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.