SOLUTION: Justin recently drove to visit his parents who live 100 miles away. On his way there his average speed was 9 miles per hour faster than on his way home (he ran into some bad weathe

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Justin recently drove to visit his parents who live 100 miles away. On his way there his average speed was 9 miles per hour faster than on his way home (he ran into some bad weathe      Log On


   



Question 992632: Justin recently drove to visit his parents who live 100 miles away. On his way there his average speed was 9 miles per hour faster than on his way home (he ran into some bad weather). If Justin spent a total of 5 hours driving, find the two rates.
Answer by stanbon(75887) About Me  (Show Source):
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Justin recently drove to visit his parents who live 100 miles away. On his way there his average speed was 9 miles per hour faster than on his way home (he ran into some bad weather). If Justin spent a total of 5 hours driving, find the two rates.
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Outward bound::
rate = x+9 ; dist = 100 m ; time = 100/(x+9) hrs
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Homeward
rate = x mph ; dist = 100 m ; time = 100/x hrs
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Equation:
time + time = 5 hrs
100/(x+9) + 100/x = 5 hrs
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100x + 100(x+9) = 5x(x+9)
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200x + 900 = 5x^2 + 45x
5x^2 - 155x - 900 = 0
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x^2 - 31x - 180 = 0
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x = 36 mph
x+9 = 45 mph
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Cheers,
Stan H.
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