At 10 AM two cars started traveling toward each other from t

At 10 A.M. two cars started traveling toward each other from towns 287 miles apart. They passed each other at 1:30 P.M. If the rate of the faster car exceeded the rate of the slower car by 6 miles per hour, find the rate, in miles per hour, of the faster car.

Solution

Faster car DATA:
time = 7/2 hr ; rate = x+6 mph ; distance = rt = (7/2)(x+6)
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Slower car DATA:
time = 7/2 hr ; rate = x mph ; distance = rt = (7/2)x
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Equation:
distance + distance = 287 miles
(7/2)(x+6) + (7/2)x = 287
7x + 21 = 287
7x = 266
x = 38 mph (speed of the slower car)
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x+6 = 44 mph (speed of the faster car)