An airplane took 4 hours to fly 2400 miles against headwind. The return trip with the wind took 3 hours. Find the speed of the plane in still air and the speed of the wind.

Use distance = rate x time

For the trip there:

2400 = rate x 4

where rate = A (airplane speed) - W (wind speed)

so:

2400 = (A - W) x 4

The return trip:

2400 = (A + W) x 3

Now solve. Since they both equal 2400, I'll set them equal to each other:

(A - W) x 4 = (A + W) x 3

4A - 4W = 3A + 3W

A = 7W

Now substitute back into one of the original equations:

2400 = (7W - W) x 4

2400 = 6W x 4

2400 = 24W

100 = W

Therefore A = 700

So airplane speed in still air is 700 mph and the wind speed is 100 mph.

Speed in still air: x

speed of wind: y

speed against headwind is x-y = 2400/4=600

speed of return trip x+y = 2400/3=800

solve the linear equation:

add the two equations to eliminate y:

2x=1400

x=700

y=100