document.write( "Question 66362: During rush hour, Fernando can drie 20 miles using the side roads in the same time it takes to travel 15 miles on the freeway. If Fernandos rate on the side roads is 9mi/h faster than his rate on the freeway, find his rate on the side roads.
\n" ); document.write( "Is the answer 29?
\n" ); document.write( "Thank you for your help :-) \n" ); document.write( "
Algebra.Com's Answer #47059 by ankor@dixie-net.com(22740)\"\" \"About 
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During rush hour, Fernando can drive 20 miles using the side roads in the same time it takes to travel 15 miles on the freeway. If Fernandos rate on the side roads is 9 mi/h faster than his rate on the freeway, find his rate on the side roads.
\n" ); document.write( "Is the answer 29?
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\n" ); document.write( "Let s = his speed on the sideroads
\n" ); document.write( "Then (s-9) = his speed on the freeway
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\n" ); document.write( "Write a time equation, time = dist/speed
\n" ); document.write( "side-road time = freeway time
\n" ); document.write( "\"20%2Fs\" = \"15%2F%28%28s-9%29%29\"
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\n" ); document.write( "Cross mult and you have:
\n" ); document.write( "20(s-9) = 15s
\n" ); document.write( "20s - 180 = 15s
\n" ); document.write( "20s - 15s = 180
\n" ); document.write( "5s = 180
\n" ); document.write( "s = 180/5
\n" ); document.write( "s = 36 mph on the side-roads
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\n" ); document.write( "Check our solution using the time eq:
\n" ); document.write( "20/36 = 15/27
\n" ); document.write( "5/9 = 5/9 confirms our solution \n" ); document.write( "
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