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

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Answer 1 · 2017-03-24T03:06:58

See the entire solution process below:

First, subtract $color(red)(B)$ from each side of the equation to isolate the term containing $P$ while keeping the equation balanced:

$S - color(red)(B) = -color(red)(B) + B + (1/2)Pl$

$S - B = 0 + (1/2)Pl$

$S - B = (1/2)Pl$

Next, multiply each side of the equation by $color(red)(2)$ to eliminate the fraction while keeping the equation balanced:

$color(red)(2)(S - B) = color(red)(2) xx (1/2)Pl$

$2(S - B) = cancel(color(red)(2)) xx (1/color(red)(cancel(color(black)(2))))Pl$

$2(S - B) = Pl$

Now, divide each side of the equation by $color(red)(l)$ to solve for $P$ while keeping the equation balanced:

$(2(S - B))/color(red)(l) = (Pl)/color(red)(l)$

$(2(S - B))/l = (Pcolor(red)(cancel(color(black)(l))))/cancel(color(red)(l))$

$(2(S - B))/l = P$

$P = (2(S - B))/l$

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