If $4 x + 6 y = 3 x y a n d 8 x + 9 y = 5 x y$ then

(a) x=2,y=3 (b) x=1,y=2 (c) x=3,y=4 (d) x=1,y=-1

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Solution

$4 x + 6 y = 3 x y a n d 8 x + 9 y = 5 x y Divide with xy \frac{6}{x} + \frac{4}{y} = 3 \frac{9}{x} + \frac{8}{y} = 5 P u t u = \frac{1}{x} a n d v = \frac{1}{y} 6 u + 4 v = 3 - - - (1) 9 u + 8 v = 5 - - - (2) (1) \times 2 12 u + 8 v = 6 - - - (3) (3) - (2) 3 u = 1 u = \frac{1}{3} \Rightarrow x = 3 6 u + 4 v = 3 4 v = 3 - 6 u 3 - 6 \times \frac{1}{3} = 3 - 2 = 1 4 v = 1 v = \frac{1}{4} \Rightarrow y = 4$

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