How do you solve for x in #ax-c=b#?

2 Answers
Apr 26, 2017

Answer:

See the solution process below:

Explanation:

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

#ax - c + color(red)(c) = b + color(red)(c)#

#ax - 0 = b + c#

#ax = b + c#

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

#(ax)/color(red)(a) = (b + c)/color(red)(a)#

#(color(red)(cancel(color(black)(a)))x)/cancel(color(red)(a)) = (b + c)/a#

#x = (b + c)/a#

Or

#x = b/a + c/a#

Apr 26, 2017

Answer:

#(b+c)/a#

Explanation:

Add 'c' in both sides, we get ax - c + c = b + c

#rArr ax = b + c# [now divide by 'a' in both sides]

#rArr (ax)/a = (b+c)/a#

#rArr x = (b+c)/a#