# Thread: How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??

1. ## How does x/(x-1) - x/(x+1) equal 1/(x+1) + 1/(x-1) ??

I'm looking at my notes and one of my professors steps goes from:

$\displaystyle \frac{x}{x-1}-\frac{x}{x+1}$

to

$\displaystyle \frac{1}{x-1}+\frac{1}{x+1}$

What happened to the numerator? I know this is not a mistake because when I enter the first line into my ti-89 it spits out the second line.

2. First, notice that $\displaystyle \frac{x}{x-1}=\frac{x-1}{x-1}+\frac{1}{x-1}=1+\frac{1}{x-1}$ and that $\displaystyle \frac{x}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$

What do you get if you put that into the given equation?

3. Thanks a lot. That explains it.

Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?

4. Originally Posted by Carbon
Thanks a lot. That explains it.

Although it seems like such an unusual (and unnecessary) step to take... I guess I just have to try to remember this if I see something like it again. Does this technique have a name?
I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.

-Dan

5. Originally Posted by topsquark
I doubt if there's a name to it. Note that his (rather clever) way of doing the problem can be done in a more direct, if somewhat longer, manner. Add the two fractions, then use a partial fraction decomposition. It takes no more than five lines.

-Dan
Can you show this? Because when I do it, I don't get 1 in the numerator.

Adding the fractions, you get 2x/[(x-1)(x+1)]
Using partial fraction I go to 2x=A(x+1) + B(x-1)
Solving for A and B, I end up getting x/(x-1) -x/(x+1)

6. Originally Posted by Carbon
Can you show this? Because when I do it, I don't get 1 in the numerator.

Adding the fractions, you get 2x/[(x-1)(x+1)]
Using partial fraction I go to 2x=A(x+1) + B(x-1)
Solving for A and B, I end up getting x/(x-1) -x/(x+1)
$\displaystyle \frac{2x}{(x + 1)(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$

2x = A(x + 1) + B(x - 1)

2x = (A + B)x + (A - B)

So we know that
2 = A + B
0 = A - B

I get A = B = 1 and the theorem follows.

-Dan

7. $\displaystyle x^1: 2 = A + B$
$\displaystyle x^0: 0 = A - B$

That gives me A = B = 1.