## Task B-1.1.

Solve the system of equations

$$
\begin{gathered}
x(y+z)=3 \\
y(x+z)=-32 \\
z(x+y)=-5
\end{gathered}
$$
## First solution.

By adding the given equations, we get $2 x y + 2 x z + 2 y z = -34$, which simplifies to $x y + x z + y z = -17$.

Since $x y + x z = 3$, it follows that $y z = -20$.

Similarly, we obtain $x z = 15$ and $x y = -12$.

Multiplying these three equations, we get $x^{2} y^{2} z^{2} = 60^{2}$, which implies $x y z = 60$ or $x y z = -60$.

From $x y z = 60$ and $y z = -20$, it follows that $x = -3$,

from $x y z = 60$ and $x z = 15$, it follows that $y = 4$,

and from $x y z = 60$ and $x y = -12$, it follows that $z = -5$.

Thus, one solution is $(-3, 4, -5)$.

From $x y z = -60$ and $y z = -20$, it follows that $x = 3$,

from $x y z = -60$ and $x z = 15$, it follows that $y = -4$,

and from $x y z = -60$ and $x y = -12$, it follows that $z = 5$.

Thus, the second solution is $(3, -4, 5)$.