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

1 Answer
Jul 5, 2017

See a solution process below:

Explanation:

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

S - color(red)(Ph) = 2B + Ph - color(red)(Ph)SPh=2B+PhPh

S - Ph = 2B + 0SPh=2B+0

S - Ph = 2BSPh=2B

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

(S - Ph)/color(red)(2) = (2B)/color(red)(2)SPh2=2B2

(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