gcd(m,n) and modular arithmetic

"If gcd(m,n) = 1, show that for any pair of integers a and b there exists an integer x such that x ≡ a mod m and x ≡ b mod n."

I'm having trouble interpreting this problem. When I look at this, I think, for example:

m = 7
n = 11
a = 6
b = 8

and

x ≡ 6 mod 7 -> 6
x ≡ 8 mod 11 -> 8

I'm not sure what I'm doing wrong. If someone could show me an example of integers m, n, a, b where the problem works I think I can finish the problem. Any response is appreciated.

>>12

"Equivalence modulo m" (or more generally "equivalence modulo an ideal" in some ring) just means that the images of two numbers in the factor ring of Z by mZ under the natural reduction homomorphism are equal.  So if \phi_m : \mathbb{Z} \rightarrow \mathbb{Z} / m\mathbb{Z} is the natural reduction homomorphism, saying " a \equiv b \pmod m " is the same as saying \phi_m (a) = \phi_m (b), where equality is understood to be in the ring which is the target of \phi_m.

So basically, some way or another, you need to keep track of which ring you're working in (equivalently the modulus), and the "≡ (mod m)" notation is at least as convenient as any other that anyone's come up with, and at this point there doesn't seem to be any reason to change.