SOLUTION: A man flies a small airplane a distance of 180 mi. Because he is flying into head wind, the trip takes him 2 hours. On the way back the wind is still blowing at the same speed so t

               SPEED     TIME       DISTANCE
GOING          r-w        t           d
RETURN         r+w        b           d

When the man covered 180 mi in 2 hours flying into head wind, the speed of the plane relative to the ground was  = 90 mph. 

This speed is the difference of the plane speed in the still air "u" and the wind's speed "v":

 = 90 = u - v.     (1)

When the man covered the same distance of 180 mi in 1 hour 12 minutes in the return trip flying with the wind, 
the speed of the plane relative to the ground was  = 150 mph. 
Here 1.2 = 1.2 hour = 1 hour and 12 minutes.  (12 minutes =  hour = 0.2 hour).

This speed is the sum of the plane speed in the still air "u" and the wind's speed "v":

 = 150 = u + v.   (2)


So you have this two equations to find u and v:

u - v =  90    (1')   
u + v = 150.   (2')

The simplest way to solve this system is to add the equations (1') and (2').
If you do, you will get

2u = 90 + 150,  or  2u = 240.  Hence,  u =  = 120.

Thus the speed of the plane in the still air is 120 mph.

Having this, you can easily find the speed of the wind v from the equation (1'):

v = u = 90 = 120 - 90 = 30 mph.

Answer.  The speed of the plane is still air is 120 mph.
         The speed of the wind is 30 ph.