Hi
So
-3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(-2A -2A -4 + 10) -A-2 = 0
-3A + -12A -12 +30 -A - 2 = 0
-16A +16 = 0
-16A = -16
Answer:
A = 1
Problem
I did all the steps correctly until this part
A+C = -2
2A+ B -2C = 10
-3A +3B + C = 0
Attempt:
I said C = -A-2
B = -2A +2C + 10
So
-3A + 3(2A + 2(-A-2) + 10) + (-A-2) = 0
-3A + 3(2A -2A -4 + 10) -A-2 = 0
-3A + -12 +30 -A - 2 = 0
-4A +16 = 0
-4A = -16
Answer:
A = 4
Correct Answer
A= 1, B= 2, C= -3
*This is related to partial fractions in Calculus.