Question

add and simplify: (x/x-1)-(2x/x^2-1)
a. x/x-1
b. x/x+1
c. x^2-x/x^2-1

Answer 1

\[\huge {x \over x-1} - {2x \over x^2-1}\]
is that the problem?

Answer 2

yes!

Answer 3

In the denominator of the second term use:

\[a^2 - b^2 = (a-b)(a+b)\]

Answer 4

(x-1)(x+1)(x/x-1)-(x-1)(x+1)(2x/x^2-1)

Answer 5

so do i multiply the whole denominator by (x-1)(x+1)?

Answer 6

(x-1)(x+1) is LCD

Answer 7

How you got (x-1)(x+1) tell me first..

Answer 8

Are you guessing it or you have done it by using the formula that I gave above???

Answer 9

i did it by the formula that you gave me

Answer 10

Yes..

Do you know about LCD ??

Answer 11

yes, I just wasn't sure what it would be for this one.

Answer 12

Can you factor out what is common in both the terms??

Answer 13

Are you getting what I said???

Can you take common factor from both the terms ?

Answer 14

is it x/x+1 ? I just factored out what was common.

Answer 15

I think x/(x-1) is common in both the terms right???

Answer 16

yes, so I factored it out. is that correct? and then I was left with x/x+1

Answer 17

Didn't you get (x-1)/(x+1) ???

Check it carefully..

Answer 18

See I show you the steps:

\[\frac{x}{x-1} - \frac{2x}{(x-1)(x+1)} \implies \frac{x}{x-1} (1 - \frac{2}{x+1}) \implies \frac{x}{x-1}(\frac{x+1 - 2}{x+1})\]

Answer 19

\[\implies (\frac{x}{x-1}) \times (\frac{x-1}{x+1}) \implies ??\]

Answer 20

Getting what I have done or not ?

If not then which step is creating problem for you??

Answer 21

I'm slightly confused as to why you multiplied x/x-1 and (1-2/x+1) in the very beginning steps..

Answer 22

I have factored out x/(x-1) out of the terms by using distributive property..

Getting??

Answer 23

ok I get it

Answer 24

Sure no???

Answer 25

I didn't see it get explained this way, so maybe this will offer some clarity. The formula for adding or subtracting two fractions without a common denominator is this:
\[\huge {a \over b} \pm {c \over d} = {a*d \pm c*b \over b*d}\]
where, a, b, c and d are just whatever is in your fractions.