document.write( "Question 784848: please help me solve:\r
\n" ); document.write( "\n" ); document.write( "Kevin drove 320 miles to a resort. He return trip took 20 min longer because his speed returning was 4 mi/h slower than going
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Algebra.Com's Answer #477294 by ankor@dixie-net.com(22740)\"\" \"About 
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Kevin drove 320 miles to a resort.
\n" ); document.write( " He return trip took 20 min longer because his speed returning was 4 mi/h slower than going
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\n" ); document.write( "Let (s+4) = speed of the outbound trip
\n" ); document.write( "then
\n" ); document.write( "s = speed of the return trip
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\n" ); document.write( "Change 20 min to 1/3 hr
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\n" ); document.write( "Write a time equation; time = dist/speed
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\n" ); document.write( "return time - outbound time = 20 min
\n" ); document.write( "\"320%2Fs\" - \"320%2F%28%28s%2B4%29%29\" = \"1%2F3\"
\n" ); document.write( "multiply by 3s(s+4) to clear the denominators, resulting in:\
\n" ); document.write( "3(s+4)(320) - 3s(320) = s(s+4)
\n" ); document.write( "960s + 3840 - 960s = s^2 + 4s
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\n" ); document.write( "Arrange as a quadratic equation on the right
\n" ); document.write( "0 = s^2 + 4s - 3840
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\n" ); document.write( "You can use the quadratic formula here but this will factor to
\n" ); document.write( "(s+64)(s-60) = 0
\n" ); document.write( "the positive solution
\n" ); document.write( "s = 60 mph on the return trip
\n" ); document.write( "and obviously 64 mph on the outbound trip
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\n" ); document.write( "Confirm this by finding the actual times
\n" ); document.write( "320/60 = 5.33
\n" ); document.write( "320/64 = 5.00
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\n" ); document.write( "dif: .33 hrs which is 20 min\r
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