<s>[INST] Solve for real numbers $x$ and $y$, simultaneously the equations given by $xy^2 = 15x^2 + 17xy + 15y^2$ and $x^2y = 20x^2 + 3y^2$. 
Solve for real numbers $x$ and $y$, simultaneously the equations given by
$$
\left\{
\begin{array}{l}
xy^2 &= 15x^2 + 17xy + 15y^2 \\
x^2y &= 20x^2 + 3y^2
\end{array}
\right.
$$

by taking $y = xt$, I get $x=0$ and $y=0$. Is there some other way or some other solution? Thanks in advance.
 [/INST] Subbing $y = x t$ is a good idea. Since you've already found $x = y = 0$, let's assume $x \ne 0$. Doing the substitution and simplifying gives
$$
xt^2 = 15 t^2 + 17 t + 15 \\
x t = 20 + 3t^2.
$$
Multiply the bottom equation by $t$ and subtract to get
$$
0 = t(3t^2 + 20) - (15t^2 + 17 t + 15)= 3t^3 -15 t^2 + 3t - 15 = 3(t-5)(t^2+1)
$$
which has the real solution $t = 5$. From this, we find the solution $(x, y) = (19, 95)$.
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