document.write( "Question 300475: A man leaves for work at the same time each morning. If his travel speed averages 30 miles per hour, he will be 18 minutes late. If his travel speed averages 45 miles per hour, he will arrive 8 minutes early. What average travel speed (in miles per hour) will put him at work precisely on time? \n" ); document.write( "
Algebra.Com's Answer #215680 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A man leaves for work at the same time each morning. \n" ); document.write( "If his travel speed averages 30 miles per hour, he will be 18 minutes late. \n" ); document.write( " If his travel speed averages 45 miles per hour, he will arrive 8 minutes early. \n" ); document.write( " What average travel speed (in miles per hour) will put him at work precisely on time? \n" ); document.write( ": \n" ); document.write( "Let s = av speed to arrive on time \n" ); document.write( ": \n" ); document.write( "let d = distance \n" ); document.write( ": \n" ); document.write( "Write two time equations; time = dist/speed \n" ); document.write( "The late equation \n" ); document.write( " \n" ); document.write( "The early equation \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "clear the denominators, multiply both equations by 180s and arrange like this: \n" ); document.write( " 6ds - 180d = 54s \n" ); document.write( "-4ds + 180d = 24s \n" ); document.write( "--------------------addition eliminates 180d \n" ); document.write( "2ds = 78s \n" ); document.write( "divide both sides by s \n" ); document.write( "2d = 78 \n" ); document.write( "d = 39 mi is the distance \n" ); document.write( ": \n" ); document.write( "Find s \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "using a calc \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "s = 39 mph to arrive on time (in 1 hr) \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution using the late equation \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "using a calc \n" ); document.write( "1.3 - 1 = .3; confirms our solution 39 miles and 39 mph\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |