document.write( "Question 1014642: A biker traveling at his usual speed can go from his house to work in 4 hours. When the weather is bad, he travels 6 miles per hour slower, and it takes him half an hour more. How far is his work from his house?
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Algebra.Com's Answer #630946 by ankor@dixie-net.com(22740)\"\" \"About 
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A biker traveling at his usual speed can go from his house to work in 4 hours.
\n" ); document.write( " When the weather is bad, he travels 6 miles per hour slower, and it takes him half an hour more.
\n" ); document.write( " How far is his work from his house?
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\n" ); document.write( "let d = the distance from his house
\n" ); document.write( "From the information we can say \"He takes 4hr normally, but 4.5 hrs in bad weather\"
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\n" ); document.write( "Write a speed equation; speed = dist/time
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\n" ); document.write( "normal speed - slower speed= 6 mph
\n" ); document.write( "\"d%2F4\" - \"d%2F4.5\" = 6
\n" ); document.write( "multiply equation by 36 to get rid of the denominators
\n" ); document.write( "36*\"d%2F4\" - 36*\"d%2F4.5\" = 36(6)
\n" ); document.write( "Cancel the denominators
\n" ); document.write( "9d - 8d = 216
\n" ); document.write( "d = 216 mi to his house
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\n" ); document.write( "See if that checks out, find the actual speeds
\n" ); document.write( "216/4 = 54 mph
\n" ); document.write( "216/4.5= 48 mph
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\n" ); document.write( "difference 6 mph
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