NOTE.-The student will see that the pairs of values which satisfy Now (xy) is a factor of each side, and hence, striking it out, we have Equations (3.) and (4.) may not be used as simultaneous equations, but each of them may be used in turn with either of the given equations. Thus, taking equations (3.) and (1.), we have - (1.) – (3.), x2 from which x = 6 ± 2 √√6; and hence from (3.), by substitution, we easily get Hence, the pairs of values satisfying the given equations NOTE. It is worth while remarking that when each of the terms of the given equations contain at least one of the unknown quantities, the values x = 0, y = 0 will always satisfy. Each of these equations taken in turn with either (1.) or (2) will easily give the required values of x and y. From (2), raising each side to the fourth power, we have— x2 + 4x3y + 6 x2y2 + 4 xy3 + y2 = 2401 ;........ (3) - 4x3y (1), then 4 xy + 6 x22 + 4 xy3 = 2064; or, 2 x3y + 3x2y2 + 2 xy3 1032; or, arranging, 2 xy(x + y)2 - x2y2 = 1032; and hence, 2 xy(49) - x2y3 or, x2y2 98 xy + 1032 11=7 49, = 1032; .(3.) from which xy = 12 or 86,......(4.) From (2) and (4), ay may now be easily obtained, and hence also the required values of x and y. 44. x + y + ≈ + U = 4a + 4b, xy + xz + xu + yz + yu + zu = 6a2 + 12 ab +663, xyz + xyu + xzu + yzu 45. x2y2 + xyz + x2yz = a, 4a3 +12a2b+ 12ab2 + 4b3, + 4 ab3 + ba. A 48. Given R = 1 + r, P = 4 (1 − R−"), M = show that R = A 2 PR", CHAPTER II. Problems Producing Quadratic Equations. 7. We shall now discuss one or two problems whose solutions depend upon quadratic equations. Ex. 1. A person raised his goods a certain rate per cent., and found that to bring them back to the original price he must lower them 3 less per cent. than he had raised them. Find the original rise per cent. Let x then = x the original rise per cent., the original price. Hence, by problem the fall per cent. to bring them to The value x = 20 is alone applicable to the problem. Remembering, however, the algebraical meaning of the negative sign, it is easy to see that the second value, x gives us the solution of the following problem: = 16%, A person lowered his goods a certain rate per cent., and found that to bring them back to the original price he must raise them 3 more per cent. than he had lowered them. Find the original fall per cent. The above solution tells us that the fall required is 163 per cent. Had we worked the latter problem first, we should have obtained x = 16 or 20, the value x = 20 indicating the solution of the former problem. |