Identify and name what you are looking for. Make a table, similar to the one we used in Example 5.4.16 to help you organize the information. We are given the rates of both Charlie and Sally, and so you can enter them into your table. We don’t know how long they have been driving at those rates, so you can let $c$ represent Charlie’s time, in hours, and let $s$ represent Sally’s time, in hours, and enter that information into the table as well. Then, you can multiply the rates and times together to obtain an expression for each person’s distance. You should end up with $36c$ for Charlie’s distance and $48s$ for Sally’s distance. Next you need to translate into a system of equations. To make the system of equations, you should recognize that Charlie and Sally will have driven the same distance once Sally has caught up to Charlie. So, $36c=48s\text{.}$ Also, since Sally left $\frac{1}{4}$ hour later, her time will be $\frac{1}{4}$ hour less than Charlie’s time. So, $s=c-\frac{1}{4}\text{.}$ Now we have a system: $$\begin{cases} 36c = 48s \\ s = c - \frac{1}{4} \end{cases}$$ Now solve the system of equations, either by substitution or elimination. This system, lends itself easily to substitution since $s$ is already isolated in the second equation. Thus, we can substitute the expression $c-\frac{1}{4}$ in place of $s$ in the first equation: $$\begin{gathered} 36c=48\left(\substitute{c-\frac{1}{4}}\right)\\ 36c=48c-12\\ -12c=-12\\ c=1 \end{gathered}$$ Note that all we were asked to find is how long Charlie will have been driving before Sally catches up, so we don’t need to go any further. We now know that Charlie had been driving for $1$ hour before Sally caught up with him.