22. Required that number, one-fourth of which is equal to the difference of the squares of its digits; and if the number be added to three-fourths of itself, the digits will be inverted. Ans. 12. 23. Required three numbers, such, that if one-half of the first, one-third of the second, and one quarter of the third be added together, their sum will be 62; also, if one-third of the first, one-fourth of the second, and one-fifth of the third be added together, their sum will be 47; but if one-fourth of the first, one-fifth of the second, and one-sixth of the third be added together, their sum will be 38. Ans. The first 24, the second 60, and the third 120. 24. Three workmen, A, B, and C, being employed to build a house, it is found that A and B together can earn £2 in six days, A and C together can earn £2. 14s. in nine days, and B and C together can earn £4 in fifteen days. How much does each person receive per day? Ans. A 3s. 8d., B 3s., and C 2s. 4d. N. B.-Questions 23 and 24 are solved by three unknown quantities. PURE QUADRATICS. A Pure Quadratic Equation is that which contains the second power of the unknown quantity, without any other power of it. Thus, x=9, y2=16, x2=a, &c. are pure quadratic equations, the value of the unknown quantity, in each case, being determined by extracting the square root of both sides of the equation, and they become as follows: x=±3, y=±4, x=±√a. Here the sign of the value of the unknown quantity is either or —. Cor. The square root of any quantity being either + or -, pure quadratic equations admit of two answers. Thus, +3x+3 make 9, and by Note 4th, Case I. in Multiplication, -3x-3 make 9, and each value, when substituted for x in the given equation, answering the conditions of the question. If the square of the unknown quantity have a coefficient, or be connected with any known quantity by the signor —, such known quantity must be transposed to the other side of the equation, as in example first, then divide by the coefficient of the unknown quantity, and extract the square root of each side of the equation in order to find its value. Pure Quadratics and others solved without completing the square. 1. Given 4x2-16=48, to find the values of x. By transposition, 4x2=48+16=64, or x2=16; .. x=±4. From the second equation, x=3y; substituting this value of x in the first equation, 3y2=48, or y2=16; .. y=±4, and x=±12. 3. Given x-y y 1: 4, to find the values of x and x2 y2=8I, and y. By Theorem 3rd, from the first equation, x: y:: 5:4; 4.x 16x2 by Theorem 2nd, 4x=5y; .. y=, or y2=25; substituting this value of y2 in the second equation, 4. Given x+y: y :: 5 : 2, to find the values of and x2+y2=117, x and y. substituting this value of y2 in the second equation, 4.x2 .. 9x+4x2-1053, or a2=81; .. x=9, and y=±6. 5. Given x+y: x—y :: 7 : 1,2 to find the values and x+y=100, } of x and y. By Theorem 5th, from the first equation, 2x 2y86; dividing the proportion by 2, (Theorems 9 and 10,) xy: 43; 3x 9x2 ..4y=3x, or y=, or y2=16; substituting the value of y2 in the second equation, 9.x2 16 =100, or 16x2+9x2=1600, or x2=64; By Theorem 6, 2y: 2x :: 2:6; dividing by 2, y:x::1:3; .. x=3y; substituting this value of x in the second equation, 3y=48, or y2=16; .. y=±4, and x=3y=±12. 7. Given x:x+y:: 2 : 5, to find the values of a and x2+xy=40, and y. From equation first, by Theorem 7, x+y:x::5:2; Theorem 4, y:x:: 3 : 2; substituting this value of y in the second equation, From the first equation, by Theorem 12, x+y:g::6:1; substituting this value of y2 in the second equation, 25x2 x2 -=24; .. 49x2—25x2=1176, 49 or 24x2=1176; .. x2=49; whence, x=17, and y=±5. 1 1 x2—2xy+y2=‚—‚ or(x—y)2=z—y; .. (xy)-1, or x-y=1; substituting this value of x-y, in the equations (A and B), but by subtraction, 2y=10, or -12; .. y= 5, or 6. |