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Canceling a Common Factor

  1. Sep 20, 2013 #1

    Drakkith

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    1. The problem statement, all variables and given/known data

    I've got a fraction here:

    [itex]\frac{14+14\sqrt{3}}{-8}[/itex]

    Why is it you can take a 2 out of the bottom and top to make it the following?

    [itex]\frac{7+7\sqrt{3}}{-4}[/itex]

    I'm lost in figuring out how this works. I thought the top was like having (14+14x), where you can take a 14 out of each term and make it 14(1+x).
     
    Drakkith, Sep 20, 2013
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  3. Sep 20, 2013 #2

    Mentallic

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    What does
    [tex]\frac{14y}{2x}[/tex]
    equal to?

    Also, you can move the negative sign from the denominator to the numerator by "taking out" a -1 from both the numerator and denominator.
     
    Mentallic, Sep 20, 2013
  4. Sep 20, 2013 #3

    SteamKing

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    Really?

    Code (Text):
    Lookit:
    14 + 14*Sqrt(3)      2*7 + 2*7 * Sqrt(3)     2*[7 + 7 * Sqrt (3)]     7 + 7 * Sqrt (3)
    ---------------- =   ------------------- =   -------------------- = ----------------
         -8                     2*(-4)                   2 * (-4)               -4


     
     
    SteamKing, Sep 20, 2013
  5. Sep 20, 2013 #4

    Drakkith

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    I understand that 14y/2x = 7y/x.

    I thought you couldn't divide the original fraction that way because it's still adding up there and you had to take a factor out or something first.
     
    Drakkith, Sep 20, 2013
  6. Sep 20, 2013 #5

    SteamKing

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    Multiplication distributes over addition, so a*(b + d) = a*b + a*d
     
    SteamKing, Sep 20, 2013
  7. Sep 20, 2013 #6

    Mentallic

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    Yes, that's true, but look at what you said earlier

    So in this case, the numerator is [itex]14(1+\sqrt{3})[/itex] so we can now let [itex]y=1+\sqrt{3}[/itex].
     
    Mentallic, Sep 20, 2013
  8. Sep 20, 2013 #7

    Drakkith

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    Arrghh... I had my answer as [itex]\frac{7(1+\sqrt{3})}{4}[/itex] , which was apparently wrong, whereas [itex]\frac{7+7\sqrt{3})}{4}[/itex] was correct.
     
    Drakkith, Sep 20, 2013
  9. Sep 20, 2013 #8

    Office_Shredder

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    Those are both the exact same number, and are equally correct actually (well, except for the missing minus sign)
     
    Office_Shredder, Sep 20, 2013
  10. Sep 20, 2013 #9

    Drakkith

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    Ah yes, forgot the negative.
    At least I got it figured out. I was so confused...
     
    Drakkith, Sep 20, 2013
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