$\displaystyle

\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 0 & 14\;\\\;0 & 0 & 1 & -13\end{vmatrix}

$

If the variables in the equations of this matrix were x, y, and z:

x = 14

z = -13

I'm just not sure what to answer for y. Does having no y value in the augmented matrix mean the system has infinitely many solutions or a unique solution?