Answer:
The rate of the jet in still air = 1030 miles per hour
And 210 miles per hour is the rate of the stream.
Explanation:
Let the speed of the jet in still air be
miles per hour.
And,the speed of the jetstream be
miles per hour.
Key:So when it is moving against the stream speed will be
and when it is moving with the stream, speed will be
![(x+y)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/y1xl6hy7ejg6ylilssi69588350b2ih8ww.png)
And we know that.
Distance=Speed multiplied with time.
Arranging the equation.
...equation 1
...equation 2
Working with the equation by using elimination method.
Dividing equation 1 with 8 and equation 2 with 9.We will have
1240=
...eqn 3
820=
...eqn 4
Now adding both the equation will decimate y.
1240+820=2
![x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3ozza40nv61jy1offmxaxutrb6y1c3ly5.png)
2060=2
![x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3ozza40nv61jy1offmxaxutrb6y1c3ly5.png)
Dividing with 2 both sides.
1030=
![x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3ozza40nv61jy1offmxaxutrb6y1c3ly5.png)
By plugging the value of
we can find y in equation 3.
1240=1030 +
![y](https://img.qammunity.org/2020/formulas/mathematics/college/uw0b7dbqmfpakodpw1nh8u5h9nrcutx8vw.png)
Subtracting 1030 from both sides.
1240-1030=
![y](https://img.qammunity.org/2020/formulas/mathematics/college/uw0b7dbqmfpakodpw1nh8u5h9nrcutx8vw.png)
=210 miles per hour
So the rate of the jet in still air = 1030 miles per hour
And 210 miles per hour is the rate of the jet stream.