document.write( "Question 694124: a boat traveled 30 miles down stream in 3 hours and made a return trip in 5 hours. find the speed of a boat in still water and the speed of the current. \n" ); document.write( "

Algebra.Com's Answer #427749 by mananth(16946) $\"\"$ $\"About$

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Boat speed speed =x mph
\n" ); document.write( "current speed =y mph
\n" ); document.write( "against current 5 hours
\n" ); document.write( "with current 3 hours
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\n" ); document.write( "Distance with current 30 miles distance against current 30
\n" ); document.write( "t=d/r against current (x-y)
\n" ); document.write( "30.00 / ( x - y )= 5.00
\n" ); document.write( "
\n" ); document.write( "5.00 x - -5.00 y = 30.00 ....................1
\n" ); document.write( "with current (x+y)
\n" ); document.write( "30.00 / ( x + y )= 3.00
\n" ); document.write( "3.00 ( x + y ) = 30.00
\n" ); document.write( "3.00 x + 3.00 y = 30.00 ...............2
\n" ); document.write( "Multiply (1) by 3.00
\n" ); document.write( "Multiply (2) by 5.00
\n" ); document.write( "we get y
\n" ); document.write( "15.00 x + -15.00 y = 90.00
\n" ); document.write( "15.00 x + 15.00 = 150.00
\n" ); document.write( "30.00 x = 240.00
\n" ); document.write( "/ 30.00
\n" ); document.write( "x = 8 mph
\n" ); document.write( "
\n" ); document.write( "plug value of x in (1) y
\n" ); document.write( "5 x -5 y = 30
\n" ); document.write( "40 -5 -40 = 30
\n" ); document.write( "-5 y = 30
\n" ); document.write( "-5 y = -10 mph
\n" ); document.write( " y = 2
\n" ); document.write( "Boat speed 8 mph
\n" ); document.write( "current 2 mph
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "x+y= 10
\n" ); document.write( "X-y= 6 +
\n" ); document.write( "30.00 / ( 10.00 )= 3
\n" ); document.write( "30 / ( 6 ) = 5
\n" ); document.write( " \n" ); document.write( "

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