document.write( "Question 909780: After walking a distance of 6 miles at a certain rate, a man decided to increase his rate per hour by 1 mile and walked 5 miles farther. Had he walked the entire distance of 11 miles at his former rate, his time would have been 15 minutes longer. Find his former rate. \n" ); document.write( "
Algebra.Com's Answer #552028 by lwsshak3(11628)\"\" \"About 
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After walking a distance of 6 miles at a certain rate, a man decided to increase his rate per hour by 1 mile and walked 5 miles farther. Had he walked the entire distance of 11 miles at his former rate, his time would have been 15 minutes longer. Find his former rate.
\n" ); document.write( "***
\n" ); document.write( "let x=former rate
\n" ); document.write( "x+1=increased rate
\n" ); document.write( "15 min=1/4 hr
\n" ); document.write( "travel time=distance/rate
\n" ); document.write( "..
\n" ); document.write( "\"11%2Fx-%286%2Fx%2B5%2F%28x%2B1%29%29=1%2F4\"
\n" ); document.write( "\"11%2Fx-6%2Fx-5%2F%28x%2B1%29%29=1%2F4\"
\n" ); document.write( "\"%285%2Fx%29-%285%2F%28x%2B1%29%29=1%2F4\"
\n" ); document.write( "lcd:x(x+1)
\n" ); document.write( "5x+5-5x=(x^2+x)/4
\n" ); document.write( "20x+20-20x=x^2+x
\n" ); document.write( "x^2+x-20=0
\n" ); document.write( "(x+5)(x-4)=0
\n" ); document.write( "x=4
\n" ); document.write( "x+1=5
\n" ); document.write( "former rate=4 mph
\n" ); document.write( "
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