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$f:\mathbb R\to \mathbb R$ monotonically increasing.

Solve the system:

$$\begin{cases} f(x) + x = f(y) + y\\ x^2 + xy + y^2 = 12\end{cases}$$

Since $f$ is monotonically increasing and $f:\mathbb R\to \mathbb R$, isn't $f$ one-to-one?

But, I can't get it past that..

Can I replace $f(x)$ with $y$ so as to reduce the unknowns?

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Since $f(x)+x$ is injective, as you already figured out, you can conclude that $x=y$. This makes the second equation trivial to solve.

You may not replace $f(x)$ with $y$, $y$ is a different, unrelated variable.

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  • $\begingroup$ If I take x = y for granted, and continue with f(x) = f(y) f(x) + x = f(y) + y which is true based on the hypothesis, is it an acceptable solution? $\endgroup$
    – AQUATH
    Aug 2 '15 at 10:03
  • $\begingroup$ You already used the first equation for figuring out that x=y is the only option. You basically said $f(x)+x=f(y)+y \rightarrow f(x)=f(y)\rightarrow x=y $. $\endgroup$
    – orion
    Aug 2 '15 at 12:20

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