Question 44686
A 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?

With the wind DATA:
distance=720 mi;rate=plane speed + 30mph; time=d/r=720/(p+30)

Against the wind DATA:
distance=720 mi; rate=plane speed -30 mph; time=720/(p-30)

EQUATION:
time with + time against = 10 hr.
720/(p+30) + 720/(p-30) = 10

Divide thru by 10 to get:

72/(p+30) + 72/(p-30)=1
LCM=p^2-900

Multiply thru by the LCM to get:
72(p-30) + 72(p+30)=p^2-900
72p-72*30+72p+72*30=p^2-900
144p=p^2-900
p^2-144p-900=0
(p+6)(p-150)=0
p=-8 or p=150

plane speed in still air = 150 mph is the only reasonable answer.

Cheers,
Stan H.