Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, then $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=$ $3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$.
Also, $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$,
so $1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$, we can take $M_{1}^{-1}=1$.
By $M_{2} \equiv(-2) \cdot(2) \cdot(1) \equiv 1(\bmod 5)$,
so $1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$, we can take $M_{2}^{-1}=1$.
By $M_{3} \equiv 3 \times 5 \times 4 \equiv 4(\bmod 7)$,
so $1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$, we can take $M_{3}^{-1}=2$.
By $M_{4} \equiv 3 \times 5 \times 7 \equiv 4 \times 7 \equiv 6(\bmod 11)$,
so $1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$, we can take $M_{4}^{-1}=2$.
By the Chinese Remainder Theorem, the solution to the system of congruences is
$$x=(5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2+(3 \cdot 5 \cdot 7)$$
- $2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11)$,
which is $x \equiv 394(\bmod 1155)$.

Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, satisfying the conditions of Theorem 1. In this case, $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$. We will find $M_{j}^{-1}$. Since $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$, we have
$$1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$$

Therefore, we can take $M_{1}^{-1}=1$. From $M_{2} \equiv(-2) \cdot(2) \cdot 1 \equiv 1(\bmod 5)$, we have
$$1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$$

Therefore, we can take $M_{2}^{-1}=1$. From $M_{3} \equiv 3 \cdot 5 \cdot 4 \equiv 4(\bmod 7)$, we have
$$1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$$

Therefore, we can take $M_{3}^{-1}=2$. From $M_{4} \equiv 3 \cdot 5 \cdot 7 \equiv 4 \cdot 7 \equiv 6(\bmod 11)$, we have
$$1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$$

Therefore, we can take $M_{4}^{-1}=2$. Thus, by Theorem 1, the solution to the system of congruences is
$$\begin{aligned} x \equiv & (5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2 \\ & +(3 \cdot 5 \cdot 7) \cdot 2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11), \end{aligned}$$

i.e., $\square$
$$x \equiv 385-231+660-420 \equiv 394(\bmod 1155)$$

Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, then $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=$ $3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$

From $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$,
so $1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$, we can take $M_{1}^{-1}=1$.
From $M_{2} \equiv(-2) \cdot(2) \cdot(1) \equiv 1(\bmod 5)$,
so $1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$, we can take $M_{2}^{-1}=1$.
From $M_{3} \equiv 3 \times 5 \times 4 \equiv 4(\bmod 7)$,
so $1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$, we can take $M_{3}^{-1}=2$.
From $M_{4} \equiv 3 \times 5 \times 7 \equiv 4 \times 7 \equiv 6(\bmod 11)$,
so $1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$, we can take $M_{4}^{-1}=2$.
By the Chinese Remainder Theorem, the solution to the system of congruences is
$$x=(5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2+(3 \cdot 5 \cdot 7)$$
- $2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11)$

which is $x \equiv 394(\bmod 1155)$.