SOLUTION: Preston rode his bike 4 mi. then he got a flat tire and had to walk back 4mi. it took him 1 hr longer than it did to ride. if his rate walking was 9mph less than his rate riding ,

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Question 887486: Preston rode his bike 4 mi. then he got a flat tire and had to walk back 4mi. it took him 1 hr longer than it did to ride. if his rate walking was 9mph less than his rate riding , find the two rates.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Preston rode his bike 4 mi. then he got a flat tire and had to walk back 4mi. it took him 1 hr longer than it did to ride. if his rate walking was 9mph less than his rate riding , find the two rates.
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Bike DATA:
dist = 4 miles ; time = x hrs ; rate = 4/x mph
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Walking DATA:
dist = 4 miles ; time = x+1 hrs ; rate = 4/(x+1)
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Equation
bike rate - walk rate = 9 mph
4/x - 4/(x+1) = 9
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4(x+1) - 4x = 9x(x+1)
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4 = 9x^2+9x
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9x^2 + 9x - 4 = 0
x = [-9+- sqrt(81-4*9*-4)]/18
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x = [-9 +-sqrt(225)]/18
x = [-9+15]/18
x = [1/3]
Bike rate = 4/x = 12 mph
Walk rate = 4(x+1) = 4/(4/3) = 3 mph
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Cheers,
Stan H.
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