recall your d = rt, distance = rate * time
thus [tex]\bf \begin{array}{lccclll}
&distance&rate&time(hrs)\\
&-----&-----&-----\\
\textit{first car}&d&55&x\\
\textit{second car}&380-d&75&x+1
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
\boxed{d}=(55)(x)\\\\
380-d=(75)(x+1)\\
----------\\
380-\left( \boxed{(55)(x)} \right)=(75)(x+1)
\end{cases}[/tex]
notice, the first car leaves at "x" time, the other leaves on hour later, or x + 1
the first car travels some distance "d", whatever that is, thus
the second car, picks up the slack, or the difference, they're 380 miles
apart, thus the difference is 380-d
