Verified answer

The difficulty here is that 12 is not prime, so the integers mod 12 are not a field. Solving systems of linear equations over arbitrary commutative rings is more of a pain because you can only divide by a coefficient when it is a unit (has a multiplicative inverse) and only numbers relatively prime to the modulus are units. Still you can try using elimination mod 12:

Subtract 4 times the second equation from the first. The x terms cancel, and you get (for the moment not reducing mod 12) -15y = -30, or equivalently 15y = 30. Reducing mod 12 gives 3y = 6, which has solutions mod 12: besides the obvious 2, there are also 6 and 10. (I'm assuming you mean solutions in the integers mod 12 -- if you mean integer solutions to the congruence, then all numbers of the form 2 + 4n for n an integer are solutions.)

Now plug these into the second equation and solve for x.