document.write( "Question 73692: Hi - could someone please provide some help on this one? Thanks so very much in advance!\r
\n" ); document.write( "\n" ); document.write( "One train leaves a station heading due west. Two hours later a second train leaves the same station heading due east. The second train is traveling 15mi/h faster than the first. Six hours after the second train leaves, the two trains are 580 miles apart. Find the rate at which each train is traveling.\r
\n" ); document.write( "\n" ); document.write( "(I think I am getting confused by the fact that the second train leaves 2 hours after the first rather than at the same time)\r
\n" ); document.write( "\n" ); document.write( "Thanks again for any help on this one :>) \n" ); document.write( "
Algebra.Com's Answer #52728 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
One train leaves a station heading due west. Two hours later a second train leaves the same station heading due east. The second train is traveling 15mi/h faster than the first. Six hours after the second train leaves, the two trains are 580 miles apart. Find the rate at which each train is traveling.
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\n" ); document.write( "Let s = speed of train 1
\n" ); document.write( "(s+15) = speed of train 2
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\n" ); document.write( "Since train 1 started two hours earlier, it has been traveling for 8 hrs;
\n" ); document.write( "While the train 2 has been traveling 6 hrs:
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\n" ); document.write( "Dist = time * speed
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\n" ); document.write( "Train 1 dist + train 2 dist = 580
\n" ); document.write( "8s + 6(s+15) = 580
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\n" ); document.write( "8s + 6s + 90 = 580
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\n" ); document.write( "14s = 580 - 90
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\n" ); document.write( "14s = 490
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\n" ); document.write( "s = 490/14
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\n" ); document.write( "s = 35 mph, train 1's speed
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\n" ); document.write( "Train 2's speed = 35 + 15 = 50 mph
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\n" ); document.write( "Check:
\n" ); document.write( "8*35 + 6*50 =
\n" ); document.write( "280 + 300 = 580 \n" ); document.write( "
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