document.write( "Question 582334: A motorcycle breaks down and the rider must walk the rest of the way to work. The motorcycle was being driven at 45mph, and the rider walks at a speed of 6mph. The distance from home to work is 25miles, and the time for the trip was 2hr. How far did the motorcyle go before it broke down? \n" ); document.write( "
Algebra.Com's Answer #372032 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A motorcycle breaks down and the rider must walk the rest of the way to work. \n" ); document.write( " The motorcycle was being driven at 45mph, and the rider walks at a speed of 6mph. \n" ); document.write( " The distance from home to work is 25miles, and the time for the trip was 2hr. \n" ); document.write( " How far did the motorcycle go before it broke down? \n" ); document.write( ": \n" ); document.write( "Let m = distance rode on the motorcycle \n" ); document.write( "then \n" ); document.write( "(25-m) = distance he walked \n" ); document.write( ": \n" ); document.write( "Write a time equation: time = dist/speed \n" ); document.write( ": \n" ); document.write( "ride time + walk time = 2 hrs \n" ); document.write( " \n" ); document.write( "multiply by LCM of the denominators, 90 will do it, results: \n" ); document.write( "2m + 15(25-m) = 90(2) \n" ); document.write( ": \n" ); document.write( "2m + 375 - 15m = 180 \n" ); document.write( "2m - 15 m = 180 - 375 \n" ); document.write( "-13m = -195 \n" ); document.write( "13m = 195 \n" ); document.write( "m = 195/13 \n" ); document.write( "m = 15 mi on the motorcycle \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the actual times \n" ); document.write( "15/45 = .33 \n" ); document.write( "10/6 = 1.67 \n" ); document.write( "------------ \n" ); document.write( "tot time: 2 hrs \n" ); document.write( " |