Question

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

Answer 1

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`

Answer 2

`(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`