Respuesta :
We will denote by x the rate of the airplane, assuming its constant, and by y, the speed of the wind.
This means:
• x+y is the speed of the airplane flying with the wind.
,• x-y is the speed of the airplane flying against the wind.
Remembering that when the speed is constant, speed equals to distance over time, then distance equals speed by time. Having in mind that when the airplane against the wind, it travels 2490 kilometers in 3 hours, we get:
[tex](x-y)\cdot3=2490[/tex]And similarly, as when flies with the wind, it travels 10320 kilometers in 8 hours:
[tex](x+y)\cdot8=10320[/tex]We obtain the following system:
[tex]\mleft\{\begin{aligned}x-y=\frac{2490}{3} \\ x+y=\frac{10320}{8}\end{aligned}\mright.[/tex]This is,
[tex]\mleft\{\begin{aligned}x-y=830 \\ x+y=1290\end{aligned}\mright.[/tex]We will solve the system by substitution. On the first equation:
[tex]x=830+y[/tex]And replacing onto the second one:
[tex]\begin{gathered} (830+y)+y=1290 \\ 2y=1290-830 \\ 2y=460 \\ y=\frac{460}{2}=230 \end{gathered}[/tex]And, thus, the value of x will be:
[tex]x=y+830=230+830=1060[/tex]This means that the rate of the plane in still air is 1060 km/h, and the rate of the wind is 230km/h.
