How do you simplify 2/(x-1)+3/(x+1)-(4x-2)/(x^2-1)?

May 18, 2017

$\frac{1}{x - 1}$; $x \ne \pm 1$

Explanation:

We have: $\frac{2}{x - 1} + \frac{3}{x + 1} - \frac{4 x - 2}{{x}^{2} - 1}$

Let's factorise the denominator of the third fraction:

$= \frac{2}{x - 1} + \frac{3}{x + 1} - \frac{4 x - 2}{\left(x + 1\right) \left(x - 1\right)}$

Then, let's collect the first two fractions:

$= \frac{2 \left(x + 1\right)}{\left(x - 1\right) \left(x + 1\right)} + \frac{3 \left(x - 1\right)}{\left(x + 1\right) \left(x - 1\right)} - \frac{4 x - 2}{\left(x + 1\right) \left(x - 1\right)}$

$= \frac{2 \left(x + 1\right) + 3 \left(x - 1\right)}{\left(x + 1\right) \left(x - 1\right)} - \frac{4 x - 2}{\left(x + 1\right) \left(x - 1\right)}$

$= \frac{2 x + 2 + 3 x - 3}{\left(x + 1\right) \left(x - 1\right)} - \frac{4 x - 2}{\left(x + 1\right) \left(x - 1\right)}$

$= \frac{5 x - 1}{\left(x + 1\right) \left(x - 1\right)} - \frac{4 x - 2}{\left(x + 1\right) \left(x - 1\right)}$

Now, let's collect these two fractions:

$= \frac{5 x - 1 - \left(4 x - 2\right)}{\left(x + 1\right) \left(x - 1\right)}$

$= \frac{5 x - 1 - 4 x + 2}{\left(x + 1\right) \left(x - 1\right)}$

$= \frac{x + 1}{\left(x + 1\right) \left(x - 1\right)}$

Finally, let's cancel $x - 1$ from the numerator and the denominator:

$= \frac{1}{x - 1}$; $x \ne \pm 1$

It is important to notice the denominators of the original fractions.

They must never be equal to zero, so that's why $x \ne \pm 1$.