<s>[INST] How to solve a system of non linear equations of the form $2ax-2x(ax^2+by^2) = 0$ Solve this system for $x,y$: 
$$2ax-2x(ax^2+by^2) = 0\tag{1}$$
$$2by-2y(ax^2+by^2) = 0\tag{2}$$
where $a$ and $b$ are constants such that $a>b>0$.
I re-arranged (1) to $x = \sqrt{\frac{a-by^2}{a}}$ and tried substituting it into (2)  but I did not know where to go from there. I'm looking for real solutions.
 [/INST] If $x = 0$, the first equation is ok, and:
$$
2by-2y(ax^2+by^2) = 0 \\
2by-2y(by^2) = 0 \\
by-by^3 = 0 \\
y-y^3 = 0 \\
y(1-y^2) = 0
$$
Thus, either $y=0$ or $y=1$ or $y=-1$. Here we have three solutions.
If $x \ne 0$:
$$
2ax-2x(ax^2+by^2) = 0 \\
ax = x(ax^2+by^2) \\
a = ax^2+by^2
$$
Replacing in the second equation, we have:
$$
2by-2y(ax^2+by^2) = 0 \\
by - y(ax^2+by^2) = 0\\
by - ya = 0 \\
y(b-a) = 0
$$
As $a > b$, the only solution is $y=0$. So:
$$
a = ax^2+by^2 \\
a = ax^2 \\
1 = x^2 \\
$$
Thus, either $x=1$ or $x=-1$.
Therefore, the solutions $(x, y)$ are: $\{ (0, 0), (0, 1), (0, -1), (1, 0), (-1, 0) \}$.
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