Let x be the velocity (rate of change ) of one of the buses. We know that the other one travels 16 km/h slower; this means that the second velocity is:
![x-16](https://img.qammunity.org/qa-images/2023/formulas/mathematics/college/9yqbcm040br2hnnejp5y.png)
Now the combined velocity would be:
![2x-16](https://img.qammunity.org/qa-images/2023/formulas/mathematics/college/jn79p9kj63yz0xvg5g0f.png)
We know that the distance is equal to time by velocity, then we have that:
![4(2x-16)=1040](https://img.qammunity.org/qa-images/2023/formulas/mathematics/college/emkmi5tnef4hzi0itu32.png)
Solving for x we have:
![\begin{gathered} 4(2x-16)=1040 \\ 2x-16=(1040)/(4) \\ 2x-16=260 \\ 2x=260+16 \\ 2x=276 \\ x=(276)/(2) \\ x=138 \end{gathered}](https://img.qammunity.org/qa-images/2023/formulas/mathematics/college/frtnl54g1gm11rt3f4z1.png)
Therefore the rate of the faster bus is 138 km/h and the rate for the slower bus is 122 km/h.