2. Solve the system of equations

$$
\begin{aligned}
x+y^{2} & =1 \\
x^{2}+y^{3} & =1
\end{aligned}
$$
II/2. Since $x=1-y^{2}$, it follows that $\left(1-y^{2}\right)^{2}+y^{3}=1$. Thus, $1-2 y^{2}+y^{4}+y^{3}=1$, which gives us $y^{2}\left(y^{2}+y-2\right)=y^{2}(y+2)(y-1)=0$. The solutions are $y=0, y=-2$ and $y=1$, with corresponding $x$ values $x=1, x=-3$ and $x=0$.

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-08.jpg?height=51&width=1627&top_left_y=1825&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_a8720c28cfa2c07912d2g-08.jpg?height=57&width=1631&top_left_y=1873&top_left_x=218)

Each pair of solutions $(x, y) \in\{(1,0),(-3,-2),(0,1)\} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ is worth 1 point