document.write( "Question 988383: If the speed were increased by 10mph, a 420-mile trip would take 1 hour less time. How long will the trip take at the slower speed? \n" ); document.write( "

Algebra.Com's Answer #609032 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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If the speed were increased by 10mph, a 420-mile trip would take 1 hour less time.
\n" ); document.write( " How long will the trip take at the slower speed?
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\n" ); document.write( "let s = the slower speed
\n" ); document.write( "then
\n" ); document.write( "(s+10) = the faster speed
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\n" ); document.write( "Write a time equation; time = dist/speed
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\n" ); document.write( "slo speed time - fast speed time = 1 hr
\n" ); document.write( " $\"420%2Fs\"$ - $\"420%2F%28%28s%2B10%29%29\"$ = 1
\n" ); document.write( "multiply equation by s(s+10), cancel the denominators
\n" ); document.write( "420(s+10) - 420s = s(s+10)
\n" ); document.write( "420s + 4200 - 420s = s^2 + 10s
\n" ); document.write( "Combine to form a quadratic equation
\n" ); document.write( "s^2 + 10s - 4200 = 0
\n" ); document.write( "You can use the quadratic formula to find s, but this will factor to
\n" ); document.write( "(s+70)(s-60) = 0
\n" ); document.write( "The positive solution is what we want here
\n" ); document.write( "s = 60 mph is the slower speed
\n" ); document.write( "\"How long will the trip take at the slower speed?\"
\n" ); document.write( " $\"420%2F60\"$ = 7 hrs
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\n" ); document.write( "We can confirm this solution by finding the time at the faster speed
\n" ); document.write( "420/70 = 6 hrs, one hr less \n" ); document.write( "

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