document.write( "Question 775661: a man left place A at 6:00 am expecting to reach place B at 9:00 am. but after walking one hour, he was delayed for half an hour, and so he had to increase his rate 1 mile per hour to reach place B at 9:00 am. find his speed before the delay and the distance between place A and place B. \n" ); document.write( "
Algebra.Com's Answer #473056 by ankor@dixie-net.com(22740)\"\" \"About 
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a man left place A at 6:00 am expecting to reach place B at 9:00 am.
\n" ); document.write( " but after walking one hour, he was delayed for half an hour, and so he had to increase his rate 1 mile per hour to reach place B at 9:00 am.
\n" ); document.write( " find his speed before the delay and the distance between place A and place B.
\n" ); document.write( ":
\n" ); document.write( "let s = his normal walking speed
\n" ); document.write( "then
\n" ); document.write( "(s+1) = his faster walking speed
\n" ); document.write( ":
\n" ); document.write( "dist = 3s
\n" ); document.write( "Because of the delay, actual walking time only 2.5 hrs
\n" ); document.write( ":
\n" ); document.write( "dist at normal speed + dist at faster speed = total dist
\n" ); document.write( "1s + 1.5(s+1) = 3s
\n" ); document.write( "s + 1.5s + 1.5 = 3s
\n" ); document.write( "2.5s + 1.5 = 3s
\n" ); document.write( "1.5 = 3s - 2.5s
\n" ); document.write( "1.5 = .5s
\n" ); document.write( "s = 1.5 /.5
\n" ); document.write( "s = 3 mph is his normal walking speed
\n" ); document.write( "then
\n" ); document.write( "3(3) = 9 mi is the distance
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check
\n" ); document.write( "3 + 1.5(4) = 9
\n" ); document.write( "3 + 6 = 9 \n" ); document.write( "
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