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

1 Answer
Mar 24, 2017

Answer:

See the entire solution process below:

Explanation:

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#