Problem 8. Solve the system

$$
\left\{\begin{aligned}
x+y+z & =13 \\
x^{2}+y^{2}+z^{2} & =61 \\
x y+x z & =2 y z
\end{aligned}\right.
$$
Answer. $(4 ; 3 ; 6),(4 ; 6 ; 3)$.

Solution. Let $y+z$ be denoted by $p$, and $yz$ by $q$. Then the system becomes

$$
\left\{\begin{aligned}
x+p & =13 \\
x^{2}+p^{2}-2 q & =61 \\
x p & =2 q
\end{aligned}\right.
$$

From the last two equations, $(x+p)^{2}-3 x p=61$, and considering the first equation, $x p=36$. Solving the system

$$
\left\{\begin{aligned}
x+p & =13 \\
x p & =36
\end{aligned}\right.
$$

we find that either $x=4, p=9$, or $x=9, p=4$. The second case is impossible, as the system

$$
\left\{\begin{aligned}
y+z & =4 \\
y z & =18
\end{aligned}\right.
$$

has no solutions, while the first case gives the above answer.