How do you solve for x in #C=b-bx#?
2 Answers
Apr 27, 2018
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
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.
NOW, given any C and b, x can be calculated directly.