document.write( "Question 532320: A cyclist and a jogger start from a town at the same time and head for a destination 19 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.9 h before the jogger. Find the rate of the cyclist. \n" ); document.write( "
Algebra.Com's Answer #350810 by ankor@dixie-net.com(22740)\"\" \"About 
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A cyclist and a jogger start from a town at the same time and head for a destination 19 mi away.
\n" ); document.write( "The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.9 h before the jogger.
\n" ); document.write( " Find the rate of the cyclist.
\n" ); document.write( ":
\n" ); document.write( "let r = the rate of the jogger
\n" ); document.write( "then
\n" ); document.write( "2r = rate of the cyclist
\n" ); document.write( ":
\n" ); document.write( "Write a time equation, time = dist/speed
\n" ); document.write( ":
\n" ); document.write( "jogger time - cyclist time = 1.9 hrs
\n" ); document.write( "\"19%2Fr\" - \"19%2F%282r%29\" = 1.9
\n" ); document.write( "multiply by 2r, results:
\n" ); document.write( "2(19) - 19 = 2r(1.9)
\n" ); document.write( "38 - 19 = 3.8r
\n" ); document.write( "19 = 3.8r
\n" ); document.write( "r = \"19%2F3.8\"
\n" ); document.write( "r = 5 mph is the rate of the jogger
\n" ); document.write( "then
\n" ); document.write( "2(5) = 10 mph is the rate of the cyclist
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check this find the actual time of each
\n" ); document.write( "19/5 = 3.8 hrs
\n" ); document.write( "19/10= 1.9 hrs
\n" ); document.write( "---------------
\n" ); document.write( "time dif: 1.9 hrs
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