document.write( "Question 60881: A slow train leaves the station, 2 hrs later a fast train, with double the speed of the slow train, leaves the station and catches up with the slow train after traveling 180 mi. Find the speed of both trains. \n" ); document.write( "
Algebra.Com's Answer #41875 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
A slow train leaves the station, 2 hrs later a fast train, with double the speed of the slow train, leaves the station and catches up with the slow train after traveling 180 mi. Find the speed of both trains.
\n" ); document.write( ":
\n" ); document.write( "When the two trains meet, they both will have traveled 180 miles
\n" ); document.write( ":
\n" ); document.write( "Let s = speed of the slow train
\n" ); document.write( "Let 2x = speed of the fast train
\n" ); document.write( ":
\n" ); document.write( "Write a time equation:
\n" ); document.write( "t = dist/ speed
\n" ); document.write( ":
\n" ); document.write( "slow train time = fast train time + 2 hr
\n" ); document.write( "180/s = 180/2s + 2
\n" ); document.write( "180/s = 90/s + 2
\n" ); document.write( ":
\n" ); document.write( "Get rid of the denominators, mult equation by 2s
\n" ); document.write( "180 = 90 + 2s
\n" ); document.write( "180 - 90 = 2s
\n" ); document.write( "2s = 90
\n" ); document.write( "s = 45 mph is the slow train.
\n" ); document.write( ":
\n" ); document.write( "You find the speed of the fast train \n" ); document.write( "
\n" );