How do I equate coefficients?

Question

I got the problem into this form:

$A+B=0$

$C-B=1$

$8A+4B-C+D=10 $

$-4B+4C-D+E=3$

$16A-4C-E=36$

I really don't know what to do after that. I suck at equating coefficients. It never makes sense to me. Please help and provide steps and a good explanation. Thank you.

Answer 1

$$A+B=0\tag{1}$$ $$C-B=1\tag{2}$$ $$8A+4B-C+D=10\tag{3}$$ $$-4B+4C-D+E=3\tag{4}$$ $$16A-4C-E=36\tag{5}$$

It is best to solve these types of problems by getting all of the variables in terms of one variable. In this case we choose $B$ because it looks like it will be easier.

From $(1)$, we get that $A=-B$.

From $(2),$ we get that $C=1+B$

Plugging this into $(3):$

$8(-B)+4B-(1+B)+D=10\implies-5B+D=11\implies D=11+5B$

Plugging this into $(4):$

$-4B+4(1+B)-(11+5B)+E=3\implies-5B+E=10\implies E=10+5B$

Plugging this into $(5):$

$16(-B)-4(1+B)-(10+5B)=36\implies-25B=50\implies B=-2$

Then we can calculate all the other variables which are in terms of $B$:

$A=2,\quad C=-1,\quad D=11-10=1,\quad E=10-10=0$

You can plug the variables back in to any of the equations to check that the answer is correct.

Answer 2

Adding equation (1) and (2),

$A+C=1$

$C=1-A$

Also from equation (1),

$B=-A$

From equation (3),

$8A+4(-A)-(1-A)+D=10$

$8A-4A-1+A+D=10$

$5A+D=10+1$

$5A+D=11$

$D=11-5A$

From equation (4),

$4(-B+C)-D+E=3$

$4(1)-D+E=3$

$E-D=-1$

$E=D-1$

$E=11-5A-1$

$E=10-5A$

Now each variable has value in terms of A. Hope now you can proceed.