## anonymous 5 years ago ((x-2)/(x-3))-((x+1)(x+3))+((x-9(/x^2-9))

1. anonymous

I think the best way to do this is to just multiply out the first two terms, and for the third term, factor the denominator to (x-3)(x+3).

2. anonymous

can you multiply it? I dont understand this. it looks like subtraction to me

3. anonymous

oh sorry, I didn't see the / in the first term. Ok, for the second term, (x+1)(x+3), multiply this out Then for the third term, (x-9)/(x^2-9), factor the denominator

4. anonymous

uh oh i typed this out wrong its supp((x-2)/(x-3))-((x+1)/(x+3))+((x-9)/(x^2-9))osed to be

5. anonymous

$\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2 -9}$ Recognize that $$x^2 - 9 = (x+3)(x-3)$$ and put them all under a common denominator of $$x^2-9$$

6. anonymous

Over a common denominator rather.

7. anonymous

so i need to multiply the first two fractions by (x+3) or x-3?

8. anonymous

Multiply the first by (x+3) on top and bottom, and the second by x-3 on top and bottom.

9. anonymous

Oh, was the second term also a fraction? Remember when you learned how to add fractions with different denominators? Use the same logic. Multiply the first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3) so that they all have a common denominator. Basically what polpak is saying but it is worth noting that this is just like what you do when you are adding normal fractions with different denominators.

10. anonymous

Oh. I may have mistyped it. Was the middle term a fraction? If not it'd be $\frac{x-2}{x-3} +(x+1)(x+3) +\frac{x-9}{x^2-9}$

11. anonymous

But you still want to get them over a common denominator of $$x^2-9$$

12. anonymous

polpak, she posted that they were all fractions a few minutes ago. I missed it too.

13. anonymous

you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated

14. anonymous

so i came up with 0(x+3)(x-3) is this correct? i somehow dont think it is

15. anonymous

im sorry 0/(x-3)(x+3)

16. anonymous

I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.

17. anonymous

no i subtracted it but it all seemed to cancel out. what am i doing wrong?

18. anonymous

$\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2-9}$ $= \frac{(x-2)(x+3)}{x^2-9} - \frac{(x+1)(x-3)}{x^2-9} + \frac{x-9}{x^2-9}$ $= \frac{[x^2+x-6] - [x^2-2x -3] + [x-9]}{x^2-9}$ $= \frac{x^2-x^2+x + 2x + x -6+3 -9}{x^2-9}$ $= \frac{4x-12}{x^2-9}$ $= \frac{4(x-3)}{x^2-9}$ $= \frac{4(x-3)}{(x+3)(x-3)}$ $= \frac{4}{x+3}$

19. anonymous

Where'd you go wrong?

20. anonymous

And hopefully you can follow that mess.

21. anonymous

i didnt change the signs in the second fraction AND i can follow it

22. anonymous

Thank you so much!

23. anonymous

Cool, np =)

24. anonymous

4/( x+3)