It may help to draw a diagram and a table similar to those found in Example 5.4.26 to help visualize the problem and organize the information that is given. We don’t know the jet’s speed in still air or the speed of the wind, so we will let $j$ represent the jet’s speed in still air and let $w$ represent the speed of the wind. Thus, the rate at which the jet is traveling with a tailwind will be represented by $j+w\text{,}$ and the rate at which the jet is traveling into a headwind will be represented by $j-w\text{.}$ 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, which are $1728$ miles and $1536$ miles, respectively. This will give us our two equations, making up our system: $$\begin{cases} \multiplyleft{4}(j+w)=1728\\\multiplyleft{4}(j-w)=1536 \end{cases}$$ 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: $$\begin{cases} 4j+4w=1728\\4j-4w=1536 \end{cases}$$ The best method to use to solve this system is the elimination method since the $w$ terms of the two equations add to zero. Add the first equation to the second equation. Then, solve for $j\text{.}$ $$\begin{aligned} \begin{aligned}4j+4w\\{}+4j-4w\end{aligned}\amp=\begin{aligned}1728\\{}+1536\end{aligned} \end{aligned}$$ So we have $$\begin{gathered} 8j=3264\\ j=408 \end{gathered}$$ Substitute $408$ for $j$ in the first original equation and solve for $w\text{:}$ $$\begin{gathered} 4(\substitute{408}+w)=1728\\ 1632+4w=1728\\ 4w=96\\ w=24 \end{gathered}$$ Therefore, the jet’s speed in still air is $408$ miles per hour and the speed of the wind is $24$ miles per hour. The check is left to the reader.