document.write( "Question 44544: A car leaves a town at 40 miles per hour. Two hours later a second car leaves the town traveling the same route and overtakes the first car in 5 hours and 20 minutes. how fast was the second car traveling? \n" ); document.write( "
Algebra.Com's Answer #29427 by Nate(3500)\"\" \"About 
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Since the first car was given a two-hour head start, it traveled 2(40 mi/h) = 80 miles ....
\n" ); document.write( "First Car's Rate: 40t + 80
\n" ); document.write( "Second Car's Rate: xt
\n" ); document.write( "The different of the two car's traveled distance is equal to zero:
\n" ); document.write( "40t + 80 - xt = 0
\n" ); document.write( "40(16/3) + 80 - (16/3)x = 0
\n" ); document.write( "-(16/3)x = -240/3 - 640/3
\n" ); document.write( "-(16/3)x = -880/3
\n" ); document.write( "x = (880/3)(3/16) = 55
\n" ); document.write( "The second car was traveling 55 miles per hour.
\n" ); document.write( "Check:
\n" ); document.write( "First Car: y = 40x + 80
\n" ); document.write( "Second Car: y = 55x
\n" ); document.write( "Where 'y' is the distance and 'x' is the time:
\n" ); document.write( "-y + 40x = -80
\n" ); document.write( "+
\n" ); document.write( "y - 55x = 0
\n" ); document.write( "-15x = -80
\n" ); document.write( "x = 16/3
\n" ); document.write( "5 hours and 20 minutes \n" ); document.write( "
\n" );