This is a linear congruence equation, which can of course be solved using the method in §2. Here, we transform it into a system of linear congruence equations with smaller moduli, which is sometimes more convenient. Since $1155 = 3 \cdot 5 \cdot 7 \cdot 11$, by the property [X] in Chapter 3 §1, this congruence equation has the same solution as the system of congruence equations

Using the equivalent transformation I in §1, this system of congruence equations becomes
$$\begin{aligned} x \equiv 1(\bmod 3), & & -x \equiv 1(\bmod 5) \\ -2 x \equiv 3(\bmod 7), & & -3 x \equiv 6(\bmod 11) \end{aligned}$$

Furthermore, using the equivalent transformations III and I of 11 (i.e., solving the second, third, and fourth equations in the above system of congruence equations), the system of congruence equations becomes
$$\begin{array}{ll} x \equiv 1(\bmod 3), & x \equiv -1(\bmod 5) \\ x \equiv 2(\bmod 7), & x \equiv -2(\bmod 11) \end{array}$$

This system of congruence equations can be solved using the method in Theorem 1. In fact, this is the system of congruence equations in our Example 1, and its solution is
$$x \equiv 394(\bmod 1155)$$

This is the solution to the original congruence equation.
$$\begin{array}{l} 19 x \equiv 556(\bmod 3), \quad 19 x \equiv 556(\bmod 5), \\ 19 x \equiv 556(\bmod 7), \quad 19 x \equiv 556(\bmod 11) \end{array}$$