How to solve following system of non-linear algebraic equations I have a system
$$
5x^2 - 5y^2 - 3x + 9y = 0,
$$
$$
5x^3 + 5y^3 - 13xy - 15x^2 - y^2 = 0.
$$
How to solve it?

First, you can observe that if $y=0$, then $x=0$, which gives out a solution.
Let us assume $y\neq 0$ now.
Take $\dfrac{x}{y}=\lambda$.
Then the first equation is:
$$(5\lambda^2-5)y^2+(9-3\lambda)y=0$$
which is
$$(5\lambda^2-5)y+(9-3\lambda)=0$$
So
$$y = \dfrac{3\lambda-9}{5\lambda^2-5}$$
The second one is:
$$(5\lambda^3+5)y^3-(15\lambda^2+13\lambda+1)y^2=0$$
which is
$$(5\lambda^3+5)y-(15\lambda^2+13\lambda+1)=0$$
Thus
$$y = \dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$,
According to these, we have
$$\dfrac{3\lambda-9}{5\lambda^2-5}=\dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$
which is equal to:
$$6\lambda^4+11\lambda^3-7\lambda^2-8\lambda+4=0$$
Factorize it:
$$(\lambda+1)(\lambda+2)(2\lambda-1)(3\lambda-2)=0$$
plug in  $y = \dfrac{3\lambda-9}{5\lambda^2-5}$.
And $\lambda\neq -1$.
