Respuesta :
recall your d = rt, distance = rate * time
so... one train goes west and the goes the opposite way... alrite... so... notice, by the time 338 miles have been covered by both, it will have been "t" hours, and whatever "t" is, is the same amount of time the westbound train has been running as well as the eastbound train has been running.
now, let's say, since by "t" hours they've covered 338 altogether, so, if the westbound train has covered say "d" miles, then the eastbound train would have covered the slack from 338 and d, that is, "338 - d".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Westbound&d&55&t\\ Eastbound&338-d&75&t \end{array} \\\\\\ \begin{cases} \boxed{d}=55t\\ 338-d=75t\\ ----------\\ 338-\boxed{55t}=75t \end{cases} \\\\\\ 338=75t+55t\implies 338=130t\implies \cfrac{338}{130}=t\implies \stackrel{hours}{2.6}=t[/tex]
so, 2hours and 36 minutes.
so... one train goes west and the goes the opposite way... alrite... so... notice, by the time 338 miles have been covered by both, it will have been "t" hours, and whatever "t" is, is the same amount of time the westbound train has been running as well as the eastbound train has been running.
now, let's say, since by "t" hours they've covered 338 altogether, so, if the westbound train has covered say "d" miles, then the eastbound train would have covered the slack from 338 and d, that is, "338 - d".
[tex]\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Westbound&d&55&t\\ Eastbound&338-d&75&t \end{array} \\\\\\ \begin{cases} \boxed{d}=55t\\ 338-d=75t\\ ----------\\ 338-\boxed{55t}=75t \end{cases} \\\\\\ 338=75t+55t\implies 338=130t\implies \cfrac{338}{130}=t\implies \stackrel{hours}{2.6}=t[/tex]
so, 2hours and 36 minutes.
