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Question 1077582: The two engine plane and a Boeing jet take off from an airport at the same time for a city 720 miles away. The rate of the two engine plane is 180 mi/h less than that of the Boeing jet. If the Boeing jet arrives 2 hours ahead of the two engine plane, find each planes rate.
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! Let n be the speed of the small plane. Then the jet flies at n+180. So:
720/n=720/n+180 +2
720(n+180)=720n+2nē+360n
2nē+360n-129600=0
nē+180n-64800=0
(n+360)(n-180)=0
n=-360 or 180
Using the positive value, we get the speed of the two engine plane to be 180 mph, and that of the jet to be 360 mph. ☺☺☺☺
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