document.write( "Question 79762: 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? \n" ); document.write( "
Algebra.Com's Answer #57236 by ankor@dixie-net.com(22740)\"\" \"About 
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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?
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\n" ); document.write( "Let t = time the 2nd cyclist is on the road when he catches up
\n" ); document.write( "Then (t+3) = time the 1st cyclist is on the road.
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\n" ); document.write( "When the 2nd cyclist catches up with the 1st, we know they will have traveled the same distance.
\n" ); document.write( ":
\n" ); document.write( "Write a distance equation: Distance = speed * time
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\n" ); document.write( "10t = 6(t+3)
\n" ); document.write( "10t = 6t + 18
\n" ); document.write( "10t - 6t = 18
\n" ); document.write( "4t = 18
\n" ); document.write( "t = 18/4
\n" ); document.write( "t = 4.5 hrs (time of the 2nd cyclist:
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\n" ); document.write( "Then the 1st cyclist time = 7.5 hrs
\n" ); document.write( "Check to see if both traveled the same distance
\n" ); document.write( "4.5(10) = 45 mi
\n" ); document.write( "7.5(6) = 45 mi also
\n" ); document.write( ":
\n" ); document.write( "Make sense to you, not that hard, right?
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