If you put this info into a distance/rate/time table, you would have "against the wind" and "with the wind" as your 2 situations. If the distance from PA to AZ is 1800 miles, then the distance backwards, from AZ to PA is 1800 miles also. That number would go into the distance category for both situations. The rate is what we are looking for. The rate for "against the wind" is the speed of the plane minus the speed of the wind since the wind is blowing into the plane, slowing it down a bit. That looks like this: p - w where p is the speed of the plane and w is the speed of the wind. The rate for "with the wind" is p + w since the wind is pushing the plane along with it, making it fly faster. It took 3.6 hours to get from PA to AZ (you have to convert the 36 minutes into hours) and it took 3 hours to go from AZ to PA. In table form that looks like this:
distance = rate x time
against wind 1800 = p - w x 3.6
with wind 1800 = p + w x 3
Since distance = rate times time, then the first equation going along in the first row is 1800 = 3.6(p - w) and
1800 = 3.6p - 3.6w
Going along in the second row we have 1800 = 3p + 3w. Let's solve this equation for p: 1800 - 3w = 3p and dividing everything by 3 we have
600 - w = p. Now we have a viable substitution to make into the first equation we wrote. If p = 600 - w, then 1800 = 3.6(600 - w) - 3.6w
and 1800 = 2160 - 3.6w - 3.6w. Combining like terms we have
-360 = -7.2w and solving for w we have that w = 50. That means that the speed of the wind is 50 mph. Now we can sub that value in for w to solve for p. 600 - w = p so 600 - 50 = p. That means that p = 550mph. The wind blows at a rate of 50 mph while the plane travels at a rate of 550 mph. There you go!
