Why $(a+d)(b+c)$ equals $-16$?

If $a,b,c$ and $d$ satisfy the equations:$$a+7b+3c+5d=0$$$$8a+4b+6c+2d=-16$$$$2a+6b+4c+8d=16$$$$5a+3b+7c+d=-16$$ then $(a+d)(b+c)$ equals $-16$.

I can't understand why $(a+d)(b+c)$ equals $-16$?

Solution 1:

Summing the second and third equations we get $a+b+c+d=0$. Summing the first and fourth, $6(a+d)+10(b+c)=-16$. This is a system in two variables .

Solution 2:

Subtracting equations (1 from 3) and (4 from 2) gives $$a-b+c+3d=16$$ $$3a+b-c+d=0$$ Adding these two gives $$4a+4d=16$$ $$a+d=4$$ Adding the second and third equations, $$10(a+b+c+d)=0$$ so $$(b+c)=-(a+d)=-4$$ Therefore $$(a+d)(b+c)=(4)(-4)=-16$$