5. Solve the system of equations:

$$
\left\{\begin{array}{l}
x y-2 y=x+106 \\
y z+3 y=z+39 \\
z x+3 x=2 z+438
\end{array}\right.
$$
Solution. Note that $y \neq 1$ (since substituting into the first equation would then yield $x-2=x+106$). Then from the first two equations, we can express $x$ and $z$:

$$
\begin{aligned}
& x=\frac{106+2 y}{y-1} \\
& z=\frac{39-3 y}{y-1}
\end{aligned}
$$

Substitute into the last equation, multiply by $(y-1)^{2}$, and expand the brackets:

$$
\begin{aligned}
-432 y^{2}+864 y+3456 & =0 \\
-432(y-4)(y-2) & =0
\end{aligned}
$$

Thus, the solutions are $(38,4,9)$ and $(-34,-2,-15)$.