Respuesta :
recall your d = rt, distance = rate * time
now, if say, by the time they meet, Mr Cunningham has travelled "d" miles, that means Mrs Cunningham must also had travelled "d" miles as well.
However, he left 3 hours earlier, so by the time he travelled "d" miles, and took say "t" hours, for her it took 3 hour less, because she started driving 3 hours later, so, she's been on the road 3 hours less than Mr Cunningham, so by the time they meet, Mrs Cunningham has travelled then "t - 3" hours.
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{Mr Cunningham}&d&40&t\\ \textit{Mrs Cunningham}&d&80&t-3 \end{array} \\\\\\ \begin{cases} \boxed{d}=40t\\ d=80(t-3)\\ ----------\\ \boxed{40t}=80(t-3) \end{cases} \\\\\\ \cfrac{40t}{80}=t-3\implies \cfrac{t}{2}=t-3\implies t=2t-6\implies 6=2t-t \\\\\\ 6=t[/tex]
now, if say, by the time they meet, Mr Cunningham has travelled "d" miles, that means Mrs Cunningham must also had travelled "d" miles as well.
However, he left 3 hours earlier, so by the time he travelled "d" miles, and took say "t" hours, for her it took 3 hour less, because she started driving 3 hours later, so, she's been on the road 3 hours less than Mr Cunningham, so by the time they meet, Mrs Cunningham has travelled then "t - 3" hours.
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{Mr Cunningham}&d&40&t\\ \textit{Mrs Cunningham}&d&80&t-3 \end{array} \\\\\\ \begin{cases} \boxed{d}=40t\\ d=80(t-3)\\ ----------\\ \boxed{40t}=80(t-3) \end{cases} \\\\\\ \cfrac{40t}{80}=t-3\implies \cfrac{t}{2}=t-3\implies t=2t-6\implies 6=2t-t \\\\\\ 6=t[/tex]
