Example 12 Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} .
\end{array}\right.
$$
(1996, Beijing Junior High School Mathematics Competition)
Solving: By taking the reciprocal and finding a common denominator, we get
$$
\left\{\begin{array}{l}
\frac{x(y+z)}{x+y+z}=2, \\
\frac{y(z+x)}{x+y+z}=3, \\
\frac{z(x+y)}{x+y+z}=4 .
\end{array}\right.
$$

Adding the three equations yields $\frac{x y+y z+z x}{x+y+z}=\frac{9}{2}$.
(4) Subtracting (1), (2), and (3) respectively, we get
$$
\left\{\begin{array}{l}
\frac{y z}{x+y+z}=\frac{5}{2}, \\
\frac{z x}{x+y+z}=\frac{3}{2}, \\
\frac{x y}{x+y+z}=\frac{1}{2} .
\end{array}\right.
$$
(5) ÷ (6) gives $\frac{y}{x}=\frac{5}{3}, y=\frac{5}{3} x$;
(6) ÷ (7) gives $\frac{z}{y}=3, z=3 y=5 x$.
Thus, $\frac{1}{x}+\frac{1}{\frac{5}{3} x+5 x}=\frac{1}{2}, \frac{1}{x}+\frac{3}{20 x}=\frac{1}{2}$.
Therefore, $x=\frac{23}{10}, y=\frac{23}{6}, z=\frac{23}{2}$ is the solution to the system of equations.