Answer :
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
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