## Cubi-Cal Group Title Add and simplify... x/(x-1) - 2x/(x^2-1) one year ago one year ago

Where are you stuck?

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

3. Cubi-Cal

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

4. tcarroll010

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. Cubi-Cal

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

6. tcarroll010

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

7. tcarroll010

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. tcarroll010

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

9. tcarroll010

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

10. tcarroll010

Think of 1 as 1^2

11. tcarroll010

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

12. Cubi-Cal

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

13. tcarroll010

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

14. Cubi-Cal

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

15. Cubi-Cal

Or would it be +1?

16. tcarroll010

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

17. Cubi-Cal

Ohhhh! Okay thanks!

18. tcarroll010

uw!

19. Cubi-Cal

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

20. tcarroll010

ok, see you in another question!

21. Cubi-Cal

Okay!