document.write( "Question 596604: the speed of joe's boat in still water is 5miles per hour. if joe can go upstream 36 miles and return downstream in 40 hours, what is the speed of the current? \n" ); document.write( "

Algebra.Com's Answer #377810 by llorton(14) $\"\"$ $\"About$

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Speed of Joe's boat going upstream = 5 - x where x is the speed of the current
\n" ); document.write( "Speed of Joe's boat going downstream = 5 + x\r
\n" ); document.write( "\n" ); document.write( "time to go upstream is distance/rate = 36/5-x
\n" ); document.write( "time to go downstream is 36/5+x
\n" ); document.write( "total time for trip is 40 hours\r
\n" ); document.write( "\n" ); document.write( "[36/(5-x)] + [36/(5+x)] = 40\r
\n" ); document.write( "\n" ); document.write( "If we multiply both sides of the above equation by (25-x^2)we get:\r
\n" ); document.write( "\n" ); document.write( "36(5+x) + 36(5-x) = 40(25-x^2)\r
\n" ); document.write( "\n" ); document.write( "Simplify:\r
\n" ); document.write( "\n" ); document.write( "180 + 36x + 180 - 36x = 40(25-x^2)
\n" ); document.write( "360 = 40(25-x^2)
\n" ); document.write( "Divide both sides of the equation by 40
\n" ); document.write( "9 = 25 - x^2
\n" ); document.write( "x^2 -25 + 9 = 25-x^2-25+x^2
\n" ); document.write( "x^2-16=0
\n" ); document.write( "x-4=0 x+4=0
\n" ); document.write( "x=4 x=-4\r
\n" ); document.write( "\n" ); document.write( "The speed of the current is 4 mph.\r
\n" ); document.write( "\n" ); document.write( "check:
\n" ); document.write( "net speed going upstream is 5-4=1mph Upstream trip takes 36 hours
\n" ); document.write( "net speed going downstream is 5+4=9mph Upstream trip takes 4 hours
\n" ); document.write( "Yotal time = 36 + 4 + 60 hours\r
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