3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{6}{5} \text {, } \\ \frac{1}{y}+\frac{1}{x+z}=\frac{3}{4} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{2}{3}\end{array}\right.$,
Answer. $(1 ; 2 ; 3)$.

Solution. Multiplying both sides of the first equation by $\frac{5}{6} x(y+z)$, both sides of the second by $\frac{4}{3} y(x+z)$, and both sides of the third by $\frac{3}{2} z(x+y)$, we obtain the system

$$
\left\{\begin{array}{l}
\frac{5}{6}(x+y+z)=x y+x z \\
\frac{4}{3}(x+y+z)=x y+y z \\
\frac{3}{2}(x+y+z)=x z+y z
\end{array}\right.
$$

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

$$
x y+x z+y z=\frac{11}{6}(x+y+z)
$$

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

$$
\left\{\begin{array}{l}
\frac{1}{3}(x+y+z)=x y \\
(x+y+z)=y z \\
\frac{1}{2}(x+y+z)=x z
\end{array}\right.
$$

Dividing the first equation by the second (this is possible because from the domain of definition of the original system it follows that $x y z \neq 0$), we get that $x=\frac{1}{3} z$, and dividing the first by the third - that $y=\frac{2}{3} z$.

Then the second equation takes the form $2 z=\frac{2}{3} z^{2}$, from which $z=3, x=1, y=2$.