instead of solving $A\vec{x} = \vec{b}$, you can solve the linear least squares formulation of the problem, $$ A^T A \vec{x} = A^t \vec{b}, $$ which is guaranteed to have the solution. In the geometric projection sense, that will be the closest possible answer to the true solution of the system.
Meaning to say, geometrically, the solutions of $A \vec{x} = \vec{b}$ are representations of $\vec{b}$ in the column space of $A$, which is not always possible. However, you can always orthogonally project $\vec{b}$ onto the column space of $A$ and the linear least squares problem will get you the corresponding solution.
A typical technique for doing this numerically is the singular-value decomposition.