<s>[INST] solve for real x and y: $x(x^2-3y^2)=2, y(3x^2-y^2)=11$ Solve for real x and y:
$x(x^2-3y^2)=2$
$y(3x^2-y^2)=11$
My attempt: I got $(x-y)^3=13$ but this doesn't always hold, I got a solution $(2,1)$.
How to proceed?
 [/INST] NOTE: You made a mistake in getting $(x-y)^3=13$. Once you add the two equations, the left side is not $(x-y)^3$.
Idea:
\begin{align*}
\frac{x(x^2-3y^2)}{y(3x^2-y^2)}&=\frac{2}{11}.
\end{align*}
Let $y=mx$, then we have
\begin{align*}
\frac{(1-3m^2)}{m(3-m^2)}&=\frac{2}{11}.
\end{align*}
This gives
$$2m^3-33m^2-6m+11=0 \implies (2m-1)(m^2-16m-11)=0.$$
One of the solutions for this is $m=1/2$, now others can be found as well and they are $8 \pm 5\sqrt{3}$.
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