Solving 4 Quadratic Simultaneous Equations

Take $\#1 - \#2 + \#3 - \#4$ to get $(b-d)(a-c)=-6$

$\#1 + \#2 - \#3 - \#4$ gives $(b-d)(a+c-2) = 0$

$\#1 - \#2 - \#3 + \#4$ gives $(b+d-2)(a-c)=2$.

So now $a+c-2 = 0$, and $$\frac{1}{a-c} = \frac{b-d}{-6} = \frac{b+d-2}{2}$$

That gives us $d=3-2b$ and $a-c = \dfrac{-2}{b-1}$ (in particular, $b \ne 1$).
Together with $a+c-2=0$ we get $a = \dfrac{b-2}{b-1}$, $c = \dfrac{b}{b-1}$. Finally, substitute into $\#1$ to get $-b+3=3$, or $b=0$, and thus $a=2$, $c=0$, $d=3$.