Example 6 Solve the system of congruences
$$x \equiv 3(\bmod 7), \quad 6 x \equiv 10(\bmod 8) .$$
This is not in the form of the system of congruences in Theorem 1. It is easy to see that the second congruence has solutions and the number of solutions is 2 (the specific solution is left to the reader):
$$x \equiv -1, 3 \pmod{8}$$

Therefore, the solutions to the original system of congruences are the solutions to the following two systems of congruences:
$$x \equiv 3 \pmod{7}, \quad x \equiv -1 \pmod{8}$$

and
$$x \equiv 3 \pmod{7}, \quad x \equiv 3 \pmod{8}$$

It is easy to find (left to the reader) that the solution to the system of congruences (21) is $x \equiv 31 \pmod{56}$; the solution to the system of congruences (22) is $x \equiv 3 \pmod{56}$. Therefore, the number of solutions to the original system of congruences is 2, and the solutions are
$$x \equiv 3, 31 \pmod{56}$$