Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $-16$?

share|improve this question
1  
Why don't you create a 4x5 augmented matrix and row reduce it as far as you can. –  user60887 Mar 25 at 3:21
    
@user60887, can't we solve this using elementary algebra tricks? I found out this question from a high-school book. –  Sush Mar 25 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 at 3:23
    
Yes of course just like the answer below. –  user60887 Mar 25 at 3:23

4 Answers 4

up vote 7 down vote accepted

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 .

share|improve this answer
1  
You gave me the general trick! I can apply it on n equations with n unknowns! –  Sush Mar 25 at 3:31
1  
@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 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 $$(a+d)(b+c)=(4)(-4)=-16$$

share|improve this answer

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 $4a+4d=16$.

share|improve this answer

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 ?

share|improve this answer
    
Yes, really it is very simple. –  Sush Mar 25 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 at 8:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.