You can always just solve it as you would without the modulo, then transfer your answers into modulo notation (e.g. 1/3 = 7 mod 10 because 7*3=1 mod 10).
Or, you can try and use the modulo at the same time.
For example, I have the following two equations. They have solutions x=11, y=12 (so mod 10 x=1 and y=2),
Add them, to get,
We now have the question `what is '? In this case, it is obviously 1. However, it won't always be so simple. For example, if 3x=8 mod 10, how would you proceed?
Well, 3*7=21=1 mod 10, so we have found the multiplicative inverse of 3 mod 10. Therefore, 3*(7*8)=8 mod 10. 7*8=56=6 mod 10. Therefore, 3*6=8 mod 10, and so x=6. This is the `standard' way of solving for such an x. You find the z such that az=1 mod n, and then you have a*(zm)=m mod n.
Anyway, I deviate. We have x=1 and the two equations,
Substituting into the second equation, we have y=3, as required.
Does that help?