## anonymous 3 years ago Add and simplify... x/(x-1) - 2x/(x^2-1)

Where are you stuck?

$\frac{ x }{ (x-1) }-\frac{2x}{x^2-1}$

3. anonymous

I'm pretty sure there is some factoring int there, but I'm not sure.

4. anonymous

You will want to factor the denominator of that second term. Then you will want to get the LCM (Least Common Multiple) for both terms.

5. anonymous

Right! So what do I multiply to get 1 and adds up to be 1?

6. anonymous

Hint: the denominator of the second term is the difference of 2 perfect squares.

7. anonymous

Once you factor that denominator of that second term, you will see one of the factors as the denominator in the first term. Take the OTHER factor of the denominator of the second term and make a factor of "1" out of it (by putting the factor in the top and bottom) and multiply the first term by that. Then you will have a common denominator.

8. anonymous

So to start, can you factor x^2 - 1 ?

9. anonymous

Another hint:$a ^{2} - b ^{2} = (a + b)(a - b)$

10. anonymous

Think of 1 as 1^2

11. anonymous

If you substitute a=x and b=1 into the equation I wrote out, you should be able to factor:$x ^{2} - 1$

12. anonymous

Okay, so my final answer would be x^2-x/x^2-1?

13. anonymous

No, but that's a good intermediate answer. You need to factor out "x-1" from the top and bottom.

14. anonymous

uh.. x/x-1?._.

15. anonymous

Or would it be +1?

16. anonymous

$\frac{ x }{ x - 1 } - \frac{ 2x }{ x ^{2} - 1 } = \frac{ x }{ x - 1 } - \frac{ 2x }{ (x - 1)(x + 1) }$ $\frac{ x(x + 1) }{ (x - 1)(x + 1) } - \frac{ 2x }{ (x - 1)(x + 1) } = \frac{ x ^{2} + x - 2x }{ (x - 1)(x + 1) }$ $\frac{ x ^{2} - x }{ (x - 1)(x + 1) } = \frac{ x\cancel{(x - 1)} }{ \cancel{(x - 1)}{(x + 1)} }$

17. anonymous

Ohhhh! Okay thanks!

18. anonymous

uw!

19. anonymous

Would you mind helping me with some other problems? My teacher isn't available.

20. anonymous

ok, see you in another question!

21. anonymous

Okay!