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

1 Answer
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)((x + 1)(x - 1))#

Then, let's collect the first two fractions:

#= frac(2(x + 1))((x - 1)(x + 1)) + frac(3(x - 1))((x + 1)(x - 1)) - frac(4 x - 2)((x + 1)(x - 1))#

#= frac(2(x + 1) + 3(x - 1))((x + 1)(x - 1)) - frac(4 x - 2)((x + 1)(x - 1))#

#= frac(2 x + 2 + 3 x - 3)((x + 1)(x - 1)) - frac(4 x - 2)((x + 1)(x - 1))#

#= frac(5 x - 1)((x + 1)(x - 1)) - frac(4 x - 2)((x + 1)(x - 1))#

Now, let's collect these two fractions:

#= frac(5 x - 1 - (4 x - 2))((x + 1)(x - 1))#

#= frac(5 x - 1 - 4 x + 2)((x + 1)(x - 1))#

#= frac(x + 1)((x + 1)(x - 1))#

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#.