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Question

Solve for x and y: $$\dfrac{a^2}{x}-\dfrac{b^2}{y}=0,\dfrac{a^2b}{x}+\dfrac{b^2a}{y}=a+b$$


Solution

$$ \dfrac{a^2}{x}-\dfrac{b^2}{y} = 0, \dfrac{a^2b}{x}+\dfrac{b^2 a}{y} = a+b $$
$$ \dfrac{a^2 y-b^2 x}{xy} = 0$$
$$ a^2y-b^2x = 0 ...(1)$$
$$ \dfrac{a^2 by +b^2 ax}{xy} = a+b $$
$$ a^2by+b^2ax = axy +bxy $$
$$ a^2by +b^2ax - axy -bxy = 0 ...(2)$$
So equation (1) and (2)
$$ a^2y-b^2x = 0 \times b $$
$$ a^2 by - b^3 x = 0$$
$$ a^2 by + b^2ax - axy - bxy = 0 $$
________________________
$$ -b^3 x-b^2ax+axy+bxy = 0$$
$$ -b^2x (b+a)+xy (a+b) = 0$$
$$ (a+b) (xy-b^2x) = 0$$
$$ a+b = 0 $$
$$ xy - b^2x = 0$$
$$ xy = b^2 x$$
$$ y = b^2 $$
equation (1)
$$ a^2 y -b^2 x = 0$$
$$ a^2b^2-b^2x = 0$$
$$ a^2 b^2 = b^2 x $$
$$ \boxed {x = a^2}$$

1140479_1139031_ans_ffd8c6a1057d43bda136140f26c60ec5.jpg

Mathematics

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