document.write( "Question 182430This question is from textbook prealgebra and introductory algebra
\n" ); document.write( ": A motorcycle breaks down and the rider must walk the rest of the way to work. The motorcycle was being driven at 45 mph, and the rider walks at a speed of 6 mph. The distance from home to work is 25 mi, and the total time for the trip was 2 hrs. How far did the motorcycle go before it broke down? \n" ); document.write( "
Algebra.Com's Answer #136917 by stanbon(75887)\"\" \"About 
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A motorcycle breaks down and the rider must walk the rest of the way to work. The motorcycle was being driven at 45 mph, and the rider walks at a speed of 6 mph. The distance from home to work is 25 mi, and the total time for the trip was 2 hrs. How far did the motorcycle go before it broke down?
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\n" ); document.write( "Walking DATA:
\n" ); document.write( "distance = x miles ; rate = 6 mph ; time = x/6 hrs
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\n" ); document.write( "Motocycle DATA:
\n" ); document.write( "rate = 45 mph ; distance = (25-x) mi ; time = (25-x)/45 hrs
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\n" ); document.write( "Equation :
\n" ); document.write( "time + time = 2 hrs
\n" ); document.write( "x/6 + (25-x)/45 = 2\r
\n" ); document.write( "\n" ); document.write( "45x + 6(25-x) = 2*6*45\r
\n" ); document.write( "\n" ); document.write( "45x + 150 - 6x = 6*90
\n" ); document.write( "39x = 540 - 150\r
\n" ); document.write( "\n" ); document.write( "39x = 390
\n" ); document.write( "x = 10 miles (distance walked)
\n" ); document.write( "25-x = 15 miles (distance ridden)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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