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Thread: Algebra/Simplifcation

  1. #1
    Member FalconPUNCH!'s Avatar
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    Algebra/Simplifcation

    Hi I was wondering how one can get from

    $\displaystyle 2x = \lambda(2x-2) $ to $\displaystyle x = \frac{\lambda}{\lambda - 1}$

    and

    $\displaystyle 2y = \lambda(2y-4) $ to $\displaystyle y = \frac{2\lambda}{\lambda - 1}$

    thanks
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  2. #2
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    Hi FalconPUNCH!,

    We have $\displaystyle 2x=\lambda(2x-2)$ . Divide throughout by $\displaystyle 2$ hence $\displaystyle x=\lambda x-\lambda$ . Take the $\displaystyle x$ and the $\displaystyle \lambda$ over therefore $\displaystyle \lambda = x(\lambda-1)$ and thus dividing throughout by $\displaystyle (\lambda -1)$ gives the required results.

    The seconds is similar. Hope this helps.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by FalconPUNCH! View Post
    Hi I was wondering how one can get from

    $\displaystyle 2x = \lambda(2x-2) $ to $\displaystyle x = \frac{\lambda}{\lambda - 1}$

    and

    $\displaystyle 2y = \lambda(2y-4) $ to $\displaystyle y = \frac{2\lambda}{\lambda - 1}$

    thanks
    $\displaystyle 2x = \lambda(2x-2) $

    $\displaystyle 2x=2x\lambda-2\lambda$

    $\displaystyle x-\lambda x=-\lambda$

    $\displaystyle x(1-\lambda)=-\lambda$

    $\displaystyle x=\frac{-\lambda}{1-\lambda}$

    $\displaystyle x=\frac{\lambda}{\lambda-1}$

    The other one is done in a similar manner.
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  4. #4
    Member FalconPUNCH!'s Avatar
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    Thank you guys
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