SOLUTION: Elizabeth drove to Thomas' house at 57 mph.Thomas' house is seventy-one miles away.Elizabeth arrived at Thomas' house at 4:27 p.m. What time did she leave?

You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let t=amount of time required for Elizabeth to make the trip to Thomas' house
Now we know that t=d/r=71/57; t=1.25 hrs or 1 hr 15min
So 4:27 p.m. minus 1hr 15min=3:12 p.m.
Hope this helps---ptaylor