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Simple algebra question

  1. Sep 29, 2005 #1

    DB

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    (another) simple algebra question

    [tex]\frac{300}{5v_w}+\frac{300}{3v_w}=8[/tex]

    can some1 give me a hand on the algebra here? thanks
     
    Last edited: Sep 29, 2005
  2. jcsd
  3. Sep 29, 2005 #2

    TD

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    Homework Helper

    Multiply both sides with [itex]v_w[/itex] to get:

    [tex]\frac{{300}}
    {{5v_w }} + \frac{{300}}
    {{3v_w }} = 8 \Leftrightarrow v_w \left( {\frac{{300}}
    {{5v_w }} + \frac{{300}}
    {{3v_w }}} \right) = 8v_w \Leftrightarrow \frac{{300}}
    {5} + \frac{{300}}
    {3} = 8v_w [/tex]
     
  4. Sep 29, 2005 #3

    DB

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    okay but i only multiply the numerators by v_w? otherwise i would get v_w^2 in the denominators...right?
     
  5. Sep 29, 2005 #4

    DB

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    i get 20 btw
     
  6. Sep 29, 2005 #5

    TD

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    Yes, that wouldn't help anything. You just multiply both sides with it, and when you multiply a fraction with a factor, it's always in the nominator. c * a/b = ca/b.

    You do this to get rid of the v_w in the denominators since you can cancel them out when they're also in the numerators.

    Correct :smile: You can check it by plugging it in the initial equation.
     
  7. Sep 29, 2005 #6

    DB

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    thanks TD, i see it perfectly now
     
  8. Sep 29, 2005 #7

    TD

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    No problem, glad I could help :smile:
     
  9. Sep 29, 2005 #8

    DB

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    just out of curiosity...how would i approach:
    [tex]\frac{300}{5v_w+10}+\frac{300}{3v_w}=8[/tex]
    ?
    thanks
     
  10. Sep 30, 2005 #9

    VietDao29

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    You can try simplify the equation a bit:
    [tex]\frac{300}{5v_w + 10} + \frac{300}{3v_w} = 8[/tex]
    [tex]\Leftrightarrow \frac{60}{v_w + 2} + \frac{100}{v_w} = 8[/tex]
    Then, you can try multiplying both sides with vw(vw + 2) to get rid of the denominator. And you will have:
    [tex]60v_w + 100(v_w + 2) = 8v_w(v_w + 2)[/tex]
    You can rearrange a bit, and you will have a quadratic equation.
    Can you go from here?
    Viet Dao,
     
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