How do you simplify #(2/(x^2-1)-1/(x+1))/(1/(12x^2-3)#?

1 Answer
Jan 29, 2018

Given #(2/(x^2-1)-1/(x+1))/(1/(12x^2-3))#

Division by a fraction is the same as multiplication by its reciprocal:

#(2/(x^2-1)-1/(x+1))(12x^2-3)/1#

Use the distributive property:

#(2(12x^2-3))/(x^2-1)-(1(12x^2-3))/(x+1)#

Twice:

#(24x^2-6)/(x^2-1)+(-12x^2+3)/(x+1)#

Add #-24+24# to the first term and #-12x+12x# to the second term:

#(24x^2-24+24-6)/(x^2-1)+(-12x^2-12x+12x+3)/(x+1)#

Separate into 4 fractions:

#(24x^2-24)/(x^2-1)+(18)/(x^2-1)+(-12x^2-12x)/(x+1)+

(12x+3)/(x+1)#

Two of the fractions reduce:

#-12x+24+(18)/(x^2-1)+(12x+3)/(x+1)#

Add +12-12 to the second fraction:

#-12x+24+(18)/(x^2-1)+(12x+12-12+3)/(x+1)#

Separate the second fraction into two fractions:

#-12x+24+(18)/(x^2-1)+(12x+12)/(x+1)+(-9)/(x+1)#

One of the fractions reduces:

#-12x+24+(18)/(x^2-1)+12+(-9)/(x+1)#

Combine like terms:

#-12x+36+(18)/(x^2-1)+(-9)/(x+1)#

The fraction #(18)/(x^2-1)# decomposes:

#18/(x^2-1) = A/(x-1)+B/(x+1)#

#18 = A(x+1)+B(x-1)#

Let x = 1:

#18 = A(1+1)+B(1-1)#

#A = 9#

Let x = -1:

#B = -9#

#18/(x^2-1) = 9/(x-1)+(-9)/(x+1)#

Substitute back into the expression:

#-12x+36+9/(x-1)+(-9)/(x+1)+(-9)/(x+1)#

Combine like terms:

#-12x+36+9/(x-1)-18/(x+1)#

Done.