Question 1120070
A man flies a small airplane from Fargo to Bismarck, North Dakota a distance of 180 miles.
 Because he is flying into a headwind, the trip takes him 2 hours.
 On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour 12 minutes.
 Change 1 hr 12 min to 1.2 hrs
 What is his speed in still air, and how fast is the wind blowing?
:
let s = aircraft speed in still air
let w = speed of the wind
then
(s-w) = effective speed into a headwind
and
(s+w) = effective speed with the wind
:
Write dist equation for each way; (dist = time * speed
2(s-w) = 180
1.2(s+w) = 180
:
Simplify, divide the 1st equatio by 2 and the 2nd equation by 1.2
s - w = 90
s + w = 150
--------------addition eliminates w, find s
2s + 0 = 240
s = 240/2
s = 120 mph the speed in still air
and 
12 + w = 150
w = 30 mph is the wind
:
;
Check this; find the distance with these values
2(120-30) = 180
1.2(120+30) = 180