document.write( "Question 243043: a truck can travel 120 miles is the same time that it takes a care to travel 180 miles. if the truck's rate is 20 mph slower than the car's, find the adverage rate of each \n" ); document.write( "
Algebra.Com's Answer #178052 by oberobic(2304)\"\" \"About 
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This problem is nicely set up for you because you are told they two vehicles take equal time. And we know that D = RT, which means D/R = T.
\n" ); document.write( "We also are told the two distances:
\n" ); document.write( "The distance the truck travels in 'T' time = 120 miles
\n" ); document.write( "The distance the car travels in 'T' time = 180 miles
\n" ); document.write( "So we can pose the equation as:
\n" ); document.write( "120/x = 180/y
\n" ); document.write( "where x=truck's speed and y=car's speed.
\n" ); document.write( "We also are told x is 20 mph less than y, so we can express this as either x + 20 = y OR x = y - 20.
\n" ); document.write( "We're now setup to solve it.
\n" ); document.write( "Let's substitute: x = y -20
\n" ); document.write( "120/(y -20) = 180/y
\n" ); document.write( "Cross multiplying, we have:
\n" ); document.write( "120y = 180(y -20) = 180y - 3600
\n" ); document.write( "Subtracting 180y from both sides
\n" ); document.write( "-60y = -3600
\n" ); document.write( "Dividing by -60
\n" ); document.write( "y = 60
\n" ); document.write( "Given that x = y - 20, we expect
\n" ); document.write( "x = 40
\n" ); document.write( "But we have to check our work to see if this answer 'works'...
\n" ); document.write( "120/x = 3, which means the truck takes 3 hrs to go 120 miles
\n" ); document.write( "180/y = 3, which means the car takes 3 hrs to go 180 miles
\n" ); document.write( "That certainly meets the problems stated requirements, so we're done. \n" ); document.write( "
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