jmc

Because $B+A=A$, $B$ must be $0$. Next, because $B-A=A⟹0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case. Now we can rewrite as . Therefore, $D=5+C$ Finally, $A-1-C=0⟹A=C+1⟹C=4$, So we have $D=5+C⟹D=5+4=\boxed{9}$.