2. Solve the system:

$$
\begin{aligned}
& \frac{x y}{x+y}=\frac{2}{3} \\
& \frac{y z}{y+z}=\frac{6}{5} \\
& \frac{z x}{z+x}=\frac{3}{4}
\end{aligned}
$$
2. The first equation is equivalent to $\frac{x+y}{x y}=\frac{3}{2}$, i.e., $\frac{1}{y}+\frac{1}{x}=\frac{3}{2}$. Similarly, the second and third equations are equivalent to $\frac{1}{z}+\frac{1}{y}=\frac{5}{6}$ and $\frac{1}{x}+\frac{1}{z}=\frac{4}{3}$. By substituting $a=\frac{1}{x}$, $b=\frac{1}{y}$, and $c=\frac{1}{z}$, we obtain the linear system $b+a=\frac{3}{2}, c+b=\frac{5}{6}$, and $a+c=\frac{4}{3}$. By adding all three equations, we get $2(a+b+c)=\frac{3}{2}+\frac{5}{6}+\frac{4}{3}=\frac{9+5+8}{6}=\frac{22}{6}$, i.e., $a+b+c=\frac{11}{6}$. From there, we calculate $c=\frac{11}{6}-\frac{3}{2}=\frac{1}{3}, a=\frac{11}{6}-\frac{5}{6}=1$, and $b=\frac{11}{6}-\frac{4}{3}=\frac{1}{2}$, so it follows that $x=1, y=2$, and $z=3$.