Proof involving modulus and CRT Let m,n be natural numbers where gcd(m,n) = 1. Suppose x is an integer which satisfies
x ≡ m (mod n)
x ≡ n (mod m)
Prove that x ≡ m+n (mod mn).
I know that since gcd(m,n)=1 means they are relatively prime so then given x, gcd(x,n)=m and gcd(x,m)=n. I have trouble getting to the next steps in proving x ≡ m+n (mod mn). Which is gcd(x, mn)=m+n

From the Chinese Remainder Theorem (uniqueness part) you know that, since $m$ and $n$ are relatively prime, the system of congruences has a unique solution modulo $mn$. 
So we only need to check that $m+n$ works. For that, we only need to verify that $m+n\equiv n\pmod{m}$, and that $m+n\equiv m\pmod{n}$.  That is very easy! 
