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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

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  1. anonymous
    • 5 years ago
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    Is your question,\[\frac{v}{v^2}-5v-6-\frac{1}{v}+1\]?

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    how do you get the fraction

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Okay, you need to use brackets to group things.

  6. anonymous
    • 5 years ago
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    i do not know how to write equations on here... i do not understand i type it in the best i can

  7. anonymous
    • 5 years ago
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    [(v \ v ^{2}-5v-6) (1 \ v+1)]

  8. anonymous
    • 5 years ago
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    in the middle it is subtract.. between the two fractions

  9. anonymous
    • 5 years ago
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    nvm it is add. so ill re do it

  10. anonymous
    • 5 years ago
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    \[(v \ v^2-5v-6)+(1 \ v+1)\]

  11. anonymous
    • 5 years ago
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    lol, you're really struggling with this editor

  12. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    okay...wait...

  14. anonymous
    • 5 years ago
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    i get it

  15. anonymous
    • 5 years ago
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    \[\frac{v}{v^2-5v-6}+\frac{1}{v+1}\]is this it?

  16. anonymous
    • 5 years ago
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    abarnzy, still there? Is that expression right before I go on?

  17. anonymous
    • 5 years ago
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    yes that is right

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

  19. anonymous
    • 5 years ago
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    \[\frac{v}{(v-6)(v+1)}+\frac{1}{v+1}=\frac{1}{v+1}\left[ \frac{v}{v-6}+1 \right]\]\[=\frac{1}{v+1}\left[ \frac{v+(v-6)}{v-6} \right]=\frac{2v-6}{(v+1)(v-6)}=\frac{2(v-3)}{(v+1)(v-6)}\]

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