# How do you solve for f in 2/3f + 5/12g=1- fg?

Aug 28, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{\frac{5}{12} g}$ and add $\textcolor{b l u e}{f g}$ to both sides of the equation to isolate the $f$ terms on the left side of the equation:

$\frac{2}{3} f + \textcolor{b l u e}{f g} + \frac{5}{12} g - \textcolor{red}{\frac{5}{12} g} = 1 - \textcolor{red}{\frac{5}{12} g} - f g + \textcolor{b l u e}{f g}$

$\frac{2}{3} f + f g + 0 = 1 - \frac{5}{12} g - 0$

$\frac{2}{3} f + f g = 1 - \frac{5}{12} g$

Next, factor an $f$ out of each term on the left side of the equation:

$f \left(\frac{2}{3} + g\right) = 1 - \frac{5}{12} g$

Now, divide each side of the equation by $\textcolor{red}{\frac{2}{3} + g}$ to solve for $f$ while keeping the equation balanced:

$\frac{f \left(\frac{2}{3} + g\right)}{\textcolor{red}{\frac{2}{3} + g}} = \frac{1 - \frac{5}{12} g}{\textcolor{red}{\frac{2}{3} + g}}$

$\frac{f \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(\frac{2}{3} + g\right)}}}}{\cancel{\textcolor{red}{\frac{2}{3} + g}}} = \frac{1 - \frac{5}{12} g}{\frac{2}{3} + g}$

$f = \frac{1 - \frac{5}{12} g}{\frac{2}{3} + g}$