Solve the following system of equations:

$$
4 x^{2}-3 y=x y^{3}
$$

$$
x^{2}+x^{3} y^{2}=2 y
$$
Solve the following system of equations:

$$
\begin{aligned}
4 x^{2}-3 y & =x y^{3} \\
x^{2}+x^{3} y^{2} & =2 y
\end{aligned}
$$

Solution. The sum of the two equations can be factored after rearrangement.

Indeed, $5 x^{2}-3 y+x^{3} y^{2}=x y^{3}+2 y$, from which

$$
5 x^{2}-5 y+x^{3} y^{2}-x y^{3}=\left(x^{2}-y\right)\left(5+x y^{2}\right)=0
$$

Since the system of equations (1) and (3) has the same roots as the original system of equations, we will examine this in the following.

From (3), we get either $x^{2}=y$ or $x y^{2}=-5$.

In the first case, substituting $x^{2}=y$ into (1) gives $x^{2}=x^{7}$, or $x^{2}\left(x^{5}-1\right)=0$. Thus, either $x=0$ and in this case $y=0$, or $x=1$ and in this case $y=1$.

In the second case, $y \neq 0$, so $x=\frac{-5}{y^{2}}$. Substituting this into (1) gives

$$
\begin{gathered}
\frac{100}{y^{4}}-3 y=-5 y, \quad \text { from which } \\
-50=y^{5}, \quad \text { or } \\
y=\sqrt[5]{-50}
\end{gathered}
$$

From here, using $x=\frac{-5}{y^{2}}$, we get $x=\frac{-5}{\sqrt[5]{50^{2}}}=-\sqrt[5]{\frac{5}{4}}$. Therefore, the system of equations (1) and (3) has three solutions: $x_{1}=y_{1}=0 ; x_{2}=y_{2}=1$ and $x_{3}=-\sqrt[5]{\frac{5}{4}} ; y_{3}=-\sqrt[5]{50}$. According to our note, these are also solutions to the original system of equations.