document.write( "Question 242666: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?... They tell me the answer is 4.5 hours, but I can't figure out how to get there. \n" ); document.write( "
Algebra.Com's Answer #177637 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Two cyclists start biking from a trail's start 3 hours apart.
\n" ); document.write( "The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour.
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\n" ); document.write( "1st cyclist DATA:
\n" ); document.write( "rate = 6 mph ; time = t hrs. ; distance = 6t miles
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\n" ); document.write( "2nd cyclist DATA:
\n" ); document.write( "rate = 10 mph ; time = t-3 hrs ; distance = 10(t-3) miles
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\n" ); document.write( "\n" ); document.write( "How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
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\n" ); document.write( "Equation:
\n" ); document.write( "distance = distance
\n" ); document.write( "6t = 10(t-3)
\n" ); document.write( "6t = 10t - 30
\n" ); document.write( "4t = 30
\n" ); document.write( "t = 7.5 hrs (time for cyclist # 1)
\n" ); document.write( "t-3 = 4.5 hrs (time for cyclist # 2)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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