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