## anonymous 5 years ago add: v/v^2-5v-6 - 1/v+1.. i know that if you mulitply the second fraction by v-6 it will give you the same denomitator i think.. but i need help please

1. anonymous

Is your question,$\frac{v}{v^2}-5v-6-\frac{1}{v}+1$?

2. anonymous

I'm going to type out for this question. If my interpretation is wrong, you're best to re-post. $\frac{v}{v^2}-5v-6-\frac{1}{v}+1$$=\frac{1}{v}-5v-6-\frac{1}{v}+1$$=-5v-5$(since the 1/v's cancel)$=-5(v+1)$

3. anonymous

how do you get the fraction

4. anonymous

on top of the first fraction is v on the bottom is v^2-5v-6 + on top of the second is 1 on bottoom is v+1

5. anonymous

Okay, you need to use brackets to group things.

6. anonymous

i do not know how to write equations on here... i do not understand i type it in the best i can

7. anonymous

[(v \ v ^{2}-5v-6) (1 \ v+1)]

8. anonymous

in the middle it is subtract.. between the two fractions

9. anonymous

nvm it is add. so ill re do it

10. anonymous

$(v \ v^2-5v-6)+(1 \ v+1)$

11. anonymous

lol, you're really struggling with this editor

12. anonymous

between the v and v^2 the V^2 part should be on bottom and the v+1 should be on the bottom

13. anonymous

okay...wait...

14. anonymous

i get it

15. anonymous

$\frac{v}{v^2-5v-6}+\frac{1}{v+1}$is this it?

16. anonymous

abarnzy, still there? Is that expression right before I go on?

17. anonymous

yes that is right

18. anonymous

You can first factor the denominator of the first fraction as$v^2-5v-6=(v-6)(v+1)$to get

19. anonymous

$\frac{v}{(v-6)(v+1)}+\frac{1}{v+1}=\frac{1}{v+1}\left[ \frac{v}{v-6}+1 \right]$$=\frac{1}{v+1}\left[ \frac{v+(v-6)}{v-6} \right]=\frac{2v-6}{(v+1)(v-6)}=\frac{2(v-3)}{(v+1)(v-6)}$