How do you solve \frac { x } { x - 1} + \frac { 4} { x + 1} = \frac { 4x - 2} { x ^ { 2} - 1}?

1 Answer
May 30, 2018

Ok, firstly, you have x-1, x+1, and x^2-1 as the denominator in your question. Thus, I will take it as the question implicitly assumes that x != 1 or -1. This is actually pretty important.

Let's combine the fraction on the right into a single fraction,
x/(x-1) + 4/(x+1) = (x(x+1))/((x-1)(x+1)) + (4(x-1))/((x-1)(x+1)) = (x^2 + x + 4x - 4)/(x^2-1) = (x^2 + 5x -4)/(x^2 -1)

Here, note that (x-1)(x+1) = x^2 - 1 from difference of two squares.

We have:

(x^2 + 5x -4)/(x^2 -1) = (4x-2)/(x^2-1)

Cancel out the denominator (multiply both sides by x^2-1),

x^2 + 5x -4 = 4x-2

Please note that this step is only possible due to our assumption at the start. Cancelling (x^2-1)/(x^2-1) = 1 is only valid for x^2-1 != 0.

x^2 + x -2 = 0

We can factorise this quadratic equation:
x^2 + x - 2 = (x - 1)(x + 2) = 0

And thus, x = 1, or x = -2.

But we're not done yet. This is the solution to the quadratic equation, but not the equation in the question.

In this case, x = 1 is an extraneous solution, which is an extra solution that is generated by the way we solve our problem, but is not an actual solution.

So, we reject x = 1, from our assumption earlier.

Therefore, x = -2.