Question

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, 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 1.5 hours. What is the plane's speed in still air, and how fast is the wind blowing?

Answer 1

Now, you have two choices, you can use elimination or substitution, in order to address the multiple variables. I will use substitution. Since both flights traveled the same distance, we can set the rt portions of our equations equal to each other and solve for one variable: rt=rt (r-w)2=(r+w)1.2 2r-2w=1.2r+1.2w 0.8r=3.2w r=4w Now, we can plug in 4w for r, in either equation: 180=(r+w)1.2 180=(4w+w)1.2 180=5w(1.2) 150=5w 30=w Since we know that r=4w, we can plug in w, to solve for r: r=4w r=4(30) r=120 Now we know our plane speed and wind speed. If we want, we can plug both into our second equation, in order to check the values: 180=(r-w)2 180=(120-30)2 180=90(2) 180=180