document.write( "Question 1034135: solve. Round to the nearest tenth of necessary\r
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document.write( "Me. Truman drove to his cottage in the mountains at a rate of 45 mi/hr. On the way back he drove at a rate of 50 mi/hr but took a route 15 mo shorter than the route by which he traveled to the cottage. How many total miles did he travel if it took him 1/2 hr longer to go than to return? \n" );
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Algebra.Com's Answer #648854 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Truman drove to his cottage in the mountains at a rate of 45 mi/hr. \n" ); document.write( " On the way back he drove at a rate of 50 mi/hr but took a route 15 mi shorter than the route by which he traveled to the cottage. \n" ); document.write( " How many total miles did he travel if it took him 1/2 hr longer to go than to return? \n" ); document.write( ": \n" ); document.write( "let m = the distance driven to the cottage \n" ); document.write( "The return was 15 mi shorter, therefore; \n" ); document.write( "(m-15) = the return distance \n" ); document.write( ": \n" ); document.write( "Write a time equation; time = dist/speed \n" ); document.write( "To time - return time = .5 hrs \n" ); document.write( " \n" ); document.write( "get rid of the denominator, multiply eq by 450, cancel the denominators and you have. \n" ); document.write( "10m - 9(m-15) = 450(.5) \n" ); document.write( "10m - 9m + 135 = 225 \n" ); document.write( "m = 225 - 135 \n" ); document.write( "m = 90 mi to the cottage \n" ); document.write( "then \n" ); document.write( "90 - 15 = 75 mi return \n" ); document.write( ": \n" ); document.write( "Total dist driven: 90 + 75 = 165 mi \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the actual time each way \n" ); document.write( "90/45 = 2.0 hr to the cottage \n" ); document.write( "75/50 = 1.5 hr return \n" ); document.write( "------------------------- \n" ); document.write( "Time dif: .5 hrs \n" ); document.write( " |