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Thread: Solve for P $S=B+\frac 12PS$

  1. #1
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    Solve for P $S=B+\frac 12PS$

    $S=B+\frac 12PS$


    $2S=2(B+\frac 12PS)$


    $2S=2B+PS$


    $2S-2B=PS$


    $\frac {2S-2B}{S}=P$


    $\frac {2(S-B)}{S}=P$

    I am unsure as to whether or not multiplying both sides by 2 was the correct thing to do. Can somebody confirm my answer? If it is wrong, how should I change my approach?
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  2. #2
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    Re: Solve for P $S=B+\frac 12PS$

    I would have done it slightly differently but come up with the same answer

    $S = B + \dfrac 1 2 P S$

    $S - B = \dfrac 1 2 P S$

    $P = \dfrac{2(S-B)}{S}$
    Thanks from jschoeppler
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