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

1 Answer
Feb 13, 2017

Answer:

#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.