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2007/08 AHSMC Part 1 Problem 11

Problem 11 of 16

The real numbers xx and yy are such that x+2y=83x + \dfrac{2}{y} = \dfrac{8}{3} and y+2x=3y + \dfrac{2}{x} = 3. The value of xyxy is

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Solution

Multiplying one equation by the other, we have xy+4+4xy=8xy + 4 + \dfrac{4}{xy} = 8. This may be rewritten as 0=(xy)2−4xy+4=(xy−2)20 = (xy)^2 - 4xy + 4 = (xy-2)^2. Hence xy=2xy = 2. The answer is (C).

University of Alberta