Note that, this can be brought down to the following square system since 1
is not a variable hence can be included at the left hand side. (Just multiply everything to see this.)
$\pmatrix{ 122.413307-122.366667 \\ -37.632195 + 37.61666667 } = \pmatrix{0.046640\\ -0.01552833}=\begin{bmatrix} 0.000046 & 0.000032\\ 0.000025 & -0.000036 \end{bmatrix}\pmatrix{ x \\ y} $ Now you are back at the $b=Az$ form which can be solved by multiplying both sides by the matrix $A^{-1}$ to obtain $z=A^{-1}b$. Again, this only makes sense if your matrix $A$ is invertible if not you have to apply further steps.
$ z = A^{-1}b = \pmatrix{ 14657.9804560261 & 13029.3159609121\\ 10179.1530944625 &-18729.6416938111 }\pmatrix{0.046640\\ -0.01552833} \approx \pmatrix{481.3 \\765.6} $
I did not pay attention to the significant digits (and it felt good!) hence you have to take care of that.