Note: This isn't a direct answer to the question. I've covered all the possible cases I could. Don't read this if you want a straight answer to the given question.
The number of equations is $m$.
Hence, there can be a $m*m$ matrix.
Hence the maximum rank can be $m$.
Hence, $m$ is the upper bound of $r$.
We conclude that $r \leq m$ ALWAYS.
For homogeneous system of equations; if $m < n$, then $r < n$. Because $m$ is the upper bound of $r$. We'll have infinite solutions.
For non-homogeneous system of equations; maybe $r(A) \neq r(A|B)$. We'll have no solutions in this case.
So, all such systems have a solution? No. This statement is false.
$m > n$. The only thing we know is $r$ lies below $m$. It can be between $m$ and $n$. Or it can be lower than $n$.
- $m > r > n$ is possible.
- $m > n > r$ is possible.
For case 2, $r<n$. It's good.
For case 1, $r > n$. Which is inconsistent and we'll have no solutions in case of non-homogeneous equations.
We always want $r \leq n$ be it any system of equations.
So, this statement is false, because we have a favourable case.
When $m = n$
$r$ could be $= m$, and hence, $r = n$.
$r$ could be $< m$, and hence $r < n$.
So, yeah, there exists favourable cases. This statement is correct.
Suppose $r$ is such that $r(A) \neq r(A|B)$. Then, we'll have no solution. But we need just one favourable case to make this statement True, and we did it.