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 $-16$?
 A: 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 .
A: 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 $$(a+d)(b+c)=(4)(-4)=-16$$
A: Add equations $(2)$ and $(3)$ to get: $10a+10b+10c+10d=0$.
Add equations $(1)$ and $(4)$ to get: $6a+10b+10c+6d=-16$.
The first equation tells you $a+d=-(b+c)$, so that $(a+d)(b+c)=-(a+d)^2$.
To find $(a+d)$, observe that the coefficients of $b$ and $c$ in the above two equations are the same, so you may subtract them to obtain $4a+4d=16$.
A: Let us try using elementary algebra tricks.  
You have four linear equations for four unknowns (which is not much). Eliminate $a$ from the first equation and put its expression (which is a linear combination of $b,c,d$) in the other equations; from the second equation, extract $b$ (which is now a linear combination of $c,d$) and put its expression in the other equations; from the third equation, extract $c$ (which is now a linear combination of $d$) and put its expression in the last equation which is linear in $d$; solve it for $d$ and go backwards.  
For illustration purposes, let me take your system putting letters in the rhs's. So, we have $$a+7b+3c+5d=A$$ $$8a+4b+6c+2d=B$$ $$2a+6b+4c+8d=C$$ $$5a+3b+7c+d=D$$ Doing what was described above, we have successively $$a=A-7 b-3 c-5 d$$ $$b=\frac{1}{52} (8 A-B-18 c-38 d)$$ $$c=\frac{1}{10} (-10 A-2 B+13 C-50 d)$$ $$d=\frac{1}{16} (-3 A+4 C-D)$$ Replacing $A,B,C,D$ by their values, you obtain $d=5,c=-1,b=-3,a=-1$ and then wathever you want which could be expressed as any function of $a,b,c,d$.  
Is this what you are looking for as a very simple method ?
