document.write( "Question 387690: A motorboat takes 4 hours to travel 320 km going upstream. the return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current? \n" ); document.write( "

Algebra.Com's Answer #275548 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
Rate*Time=Distance
\n" ); document.write( " $\"R%2At=D\"$
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\n" ); document.write( "Let the boat rate be $\"B\"$ , the current rate $\"C\"$ .
\n" ); document.write( "With the current,
\n" ); document.write( " $\"%28B%2BC%292=320\"$
\n" ); document.write( "1. $\"B%2BC=160\"$
\n" ); document.write( "Against the current,
\n" ); document.write( " $\"%28B-C%294=320\"$
\n" ); document.write( "2. $\"B-C=80\"$
\n" ); document.write( "Add the equations together,
\n" ); document.write( " $\"B%2BC%2BB-C=160%2B80\"$
\n" ); document.write( " $\"2B=240\"$
\n" ); document.write( " $\"highlight%28B=120%29\"$ kph
\n" ); document.write( "Then use either equation to solve for $\"C\"$ .
\n" ); document.write( " $\"120%2BC=160\"$
\n" ); document.write( " $\"highlight%28C=40%29\"$ kph \n" ); document.write( "

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