Question 1005093
speed of the plane = 200 miles per hour.


let w = the speed of the wind.


with the wind, the combined speed of the plane and the wind is 200 + w.


against the wind, the combined speed of the plane and the wind is 200 - w.


the plane travels 330 miles with the wind and 330 miles against the wind.


total flying time is 3 and 1/3 hours = 10/3 hours.


let T1 = the amount of time it takes flying with the wind.


let T2 = the amount of time it takes flying against the wind.


the rate * time = distance equations become:


(200 + w) * T1 = 330


(200 - w) * T2 = 330


solve for T1 and T2 to get:


T1 = 330 / (200 + w)


T2 = 330 / (200 - w)


since T1 + T2 is equal to 3 and 1/3 hours = 10/3 hours, we get:


T1 + T2 = 10/3 hours.


after your replace T1 and T2 with their equivalents, the equation becomes:


330 / (200 + w) + 330 / (200 - w) = 10/3


multiply both sides of this equation by 3 and divide both sides of this equation by 10 to get:


99 / (200 + w) + 99 / (200 - w) = 1


multiply both sides of this equation by (200 + w) * (200 - w) to get:


99 * (200 - w) + 99 * (200 + w) = (200 + w) * (200 - w)


the algebra  gets a little ugly, but, when you solve this equation for w, you get:


w = 20 miles per hour.


you can confirm this solution is correct by replacing w in the original equations to see what you get:


the original equations to work with are:


T1 = 330 / (200 + w)


T2 = 330 / (200 - w)


solving for T1 and T2, using w = 20 miles per hour.


T1 = 1.5 hours


T2 = 1.833333 hours


T1 + T2 = 3.33333333..... which is the same as 3 and 1/3 which is the same as 10/3.


if you need help with the ugly, let me know.


otherwise i'll assume you can handle it.