~~Since 4 and 3 are coprime, you can obtain every integer as 4a-3b for some a and b, and thus the image is isomorphic to Z/(9) x Z~~.

In general, for each factor you get the quotient of Z by the ideal generated by the gcd of the coefficients in your expression.

**EDIT** Sorry for the confusion, wrote too quickly, hope this clarifies better:

Imagine that you are working on R^3 (real vectors). If you take *any* nonzero vector v and quotient our R^3/(v) you always get something that is isomorphic to R^2, right?
Well, sort of the same thing is true for Z, but now you need to care about gcd's; take your vector v=(9,12); you cannot extend it to a basis of Z^2 because it has a nontrivial gcd, so write it as 3(3,4). Now, take a vector extending (3,4) to a basis of Z^2, for instance ~~(0,1)~~ (1,1). Now, *every* element v in Z^2 can be written in a unique way as v = a(3,4) + b(1,1); if you quotient out by (3,4), you'd only have the 'b' term, getting a free part of rank one. but you have the 3 multiplying , so the image ox v under the quotient is (a mod 3, b), and thus you get as a quotient Z/(3) x Z. In general, if you take Z^n/(w) the result will be Z^(n-1) x Z/(gcd(w)).

For the case in which you take quotient by the submodule spanned by more than one vector, ref the answer by Armin and the reference to Smith form.