3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=-\frac{2}{15} \\ \frac{1}{y}+\frac{1}{x+z}=-\frac{2}{3} \\ \frac{1}{z}+\frac{1}{x+y}=-\frac{1}{4}\end{array}\right.$,
Answer. $(5 ;-1 ;-2)$.

Solution. Multiplying both sides of the first equation by $-\frac{15}{2} x(y+z)$, both sides of the second by $-\frac{3}{2} y(x+z)$, and the third by $-4 z(x+y)$, we obtain the system

$$
\left\{\begin{array}{l}
-7.5(x+y+z)=x y+x z \\
-1.5(x+y+z)=x y+y z \\
-4(x+y+z)=x z+y z
\end{array}\right.
$$

Adding all three equations term by term and dividing the resulting equality by two, we get the equality

$$
x y+x z+y z=-6.5(x+y+z)
$$

Subtracting each of the equations from the last system from this equality, we find that

$$
\left\{\begin{array}{l}
-2.5(x+y+z)=x y \\
(x+y+z)=y z \\
-5(x+y+z)=x z
\end{array}\right.
$$

Dividing the first equation by the second (this is possible because from the domain of the original system it follows that $x y z \neq 0$), we get that $x=-2.5 z$, and dividing the first by the third - that $y=0.5 z$.

Then the second equation takes the form $-z=0.5 z^{2}$, from which $z=-2, x=5, y=-1$.