How do you express # (x + 2)/((2x+7)(x+1))# in partial fractions?

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Answer 1 · 2016-11-12T02:41:24

#A/(2x+ 7) + B/(x + 1) = (x+ 2)/((2x + 7)(x + 1))#

Put on a common denominator.

#(A(x + 1))/((2x+ 7)(x + 1)) + (B(2x + 7))/((2x + 7)(x + 1)) = (x+2)/((2x+ 7)(x + 1))#

#Ax + A + 2Bx + 7B = x+ 2#

#(A + 2B)x + (A + 7B) = x + 2#

So, #A + 2B = 1# and #A + 7B = 2#.

Solve.

#A = 1 - 2B -> 1 - 2B + 7B = 2#

#5B = 1#

#B = 1/5#

#:.A = 1 - 2(1/5) = 3/5#

Hence, the partial fraction decomposition is #3/(5(2x + 7)) + 1/(5(x+ 1)) = (x + 2)/((2x + 7)(x + 1))#.

Hopefully this helps!