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

Jun 17, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{a}$ from each side of the equation:

$\left[\frac{a - c x}{b + \mathrm{dx}}\right] + a - \textcolor{red}{a} = 0 - \textcolor{red}{a}$

$\left[\frac{a - c x}{b + \mathrm{dx}}\right] + 0 = - a$

$\frac{a - c x}{b + \mathrm{dx}} = - a$

Next, multiply each side of the equation by $\textcolor{red}{\left(b + \mathrm{dx}\right)}$ to eliminate the fraction while keeping the equation balanced:

$\textcolor{red}{\left(b + \mathrm{dx}\right)} \times \frac{a - c x}{b + \mathrm{dx}} = - a \textcolor{red}{\left(b + \mathrm{dx}\right)}$

$\cancel{\textcolor{red}{\left(b + \mathrm{dx}\right)}} \times \frac{a - c x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{b + \mathrm{dx}}}}} = \left(- a \times \textcolor{red}{b}\right) - \left(a \times \textcolor{red}{\mathrm{dx}}\right)$

$a - c x = - a b - a \mathrm{dx}$

Then, add $\textcolor{red}{c x}$ and add $\textcolor{b l u e}{a b}$ to each side of the equation to isolate the $x$ terms while keeping the equation balanced:

$a + \textcolor{b l u e}{a b} - c x + \textcolor{red}{c x} = - a b + \textcolor{b l u e}{a b} + \textcolor{red}{c x} - a \mathrm{dx}$

$a + a b - 0 = 0 + c x - a \mathrm{dx}$

$a + a b = c x - a \mathrm{dx}$

$a + a b = \left(c - a d\right) x$

Now, divide each side of the equation by $\textcolor{red}{c - a d}$ to solve for $x$ while keeping the equation balanced:

$\frac{a + a b}{\textcolor{red}{c - a d}} = \frac{\left(c - a d\right) x}{\textcolor{red}{c - a d}}$

$\frac{a + a b}{c - a d} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(c - a d\right)}}} x}{\cancel{\textcolor{red}{c - a d}}}$

$\frac{a + a b}{c - a d} = x$

$x = \frac{a + a b}{c - a d}$

Or

$x = \frac{a \left(1 + b\right)}{c - a d}$