Solving congruences using PARI
Pari's "chinese" function (as in Chinese Remainder Theorem) does the solution for you, for two congruences. You can fold in more congruences via induction.

? chinese(Mod(42,91), Mod(11, 37))
%1 = Mod(3045, 3367)


The previous example solves for $x \equiv 42 \pmod{91}$ and $x \equiv 11 \pmod{37}$, which is unique modulo $91 \cdot 37 = 3367$. You can use "raise" to promote a Mod object to the integers.