kelprofessor said:
An airplane can fly 360 miles into the wind in 3 hours. If the plane reverses the direction and thewind conditions remain the same, the return trip takes only two hours. Find the speed of the plane in still air.
I am unable to set up the equations.
Let x = speed of plane in still air
Let y = wind speed
Flying INTO the wind, the speed of the plane in still air is REDUCED by the wind speed. So, flying into the wind, the plane makes (x - y) miles per hour.
To fly 360 miles at (x - y) mph, it will take 360/(x - y) hours. This comes from the basic relationship d=r/t, or d/r = t. And we know it takes the plane 3 hours to do this....so,
360 / (x - y) = 3
or, 360 = 3(x - y)
Or, if you divide both sides of the equation by 3,
120 = x - y
<----here is one of your equations
Flying WITH the wind, the speed of the wind increases the speed of the plane in still air. If the plane can fly x mph in still air, WITH a tailwind of y mph, the plane's speed will be (x + y) mph. If the plane flies 360 miles at (x + y) mph, the time it takes will be 360/(x + y) hours.
We're told that the time for this part of the trip is 2 hours, so
360 / (x + y) = 2
or,
360 = 2(x + y)
or, if you divide both sides of the equation by 2,
180 = x + y
<----this is your second equation
Ok...
x - y = 120
x + y = 180
There's your system. And remember that the question asks you for the speed of the plane in still air. Look at how we defined the variables...x is the speed of the plane in still air.