Example 1 Solve the system of congruences $\left\{\begin{array}{l}x \equiv 1(\bmod 7), \\ x \equiv 1(\bmod 8), \\ x \equiv 3(\bmod 9) .\end{array}\right.$
Since $7, 8, 9$ are pairwise coprime, the system of congruences has a solution.
$$
\begin{array}{l}
M=7 \cdot 8 \cdot 9=504, M_{1}=\frac{504}{7}=72, M_{2}=\frac{504}{8}=63, M_{3}=\frac{504}{9}=56 . \text { Thus } \\
M_{1}^{\prime} \equiv \frac{1}{72} \equiv \frac{1}{7 \cdot 10+2} \equiv \frac{1}{2} \equiv \frac{4}{8} \equiv \frac{4}{7+1} \equiv 4(\bmod 7), \\
M_{2}^{\prime} \equiv \frac{1}{63} \equiv \frac{1}{64-1} \equiv \frac{1}{-1} \equiv-1 \equiv 7(\bmod 8), \\
M_{3}^{\prime} \equiv \frac{1}{56} \equiv \frac{1}{6 \cdot 9+2} \equiv \frac{1}{2} \equiv \frac{5}{10} \equiv \frac{5}{9+1} \equiv 5(\bmod 9) .
\end{array}
$$

Therefore, $x \equiv 4 \cdot 72 \cdot 1+7 \cdot 63 \cdot 1+5 \cdot 56 \cdot 3 \equiv 57(\bmod 504)$ is the solution.