3. (3 points) Solve the system of equations: $\left\{\begin{array}{l}x y=5(x+y) \\ x z=4(x+z) \\ y z=2(y+z)\end{array}\right.$
Answer: $x=y=z=0$ or $x=-40, y=\frac{40}{9}, y=\frac{40}{11}$

Solution: It is easy to verify that if one of the variables is 0, then the others are also 0. Otherwise, the system can be transformed into

$$
\left\{\begin{array}{l}
\frac{1}{5}=\frac{1}{x}+\frac{1}{y} \\
\frac{1}{4}=\frac{1}{x}+\frac{1}{2} \\
\frac{1}{2}=\frac{1}{z}+\frac{1}{y}
\end{array}\right.
$$

By adding the first two equations and subtracting the last one, we get $\frac{1}{5}+\frac{1}{4}-\frac{1}{2}=\frac{2}{x}$, from which $-\frac{1}{20}=\frac{2}{x}$, that is, $x=-40$. Similarly, we find the other unknowns.