document.write( "Question 155108: 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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Algebra.Com's Answer #114205 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "Let t=amount of time that passes before the second cyclist catches up with the first from the time the second cyclist starts
\n" ); document.write( "distance first cyclist travels=6*3+6*t(3 hour head start)
\n" ); document.write( "distance second cyclist travels=10*t
\n" ); document.write( "Now when the above two distances are equal, the second cyclist will have caught up with the first cyclist, so our equation to solve is:
\n" ); document.write( "10t=18+6t subtract 6t from each side
\n" ); document.write( "10t-6t=18-6t
\n" ); document.write( "4t=18
\n" ); document.write( "t=4.5 hours \r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "3*6+6*4.5=10*4.5
\n" ); document.write( "18+27=45
\n" ); document.write( "45=45\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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