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$$x+2^x=y+2^y$$

$$x^2+xy+y^2=12$$ I'm having trouble solving this problem, please do not solve the entire problem, I just want a hint. I don't have any good idea.

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    $\begingroup$ As a first guess, try $x=y$. It is easy to see that $x=y=\pm 2$ is a solution. $\endgroup$ – Paul May 9 '16 at 13:41
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The function $f(t)=t+2^t$ is strictly increasing in $\mathbb{R}$.
So from $x+2^x=y+2^y$ it directly follows that $x=y$.
And once you find that, it's easy to proceed.

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To add to the answer by peter petrov, $f(t)=t+2^2$ is strictly increasing in $\mathbb R$. This can be checked by considering $x+2^x = y + 2^y$. Rearrange to get $$x-y = 2^y - 2^x$$ Now consider $x>y$. This means that the LHS is positive and the RHS is negative, hence equality is impossible. Similar results for $x<y$. Hence, $x = y$.
It then follows that the result has to be $x=y=\pm2.$

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It's very simple in fact. If you notice that the implicit region defined in $\mathbb{R}^2$ by $$x + 2^x == y + 2^y$$ is a straight line passing on $x=y$, and the other curve, $$x^2+x y+y^2=12$$ is a rotated ellipse, with center in $0$, you're home.

Basically, the solutions mentioned by @Paul, $\pm(2,2)$ are, therefore, the only solutions for your system in $\mathbb{R}^2$, since any straight line just intercepts a ellipse in 2 points (if not tangent!)

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  • $\begingroup$ "Please do not solve the entire problem, I just want a hint." $\endgroup$ – Théophile May 9 '16 at 14:46
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    $\begingroup$ This is not solving it! It's insightful set of observations that can be done to solve this problem. Would you have given full marks to my answer in a exam? $\endgroup$ – Guilherme Thompson May 9 '16 at 14:54

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