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

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

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

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

$$
x y+x z+y z=\frac{1}{4}(x+y+z)
$$

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

$$
\left\{\begin{array}{l}
\frac{3}{2}(x+y+z)=x y \\
-\frac{3}{4}(x+y+z)=y z \\
-\frac{1}{2}(x+y+z)=x z
\end{array}\right.
$$

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 $x=-2 z$, and dividing the first by the third - that $y=-3 z$.

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