How do I equate coefficients? 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. 
 A: $$A+B=0\tag{1}$$
$$C-B=1\tag{2}$$
$$8A+4B-C+D=10\tag{3}$$
$$-4B+4C-D+E=3\tag{4}$$
$$16A-4C-E=36\tag{5}$$
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.
A: 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.
