# How do you solve for P in S=B+(1/2)Pl?

Mar 24, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{B}$ from each side of the equation to isolate the term containing $P$ while keeping the equation balanced:

$S - \textcolor{red}{B} = - \textcolor{red}{B} + B + \left(\frac{1}{2}\right) P l$

$S - B = 0 + \left(\frac{1}{2}\right) P l$

$S - B = \left(\frac{1}{2}\right) P l$

Next, multiply each side of the equation by $\textcolor{red}{2}$ to eliminate the fraction while keeping the equation balanced:

$\textcolor{red}{2} \left(S - B\right) = \textcolor{red}{2} \times \left(\frac{1}{2}\right) P l$

$2 \left(S - B\right) = \cancel{\textcolor{red}{2}} \times \left(\frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) P l$

$2 \left(S - B\right) = P l$

Now, divide each side of the equation by $\textcolor{red}{l}$ to solve for $P$ while keeping the equation balanced:

$\frac{2 \left(S - B\right)}{\textcolor{red}{l}} = \frac{P l}{\textcolor{red}{l}}$

$\frac{2 \left(S - B\right)}{l} = \frac{P \textcolor{red}{\cancel{\textcolor{b l a c k}{l}}}}{\cancel{\textcolor{red}{l}}}$

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

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