From Wikipedia: en.wikipedia.org/wiki/Chinese_remainder_theorem
Suppose
$x \equiv a_1 \pmod{n_1} \\ x \equiv a_2 \pmod{n_2}$
where $n_1$ and $n_2$ are coprime to each other.
How does this result in the following:
$n_2 [n_2^{-1}]_{n_1} + n_1 [n_1^{-1}]_{n_2} = 1$
where the notation $ [a^{-1}]_b$ denote the multiplicative inverse of $a \pmod{b}$