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

2 Answers
Apr 27, 2018

Answer:

#x = 1 - C/b#

Explanation:

Let's start by rewriting as

#bx - b = -C#

Now factor

#b(x- 1) = -C#

#x - 1 = -C/b#

#x = 1 - C/b#

Hopefully this helps!

Apr 27, 2018

Answer:

#x = 1 - C/b#

Explanation:

You can separate the constants from the variable with algebraic operations. In this case we will still end up with an expression in C and b, as they are not given values.

#C = b - bx = b(1 - x)# (distributive property)
#C/b = 1 - x# (divide both sides by b)
#C/b - 1 = - x# (subtract 1 from both sides)
#x = 1 - C/b# (multiply both sides by -1 and rearrange for neatness)

NOW, given any C and b, x can be calculated directly.