document.write( "Question 239415: A biker completed a trip from town A to town B. If he rode 3 km per hour faster, then he could spend one hour less for this trip. If he rode 2 km per hour slower, he would be one hour late.
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Algebra.Com's Answer #175836 by ankor@dixie-net.com(22740)\"\" \"About 
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A biker completed a trip from town A to town B.
\n" ); document.write( " If he rode 3 km per hour faster, then he could spend one hour less for this trip.
\n" ); document.write( "If he rode 2 km per hour slower, he would be one hour late.\r
\n" ); document.write( "\n" ); document.write( "Find the distance between towns, speed of the biker, and time of traveling.
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\n" ); document.write( "Let s = the biker's speed
\n" ); document.write( "then
\n" ); document.write( "(s+3) = 3 km faster speed
\n" ); document.write( "and
\n" ); document.write( "(s-2) = 2 km slower speed
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\n" ); document.write( "Let d = dist from a to b
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\n" ); document.write( "Write two time equations: time = dist/speed
\n" ); document.write( "\"d%2Fs\" - \"d%2F%28s%2B3%29\" = 1 (he went 3 km/h faster)
\n" ); document.write( "and
\n" ); document.write( "\"d%2F%28s-2%29\" - \"d%2Fs\" = 1 (he went 2 km/hr slower)
\n" ); document.write( ":
\n" ); document.write( "Both equations = 1, so we can write it:
\n" ); document.write( "\"d%2Fs\" - \"d%2F%28s%2B3%29\" = \"d%2F%28s-2%29\" - \"d%2Fs\"
\n" ); document.write( "simplify, divide each term by d
\n" ); document.write( "\"1%2Fs\" - \"1%2F%28s%2B3%29\" = \"1%2F%28s-2%29\" - \"1%2Fs\"
\n" ); document.write( "combine like terms
\n" ); document.write( "\"1%2Fs\" + \"1%2Fs\" - \"1%2F%28s%2B3%29\" = \"1%2F%28s-2%29\"
\n" ); document.write( ":
\n" ); document.write( "\"2%2Fs\" - \"1%2F%28s%2B3%29\" = \"1%2F%28s-2%29\"
\n" ); document.write( ":
\n" ); document.write( "Multiply by s(s+3)(s-2) to get rid of the denominators:
\n" ); document.write( "2(s+3)(s-2) - s(s-2) = s(s+3)
\n" ); document.write( ":
\n" ); document.write( "2(s^2 + s - 6) - s^2 + 2s = s^2 + 3s
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\n" ); document.write( "2s^2 + 2s - 12 - s^2 + 2s = s^2 + 3s
\n" ); document.write( "Combine like terms on the left
\n" ); document.write( "2s^2 - s^2 - s^2 + 2s + 2s - 3s - 12 = 0
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\n" ); document.write( "s - 12 = 0
\n" ); document.write( "s = 12 km/hr is the speed
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\n" ); document.write( "Find the dist between towns using
\n" ); document.write( "\"d%2Fs\" - \"d%2F%28s%2B3%29\" = 1
\n" ); document.write( ":
\n" ); document.write( "\"d%2F12\" - \"d%2F%2815%29\" = 1
\n" ); document.write( "Multiply by 60
\n" ); document.write( "5d - 4d = 60
\n" ); document.write( "d = 60 km is the distance
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\n" ); document.write( "Find the time: \"60%2F12\" = 5 hrs, time of travel
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\n" ); document.write( "Check solutions in:
\n" ); document.write( "\"d%2F%28s-2%29\" - \"d%2Fs\" = 1
\n" ); document.write( ":
\n" ); document.write( "\"60%2F10\" - \"60%2F12\" =
\n" ); document.write( "6 - 5 = 1 hr
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