recall d = rt. distance = rate * time.
[tex]\bf \boxed{A}\stackrel{8~mph}{\rule[0.35em]{15em}{0.25pt}} \\\\\\ \boxed{B}\stackrel{7~mph}{\rule[0.35em]{13em}{0.25pt}}[/tex]
so say we have two folks, A and B, both going the same direction, one slower than the other.
after a certain "t" hours, they both will be 0.25 or 1/4 of a mile apart, since A is the faster one, then A will be ahead.
by the time "t" hours hit, A has been walking "t" hours and B has also been walking "t" hours. If by then say B has covered "d" miles, then since A is ahead, A must have covered by then "d + ¼" miles.
[tex]\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ A&d+\frac{1}{4}&8&t\\ B&d&7&t \end{array}~\hspace{4em} \begin{cases} d+\cfrac{1}{4}=8t\\[1em] \boxed{d}=7t \end{cases} \\\\\\ \boxed{7t}+\cfrac{1}{4}=8t\implies \cfrac{1}{4}=t\impliedby \begin{array}{llll} \textit{after }\frac{1}{4}\textit{ of an hour}\\\\ \textit{or 15 minutes} \end{array}[/tex]
