document.write( "Question 67731: Steve leaves Nashville to visit his cousin David in Napa 80 miles away. He travels at an average speed of 50 m/hr. Half an hour later, David leaves to visit Steve traveling at an average speed of 60 m/hr. How long after David leaves will they meet?\r
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\n" ); document.write( "\n" ); document.write( "(I would really appreciate it if you would be able to answer it for me because I had been trying to make up an equation this whole entire day. Thanks in advance.) \n" ); document.write( "
Algebra.Com's Answer #48183 by ankor@dixie-net.com(22740)\"\" \"About 
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Steve leaves Nashville to visit his cousin David in Napa 80 miles away. He travels at an average speed of 50 m/hr. Half an hour later, David leaves to visit Steve traveling at an average speed of 60 m/hr. How long after David leaves will they meet?
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\n" ); document.write( "\n" ); document.write( "When they do meet they will have traveled a total of 80 mi
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\n" ); document.write( "Let t = David's time
\n" ); document.write( "Then (t+.5) = Steve's time (he left a half hour earlier)
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\n" ); document.write( "Distance = speed * time
\n" ); document.write( "Steve's distance: 50(t+.5)
\n" ); document.write( "David's distance: 60t
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\n" ); document.write( "Steve's distance + David's distance = 80
\n" ); document.write( "50(t+.5) + 60t = 80
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\n" ); document.write( "50t + 25 + 60t = 80
\n" ); document.write( "110t = 80 - 25
\n" ); document.write( "110t = 55
\n" ); document.write( "t = 55/110
\n" ); document.write( "t = .5 or half an hour
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\n" ); document.write( "Check our solution
\n" ); document.write( "Steve would be driving for 1 hr and david for a half hour:
\n" ); document.write( "(50 * 1) + (60 * .5) = 80
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