document.write( "Question 1034135: solve. Round to the nearest tenth of necessary\r
\n" ); document.write( "\n" ); 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?
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Algebra.Com's Answer #648854 by ankor@dixie-net.com(22740)\"\" \"About 
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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?
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\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
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\n" ); document.write( "Write a time equation; time = dist/speed
\n" ); document.write( "To time - return time = .5 hrs
\n" ); document.write( "\"m%2F45\" - \"%28%28m-15%29%29%2F50\" = .5
\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
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\n" ); document.write( "Total dist driven: 90 + 75 = 165 mi
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\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
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