I notice that the second equation involves a difference of squares, which can be factored as $(x+y)(x-y)$. This means I can substitute $x+y = 6$ into the second equation and get $6(x-y) = 12$. Dividing both sides by 6, I get $x-y = 2$. 
If $x+y = 6$ and $x^2-y^2 = 12$, then what is $x-y$?