SOLUTION: I do not know if I chose the right category, but this is problem solving word problem. John rides a bike going 8mph on a route that is 4miles less than the route on the return

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Question 1139823: I do not know if I chose the right category, but this is problem solving word problem.
John rides a bike going 8mph on a route that is 4miles less than the route on the return trip. He travels 7mph on the return trip. If it takes him an hour longer to return then go, how much total time does he spend riding his bike?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
                SPEED     TIME       DISTANCE

Trip Going         8      (d-4)/8      d-4

Return Trip        7        d/7        d

DIFFERENCE                 1

d%2F7-%28d-4%29%2F8=1
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Solve for d, and evaluate find sum %28d-4%29%2F8%2Bd%2F7.

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8d-7%28d-4%29=56
87-7d%2B28=56
d=56-28
d=28
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Total_Time_both_ways24%2F8%2B4=3%2B4=highlight%287%29hours

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The trip going is 4 miles shorter than the trip returning:

let x = length (miles) of trip returning
then x-4 = length of trip going

His speed going is 8mph; his speed returning is 7mph. So

x/7 = time returning
(x-4)/8 = time going

His time returning is 1 hour more than his time going:

x%2F7+=+%28x-4%29%2F8%2B1

That turns into a linear equation that is easily solved....