document.write( "Question 136647This question is from textbook Algebra Structure and Method
\n" ); document.write( ": Jacqui commutes 30 mi to her job each day. She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work. Find her new speed. \n" ); document.write( "
Algebra.Com's Answer #100068 by ankor@dixie-net.com(22740)\"\" \"About 
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Jacqui commutes 30 mi to her job each day. She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work. Find her new speed
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\n" ); document.write( "Let s = new speed
\n" ); document.write( "then
\n" ); document.write( "(s-10) = original speed
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\n" ); document.write( "Convert 6 min to hrs: 6/60 = .1 hrs
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\n" ); document.write( "Write a time equation: Time = dist/speed
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\n" ); document.write( "Fast trip + 6 min = slow trip
\n" ); document.write( "\"30%2Fs\" + .1 = \"30%2F%28%28s-10%29%29\"
\n" ); document.write( "Multiply equation by s(s-10) to get rid of the denominators:
\n" ); document.write( "s(s-10)*\"30%2Fs\" + .1(s(s-10) = s(s-10)*\"30%2F%28%28s-10%29%29\"
\n" ); document.write( "Results
\n" ); document.write( "30(s-10) + .1s^2 - 1s = 30s
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\n" ); document.write( "30s - 300 + .1s^2 - s - 30s = 0
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\n" ); document.write( ".1s^2 - s - 300 = 0; a quadratic equation
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\n" ); document.write( "s^2 - 10s - 3000 = 0; mult by 10 to get rid of the decimal
\n" ); document.write( "Factors to:
\n" ); document.write( "(s-60)(s+50) = 0
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\n" ); document.write( "S = 60 mph is the new speed
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\n" ); document.write( "Check solution by finding the times of each trip:
\n" ); document.write( "30/50 = .6 hrs or 36 min
\n" ); document.write( "30/60 = .5 hrs or 30 min
\n" ); document.write( "------------------------
\n" ); document.write( "difference time = 6 min\r
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\n" ); document.write( "\n" ); document.write( " =
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