1. Find all real numbers $x, y$, and $z$ that solve the system of equations

$$
\begin{aligned}
x+y+2 z & =0 \\
x y-z^{2} & =0 \\
y^{2}+5 z+6 & =0
\end{aligned}
$$
I/1. From the first equation, we express $x=-2z-y$ and substitute it into the second. We get $(-2z-y)y-z^2=0$ or $-(z+y)^2=0$. From this, it follows that $z=-y$. We substitute into the third equation. We get $y^2-5y+6=0$ or $(y-2)(y-3)=0$. Thus, $y=2$ or $y=3$.

The system of equations is solved by $x=2, y=2, z=-2$ and $x=3, y=3, z=-3$.

Expressing one of the variables $\mathbf{z}$ in terms of the other two (for example, $x=-2z-y$) and determining that they cannot all be positive at the same time . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point

Conclusion $z=-y$ or $x=y$...........................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_466c4f14e439a88888e4g-06.jpg?height=54&width=1633&top_left_y=881&top_left_x=220)

Factorization $(y-2)(y-3)=0$ (can also be $(z+2)(z+3)=0$) ...................... 1 point

Each of the triples $x=2, y=2, z=-2$ or $x=3, y=3, z=-3 \ldots \ldots$ 1 point each