Let $m_1, m_2, ..., m_r \in \Z$ be integers greater than 1 that are pairwise co-prime. Let $a_1, a_2, ..., a_r \in \Z$, then the system,

$x \equiv a_1 \pmod{m_1}$
$x \equiv a_2 \pmod{m_2}$
...
$x \equiv a_r \pmod{m_r}$

admits a unique solution $\pmod{m_1\cdot m_2 \cdot ... \cdot m_r}$

Example

Consider the following linear congruences:

$x \equiv 2 \pmod{3}$
$x \equiv 3 \pmod{5}$
$x \equiv 5 \pmod{7}$

Therefore, $m = m_1 \cdot m_2 \cdot m_3 = 105$.

For $b_1: \frac{m}{m_1} \cdot b_1 \equiv 1 \pmod{m_1}$
$35 \cdot b_1 \equiv 1 \pmod{3} = 2\cdot b_1 \equiv 1 \pmod{3}$
$b_1 \equiv 2 \pmod{3}$

For $b_2: \frac{m}{m_2} \cdot b_2 \equiv 1 \pmod{m_2}$
$21 \cdot b_2 \equiv 1 \pmod{5} = b_2 \equiv 1 \pmod{5}$

For $b_3: \frac{m}{m_3} \cdot b_3 \equiv 1 \pmod{m_3}$
$15 \cdot b_3 \equiv 1 \pmod{7} = b_3 \equiv 1 \pmod{7}$

$x = \frac{m}{m_1} \cdot b_1 \cdot a_1 + \frac{m}{m_2} \cdot b_2 \cdot a_2 + \frac{m}{m_3} \cdot b_3 \cdot a_3$
$x = 140 + 63 + 75 = 278$
$x = 68 \pmod{105}$