document.write( "Question 167160: Roma Sherry drove 330 miles from her hometown to Tucson. During her return trip, she was able to increase her speed by 11 mph. If her return trip took 1 hour less time, find her original speed and her speed returning home. \n" ); document.write( "
Algebra.Com's Answer #123305 by ankor@dixie-net.com(22740)\"\" \"About 
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Roma Sherry drove 330 miles from her hometown to Tucson. During her return trip, she was able to increase her speed by 11 mph. If her return trip took 1 hour less time, find her original speed and her speed returning home.
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\n" ); document.write( "Let s = original speed
\n" ); document.write( "then
\n" ); document.write( "(s+11) = return speed
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\n" ); document.write( "Write a time equation: Time = \"distance%2Fspeed\"
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\n" ); document.write( "Original time = return time + 1 hr
\n" ); document.write( "\"330%2Fs\" = \"330%2F%28%28s%2B11%29%29\" + 1
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\n" ); document.write( "Multiply equation by s(s+11) and you have:
\n" ); document.write( "330(s+11) = 330s + s(s+11)
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\n" ); document.write( "330s + 3630 = 330s + s^2 + 11s
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\n" ); document.write( "0 = 330s - 330s + s^2 + 11s - 3630
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\n" ); document.write( "A quadratic equation:
\n" ); document.write( "s^2 + 11s - 3630 = 0
\n" ); document.write( "Factor this to:
\n" ); document.write( "(s + 66)(s - 55) = 0
\n" ); document.write( "Positive solution
\n" ); document.write( "s = 55 mph is original speed.
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\n" ); document.write( "Find the time
\n" ); document.write( "330/55 = 6 hr, original time
\n" ); document.write( "and
\n" ); document.write( "330/66 = 5 hrs, faster time; confirms our solution.
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