anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Is your question,\[\frac{v}{v^2}-5v-6-\frac{1}{v}+1\]?
anonymous
  • 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)\[=-5(v+1)\]
anonymous
  • anonymous
how do you get the fraction

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anonymous
  • 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
anonymous
  • anonymous
Okay, you need to use brackets to group things.
anonymous
  • anonymous
i do not know how to write equations on here... i do not understand i type it in the best i can
anonymous
  • anonymous
[(v \ v ^{2}-5v-6) (1 \ v+1)]
anonymous
  • anonymous
in the middle it is subtract.. between the two fractions
anonymous
  • anonymous
nvm it is add. so ill re do it
anonymous
  • anonymous
\[(v \ v^2-5v-6)+(1 \ v+1)\]
anonymous
  • anonymous
lol, you're really struggling with this editor
anonymous
  • anonymous
between the v and v^2 the V^2 part should be on bottom and the v+1 should be on the bottom
anonymous
  • anonymous
okay...wait...
anonymous
  • anonymous
i get it
anonymous
  • anonymous
\[\frac{v}{v^2-5v-6}+\frac{1}{v+1}\]is this it?
anonymous
  • anonymous
abarnzy, still there? Is that expression right before I go on?
anonymous
  • anonymous
yes that is right
anonymous
  • anonymous
You can first factor the denominator of the first fraction as\[v^2-5v-6=(v-6)(v+1)\]to get
anonymous
  • anonymous
\[\frac{v}{(v-6)(v+1)}+\frac{1}{v+1}=\frac{1}{v+1}\left[ \frac{v}{v-6}+1 \right]\]\[=\frac{1}{v+1}\left[ \frac{v+(v-6)}{v-6} \right]=\frac{2v-6}{(v+1)(v-6)}=\frac{2(v-3)}{(v+1)(v-6)}\]

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