document.write( "Question 160125This question is from textbook College Algebra
\n" ); document.write( ": A boat has a speed of 5 mph in still water. The boat can travel 21 miles with the current in the same amount of time it takes to travel 9 miles against the current. Find the rate of the current. \n" ); document.write( "
Algebra.Com's Answer #118164 by ptaylor(2198)\"\" \"About 
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\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "(Note: when travelling with the current, we add the speed of the current; when travelling against the current, we subtract the speed of the current)
\n" ); document.write( "Let r=rate of the current
\n" ); document.write( "Time travelling with the current=21/(5+r)
\n" ); document.write( "Time travelling against the current=9/(5-r)\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the above times are equal, so:\r
\n" ); document.write( "\n" ); document.write( "21/(5+r)=9/(5-r) multiply each side by (5-r)(5+r) or cross multiply
\n" ); document.write( "21(5-r)=9(5+r) get rid of parens
\n" ); document.write( "105-21r=45+9r subtract 105 and also 9r from each side
\n" ); document.write( "105-105-21r-9r=45-105+9r-9r collect like terms\r
\n" ); document.write( "\n" ); document.write( "-30r=-60 divide each side by -30
\n" ); document.write( "r=2 mph---------------------------speed of the current\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "21/7=9/3
\n" ); document.write( "3=3\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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