Let \(V\), \(V_a\) and \(V_w\) be the total speed, speed of airplane on still air and speed of the wind respectively .

Let \(t\) be the time \((hrs)\) it took the plane to travel 660 miles, thus the time it took to travel againts the wind is \(t + \frac{2}{3}\).

\(V = \frac{distance}{time}\)

With the wind: \(V = V_a + Vw\)

Againts the wind: \(V = V_a - V_w\)

\(\frac{660}{t} = 200 + Vw\)

\(\frac{660}{t+\frac{2}{3}} = 200 - Vw\)

Add the previous two equations.

\(\frac{660}{t} + \frac{660}{t+\frac{2}{3}} = 400 +0\)

\(660(t+\frac{2}{3} + t) = 400t(t+\frac{2}{3})\)

\(\frac{33}{20}(2t + \frac{2}{3}) = t^2 + \frac{2t}{3}\)

\(\frac{33t}{10} + \frac{11}{10} - t^2- \frac{2t}{3} = 0\)

\(\frac{79t}{30}-t^2 +\frac{11}{10} = 0\)

\(30t^2 -79t -33 = 0\)

\(x_1 = \frac{79+\sqrt{79^2 - 4(30)(-33)}}{60}\)

\(x_1 = 3 ~hrs\)

\(x_2 = \frac{79-\sqrt{79^2 - 4(30)(-33)}}{60}\)

\(x_2 = -0.37\)

Consider the positve root and solve for the speed of the wind.

\(\frac{660}{3} = 200 + V_w\)

\(V_w = 20 ~mph\) Answer