Solve the system of equations $x+2^x=y+2^y$ and $x^2+xy+y^2=12$ $$x+2^x=y+2^y$$
$$x^2+xy+y^2=12$$
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.
 A: 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.      
A: 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 $x=y=\pm2.$
A: 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!)
