Respuesta :
Let x be the distance traveled by one of the buses (let's call it Bus A) and y be the rate of the other bus (let's call it Bus B).
Let's remember the relation that exist between the speed (rate), the distance and the time:
[tex]V=\frac{d}{t}[/tex]We know that it takes 3 hours for Bus A to travel x miles at a speed of y+15 (it is 15 faster than Bus B). This expressed as an equation is:
[tex]y+15=\frac{x}{3}[/tex]We also know that it takes 3 hours for Bus B to travel 405-x miles at a speed of y:
[tex]y=\frac{405-x}{3}[/tex]Solve the obtained system of equations by equalization:
[tex]\begin{gathered} y+15=\frac{x}{3} \\ x=3(y+15) \\ y=\frac{405-x}{3} \\ x=405-3y \end{gathered}[/tex][tex]\begin{gathered} x=x \\ 3y+45=405-3y \\ 6y=360 \\ y=\frac{360}{6} \\ y=60 \end{gathered}[/tex]The rate of Bus B is 60 and the rate of Bus A is 75 (Bus A is 15 faster than Bus B).
