How do you solve for B in #S=2B+Ph#?

Question

13738 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-05T20:58:58

See a solution process below:

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

#S - color(red)(Ph) = 2B + Ph - color(red)(Ph)#

#S - Ph = 2B + 0#

#S - Ph = 2B#

Now, divide each side of the equation by #color(red)(2)# to solve for #B# while keeping the equation balanced:

#(S - Ph)/color(red)(2) = (2B)/color(red)(2)#

#(S - Ph)/2 = (color(red)(cancel(color(black)(2)))B)/cancel(color(red)(2))#

#(S - Ph)/2 = B#

#B = (S - Ph)/2#

Or

#B = S/2 - (Ph)/2#