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:
[tex]x-16[/tex]Now the combined velocity would be:
[tex]2x-16[/tex]We know that the distance is equal to time by velocity, then we have that:
[tex]4(2x-16)=1040[/tex]Solving for x we have:
[tex]\begin{gathered} 4(2x-16)=1040 \\ 2x-16=\frac{1040}{4} \\ 2x-16=260 \\ 2x=260+16 \\ 2x=276 \\ x=\frac{276}{2} \\ x=138 \end{gathered}[/tex]Therefore the rate of the faster bus is 138 km/h and the rate for the slower bus is 122 km/h.
