The Solution:
Let the speed of the jet without the wind be x and the speed of the wind be y.
By formula,
![\text{ Speed (S) =}\frac{\text{ distance (d) }}{\text{ time (t) }}](https://img.qammunity.org/2023/formulas/mathematics/college/gp7yz0dxad5g07r56t53luj59rwyav4rqj.png)
So,
With the Tailwind:
d=distance = 2660 miles
t=time = 4 hours
s = speed = (x+y) m/h
Substituting these values in the formula above, we get
![\begin{gathered} (x+y)=(2660)/(4) \\ \\ x+y=665\ldots\text{eqn}(1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8hg63lznlh2ixb15u1mdhczp6hmyfsurr1.png)
Wind the Headwind:
d = 2492 miles
t = 4 hours
speed = (x-y) m/h
Substituting, we get
![\begin{gathered} (x-y)=(2492)/(4) \\ \\ x-y=623\ldots\text{eqn}(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4d5rp5zx0q6wsn949q8bvqzuraxsedzu6x.png)
Solving eqn(1) and eqn(2) simultaneously by the Elimination Method.
![\begin{gathered} x+y=665 \\ x-y=623 \\ -------- \\ 2x=1288 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p91bd1y2snrvzk8p97vmmr5ozkwthrfz3y.png)
Dividing both sides by 2, we get
![x=(1288)/(2)=644\text{ miles/hour}](https://img.qammunity.org/2023/formulas/mathematics/college/sqjmdbrz7o12ybz6c31x7oijsz5pgwdvfs.png)
Thus, the speed of the jet is 644 miles/hour.
The speed of the wind ( the value of y):
We shall substitute 644 for x in eqn(1).
![\begin{gathered} 644+y=665 \\ y=665-644 \\ y=21\text{ miles/hour} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3lt8hs5gjgltyg2rusofwpb9y4d0w4jxck.png)
Therefore, the speed of the wind is 21 miles/hour.