. A person bought 100 animals for 1007., namely, oxen at 57. each, sheep at 17., and fowls at 1 shilling each; how many of each kind did he purchase? Ans. 19 oxen, 1 sheep, and 80 fowls. VOL. I. N 7. I owe a person 3 shillings, and have nothing about me but guineas, and he has nothing but crowns; how must I discharge the debt? Ans. I must give 3 guineas, and receive 12 crowns. 8. A jeweller requires to mix gold of 14, 11, and 9 carats fine, so as to make a composition of 20 ounces of 12 carats fine; find the quantities of each kind of gold to form the required mixture. Ans. 8, 10, 2 ounces, or 10, 5, 5 ounces. 9. Given 5x + 7y + 11 z = 224, to find all the possible values of x, y, and z in whole numbers. Ans. The number of ways is 59. INDETERMINATE COEFFICIENTS. 135. If two series of quantities be equal to each other, as a + b x + c x2 + dx23 + ... = A + B x + C x2 + Dx3 + . . whatever be the value of x, and where the coefficients are independent of x; then will Aa, Bb, Cc, D=d, etc. For by transposition we have a- A = (B − b) x + (C − c) x2 + (D − d) x3 + = .. and if a is not equal to A, then a - A will be some constant quantity; but the second side of this equation varies as x varies, and may be made less than the fixed quantity a - A by taking a sufficiently small value for x, which is absurd, and therefore we must have a A. The proposed equation, then, separates itself into two equations, viz., A = a, and bx + c x2 + dx3 + . . . = B x + С x2 + D x3 + Dividing each side of the latter by x gives b + c x + d x2 + ex2 + . . . B+Cx + D x2 + Ex3 + which by similar reasoning separates into the two equations, Bb, and c + d x + ex2 + . . = C + D x + Ex2 + and the latter gives, as before, C = c, and so on; hence we have A = a, Bb, C = c, D = d, etc. EXAMPLES. 1. Required the development of (ax) in a series. Let (ax) = A + B x + Cx2 + D x3 + Ex1+ and equating the coefficients of the like powers of x, we have = BINOMIAL THEOREM. 136. By actual multiplication we have (1+x) = 1+2x+x2; (1+x)3 = 1 + 3x + 3 x2 + x3 ; (1 + x) = 1 + 4x + etc., and hence generally we see that the form of the expansion of (1 + x)", when n is a positive integer, is 1 + nx + A x2 + Bx3 + Сx + where the coefficient of the second term is n, and the coefficients A, B, C, etc., are entirely independent of x. m But if the exponent be a positive fraction, as — then we have n (1 + x)" = (1 + x)TM = / (1 + mx + A'x2 + B'x3 + C'x' + ...). And since (1+ y)" = 1 + ny + Ay2+By+Cy+..., if У = ax + bx2 + cx3 +. . . (1 + a x + bx2 + cx2 + ...)" Hence, conversely, the nth root of 1+nax + A1 x2 + В ̧ Ñ3 + Let nam, then will a = m ... = 1+ ax + bx2 + cx3 + ... and therefore the nth root of 1+mx + Px2 + Qx3 + is of the form ... x+p x2+q x3 +... Again, if the exponent be negative, as n, where n may be either integer or fractional, then we have (1 + x)" = 1 1 + n x + A x2+Bx3 +Сx2 + (1). by actual division; therefore generally the form of the expansion of 1 + n x + A x2 + B x3 + С x + Now if we substitute x + y for x in equation (1) we have + n (x + y) + A (x + y)2 + B (x + y)3 + ... 1 + n x + A x2 + BxCx* +...) +ny +2 Axy +3В x2y +4 Cx3y+.. (1 + x + y)" = = 1 But (1+x+y)" = {(1 + y) +x}" = (1 + y)" {1+ { 1+ y x3 +A + B + (1+ y) 2 (1+y)3 } = (1 + y)" + n x (1 + y)" ~1 -1 + A x2 (1 + y)"−2 + Bx2 (1+ y)" +Cx*(1+ y)*-* + • • nA,xy+n B1xy3 +... Ax2+A (n − 2) x2y + AA, x2 y2 +.....(3), + B+B(n-3)x3y+... + Cx+... where A1, B1, etc., A2, B2, etc., are what A, B, etc. become when n is changed into n 1, n - 2, etc. Now the series (2) and (3) must be identical; hence equating the coefficients of the same powers, or combinations of powers, of x and y, in both, we get sequently the coefficients of x y and xy in (3) are identical, each being Comparing the coefficients of xy' and xy in Finding now the values of B, and A,, we can find the coefficients of xy3, x'y', and x3y in (3) to be respectively |