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)\"\" \"About 
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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( "\"m%2F45\" + \"%28%2825-m%29%29%2F6\" = 2
\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
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