System of Equations $ 2x ^ 3 + x + 2y - y ^ 3 = 0$, $4x ^ 3 - 2xy + 3y ^ 2 = 10$

Question

I have found a system of equations which I cannot solve.

$$\begin {cases} 2x ^ 3 + x + 2y - y ^ 3 = 0 \\ 4 x ^ 2 - 2xy + 3y ^ 2 = 10 \\ \end {cases} $$

I noticed that the second equation can be written as $\frac{8x ^ 3 + y ^ 3}{2x + y} + 2y ^ 2 = 10$, so I tried to write the first equation similarly and ended up with $8x ^ 3 + y ^ 3 = 5y ^ 3 - 4x - 8y$, but, in the end, it did not really help.

Answer 1

Homogenization helps.

We obtain: $$2x^3+(x+2y)\cdot\frac{4x^2-2xy+3y^2}{10}-y^3=0$$ or $$(y-2x)(12x^2+9xy+4y^2)=0$$ or $$y=2x.$$

Can you end it now?