Answer:
1100 and 130 (km/h)
Explanation:
1. if the velocity of the wind is 'w' and the velocity of the plane in still air is 'p', then
2. it is possible to make up two equations:
the fly against the wind: (p-w)*8=7760;
the fly with the wind: (p+w)*3=3690.
3. if to solve the system made up, then:
![\left \{ {{3(p+w)=3690} \atop {8(p-w)=7760}} \right. \ => \ \left \{ {{p+w=1230} \atop {p-w=970}} \right. \ => \ \left \{ {{p=1100} \atop {w=130}} \right.](https://img.qammunity.org/2022/formulas/mathematics/college/xjh6xjprd6u21qxuvl9pm38r8r454zae06.png)
4. the required rate of the plane in still air is p=1100 km/h; the rate of the wind is w=130 km/h.