A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the plane took 4 hours. Find the wind speed and the speed of the plane in still air.

Answer:Wind speed: [tex]\rm 40\; km \cdot h^{-1}[/tex].Speed of the plane in still air: [tex]\rm 320\; km \cdot h^{-1}[/tex].Step-by-step explanation:This problem involves two unknowns:wind speed, andspeed of the plane in still air.Let the speed of the wind be [tex]x \rm \; km \cdot h^{-1}[/tex], and the speed of the plane in still air be [tex]y\rm \; km \cdot h^{-1}[/tex]. It takes at least two equations to find the exact solutions to a system of two variables.Information in this question gives two equations:It takes the plane three hours to travel [tex]\rm 960\; km[/tex] from Ottawa to with a tail wind (that is: at a ground speed of [tex]x + y[/tex].)It takes the plane four hours to travel [tex]\rm 960\; km[/tex] from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of [tex]-x + y[/tex].)Create a two-by-two system out of these two equations:[tex]\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right.[/tex].There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.Note that this system is equivalent to[tex]\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right.[/tex].[tex]\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right.[/tex].Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only [tex]1 \times 960[/tex] on the right-hand side. In comparison, adding them will give [tex]7 \times 960[/tex], which is much more complex to evaluate. Subtracting the second equation ([tex]3 \times (2)[/tex]) from the first ([tex]4 \times (1)[/tex]) will give the equation[tex](12 - (-12) x = 1 \times 960[/tex].[tex]24 x = 960[/tex].[tex]x = 40[/tex]. Substitute [tex]x[/tex] back into either equation [tex](1)[/tex] or [tex](2)[/tex] of the original system. Solve for [tex]y[/tex] to obtain [tex]y = 320[/tex].In other words, Wind speed: [tex]\rm 40\; km \cdot h^{-1}[/tex].Speed of the plane in still air: [tex]\rm 320\; km \cdot h^{-1}[/tex].