# How do you solve (x+1)/(x-1)=2/(2x-1)+2/(x-1)?

Feb 13, 2017

$x = \frac{3}{2}$

#### Explanation:

We have an equation,
$\frac{x + 1}{x - 1} = \frac{2}{2 x - 1} + \frac{2}{x - 1}$

For simplicity, which you might find to your liking, I'll be taking $x - 1$ as $y$, so $y = x - 1$, turning the equation to

$\frac{x + 1}{y} = \frac{2}{y + x} + \frac{2}{y}$
(Note, 2x-1=x+x-1=x+y)

Now, we have two terms, one on either side of the equation to have the same denominator, being $\frac{x + 1}{y}$ and $\frac{2}{y}$.
so what I'll be doing is that I'll subtract on both sides by $\frac{2}{y}$, hence removing the $\frac{2}{y}$ term from the right side of the equation and bringing it to the left. That gives us

$\frac{x + 1 - 2}{y} = \frac{2}{x + y}$
Since the terms on the left-hand side already had the same denominator, I already grouped them together.

By solving the sum on the numerator of the left-hand side, we get
$\frac{x - 1}{y} = \frac{2}{x + y}$

Now, what is the numerator of the left-hand side similar to? If you go back up to the top of the answer, you'll see that I equated $y = x - 1 \implies \frac{x - 1}{y} = 1$

So that makes the left-hand side of the equation as
$1 = \frac{2}{x + y}$

Multiplying by $x + y$, gives us
$x + y = 2$

Now, we know that $y = x - 1$, so substituting back into the equation, give us
$x + x - 1 = 2 \implies 2 x - 1 = 2$

I'm sure it'll be easy to find the answer now.