<PRE><B>Question 909780</B>
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.
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let x=former rate
x+1=increased rate
15 min=1/4 hr
travel time=distance/rate
..
{{{11/x-(6/x+5/(x+1))=1/4}}}
{{{11/x-6/x-5/(x+1))=1/4}}}
{{{(5/x)-(5/(x+1))=1/4}}}
lcd:x(x+1)
5x+5-5x=(x^2+x)/4
20x+20-20x=x^2+x
x^2+x-20=0
(x+5)(x-4)=0
x=4
x+1=5
former rate=4 mph</PRE>