Let $A$ be an $n\times m$ matrix, $x$ is an $m\times 1$ vector, and $b$ is an $n\times 1$ vector.
The system of equations $Ax=b$ has a solution iff the row rank of $A$ equals the row rank of $A|b$ ($A$ augmented with $b$).
My attempt at a proof: Let $A'|b'$ represent the row reduced form of $A|b$. Suppose that $Ax=b$ has a solution. Then $A'|b'$ has a solution if and only if the nonzero rows in $A'$ are nonzero rows in $b'$. Since row reduction doesn't affect rank, we have that $rank(A)=rank(A|b)$.
I'm stuck on this problem working backwards.