Question 90884
 plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
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Let the speed of the plane in calm air be "p".
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Against the wind DATA:
distance = 720 mi ; rate = p-30 ; time = d/r = 720/(p-30)
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With the wind DATA;
distance = 720 mi ; rate = p+30 ; time = d/r = 720/(p+30)
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EQUATION:
time + 1ime = 10 hrs.
720/(p-30) + 720/(p+30) = 10
72/(P-30)  + 72/(p+30) = 1
Multiply thru by P^2-900 to get:
72(p+30) + 72(p-30) = p^2-900
144p= p^2-900
p^2-144p-900 = 0
(p+6)(p-150)=0
p = 150 mph (speed of the plane in still air)
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Cheers,
Stan H.