Example 2 Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+\frac{9}{x}+\frac{4}{y}=10, \\
\left(x^{2}+9\right)\left(y^{2}+4\right)=24 x y .
\end{array}\right.
$$
(1993, Taiyuan Junior High School Mathematics Competition)
The original system of equations can be transformed into
$$
\left\{\begin{array}{l}
x+\frac{9}{x}+y+\frac{4}{y}=10, \\
\left(x+\frac{9}{x}\right)\left(y+\frac{4}{y}\right)=24 .
\end{array}\right.
$$

From (1), let \( x+\frac{9}{x}=5+t \) and \( y+\frac{4}{y}=5-t \). Then from (2), we get
$$
(5+t)(5-t)=24,
$$

which simplifies to \( t^{2}=1 \). Therefore, \( t= \pm 1 \).
$$
\therefore\left\{\begin{array}{l}
x+\frac{9}{x}=6, \\
y+\frac{4}{y}=4
\end{array}\right.
$$

or \(\left\{\begin{array}{l}x+\frac{9}{x}=4, \\ y+\frac{4}{y}=6 .\end{array}\right.\)
From (3), we get \( x=3, y=2 \). It is easy to see that (4) has no real solutions.
Upon verification, \( x=3, y=2 \) is a solution to the original system of equations.