Another way to write ##X=A/B \equiv AB^{-1}## is ##XB=A##. This has a unique solution X if B is not singular. You can solve for X in XB=A using Gaussian elimination.
Another way to write ##X=B \backslash A \equiv B^{-1}A## is ##BX=A##. This, too, has a a unique solution X if B is not singular. You can solve for X in BX=A using Gaussian elimination.
What if B is singular? The standard approach is to use the pseudo-inverse, and now you have but no choice to compute that inverse, typically via singular value decomposition.