How do you solve for G in AL+G/EB=R+A ?

Apr 29, 2017

See the entire solution process below:

Explanation:

First, subtract $\textcolor{red}{A L}$ from each side of the equation to isolate the term containing the $G$ while keeping the equation balanced:

$A L + \frac{G}{E} B - \textcolor{red}{A L} = R + A - \textcolor{red}{A L}$

$A L - \textcolor{red}{A L} + \frac{G}{E} B = R + A - A L$

$0 + \frac{G}{E} B = R + A - A L$

$\frac{G}{E} B = R + A - A L$

Now, multiply each side of the equation by $\frac{\textcolor{red}{E}}{\textcolor{b l u e}{B}}$ to solve for $G$ while keeping the equation balanced:

$\frac{\textcolor{red}{E}}{\textcolor{b l u e}{B}} \cdot \frac{G}{E} B = \frac{\textcolor{red}{E}}{\textcolor{b l u e}{B}} \left(R + A - A L\right)$

$\frac{\cancel{\textcolor{red}{E}}}{\cancel{\textcolor{b l u e}{B}}} \cdot \frac{G}{\textcolor{red}{\cancel{\textcolor{b l a c k}{E}}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{B}}} = \frac{\textcolor{red}{E}}{\textcolor{b l u e}{B}} \left(R + A - A L\right)$

$G = \frac{\textcolor{red}{E}}{\textcolor{b l u e}{B}} \left(R + A - A L\right)$

Or

$G = \frac{\textcolor{red}{E} \left(R + A - A L\right)}{B}$

Or

$G = \frac{E R + E A - E A L}{B}$

Or

$G = \frac{E R}{B} + \frac{E A}{B} - \frac{E A L}{B}$