document.write( "Question 192729: Jim drives from the Twin Cities to Monroe Louisiana, a distance of 1075 miles in record time. On the way back he averages 5 mph slower and the trip takes an additional 1.5 hrs. How fast did he drive to Monroe LA.? Round to the nearest tenth of a mph. \n" ); document.write( "
Algebra.Com's Answer #144817 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Jim drives from the Twin Cities to Monroe Louisiana, a distance of 1075 miles \n" ); document.write( " in record time. On the way back he averages 5 mph slower and the trip takes \n" ); document.write( " an additional 1.5 hrs. How fast did he drive to Monroe LA.? \n" ); document.write( " Round to the nearest tenth of a mph. \n" ); document.write( ": \n" ); document.write( "let s = his original speed to La \n" ); document.write( "then \n" ); document.write( "(s-5) = his return speed \n" ); document.write( "; \n" ); document.write( "Write a time equation: Time = \n" ); document.write( "; \n" ); document.write( "Time to + 1.5 hr = time to return \n" ); document.write( " \n" ); document.write( "Multiply equation by s(s-5), results \n" ); document.write( "1075(s-5) + 1.5s(s-5) = 1075s \n" ); document.write( ": \n" ); document.write( "1075s - 5375 + 1.5s^2 - 7.5s - 1075s = 0 \n" ); document.write( ": \n" ); document.write( "1.5s^2 - 7.5s - 5375 = 0 \n" ); document.write( ": \n" ); document.write( "A quadratic equation, use the quadratic formula: \n" ); document.write( " \n" ); document.write( "In this equation: x = s; a = 1.5, b = -7.5, c = -5375 \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "Two solutions but we only want the positive one: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "s = 62.4 mph to Monroe La \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the time for each trip; return speed: 62.4 - 5 = 57.4 \n" ); document.write( "1075/57.4 = 18.73 hrs \n" ); document.write( "1075/62.4 = 17.23 hrs \n" ); document.write( "----------------------- \n" ); document.write( "difference: 1.5 hrs \n" ); document.write( " |