Given the word problem, we can deduce the following information:
1. An airplane flying with the wind takes 5 hours to travel a distance of 1500 miles.
2. The return trip takes 6 hours flying against the wind.
To determine the speed of the airplane in still air and the wind speed, we follow the process as shown below. Let:
s=speed of the airplane in still air
w= speed of the wind
Next, we write the distance equation for each way:
5(s+w)=1500
6(s-w)=1500
Then, we solve for s in 5(s+w)=1500:
![\begin{gathered} 5\mleft(s+w\mright)=1500 \\ \text{Simplify and rearrange} \\ s+w=(1500)/(5) \\ s+w=300 \\ s=300-w \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/drhr8s6qf3amxgv2bpcyojw45p6h7vmevz.png)
We plug in s=300-w into 6(s-w)=1500:
![\begin{gathered} 6\mleft(s-w\mright)=1500 \\ 6(300-w-w)=1500 \\ \text{Simplify and rearrange} \\ 300-2w=(1500)/(6) \\ 300-2w=250 \\ 2w=300-250 \\ 2w=50 \\ w=(50)/(2) \\ \text{Calculate} \\ w=25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cf5oh70308tpb6of5w02bhl7cz0hqnakdz.png)
We plug in w=25 into s=300-w:
![\begin{gathered} s=300-w \\ s=300-25 \\ s=275 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4hkhpuhaiku7ww02mg1ik85xrmwwzjf98f.png)
Therefore, the answers are:
The speed of the airplane in still air is 275 miles per hour.
The wind speed is 25 miles per hour.