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I have the following equation that I don't know how to solve: $$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(x-y)} \end{cases} $$ Here's what I've tried (but my mathematical instinct tells me that I didn't solve it right): $$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(x-y)} \end{cases} \rightarrow \begin{cases} x = \frac {1}{y} \\ \frac {1}{y}^{2x-y} = y^{2(x-y)} \end{cases} \rightarrow y^{y-2x} = y^{2(x-y)} \rightarrow y-2x = 2x-2y \rightarrow 4x=3y \rightarrow \frac{4}{y} = 3y \rightarrow 3y^2-4 = 0\rightarrow \begin{cases} y_1=\frac{-2\sqrt{3}}{3} \\ y_2= \frac{2\sqrt{3}}{3}\end{cases} \rightarrow \begin{cases} y_1=\frac{-2\sqrt{3}}{3} \\ y_2= \frac{2\sqrt{3}}{3}\end{cases} \rightarrow \begin{cases} x_1=\frac{-\sqrt{3}}{2} \\ x_2= \frac{\sqrt{3}}{2}\end{cases}$$

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Your attempted solution is correct. But you missed the trivial solution $(x,y)=(1,1)$.

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  • $\begingroup$ Thanks. I didn't see that. $\endgroup$
    – T4yl0r
    Jul 25, 2016 at 14:30
  • $\begingroup$ @T4yl0r You're welcome. $\endgroup$ Jul 25, 2016 at 14:31

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