Let:
x = rate of the slower bus
y = rate of the faster bus
One bus travel 18 mi/h slower than the other, so:
[tex]x=y-18[/tex]the buses leave towns 296 miles apart at the same time and they meet in 2 hours, so:
[tex]\begin{gathered} 2x+2y=296 \\ where \\ x=y-18 \\ 2(y-18)+2y=296 \\ 2y-36+2y=296 \\ 4y-36=296 \\ 4y=296+36 \\ 4y=332 \\ y=\frac{332}{4} \\ y=\frac{83mi}{h} \end{gathered}[/tex]Replace the value of y into:
[tex]\begin{gathered} x=y-18 \\ x=83-18 \\ x=\frac{65mi}{h} \end{gathered}[/tex]Answer:
rate of slower bus:
[tex]\frac{65mi}{h}[/tex]rate of the faster bus:
[tex]\frac{83mi}{h}[/tex]