# SOLUTION: I know that this motion problem involves a system of two equations and I could solve it if I could only figure out how to set it up correctly. I have been using the Distance = Ra

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Travel -> SOLUTION: I know that this motion problem involves a system of two equations and I could solve it if I could only figure out how to set it up correctly. I have been using the Distance = Ra      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Travel Word Problems Question 5092: I know that this motion problem involves a system of two equations and I could solve it if I could only figure out how to set it up correctly. I have been using the Distance = Rate x Time formula to try and set it up, but I just can't figure it out and it's driving me nuts! Here's the problem: "An express train and a local train leave Gray's Lake at 3PM and head for Chicago 50 miles away. The express travels twice as fast as the local and arrives 1 hour ahead of the local. Find the speed of each train." Here's what I've tried to do to set this up. I've tried to set up a table so I can see what my equations should be. I'll try to approximate below. Using the Rate x Time = Distance formula. Let s=The speed the train is traveling and let t=the time that it takes to get to Chicago. Rate Time Distance Local s x t =50 Express 2(s)x t-1 =50 The above is as far as I get, but I don't think it's right and thus won't get me to the speed that each train is traveling at. Help! Karen Henry Answer by rapaljer(4667)   (Show Source): You can put this solution on YOUR website!I think your mistake is not putting parentheses around the time (t-1). As you said, let s = speed of the local 2s = speed of the express t= time of the local t-1 = time of the express Distances are equal at 50 miles, so (rate times time) = (rate times time) s*(t) = 2s *(t-1) Since s the speed of the train does not equal zero, you can divide both sides by s: t= 2(t-1) t=2t -2 -t = -2 t=2 hours for the local t-1 = 1 hour for the express After you do that, then remember that D=RT, so Rate for the local = =25 mph Rate for the express = = 50 mph. R^2 at SCC