SOLUTION: at 7 am two cars started from the same place, one traveling east and the oher traveling west. at 10:30 am they were 287 miles apar. if the rate of the fast car exceeded the rate of

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Question 165963: at 7 am two cars started from the same place, one traveling east and the oher traveling west. at 10:30 am they were 287 miles apar. if the rate of the fast car exceeded the rate of the slow car by 6 mph, find the rate of each car.
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Remember: FastCar=FC; SlowCar=SC
In the condition: FC exceeded SC by 6mph ---> FC=SC%2B6mph, EQN 1
We know that Speed=Distance%2FTime and Speed=SC%2BFC that covered 287 miles in 3.5 hours (10:30-7:00=3.5 hrs) right?
.
Putting into eqn all given via our formula:
SC%2BFC=287%2F3.5, plug in EQN 1:
SC%2BSC%2B6=287%2F3.5
%282SC%2B6%293.5=287, cross multiply
7SC%2B21=287 ----> 7SC=287-21=266
cross%287%29SC%2Fcross%287%29=cross%28266%2938%2Fcross%287%29
highlight%28SC=38mph%29, Speed of Slow Car
Via EQN 1,
highlight%28FC=38%2B6=44mph%29, Speed of Fast Car
In doubt? Both cars should travlled a Distance of 287 miles in 3.5 hrs. Let's see,
%28SC%2BFC%29%283.5%29=287mi
%2838%2B44%29%283.5%29=287
81%2A3.5=287
287mi=287mi
Thank you,
Jojo