document.write( "Question 584673: A boat travels 60 miles upstream in 5 hours. The boat travels the same distance downstream in 3 hours. Find the rate of the boat still water and rate of the current?
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Algebra.Com's Answer #372847 by mananth(16946)\"\" \"About 
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Boat speed =x mph
\n" ); document.write( "current speed =y mph
\n" ); document.write( "against wind x-y 5 hours
\n" ); document.write( "with wind x+y 3 hours
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\n" ); document.write( "Distance = same= upstream= 60 miles Downstream = 60 miles
\n" ); document.write( "t=d/r against current
\n" ); document.write( "60 / ( x - y )= 5
\n" ); document.write( "5 ( x - y ) = 60
\n" ); document.write( "5 x - 5 y = 60 ....................1
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\n" ); document.write( "60 / ( x + y )= 3
\n" ); document.write( "3 ( x + y ) = 60
\n" ); document.write( "3 x + 3 y = 60 ...............2
\n" ); document.write( "Multiply (1) by 3
\n" ); document.write( "Multiply (2) by 5
\n" ); document.write( "we get
\n" ); document.write( "15 x + -15 y = 180
\n" ); document.write( "15 x + 15 y = 300
\n" ); document.write( "30 x = 480
\n" ); document.write( "/ 30
\n" ); document.write( "x = 16 mph
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\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "5 x -5 y = 60
\n" ); document.write( "80 -5 y = 60
\n" ); document.write( "-5 y = 60 -80
\n" ); document.write( "-5 y = -20
\n" ); document.write( " y = 4 mph
\n" ); document.write( " Boat 16 mph
\n" ); document.write( " current 4 mph
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