Question

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

Answer 1

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

Answer 2

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)\[=-5(v+1)\]

Answer 3

how do you get the fraction

Answer 4

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

Answer 5

Okay, you need to use brackets to group things.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

nvm it is add. so ill re do it

Answer 10

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

Answer 11

lol, you're really struggling with this editor

Answer 12

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

Answer 13

okay...wait...

Answer 14

i get it

Answer 15

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

Answer 16

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

Answer 17

yes that is right

Answer 18

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

Answer 19

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