<PRE><B>Question 661473</B>
Two friends living 39 mi apart, leave their homes at the same time on bicycles and travel toward each other.If one person averages 2 mi/ hr more than the other,and they meet in 11/2 hour,what is each persons average rate of cycling?
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Faster cyclist:
time = 11/2 hrs ; distance =  x miles ; rate = x/(11/2) = (2x/11) mph
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Slower cyclist:
time = 11/2 hrs ; distance = (39-x) miles ; rate = (39-x)/(11/2) = (78-2x)/11 mph
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Equation:
Faster - Slower = 2 mph
(2x/11) - (78-2x)/11 = 2 mph
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2x -(78-2x) = 22
4x = 100
x = 25 miles
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Faster rate = (2x/11) = 50/11 = 4.55 mph
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Slower cyclist = (78-2x)/11 = (78-50)/11 = 28/11 = 2.55 mph
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Cheers,
Stan H.

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