Question

Decompose 8y+7/y^2+y-2 into partial fractions.

Answer 1

Factoring the denominator, we have:
y^2 + y - 2
= (y + 2)(y - 1)

Then using partial fractions:
(8y + 7)/(y^2 + y - 2)
= (8y + 7)/[(y + 2)(y - 1)] = A/(y + 2) + B/(y - 1)

Multiply both sides by (y + 2)(y - 1) to obtain:
8y + 7 = A(y - 1) + B(y + 2)

Expand the right side and collect like terms:
8y + 7 = A(y - 1) + B(y + 2)
8y + 7 = Ay - A + By + 2B
8y + 7 = Ay + By - A + 2B
8y + 7 = (A + B)y + (-A + 2B)

Equate the coefficients, so you have the following system of equations:
A + B = 8
-A + 2B = 7

The solution is A = 3, B = 5, so we have:
(8y + 7)/(y^2 + y - 2)
= (8y + 7)/[(y + 2)(y - 1)]
= A/(y + 2) + B/(y - 1)
= 3/(y + 2) + 5/(y - 1)

I hope this helps!