Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =5 \\
x^{2}+y^{2}+z^{2} & =9 \\
x y+u+v x+v y & =0 \\
y z+u+v y+v z & =0 \\
z x+u+v z+v x & =0
\end{aligned}
$$
$$
\begin{aligned}
x+y+z & =5 \\
x^{2}+y^{2}+z^{2} & =9 \\
x y+u+v x+v y & =0 \\
y z+u+v y+v z & =0 \\
z x+u+v z+v x & =0 .
\end{aligned}
$$

Solution. Subtract (4) from (3), and (5) from (3). After rearrangement, we get the relations $(y+v)(x-z)=0$ and $(x+v)(y-z)=0$. From these, we can conclude that at least two of $x, y, z$ are equal. If, for instance, $x \neq z$ and $y \neq z$, then from the last two equations we get $y+v=0$ and $x+v=0$, hence $x=y=-v$.

Let's assume, for example, that $x=y$. Then from (1) and (2) we have

$$
2 x+z=5, \quad 2 x^{2}+z^{2}=9
$$

Solving these, we get the solutions $x_{1}=y_{1}=2, z_{1}=1$ and $x_{2}=y_{2}=4 / 3, z_{2}=7 / 3$. From the relation $(y+v)(x-z)=0$, we get $v_{1}=-2, v_{2}=-4 / 3$, and the value of $u$ can be easily calculated from (3): $u_{1}=4, u_{2}=16 / 9$. Of course, instead of $x=y$, we can also have $x=z$ or $y=z$, which yields four more solutions.

In summary, the system of equations has six solutions:
$x_{1}=2$
$y_{1}=2$
$z_{1}=1$
$u_{1}=4$
$v_{1}=-2$
$x_{2}=4 / 3$
$y_{2}=4 / 3$
$z_{2}=7 / 3$
$u_{2}=16 / 9$
$v_{2}=-4 / 3$
$x_{3}=2$
$y_{3}=1$
$z_{3}=2$
$u_{3}=4$
$v_{3}=-2$
$x_{4}=4 / 3$
$y_{4}=7 / 3$
$z_{4}=4 / 3$
$u_{4}=16 / 9$
$v_{4}=-4 / 3$
$x_{5}=1$
$y_{5}=2$
$z_{5}=2$
$u_{5}=4$
$v_{5}=-2$
$x_{6}=7 / 3$
$y_{6}=4 / 3$
$z_{6}=4 / 3$
$u_{6}=16 / 9$
$v_{6}=-4 / 3$