an airplane has just enough fuel for a five hour flight how far can it fly a round trip and it flies at 225 miles per hour with the wind during the first leg of the trip and at 180 miles per hour Against the Wind or the return

To solve this we are going to use the speed equation: [tex]S= \frac{d}{t} [/tex]
where
[tex]S[/tex] is speed
[tex]d[/tex] is distance 
[tex]t[/tex] is time 

First, we are going to label the information that the problem is giving us:
Since the plane has enough fuel for a five our flight, the total time of the round trip will be five hours; therefore, [tex]t_{t}=5[/tex]. We know that the speed of the plane with the wind is 225 miles per hour, so [tex]S_{ww}=225[/tex]. We also know that the speed of the plane against the wind is 180 miles per hour, so [tex]S_{aw}=180[/tex]. Notice that Since we are dealing with a round trip, the distance against the wind and the distance with the wind will be the same. 

For the trip with the wind:
[tex]S_{ww}= \frac{d}{t_{w}} [/tex]
[tex]225= \frac{d}{t_{w}} [/tex] equation (1)

For the trip against the wind:
[tex]S_{aw}= \frac{d}{t_{a} } [/tex]
[tex]180= \frac{d}{t_{a} } [/tex] equation (2)

Since the whole trip takes 5 hours, [tex]t_{w}+t_{a}=5[/tex]
Solving for [tex]t_{w}[/tex]:
[tex]t_{w}=5-t_{a}[/tex] equation (3)

Replacing (3) in (1):
[tex]225= \frac{d}{t_{w}} [/tex]
[tex]225= \frac{d}{5-t_{a}} [/tex] equation (4)

Solving for [tex]d[/tex] in (2)
[tex]180= \frac{d}{t_{a} } [/tex]
[tex]d=180t_{a}[/tex] equation (5)

Replacing (5) in (4):
[tex]225= \frac{d}{5-t_{a}} [/tex]
[tex]225= \frac{180t_{a}}{5-t_{a}} [/tex]
[tex]225(5-t_{a})=180t_{a}[/tex]
[tex]1125-225t_{a}=180t_{a}[/tex]
[tex]405t_{a}=1125[/tex]
[tex]t_{a}= \frac{1125}{405} [/tex]
[tex]t_{a}= \frac{25}{9} [/tex] equation (6)

Replacing (6) in (5):
[tex]d=180t_{a}[/tex]
[tex]d=180( \frac{25}{9} )[/tex]
[tex]d=500[/tex] miles 

We can conclude that the plane can travel a distance of 500 miles to make a round trip if it has just enough fuel for a five hour flight.