Answer:
[tex]\large \boxed{\text{ Car = 60 mi/h: bus = 20 mi/h}}[/tex]
Step-by-step explanation:
A. Car rate
Let c = the car rate
Then c - 40 = bus rate
Distance = rate × time
Time = distance/rate
[tex]\begin{array}{rcll}\dfrac{540}{c} & = & \dfrac{180}{c - 40} & \\\\\dfrac{540(c - 40)}{c} & = & 180 &\text{Multiplied each side by 180 - c}\\\\540(c - 40) & = & 180c & \text{Multiplied each side by c}\\540c - 21600 & = & 180c & \text{Distributed the 540}\\360c -21600 & = & 0 & \text{Subtracted 180c from each side}\\\end{array}\\[/tex]
[tex]\begin{array}{rcll}360c & = &21600 &\text{Added 21600 to each side}\\c & = & \dfrac{21600}{360} & \text{Divided each side by 360}\\\\c & = & 60 &\text{Simplified}\\\end{array}\\\text{The average rate of the car is $\large \boxed{\textbf{60 mi/h}}$}[/tex]
B. Bus rate
[tex]\text{Bus rate} = c - 40 =60 - 40 = \mathbf{20}\\\text{The average rate of the bus is $\large \boxed{\textbf{20 mi/h}}$}[/tex]
Check:
[tex]\begin{array}{rcl}\dfrac{540}{60} & = & \dfrac{180}{20}\\\\9 & = & 9\\\end{array}[/tex]
OK.
