Problem 2. Solve the system of equations

$$
\left\{\begin{array}{l}
x(y+z)=5 \\
y(x+z)=10 \\
z(x+y)=13
\end{array}\right.
$$
Solution. First method. We introduce the substitution $x_{1}=x y, x_{2}=x z, x_{3}=y z$. Then the system takes the form

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=5 \\
x_{1}+x_{3}=10 \\
x_{2}+x_{3}=13
\end{array}\right.
$$

We express $x_{2}$ from the first equation and substitute it into the third,

$$
\left\{\begin{array}{l}
x_{2}=5-x_{1} \\
x_{1}+x_{3}=10 \\
5-x_{1}+x_{3}=13
\end{array}\right.
$$

then we have

$$
\left\{\begin{array}{l}
x_{2}=5-x_{1} \\
x_{1}+x_{3}=10 \\
-x_{1}+x_{3}=8
\end{array}\right.
$$

We express $x_{3}$ from the third equation and substitute it into the second.

$$
\left\{\begin{array}{l}
x_{2}=5-x_{1} \\
x_{1}+8+x_{1} \\
x_{3}=8+x_{1}
\end{array}=10 \Leftrightarrow\left\{\begin{array} { l } 
{ x _ { 2 } = 5 - x _ { 1 } } \\
{ 2 x _ { 1 } = 2 } \\
{ x _ { 3 } = 8 + x _ { 1 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x _ { 2 } = 4 } \\
{ x _ { 1 } = 1 } \\
{ x _ { 3 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x y=1 \\
x z=4 \\
y z=9
\end{array}\right.\right.\right.\right.
$$

If we divide the first two equations, we get $\frac{z}{y}=4$ or $z=4 y$.

If we substitute this into the third equation, we get $4 y^{2}=9$ or $y= \pm \frac{3}{2}$. Therefore, $z= \pm 6$ and $x= \pm \frac{2}{3}$.

Second method. If we expand the brackets in each equation of the system, we get

$$
\left\{\begin{array}{l}
x y+x z=5 \\
y x+y z=10 \\
z x+z y=13
\end{array}\right.
$$

If we add the three equations, we get $2(x y+y z+z x)=28$, i.e., $x y+y z+z x=14$. If we subtract the first equation from the last equation, then the second equation, and then the third equation, we get $y z=9, x z=4$, and $x y=1$.

If we divide the first two of the last three equations, we get $y=\frac{9}{4} x$, and substituting this into the third equation, we get $\frac{9}{4} x^{2}=1$, i.e., $x^{2}=\frac{4}{9}$. The numbers that satisfy the last equation are $x=\frac{2}{3}$ and $x=-\frac{2}{3}$. It is easy to get $y= \pm \frac{3}{2}$ and $z= \pm 6$.