We know that two jets leave Harrisburg at the same, time, one flying east, and another flying west.
We will denote the speed of the second jet by x (in km/h). Thus, the speed of the first jet is x+20. Remembering that:
![v=(d)/(t)](https://img.qammunity.org/2023/formulas/mathematics/college/7bvf02ex7prlyl84jiizv8vikm7s8zddn1.png)
where v is speed, d is distance and t is time, we know that for the first jet:
![x+20=(d_1)/(4)\Rightarrow4x+80=d_1](https://img.qammunity.org/2023/formulas/mathematics/college/qwnwd355vv5ma5g8fr07xo2tlv199mu2zf.png)
Where d₁ represents the distance of the first jet from the starting point. For the second jet:
![x=(d_2)/(4)\Rightarrow4x=d_2](https://img.qammunity.org/2023/formulas/mathematics/college/k3bpb23iept814z3qu9zpu2eqegvztlsa6.png)
Where d₂ represents the distance of the second jet from the starting point.
We also know that:
![d_1+d_2=6000](https://img.qammunity.org/2023/formulas/mathematics/college/a5hjtr43ynuii9o0y3tz7bo7iddlwbr7jt.png)
As:
Thus, we have that:
![\begin{gathered} (4x+80)+(4x)=6000 \\ \text{And solving for x, we get:} \\ 8x+80=6000 \\ 8x=5920 \\ x=(5920)/(8)=740 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5qdx04i6es9x6t65q8z7p7ma90enpag94t.png)
This means that the second jet has a speed of 740km/h, and the first jet has a speed of 760km/h (20km/h greater than the second one).