document.write( "Question 1122280: 2 LINEAR EQUATIONS IN 2 UNKNOWNS or SIMULTANEOUS LINEAR EQUATIONS IN 2 UKNOWNS\r
\n" ); document.write( "\n" ); document.write( "a swimmer going downstream takes 1 hour and 20 minutes to travel a certain distance. it takes the swimmer 4 hours to make the return trip against the current. if the river flows at the rate of 1.5 miles per hour, find the rate of the swimmer in still water and the distance traveled one way. \n" ); document.write( "

Algebra.Com's Answer #738379 by ikleyn(52790) $\"\"$ $\"About$

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document.write( "Let u be the rate of the swimmer in still water, in miles per hour,\r\n" );
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document.write( "and let d be the distance traveled one way.\r\n" );
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document.write( "Then the effective speed (rate) of the swimmer downstream is  (u+1.5)  miles per hour, while \r\n" );
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document.write( "     the effective speed (rate) of the swimmer upstream   is  (u-1.5)  miles per hour.\r\n" );
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document.write( "The equation for the effective rate downstream is\r\n" );
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document.write( "    u + 1.5 = ,       ({{4/3}}} =  hours =  1 hour and 20 minutes)\r\n" );
document.write( "or\r\n" );
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document.write( "    u + 1.5 = .\r\n" );
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document.write( "The equation for the effective rate upstream is\r\n" );
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document.write( "    u - 1.5 = .       \r\n" );
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document.write( "Thus you have this system of 2 equations in 2 unknowns\r\n" );
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document.write( "    u + 1.5 = ,     (1)\r\n" );
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document.write( "    u - 1.5 = .       (2)\r\n" );
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document.write( "Subtract eq(2) from eq(1). You will get\r\n" );
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document.write( "    1.5 - (-1.5) =  - ,    or    3 =  = ,\r\n" );
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document.write( "which implies   d = 6.   Thus the one way distance is 6 miles.\r\n" );
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document.write( "Now from eq(2),  u =  + 1.5 =  + 1.5 = 3 miles per hour.\r\n" );
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document.write( "Answer.  One way distance is 6 miles.  The swimmer rate in still water is 3 miles per hour.\r\n" );
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document.write( "Check.   Time to swim downstream is  =  =  hours.   ! Correct !\r\n" );
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document.write( "         Time to swim upstream   is  =  = 4 hours.    ! Correct !\r\n" );
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