Squaring the first equation, and multiplying the second by m2, we have, by transposing in the second, and, n2x2 = m2y2, m2x2 = m2 a2 m2y2; and then, by addition, we obtain, x2 (m2 + n2) = m2a2, Let the largest number be denoted by 2x; then the less will be denoted by x. and, or, We shall then have, (10.) Then and, Let the numbers be denoted by x and y. x: y :: m : n; or, nx = my, x2 - y2 = b2; Let the amount in dollars, placed at interest, be denoted NOTE. This, it will be seen, is an equation of the first degree, and is placed among those of the second degree, to lead the student to have confidence in his own methods, and not to rely too implicitly on the arrangements of the author. Transposing, arranging the terms, and factoring, we have, dividing by the coefficient of x2, we have, b2n2; b2n2 m2 m2, so as to reduce term of the second member, by n2 a2n4 = b2m2n2 + a2m2n2 + b2m2n2 + b2n1. Cancelling like terms, and factoring, = n2(a2m2 + b2 n2 — a2n2); then extracting the root of both members, ; Let x denote the greater number; then 11 - will denote the less. By the conditions of the question, Either root will satisfy the conditions of the problem. (5.) Let x denote the greater number; then, since the difference of the two numbers is 3, x 3 will denote the less By the conditions of the question, The plus value belongs to the problem; hence, the numbers are 8 and 5. (6.) Let x denote the number of sheep which he purchased. 1200 X 15 18000 х Then, the number of sheep sold will be represented by 15, and 2(15) the amount of profit. Now, had there been no profit, the amount received for the sheep would have been just equal to the cost, less the value of the fifteen unsold; and, consequently, the amount received, less the profit, must be just equal to this difference. That is reducing to shillings, and, by reducing and dividing by the coefficient of x2, we have, x2 + 45x = 9000; from which, by taking the positive value, we have, x 75. |