If, now, we substitute the value of xz from Equation (2), in Equation (1), we shall have, then, adding together Equations (3) and (4), we have, 13000y = 13000z + 13000; 13y = 132 + 13; (4.) or, Now, to eliminate x from Equations (2) and (3), mul. tiply the first by y, and the second by z, and we have, Now, substituting for y its value in Equation (5), we have, 13z(z+1) 36(z+1) + 492; = 13z2 + 13z = 362 + 36 + 49%; that is, 2 = The negative value of z is not applicable to the question. (18.) Let the second number be denoted by x, and the differ ence between the second and first by y. Then, and, Then, also, that is, x = 2d number, x + y + 6 = 3d number. 3x + 6 = 33, and, hence, x = 9; (x − y)2 + x2 + (x + y + 6)~= 467; 3x2+12x+12y + 2y2 431; Let x denote the digit which stands in the ten's place, and y the digit which stands in the unit's place. Substituting the value of y2 from Equation (1), we have, Let the numbers be denoted, respectively, by x, y, and z. Then, xy = 2, xz = 4, y2 + 22 = 20. From the first equation, we have, x= 2 y 22 ; and hence, = 4; or, 2 = 2y. y Substituting this value of z in the third condition, we have, whence, y2+4y2 = 20; 5y2 = 20, y2 = 4, and y = 2. Hence, the numbers are 1, 2, and 4. (23.) Let the numbers be denoted by x, y, and z. Adding the first and third equations, we have, 3y = 45; or, y = 15; from which we easily find, (24.) Let the numbers be denoted by x, y, and 2. Then, xy = a, X2 = b, y2 + z2 = c. From the first equation, we have, Let the greater number be denoted by x, and the less by y. But Equations (1) and (2) may be put under the forms, and, x2 + xy = 144, xy- y2 = 14; or, xy= 14+ y2; and subtracting, x2 + y2 = 130; or, a2 = 130 y2. Substituting in Equation (3), the value of xy = 144 - x2, and then for 2, its value, 130 y2, and we obtain, |