A plane travels from Orlando to Denver and back again. On the five-hour trip from Orlando to Denver, the plane has a tailwind of 40 miles per hour. On the return trip from Denver to Orlando, the plane faces a headwind of 40 miles per hour. This trip takes six hours. What is the speed of the airplane in still air?
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x/(y+40)=5, x/(y-40)=6 where y is the speed of the plane and x is the distance.
this gets y=440
how did you get that
First you determine your knowns and unknowns. You know the time it takes and the winds. You do not know the plane's speed(y) nor the distance(x) so assign variable to them. You have two trip with specific conditions to each so you can create two equations to model them. A tail wind means that the wind is pushing the plane so the total speed is y+40. You are given time and to get time you need to divide distance by speed. So x/(y+40)=5. In the other case a headwind causes the plane to slow down so the speed is (y-40) and again to get time you divide distance by speed. x/(y-40)=6. 2 equations with 2 unknowns is possible to solve. Thus solve for y because you dont need to know x's value.