How to find $x,y$
\begin{align} 2^{x^2+y} + 2^{x + y^2} & = 128 \\[4pt] \sqrt{x} + \sqrt{y} & = 2\sqrt{2} \end{align}
I know $x=2$, $y=2$ but if they are only solutions?
2026-03-25 20:36:06.1774470966
How to find $x,y$ : $2^{x^2+y} + 2^{x + y^2} = 128$ , $\sqrt{x} + \sqrt{y} = 2\sqrt{2}$
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i found also $$x=2,y=2$$ you can eliminate $y$ and you will get the following equation $$2^{x^2+(2\sqrt{2}-\sqrt{x})^2}+2^{x+(2\sqrt{2}-\sqrt{x})^4}=128$$ for $x$ and we will find again $x=2$