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

Jan 29, 2018

Given $\frac{\frac{2}{{x}^{2} - 1} - \frac{1}{x + 1}}{\frac{1}{12 {x}^{2} - 3}}$

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

$\left(\frac{2}{{x}^{2} - 1} - \frac{1}{x + 1}\right) \frac{12 {x}^{2} - 3}{1}$

Use the distributive property:

$\frac{2 \left(12 {x}^{2} - 3\right)}{{x}^{2} - 1} - \frac{1 \left(12 {x}^{2} - 3\right)}{x + 1}$

Twice:

$\frac{24 {x}^{2} - 6}{{x}^{2} - 1} + \frac{- 12 {x}^{2} + 3}{x + 1}$

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

$\frac{24 {x}^{2} - 24 + 24 - 6}{{x}^{2} - 1} + \frac{- 12 {x}^{2} - 12 x + 12 x + 3}{x + 1}$

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:

$- 12 x + 24 + \frac{18}{{x}^{2} - 1} + \frac{12 x + 3}{x + 1}$

Add +12-12 to the second fraction:

$- 12 x + 24 + \frac{18}{{x}^{2} - 1} + \frac{12 x + 12 - 12 + 3}{x + 1}$

Separate the second fraction into two fractions:

$- 12 x + 24 + \frac{18}{{x}^{2} - 1} + \frac{12 x + 12}{x + 1} + \frac{- 9}{x + 1}$

One of the fractions reduces:

$- 12 x + 24 + \frac{18}{{x}^{2} - 1} + 12 + \frac{- 9}{x + 1}$

Combine like terms:

$- 12 x + 36 + \frac{18}{{x}^{2} - 1} + \frac{- 9}{x + 1}$

The fraction $\frac{18}{{x}^{2} - 1}$ decomposes:

$\frac{18}{{x}^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}$

$18 = A \left(x + 1\right) + B \left(x - 1\right)$

Let x = 1:

$18 = A \left(1 + 1\right) + B \left(1 - 1\right)$

$A = 9$

Let x = -1:

$B = - 9$

$\frac{18}{{x}^{2} - 1} = \frac{9}{x - 1} + \frac{- 9}{x + 1}$

Substitute back into the expression:

$- 12 x + 36 + \frac{9}{x - 1} + \frac{- 9}{x + 1} + \frac{- 9}{x + 1}$

Combine like terms:

$- 12 x + 36 + \frac{9}{x - 1} - \frac{18}{x + 1}$

Done.
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