Question

If (x + y)/xy = 2 and (x-y)/xy = 1 , then value of x and y are​

Answer 1

$Given \: \frac{(x+y)}{xy} = 2 \\\implies \frac{x}{xy} + \frac{y}{xy} = 2$

$\implies \frac{1}{y} + \frac{1}{x} = 2 \: --(1)$

$and \: \frac{(x-y)}{xy} = 1 \\\implies \frac{x}{xy} - \frac{y}{xy} = 1$

$\implies \frac{1}{y} - \frac{1}{x} = 1 \: --(2)$

$Add \: equations\: (1)\:and \: (2) , we \:get$

$\frac{1}{y} +\cancel { \frac{1}{x} }+ \frac{1}{y} - \cancel {\frac{1}{x} }= 2 + 1$

$\implies \frac{2}{y} = 3$

$\implies \frac{2}{3} = y$

$\implies \blue{ y = \frac{2}{3} } \: --(3)$

$Subtract \: equation \:(2) \:from \: equation\:(1),\\we \:get$

$\frac{1}{y} +\frac{1}{x} + \Big(\frac{1}{y} - \frac{1}{x} \Big)= 2 - 1$

$\implies \frac{1}{y} +\frac{1}{x} - \Big(\frac{1}{y} + \frac{1}{x} \Big)= 1$

$\implies \frac{2}{x} = 1$

$\implies 2 = x$

$\blue { \implies x = 2 }$

Therefore.,

$\green { Value \: of \: x = 2 }$

$\green { Value \: of \: y = \frac{2}{3} }$

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