document.write( "Question 166425: 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 #122754 by Mathtut(3670)\"\" \"About 
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\n" ); document.write( "so lets call d1=r1*T1 1st cyclist T1=T+3 r1=6
\n" ); document.write( ".............d2=r2*T2 2nd cyclist T2=T r2=10
\n" ); document.write( ":
\n" ); document.write( "d1 has to equal d2 since they travel the same distance.
\n" ); document.write( ":
\n" ); document.write( "equations:
\n" ); document.write( "\"d1=6%28T%2B3%29\"
\n" ); document.write( "\"d1=10T\"
\n" ); document.write( "and since: \"d1=d2\"\r
\n" ); document.write( "\n" ); document.write( "then:\"6T%2B18=10T\"----->\"4T=18\" \"T=4.5\"
\n" ); document.write( ":
\n" ); document.write( "2nd cyclist was cycling \"T=4.5hours\" before he caught the 1st cyclist.\r
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