document.write( "Question 4290: A freighter leaves port steaming South. Two hours later a Coast Guard Cutter located 50 miles to the north of the port is ordered to give chase. If it takes the cutter 4 hours to overtake the ship steaming at 20 miles per hour faster, how fast were both traveling? \n" ); document.write( "
Algebra.Com's Answer #1930 by rapaljer(4671)\"\" \"About 
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Let x = rate of the freighter.
\n" ); document.write( " x + 20 = rate of the Coast Guard Cutter
\n" ); document.write( " 4 + 2 = 6 = time of the freighter
\n" ); document.write( " 4 = time of the Coast Guard Cutter\r
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\n" ); document.write( "\n" ); document.write( " D=RT
\n" ); document.write( " 6x= distance of the freighter
\n" ); document.write( " 4(x + 20) = distance of Coast Guard Cutter\r
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\n" ); document.write( "The equation is based upon the fact that:\r
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\n" ); document.write( "\n" ); document.write( "Distance traveled by the Coast Guard Cutter is 50 miles more than the distance traveled by the freighter
\n" ); document.write( "4(x + 20) = 6x + 50
\n" ); document.write( "4x + 80 = 6x + 50\r
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\n" ); document.write( "\n" ); document.write( "Subtract 4x from each side:
\n" ); document.write( "4x - 4x + 80 = 6x - 4x + 50
\n" ); document.write( "80 = 2x + 50\r
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\n" ); document.write( "\n" ); document.write( "Subtract 50 from each side:
\n" ); document.write( "80 - 50 = 2x + 50 - 50
\n" ); document.write( "30 = 2x
\n" ); document.write( "x = 15 mph = rate of the freighter
\n" ); document.write( "x+20 = 35 mph = rate of Coast Guard Cutter.\r
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\n" ); document.write( "\n" ); document.write( "Check: Freighter travels 15 mph for 6 hours, which is 90 miles. The Coast Guard Cutter travels at 35 mph for 4 hours, which is 140 miles, a difference of 50 miles. \r
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\n" ); document.write( "\n" ); document.write( "Nice problem.\r
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\n" ); document.write( "\n" ); document.write( "R^2 from SCC
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