document.write( "Question 227740: if a car averages 55 mph on a certain trip, it will arrive 2 hours early. If the car averages 35 mph it will arrive 2 hours late. How many miles is the trip? \n" ); document.write( "
Algebra.Com's Answer #169206 by stanbon(75887)\"\" \"About 
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if a car averages 55 mph on a certain trip, it will arrive 2 hours early. If the car averages 35 mph it will arrive 2 hours late. How many miles is the trip?
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\n" ); document.write( "The difference in the time is 4 hours.
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\n" ); document.write( "1st car DATA:
\n" ); document.write( "rate = 55 mph ; distance = x miles ; time = d/r = x/55 hrs
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\n" ); document.write( "2nd car DATA:
\n" ); document.write( "rate = 35 mph ; distance = x miles ; time = d/r = x/35 hrs
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\n" ); document.write( "Equation:
\n" ); document.write( "slower time - faster time = 4 hrs
\n" ); document.write( "x/35 - x/55 = 4
\n" ); document.write( "55x - 35x = 4*35*55
\n" ); document.write( "20x = 7700
\n" ); document.write( "x = 385 miles (length of the trip)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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