document.write( "Question 1127535: Leaving from the same starting line at the same time, Tracy jogs along the riverbank at a constant speed, while Joe paddles a canoe at a constant speed in the river beside her. If Joe was in still water, his speed would be the same as Tracy, but Joe is paddling against a 2 mile per hour current. If Tracy arrives at the end of her 6-mile jog 1 hour before Joe arrives at the same finish line, how fast was Tracy jogging? \n" ); document.write( "

Algebra.Com's Answer #743960 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Let x = \"how fast is Tracy jogging\".\r\n" );
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document.write( "Then Joe' speed in still water is the same x miles per hour, according to the condition,\r\n" );
document.write( "and Joe's effective speed paddling against the current is (x-2) miles per hour.\r\n" );
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document.write( "Then the time equation is\r\n" );
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document.write( " -  = 1 hour.\r\n" );
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document.write( "Thus the setup is done.\r\n" );
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document.write( "To solve equation, start multiplying both sides by x*(x-2).\r\n" );
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document.write( "6x - 6(x-2) = x*(x-2)\r\n" );
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document.write( "x^2 - 2x - 12 = 0\r\n" );
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document.write( " =  =  = .\r\n" );
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document.write( "The positive root   = 4.605 mph is the solution. \r\n" );
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document.write( "Check.   -  = 1.000   ! Correct !\r\n" );
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