document.write( "Question 204814: Two cyclists start biking from a trails 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 caches up with the first from the time the second cyclist started biking? \n" ); document.write( "
Algebra.Com's Answer #154609 by ankor@dixie-net.com(22740)\"\" \"About 
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Two cyclists start biking from a trails start 3 hours apart. the second cyclist
\n" ); document.write( " travels at 10 miles per hour and starts 3 hours after the first cyclist who is
\n" ); document.write( " traveling at 6 miles per hour. How much time will pass before the second
\n" ); document.write( "cyclist caches up with the first from the time the second cyclist started biking?
\n" ); document.write( ":
\n" ); document.write( "Let t = travel time of the 2nd cylist
\n" ); document.write( "then
\n" ); document.write( "(t+3) = travel time of the 1st cylist
\n" ); document.write( ":
\n" ); document.write( "When the 2nd catches the 1st, they will have traveled the same distance
\n" ); document.write( "Write dist equation: Dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "2nd cyclist dist = 1st cyclist 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 to catch the 1st
\n" ); document.write( ":
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\n" ); document.write( "Check solution by finding the distance of each (should be equal)
\n" ); document.write( "10 * 4.5 = 45 mi
\n" ); document.write( "6 * 7.5 = 45 mi \n" ); document.write( "
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