Question

(x)/(x-2) - (-2)/(x+2)=(8)/(4-x^2)

Answer 1

\[\frac{x}{x - 2} - \frac{-2}{x + 2} = \frac{8}{4 - x^2}\]

For convenience express the middle fraction as an addition:
\[\frac{x}{x - 2} + \frac{2}{x + 2} = \frac{8}{4 - x^2}\]

And make sure the denominators are factored:
\[\frac{x}{x - 2} + \frac{2}{x + 2} = \frac{8}{(2 + x)(2 - x)}\]

Answer 2

I know this but how do i reverse the (2-x) ?

Answer 3

Okay to reverse that, take out a negative like so:

\[\frac{x}{x - 2} + \frac{2}{x + 2} = -\frac{8}{(x+2)(x-2)}\]

Answer 4

Then I add this on both sides.

Answer 5

Here's what to do...

Answer 6

Add the fraction on the right to both sides; subtract 2/(x + 2) from both sides:

\[\frac{x}{x - 2} +\frac{8}{(x+2)(x-2)}= -\frac{2}{x + 2} \]

Answer 7

Why ?

Answer 8

Actually...

Answer 9

Instead of doing that...

Answer 10

Lets reduce x/(x - 2)

Answer 11

It's too bad you found out that way. It is better to discover such things on your own.

Answer 12

I'm going to continue posting my solution anyway.

Answer 13

Thank you.

Answer 14

You should try to figure it out on your own why no solutions exist.

Answer 15

The result is \[\frac{ x+2 }{ x-2 }\]

Answer 16

I think.

Answer 17

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