Solve the following system of equations:

$$
\begin{aligned}
2 y+x-x^{2}-y^{2} & =0 \\
z-x+y-y(x+z) & =0 \\
-2 y+z-y^{2}-z^{2} & =0
\end{aligned}
$$
Solution. Let's rearrange the equations:

$$
\begin{aligned}
& \frac{5}{4}=\left(x-\frac{1}{2}\right)^{2}+(y-1)^{2} \\
& z-x=y(x+z-1) \\
& \frac{5}{4}=(y+1)^{2}+\left(z-\frac{1}{2}\right)^{2}
\end{aligned}
$$

From $\left(1^{\prime}\right)$ and $\left(3^{\prime}\right)$:

$$
\left(x-\frac{1}{2}\right)^{2}+(y-1)^{2}=(y+1)^{2}+\left(z-\frac{1}{2}\right)^{2}
$$

Rearranging:

$$
\left(x-\frac{1}{2}\right)^{2}-\left(z-\frac{1}{2}\right)^{2}=(y+1)^{2}-(y-1)^{2}, \quad(x+z-1)(x-z)=4 y
$$

From ( $\left.2^{\prime}\right)$, we get: $x-z=-y(x+z-1)$, substituting this in:

$$
-y(x+z-1)^{2}=4 y, \quad \text { i.e. } \quad 0=y\left(4+(x+z-1)^{2}\right)
$$

Since the second factor is positive, $y=0$, so from (2) we get $z-x=0$, i.e., $z=x$. From (1), $x^{2}-x=x(x-1)=0$, i.e., $x=0$ or $x=1$.

Thus, $x_{1}=0, y_{1}=0, z_{1}=0$, and $x_{2}=1, y_{2}=0, z_{2}=1$, and both are indeed solutions to the system of equations.