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

“MIPT-2015”, 11th grade, solutions for ticket 5

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

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

$$
xy+xz+yz=-6.5(x+y+z).
$$

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

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

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

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