Example 1 Let $a, b, c$ be non-zero real numbers, and
$$
a b=2(a+b), b c=3(b+c), c a=4(a+c) \text {. }
$$

Then $a+b+c=$ $\qquad$
【Analysis】The number of unknowns is equal to the number of equations, so it can be solved directly.
Solution From the equations given in the problem, we have
$\left\{\begin{array}{l}\frac{1}{a}+\frac{1}{b}=\frac{1}{2}, \\ \frac{1}{b}+\frac{1}{c}=\frac{1}{3}, \\ \frac{1}{c}+\frac{1}{a}=\frac{1}{4}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{1}{a}=\frac{5}{24}, \\ \frac{1}{b}=\frac{7}{24} \\ \frac{1}{c}=\frac{1}{24}\end{array} \Rightarrow\left\{\begin{array}{l}a=\frac{24}{5}, \\ b=\frac{24}{7}, \\ c=24 .\end{array}\right.\right.\right.$
Therefore, $a+b+c=\frac{24}{5}+\frac{24}{7}+24=\frac{1128}{35}$.