# 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 $-16$?

• Why don't you create a 4x5 augmented matrix and row reduce it as far as you can. – user60887 Mar 25 '14 at 3:21
• @user60887, can't we solve this using elementary algebra tricks? I found out this question from a high-school book. – Silent Mar 25 '14 at 3:22
• Just solve. Summing the second and third equations, we get $a+b+c+d=0$. It gets easier thereafter. – chubakueno Mar 25 '14 at 3:23
• Yes of course just like the answer below. – user60887 Mar 25 '14 at 3:23

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 .

• You gave me the general trick! I can apply it on n equations with n unknowns! – Silent Mar 25 '14 at 3:31
• @Sush actually, this excercise seems to be designed to apply this trick. In general, systems are way nastier :) . There are general methods however, like Gaussian Elimination or determinants and matrices. – chubakueno Mar 25 '14 at 3:43

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 $$(a+d)(b+c)=(4)(-4)=-16$$

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 $4a+4d=16$.

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 ?

• Yes, really it is very simple. – Silent Mar 25 '14 at 8:29
• @Sush. Yes, it is ! For sure, if you know matrices, it is simpler but, otherwise, with few linear equations, this kind of elemination is easy. – Claude Leibovici Mar 25 '14 at 8:31