Question

How do you solve for x in `[ (a-cx)/(b+dx) ]+ a = 0`?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(a)` from each side of the equation:

`[(a - cx)/(b + dx)] + a - color(red)(a) = 0 - color(red)(a)`

`[(a - cx)/(b + dx)] + 0 = -a`

`(a - cx)/(b + dx) = -a`

Next, multiply each side of the equation by `color(red)((b + dx))` to eliminate the fraction while keeping the equation balanced:

`color(red)((b + dx)) xx (a - cx)/(b + dx) = -acolor(red)((b + dx))`

`cancel(color(red)((b + dx))) xx (a - cx)/color(red)(cancel(color(black)(b + dx))) = (-a xx color(red)(b)) - (a xx color(red)(dx))`

`a - cx = -ab - adx`

Then, add `color(red)(cx)` and add `color(blue)(ab)` to each side of the equation to isolate the `x` terms while keeping the equation balanced:

`a + color(blue)(ab) - cx + color(red)(cx) = -ab + color(blue)(ab) + color(red)(cx) - adx`

`a + ab - 0 = 0 + cx - adx`

`a + ab = cx - adx`

`a + ab = (c - ad)x`

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

`(a + ab)/color(red)(c - ad) = ((c - ad)x)/color(red)(c - ad)`

`(a + ab)/(c - ad) = (color(red)(cancel(color(black)((c - ad))))x)/cancel(color(red)(c - ad))`

`(a + ab)/(c - ad) = x`

`x = (a + ab)/(c - ad)`

Or

`x = (a(1 + b))/(c - ad)`