document.write( "Question 58754: Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longer than Smith’s. How fast was each one traveling. \r
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Algebra.Com's Answer #40365 by ankor@dixie-net.com(22740)\"\" \"About 
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Ride the peaks. Smith bicycled 45 miles going east from Durango, and Jones bicycled 70 miles. Jones averaged 5 miles per hour more than Smith, and his trip took one half hour longer than Smith’s. How fast was each one traveling.
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\n" ); document.write( "That Smith is going east from Durango does not seem to have any bearing on the
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\n" ); document.write( "Time = dist/speed;
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\n" ); document.write( "Let s = Smith's speed
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\n" ); document.write( "Smith's time = 45/s
\n" ); document.write( "Jones's time = 70/(s+5)
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\n" ); document.write( "Smith's time + 1/2 hr = Jone's time
\n" ); document.write( "\"45%2Fs++%2B+.5\" = \"70%2F%28s%2B5%29\"
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\n" ); document.write( "Mult equation by s(s+5) and eliminate the denominators, then you have:
\n" ); document.write( "45(s+5) + .5(s(s+5) = 70s
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\n" ); document.write( "45x + 225 + .5s^2 + 2.5s = 70s
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\n" ); document.write( ".5s^2 + 45s + 2.5s - 70s + 225 = 0
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\n" ); document.write( ".5s^2 - 22.5s + 225 = 0
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\n" ); document.write( "Mult eq by 2 and get rid of these decimals
\n" ); document.write( "s^2 - 45s + 450 = 0
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\n" ); document.write( "Factors to:
\n" ); document.write( "(s - 15)(s - 30) = 0
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\n" ); document.write( "s = 15; and s = 30; I'm not sure, but 30 seems fast for a bike to average for
\n" ); document.write( "that distance. I'm going to say Smith's speed = 15 mph & Jones speed = 20 mph
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\n" ); document.write( "Check it: Jones time - Smiths time = .5hr
\n" ); document.write( "70/20 - 45/15 =
\n" ); document.write( "3.5 - 3 = .5
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\n" ); document.write( "The other solution, s = 30:
\n" ); document.write( "70/35 - 45/30 =
\n" ); document.write( "2 - 1.5 = .5 also\r
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