It may help to draw a diagram and a table similar to those found in Example 5.4.22 to help visualize the problem and organize the information that is given. We don’t know the canoe’s speed in still water or the speed of the current, so we will let $b$ represent the canoe’s speed in still water and let $c$ represent the speed of the current. Thus, the rate at which the canoe is traveling upstream, against the current, will be represented by $b-c\text{,}$ and the rate at which the canoe is traveling downstream, with the current, will be represented by $b+c\text{.}$ Translate into a system of equations. To make the system of equations, we use the fact that the product of the rate and the time is equal to the distance for each trip. This will give us our two equations, making up our system: $$\begin{cases} 4(b-c)=24\\3(b+c)=24 \end{cases}$$ Solve the system of equations, either by substitution or elimination. The first thing to do is simplify each equation by using the distributive property to get rid of the parentheses: $$\begin{cases} 4b-4c=24\\3b+3c=24 \end{cases}$$ Since all the variable terms have a coefficient (not equal to one), the best method to use to solve this system is the elimination method. Notice that the $c$ terms already have opposite signs, so we’ll work to eliminate the variable $c\text{.}$ The least common mutiple of $4$ and $3$ is $12\text{,}$ so if we multiply both sides of the first equation by $3$ and both sides of the second equation by $4\text{,}$ then the new $c$ terms will be $-12c$ and $12c$ respectively, which eliminate the $c$ terms when we add the new equations together. Multiply both sides of the first equation by $3$ and both sides of the second equation by $4\text{:}$ $$\begin{gathered} \multiplyleft{3}(4b-4c)=\multiplyleft{3}(24)\\ \multiplyleft{4}(3b+3c)=\multiplyleft{4}(24) \end{gathered}$$ Now, our new system is: $$\begin{cases} 12b-12c=72\\12b+12c=96 \end{cases}$$ Add the first equation to the second equation. Then, solve for $b\text{.}$ $$\begin{aligned} \begin{aligned}12b-12c\\{}+12b+12c\end{aligned}\amp=\begin{aligned}72\\{}+96\end{aligned} \end{aligned}$$ So we have $$\begin{gathered} 24b=168\\ b=7 \end{gathered}$$ Substitute $7$ for $b$ in the first original equation and solve for $c\text{:}$ $$\begin{gathered} 4(\substitute{7}-c)=24\\ 28-4c=24\\ -4c=-4\\ c=1 \end{gathered}$$ Therefore, the canoe’s speed in still water is $7$ miles per hour and the speed of the current is $1$ mile per hour. The check is left to the reader.