Solve for real numbers $x$ and $y$, simultaneously the equations given by $xy^2 = 15x^2 + 17xy + 15y^2$ and $x^2y = 20x^2 + 3y^2$.

Question

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.

Answer 1

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 $(x, y) = (19, 95)$.

Answer 2

We can linearize this system by introducing variables for the monomials. Then there are exactly four solutions $$ (x,y)= (0,0),(19,95),(-17i,17),(17i,17). $$