Question 110068
a man rode a bycycle for 12 miles then hiked an additional 8 miles . the total time for the trip was 5 hrs . if the rate when he was riding his bicycle was 10 miles per hour faster than his rate walking , what was each rate 
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Bike DATA:
distance = 12 miles ; rate= x mph ; time = d/r = 12/x hrs
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Walk DATA:
distance = 8 miles ; rate = x-10 mph ; time = d/r = 8/(x-10) hrs
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EQUATION:
time + time = 5 hrs.
12/x + 8/(x-10) = 5
3/x + 2/(x-10) = 5/4
[3x-30+2x] = (5/4)x(x-10)
5x-30 = (5/4)x(x-10)
x-6 = [x^2-10x)/4
x^2-10x = 4x-24
x^2-14x+24=0
x^2-2x-12x+24=0
x(x-2)-12(x-2)=0
(x-2)(x-12)=0
x = 2 or x = 12
Positive answer:
x-10 = 2 mph (walking rate)
x=12 mph (biking rate)

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Cheers,
Stan H.