document.write( "Question 370418: Two cars each travel 75 miles at a constant rate. one car travels 6 mph faster than the other and arrives 5 minutes before the other arrives. Find the rates of speed of the two cars. \n" ); document.write( "
Algebra.Com's Answer #264144 by ankor@dixie-net.com(22740)\"\" \"About 
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Two cars each travel 75 miles at a constant rate.
\n" ); document.write( " one car travels 6 mph faster than the other and arrives 5 minutes before the other arrives.
\n" ); document.write( " Find the rates of speed of the two cars.
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\n" ); document.write( "Let s = speed of the slower car
\n" ); document.write( "then
\n" ); document.write( "(s+6) = speed of the faster car
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\n" ); document.write( "Convert 5 min to hrs 5/60 = 1/12 hr
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\n" ); document.write( "Write a time equation: Time = dist/speed
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\n" ); document.write( "Slow car time - faster car time = 5 min (1/12 hr)
\n" ); document.write( "\"75%2Fs\" - \"75%2F%28%28s%2B6%29%29\" = \"1%2F12\"
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\n" ); document.write( "Multiply by 12s(s+6) to get rid of the denominators, results:
\n" ); document.write( "12(s+6)*75 - 12s(75) = s(s+6)
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\n" ); document.write( "900(s+6) - 900s = s^2 + 6s
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\n" ); document.write( "900s + 5400 - 900s = s^2 + 6s
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\n" ); document.write( "5400 = s^2 + 6s
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\n" ); document.write( "A quadratic equation
\n" ); document.write( "0 = s^2 + 6s - 5400
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\n" ); document.write( "Use the quadratic formula to find s
\n" ); document.write( "\"s+=+%28-6+%2B-+sqrt%286%5E2-4%2A1%2A-5400+%29%29%2F%282%2A1%29+\"
\n" ); document.write( "do the math, you should get a positive solution s ~ 70.546 mph\r
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