Good work, Harry.
You can extend your method to the third equation, then substitute the values you found for (A+C) and (B+D).
AC(B+D)+BD(A+C)=3 becomes 4(AC) + BD=3 using your first values. Then you can solve that simultaneously with equation 4 to get values for AC and BD.
At that point there's enough information to evaluate A and C.
I don't have the time now for a full listing, but I've found these other values for A:
Posted by xdog
on 2016-08-11 18:20:34