document.write( "Question 168706: mr winkle jogs 4 miles and then turns and jogs back to his starting point.the first part of his jog he was going mostly uphill so his speed was 2 miles per hr slower than his speed returning.if the total time he spent was 1 2/3 hour find his rate going and his rate returning \n" ); document.write( "
Algebra.Com's Answer #124364 by ankor@dixie-net.com(22740)\"\" \"About 
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mr winkle jogs 4 miles and then turns and jogs back to his starting point.the first part of his jog he was going mostly uphill so his speed was 2 miles per hr slower than his speed returning.if the total time he spent was 1 2/3 hour find his rate going and his rate returning
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\n" ); document.write( "Let s = speed returning
\n" ); document.write( "then
\n" ); document.write( "(s-2) = speed up the hill
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: Time = dist/speed
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\n" ); document.write( "Time return + time out = 1 & 2/3 hrs
\n" ); document.write( "\"4%2Fs\" + \"4%2F%28%28s-2%29%29\" = 1\"2%2F3\"
\n" ); document.write( ":
\n" ); document.write( "\"4%2Fs\" + \"4%2F%28%28s-2%29%29\" = \"5%2F3\"
\n" ); document.write( "Multiply equation by 3s(s-2) to get rid of the denominators, results
\n" ); document.write( "4*3(s-2) + 4*3s = 5s(s-2)
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\n" ); document.write( "12s - 24 + 12s = 5s^2 - 10s
\n" ); document.write( ":
\n" ); document.write( "24s - 24 = 5s^2 - 10s
\n" ); document.write( ":
\n" ); document.write( "0 = 5s^2 - 10s - 24s + 24
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\n" ); document.write( "A quadratic equation, solve for s
\n" ); document.write( "5s^2 - 34s + 24 = 0
\n" ); document.write( "Factors to:
\n" ); document.write( "(5s - 4)(s - 6) = 0
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\n" ); document.write( "5s = 4
\n" ); document.write( "s = \"4%2F5\"; this solution makes no sense.
\n" ); document.write( "and
\n" ); document.write( "s = 6 mph is his return speed, then obviously 4 mph is his up-hill speed
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\n" ); document.write( "Check solution by find the times
\n" ); document.write( "4/6 + 4/4 = 1\"2%2F3\" hrs
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