A plane flies 720 mi against a steady 30mih headwind and the

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?   

Solution

distance(d)=rate(r) times time(t) or d=rt; t=d/r and r=d/t
Let r=rate (speed) of plane in still air
time going =720/(r-30)
time returning=720/(r+30)
(Note: if the plane is flying against the wind, then we must subtract the wind speed from the planes speed in still air. If it flying with the wind then we add)
Now we are told that the time going + time returning=10 h, so:
720/(r-30)+720/(r+30)=10 multiply each term by (r-30)(r+30)
720(r+30)+720(r-30)=10(r-30)(r+30) get rid of parens
720r+21600+720r-21600=10r^2-9000 subtract 10r^2 and add 9000 to both sides
-10r^2+1440r+9000=0 divide each term by -10
r^2-144r-900=0 quadratic in standard form and it can be factored
(r+6)(r-150)=0
discount negative value --in this problem rate of speed is positive
r=150---------------------rate (speed) of plane in still air

CK
720/(150-30)+720/(150+30)= 10 h
6 h + 4 h=10 h
10 h=10 h