a boat travels upstream on the allegheny river for 2 hours. the return trip only takes 1.7 hours because the boat travels 2.5 miles per hour faster downstream due to the current. how far does the boat travel one way

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] time 

We know from our problem that the upstream trip takes 2 hours, so [tex]t_{u}=2[/tex]. We also know that the downstream trip takes 1.7 hours, so [tex]t_{d}=1.7[/tex]. Notice that the distance of both trips is the same, so we are going to use [tex]d[/tex] to represent that distance.
Now, lets use our equation to relate the quantities:

For the upstream trip:
[tex]S_{u}= \frac{d}{t_{u} } [/tex]
[tex]S_{u}= \frac{d}{2} [/tex] equation (1)

For the downstream trip:
[tex]S_{d}= \frac{d}{t_{d} } [/tex]
[tex]S_{d}= \frac{d}{1.7} [/tex] equation (2)

We know that the boat travels 2.5 miles per hour faster downstream, so the speed of the boat upstream will be the speed of the boat downstream minus 2.5 miles per hour:
[tex]S_{u}=S_{d}-2.5[/tex] equation (3)

Replacing (3) in (1):
[tex]S_{u}= \frac{d}{2} [/tex]
[tex]S_{d}-2.5= \frac{d}{2} [/tex] equation (4)

Solving for [tex]d[/tex] in equation (2):
[tex]S_{d}= \frac{d}{1.7} [/tex]
[tex]d=1.7S_{d}[/tex] equation (5)

Replacing (5) in (4):
[tex]S_{d}-2.5= \frac{d}{2} [/tex]
[tex]S_{d}-2.5= \frac{1.7S_{d}}{2} [/tex]
[tex]2(S_{d}-2.5)=1.7S_{d}[/tex]
[tex]2S_{d}-5=1.7S_{d}[/tex]
[tex]0.3S_{d}=5[/tex]
[tex]S__{d}= \frac{5}{0.3} [/tex]
[tex]S_{d}= \frac{50}{3} [/tex] equation (6)

Replacing (6) in (5)
[tex]d=1.7S_{d}[/tex]
[tex]d=1.7( \frac{50}{3} )[/tex]
[tex]d= \frac{85}{3} [/tex] miles

We can conclude that the boat travel [tex] \frac{85}{3} [/tex], which is approximately 28.3 miles, in one way.