Question

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

Answer 1

`x=3/2`

Explanation:

We have an equation,
`{x+1}/{x-1}=2/{2x-1}+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

`{x+1}/y=2/{y+x}+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 `{x+1}/y` and `2/y`.
so what I'll be doing is that I'll subtract on both sides by `2/y`, hence removing the `2/y` term from the right side of the equation and bringing it to the left. That gives us

`{x+1-2}/y=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
`{x-1}/y=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=>{x-1}/y=1`

So that makes the left-hand side of the equation as
`1=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=>2x-1=2`

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