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Math Help - Can't believe I'm asking this...

  1. #1
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    Can't believe I'm asking this...

    It has been a while since I've done algebra. I need to solve this equation in terms of L.

    (Kw)/(Ln)=(Nw)/(Lr)

    Note: N and n are different variables.
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  2. #2
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    Haha ya same here ... I'll give it a shot but no guarentees


    Equation:

    (Kw)/(Ln) = (Nw)/(Lr)
    [L^2r] (Kw)/(Ln) = (Nw)/(Lr) [L^2r]
    gives us
    (L^2Krw)/(L^3nr) = LNw
    (L^2Krw)/L^3nr)/ [Nw] = (LNw)/ [Nw]
    gives us
    Kr/Lwr = L


    Not sure if something can be given in terms of L with L being on both sides ... but that is the best I got for ya =D
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  3. #3
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    Quote Originally Posted by mcox05 View Post
    Haha ya same here ... I'll give it a shot but no guarentees


    Equation:

    (Kw)/(Ln) = (Nw)/(Lr)
    [L^2r] (Kw)/(Ln) = (Nw)/(Lr) [L^2r]
    gives us
    (L^2Krw)/(L^3nr) = LNw You can't use the rule of distribution here!
    (L^2Krw)/L^3nr)/ [Nw] = (LNw)/ [Nw]
    gives us
    Kr/Lwr = L


    Not sure if something can be given in terms of L with L being on both sides ... but that is the best I got for ya =D
    This fraction ...... \dfrac{Kw} {Ln}=\dfrac{Nw}{Lr} ...... simplifies to: ...... \dfrac{K} {n}=\dfrac{N}{r}

    That means L\in \mathbb{R}\setminus\{0\}
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