[RESOLVED] solving simultaneous linear congruences

[TBeck]

I know how to do basic ones like this:
x = 2 (mod 9)
x = 3 (mod 7)

but when there are coefficients infront of the xs I'm not sure how to go about solving it, for example something like:

22x = 2 (mod 30)
5x = 7 (mod 18)

[TBeck]

I got it solved...
5x = 7 (mod 18) can be written as 5x = 25 (mod 18), and since the greatest common divisor of 5 and 18 is one we can divide both sides by 5. Leaving us with x = 5 (mod 18), and so the equation x = 5 + 18y can be substituted into the other congruence and we can solve it from there.