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This is a bit embarrassing, but I can't seem to solve for $x$ in $$2x=\frac{x}{y}-\frac{1}{1-y}.$$ Could someone please give me a hand!

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$x\left(2-\frac1{y}\right)=-\frac1{1-y}$. Do you see how to solve now? – J. M. Oct 27 '11 at 8:57
up vote 3 down vote accepted

Recall that $\frac xy = x\cdot\frac1y$, so $$ 2x = \frac xy-\frac{1}{1-y}\Leftrightarrow x\cdot\left(2-\frac1y\right) = -\frac1{1-y}\Leftrightarrow x = \frac{y}{(1-y)(1-2y)}. $$ Here you should be aware that $y\neq 0,y\neq 1$ (as it appears in the original equation). Moreover, $y\neq1/2$ which apper in the final answer does not reduce the space of solutions since the original equation does not have a solution in $x$ if $y=1/2$.

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

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

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

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

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