document.write( "Question 315060: 2. Two cyclists start biking from a trial and start 3 hours apart. The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph. How much time will pass before the 2nd cyclist catches up with the 1st from the time the 2nd cyclist started biking?\r
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Algebra.Com's Answer #225362 by ankor@dixie-net.com(22740)\"\" \"About 
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The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph.
\n" ); document.write( " How much time will pass before the 2nd cyclist catches up with the 1st from the time the 2nd cyclist started biking?
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\n" ); document.write( "Let t = travel time of the 2nd cyclist
\n" ); document.write( "then
\n" ); document.write( "(t+3) = travel time of the 1st cyclist
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\n" ); document.write( "When the 2nd overtakes the 1st, they will have traveled the same distance
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\n" ); document.write( "Write distance equation; dist = speed * time
\n" ); document.write( "2nd dist = 1st dist
\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%2F4\"
\n" ); document.write( "t = 4.5 hrs for the 2nd cyclist to catch the first
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\n" ); document.write( "Check solution by finding distance of each, should be equal:
\n" ); document.write( "10*4.5 = 45 mi
\n" ); document.write( "6 *7.5 = 45 mi
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