39.2. Solve the system of equations in integers:

$$
\left\{\begin{array}{l}
x+y+z=6 \\
x+y z=7
\end{array}\right.
$$

$$
\text { (8-10 grades) }
$$
39.2. Expressing $x$ from the second equation and substituting into the first, we get:

$$
7-y z+y+z=6, \text { from which } y=\frac{z+1}{z-1}=1+\frac{2}{z-1}
$$

It is clear that $y$ will be an integer only if $z-1$ equals one of the numbers $-2,-1,1,2$, i.e., $z$ takes one of the values $-1,0,2,3$. Then for $y$ we will get the respective values $0,-1,3,2$. Since $x=7-y z$, we will have the values $7,7,1,1$ for $x$ respectively.

Thus, the system has the following 4 solutions in integers:

$$
\begin{aligned}
& x_{1}=7, \quad y_{1}=0, \quad z_{1}=-1 ; \\
& x_{2}=7, \quad y_{2}=-1, \quad z_{2}=0 \text {; } \\
& x_{3}=1, \quad y_{3}=3, \quad z_{3}=2 \\
& x_{4}=1, \quad y_{4}=2, \quad z_{4}=3
\end{aligned}
$$