Let's say we have
$${\bf A} = {\bf B} {\bf X}$$
where $\bf A$ and $\bf B$ are known matrices, $\bf X$ is unknown. In case $\bf B$ was square, a solution can be found by ${\bf B}^{-1} {\bf A} = {\bf X}$. But how do you attempt to solve for $\bf X$ when $\bf B$ is not square, i.e., $n \neq m$?
Let $\mathbf{a}_1,\ldots,\mathbf{a}_k$ be the columns of $\mathbf{A}$ (so $\mathbf{A}$ is $n\times k$ for some $k$). Notice that we will have $\mathbf{B}$ is $n\times m$ and $\mathbf{X}$ is $m\times k$ for some integer $m$ (for $\mathbf{BX}=\mathbf{A}$ to work out). Let $\mathbf{x}_1,\ldots,\mathbf{x}_k$ be the columns of $\mathbf{X}$.
Notice that $\mathbf{a}_i$ depends only on $\mathbf{B}$ and $\mathbf{x}_i$, since $\mathbf{Bx}_i = \mathbf{a}_i$. So to determine the $i$th column of $\mathbf{X}$, it suffices to solve the system of linear equations $$\mathbf{B}\mathbf{x}_i = \mathbf{a}_i.$$ So finding $\mathbf{X}$ is equivalent to solving $k$ systems of linear equations.
In fact, you can just do them all at the same time. Simply take a matrix that is made up of $\mathbf{B}$ followed by $\mathbf{A}$: $$\left(\mathbf{B}|\mathbf{A}\right)$$ and use Gauss-Jordan elimination on $\mathbf{B}$. The solutions you find for each column corresponding to $\mathbf{A}$ yield the columns of $\mathbf{X}$.
In addition to Arturo's answer, maybe you could take a look at mine to this question.
The answer is given by the Penrose pseudoinverse:
$z=B^{pseudo}A $. While an exact solution exists only in the case $B^{pseudo}=B^{-1}$ the solution always guarantees that $||A-Bz||=minimum$
