Example 2. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \\ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} .\end{array}\right.$
First, find a common denominator for the left side of each equation, then take the reciprocal of both sides of each equation, and let $x+y+z=k$. The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
x y + x z = 2 k, \\
y z + y x = 3 k, \\
z x + z y = 4.6.
\end{array}\right.
$$
(1) + (2) + (3) gives $x y + y z + z x = \frac{9}{2} k$.
Subtracting (1), (2), and (3) from (4) respectively, we get
$$
\left\{\begin{array}{l}
x y = \frac{1}{2} k, \\
y z = \frac{5}{2} k, \\
z x = \frac{3}{2} k.
\end{array}\right.
$$
Multiplying (5) $\times$ (6) $\times$ (7) gives $x y z = \frac{k \sqrt{30 k}}{4}$.

Dividing (8) by (5), (6), and (7) respectively, we get
$$
\left\{\begin{array}{l}
x = \frac{\sqrt{30 k}}{10}, \\
y = \frac{\sqrt{30 k}}{6}, \\
z = \frac{\sqrt{30 k}}{2}.
\end{array}\right.
$$
Substituting (9), (10), and (11) into $x + y + z = k$ yields $k = \frac{529}{30}$. Substituting back into (9), (10), and (11) gives the solution to the original system of equations as
$$
x = \frac{23}{10}, y = \frac{23}{6}, z = \frac{23}{2}.
$$

Solving the system of equations using the overall elimination method, the solution process is natural, smooth, and concise.