How do you solve for x in #C=b-bx#?

1 Answer
Feb 10, 2017

See the entire solution process below:

Explanation:

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

#C - color(red)(b) = b - bx - color(red)(b)#

#C - b = b - color(red)(b) - bx#

#C - b = 0 - bx#

#C - b = -bx#

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

#(C - b)/color(red)(-b) = (-bx)/color(red)(-b)#

#(C - b)/color(red)(-b) = (color(red)(cancel(color(black)(-b)))x)/cancel(color(red)(-b))#

#(C - b)/-b = x#

#x = (C - b)/-b#

Or

#x = C/-b - b/-b#

#x = -C/b + b/b#

#x = -C/b + 1#

#x = 1 - C/b#