SOLUTION: an airplane takes 5 hours to travel a distance of 2800 miles with wind. The return trip takes 5 hours and 36 minutes against the wind. Find the speed of the plane in still air, and

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Question 996403: an airplane takes 5 hours to travel a distance of 2800 miles with wind. The return trip takes 5 hours and 36 minutes against the wind. Find the speed of the plane in still air, and the speed of the wind. ( Hint: convert all time to hours)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+36%2F60+=+.6+
Let +s+ = the speed of the plane in still air in mi/hr
Let +w+ = the speed of the wind in mi/hr
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Equation for flying with the wind:
(1) +2800+=+%28+s+%2B+w+%29%2A5+
Equation for flying against the wind:
(2) +2800+=+%28+s+-+w+%29%2A5.6+
------------------------------
(1) +2800+=+5s+%2B+5w+
and
(2) +2800+=+5.6s+-+5.6w+
------------------------
+5.6+=+.8%2A7+
(2) +2800+=+.8%2A7%2As+-+.8%2A7%2Aw+
(2) +2800%2F.8+=+7s+-+7w+
(2) +7s+-+7w+=+3500+
----------------------
Multiply both sides of (1) by +7+
and both sides of (2) by +5+
then add the equations
(1) +35s+%2B+35w+=+19600+
(2) +35s+-+35w+=+17500+
------------------------
+70s+=+37100+
+s+=+530+
and
(1) +5s+%2B+5w+=+2800+
(1) +5%2A530+%2B+5w+=+2800+
(1) +2650+%2B+5w+=+2800+
(1) +5w+=+150+
(1) +w+=+30+
the speed of the plane in still air is 530 mi/hr
the speed of the wind is 30 mi/hr
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To check, plug numbers back into equations