document.write( "Question 1059912: If a bicyclist is traveling 41 miles at a certain speed, the return speed is made 6 mph slower. The total time for the round trip is 15 hours. What is the average speed of the bicyclist on the first trip?
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Algebra.Com's Answer #674961 by Boreal(15235)\"\" \"About 
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x=speed
\n" ); document.write( "x*t=distance
\n" ); document.write( "x*t=41, so the outbound time is 41/x
\n" ); document.write( "(x-6)*t=41
\n" ); document.write( "t=41/(x-6), which is the inbound time.
\n" ); document.write( "Both times add to 15
\n" ); document.write( "41/x+{41/(x-6)}=15
\n" ); document.write( "Clear fractions by multiplying everything by x(x-6)
\n" ); document.write( "41(x-6)+41x=15x(x-6)
\n" ); document.write( "41x-246+41x=15x^2-90x
\n" ); document.write( "15x^2-172x+246=0
\n" ); document.write( "x=(1/30)(172+/-sqrt (172^2-60*246); sqrt 14824=121.75
\n" ); document.write( "x=(1/30)(293.75)=9.79 mph
\n" ); document.write( "x=(1/30)(50.25)=1.675 mph, and that can't work because x-6 would be negative
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C15%2C-300%2C300%2C15x%5E2-172x%2B246%29\"
\n" ); document.write( "At 9.79 mph, it takes 4.19 hours outbound ANSWER 9.79 mph
\n" ); document.write( "At 3.79 mph, it takes 10.82 hours inbound
\n" ); document.write( "They add to 15.01, or 15 with rounding error.
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