###### Example3.3.9

Solve for \(F\) in \(C=\frac{5}{9}(F-32)\text{.}\) (This represents the relationship between temperature in degrees Celsius and degrees Fahrenheit.)

To solve for \(F\text{,}\) we first need to see that it is contained inside a set of parentheses. To get the expression \(F-32\) on its own, we'll need to eliminate the \(\frac{5}{9}\) outside those parentheses. One way we can “undo” this multiplication is by dividing each side by \(\frac{5}{9}\text{.}\) As we learned in Section 3.2 though, a better approach is to instead multiply each side by the reciprocal of \(\frac{9}{5}\text{:}\)

\begin{align*} C\amp=\frac{5}{9}(\attention{F}-32)\\ \multiplyleft{\frac{9}{5}}C\amp=\multiplyleft{\frac{9}{5}}\frac{5}{9}(\attention{F}-32)\\ \frac{9}{5}C\amp=\attention{F}-32 \end{align*}Now that we have \(F-32\text{,}\) we simply need to add \(32\) to each side to finish solving for \(F\text{:}\)

\begin{align*} \frac{9}{5}C\addright{32}\amp=\attention{F}-32\addright{32}\\ \frac{9}{5}C+32\amp=\attention{F}\\ F\amp=\frac{9}{5}C+32 \end{align*}