document.write( "Question 547099: At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. Find the rate of the streams current? \n" ); document.write( "
Algebra.Com's Answer #356254 by mananth(16946)\"\" \"About 
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let speed of boat be x
\n" ); document.write( "speed of current be y\r
\n" ); document.write( "\n" ); document.write( "downstream speed = x+y
\n" ); document.write( "upstream speed = x-y\r
\n" ); document.write( "\n" ); document.write( "15/(x-y) -15/(x+y)=5\r
\n" ); document.write( "\n" ); document.write( "LCD = (x+y)(x-y)\r
\n" ); document.write( "\n" ); document.write( "multiply by the LCD\r
\n" ); document.write( "\n" ); document.write( "15(x+y)-15(x-y)=5(x^2-y^2)
\n" ); document.write( "15x+15y-15x+15y=5(x^2-y^2)
\n" ); document.write( "30y=5(x^2-y^2)
\n" ); document.write( "----------------
\n" ); document.write( "Speed is doubled\r
\n" ); document.write( "\n" ); document.write( "downstream speed = 2x+y
\n" ); document.write( "upstream speed = 2x-y
\n" ); document.write( "15/(2x-y)-15/(2x+y)=1
\n" ); document.write( "LCD = (2x+y)(2x-y)
\n" ); document.write( "multiply by the LCD
\n" ); document.write( "15(2x+y)-15(2x-y)=4x^2-y^2
\n" ); document.write( "30x+15y-30x+15y=4x^2-y^2
\n" ); document.write( "30y=4x^2-y^2\r
\n" ); document.write( "\n" ); document.write( "equate 30y from both equations
\n" ); document.write( "5x^2-5y^2=4x^2-y^2
\n" ); document.write( "x^2-4y^2=0
\n" ); document.write( "x=2y
\n" ); document.write( "substitute x=2y in
\n" ); document.write( "15/(x-y) -15/(x+y)=5
\n" ); document.write( "15/(2y-y)-15/(2y+y)=5
\n" ); document.write( "15/y-15/3y=5
\n" ); document.write( "45-15=15y
\n" ); document.write( "30=15y
\n" ); document.write( "30/15=y
\n" ); document.write( "y=2 the current speed
\n" ); document.write( "x=4 mph boat speed\r
\n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca
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