document.write( "Question 366457: a boat traveled 420 miles downstream and 420 miles back upstream. the trip downstream took 20 hours. the trip back took 140 hours. what is the speed of the boat in still water. what is the speed of the current \n" ); document.write( "
Algebra.Com's Answer #261130 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
let speed of boat with no current be x mph
\n" ); document.write( "speed of current be y
\n" ); document.write( "..
\n" ); document.write( "downstream
\n" ); document.write( "420/(x+y)=20
\n" ); document.write( "420 = 20x+20y
\n" ); document.write( "21=x+y.....................1
\n" ); document.write( "upstream
\n" ); document.write( "420/ x-y=140
\n" ); document.write( "420 = 140(x-y)
\n" ); document.write( "420=140x-140y
\n" ); document.write( "3=10x-10y...................2
\n" ); document.write( "multiply equation 1 by 10
\n" ); document.write( "210=10x+10y
\n" ); document.write( "add this to equation 2
\n" ); document.write( "213=20x
\n" ); document.write( "x=213/20
\n" ); document.write( "x=10.65 mph ... speed of boat in still water
\n" ); document.write( "x+y =21
\n" ); document.write( "y=21-10.65
\n" ); document.write( "y= 9.35 mph ... speed of current
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );