document.write( "Question 310467: The speed of the current in the Yankaton River is 3 kilometers per hour. If a boat travels 18 kilometers downstream in this river and then returns to its starting point all in 4 1/2 hours, what is the rate of the boat in still water? \n" ); document.write( "
Algebra.Com's Answer #221973 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
the speed of the boat in still water be x km/hr.\r
\n" ); document.write( "\n" ); document.write( "Downstream
\n" ); document.write( "the speed will be x+3 km/hr.
\n" ); document.write( "Distance traveled is 18 km.
\n" ); document.write( "So time taken = 18/x+3
\n" ); document.write( "--
\n" ); document.write( "Upstream
\n" ); document.write( "the speed of the boat is x-3 mph.
\n" ); document.write( "Distance is 18km.
\n" ); document.write( "Time 18 / x-3
\n" ); document.write( ".
\n" ); document.write( "Add up the two .\r
\n" ); document.write( "\n" ); document.write( "18/x+3 + 18/x-3 = 9/2 hours\r
\n" ); document.write( "\n" ); document.write( "18(x-3)+18(x+3)= 9/2(x^2-9)\r
\n" ); document.write( "\n" ); document.write( "18x-54+1854 = 9x^2-81 / 2\r
\n" ); document.write( "\n" ); document.write( "2(36x)= 9x^2-81\r
\n" ); document.write( "\n" ); document.write( "72x=9x^2-81\r
\n" ); document.write( "\n" ); document.write( "9x^2-72x-81=0 ( divide by 9)
\n" ); document.write( "x^2-8x-9=0
\n" ); document.write( "x^2-9x+x-9=0
\n" ); document.write( "x(x-9)+1(x-9)=0
\n" ); document.write( "(x+1)(x-9)=0
\n" ); document.write( "x= -1 Or x=9\r
\n" ); document.write( "\n" ); document.write( "Speed cannot be negative
\n" ); document.write( "so boat speed in still water = 9 km/hr\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );