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

1. 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?

2. Re: Solve for P $S=B+\frac 12PS$

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

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

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

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