Example 4 Solve the system of congruences
$$x \equiv 3(\bmod 8), \quad x \equiv 11(\bmod 20), \quad x \equiv 1(\bmod 15)$$
Here $m_{1}=8, m_{2}=20, m_{3}=15$ are not pairwise coprime, so we cannot directly apply Theorem 1. It is easy to see that the solution to this system of congruences is the same as the solution to the system of congruences
$$\left\{\begin{array}{l}
x \equiv 3(\bmod 8), \\
x \equiv 11(\bmod 4), \\
x \equiv 11(\bmod 5), \\
x \equiv 1(\bmod 5), \\
x \equiv 1(\bmod 3)
\end{array}\right.$$

It is clear that any $x$ satisfying the first equation must also satisfy the second equation, and the third and fourth equations are the same. Therefore, the original system of congruences has the same solution as the system of congruences
$$\left\{\begin{array}{l}
x \equiv 3(\bmod 8) \\
x \equiv 1(\bmod 5) \\
x \equiv 1(\bmod 3)
\end{array}\right.$$

The system of congruences (20) satisfies the conditions of Theorem 1. It is easy to solve (left to the reader) that the solution to the system of congruences (20) is
$$x \equiv-29(\bmod 120) \text {. }$$

Noting that $[8,20,15]=120$, this is also the solution to the original system of congruences, and the number of solutions is 1.