document.write( "Question 885920: Len is planning a three hour trip down a river and back to his starting point. He knows he can paddle in still water at 3 miles per hour and that the rate of the current is 2 miles per hour. How much time can he spend going downstream? How far downstream can he travel?
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Algebra.Com's Answer #535554 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+d+\" = the one-way distance traveled
\n" ); document.write( "Let \"+t%5B1%5D+\" = time in hrs to travel downstream
\n" ); document.write( "Let \"+t%5B2%5D+\" = time in hrs to travel upstream
\n" ); document.write( "Going downstream, Len can travel at a rate of:
\n" ); document.write( "\"+3+%2B+2+=+5+\" mi/hr
\n" ); document.write( "Going upstream, Len can travel at a rate of:
\n" ); document.write( "\"+3+-+2+=+1+\" mi/hr
\n" ); document.write( "------------------------
\n" ); document.write( "Going downstream:
\n" ); document.write( "(1) \"+d+=+5t%5B1%5D+\"
\n" ); document.write( "Going upstream:
\n" ); document.write( "(2) \"+d+=+1%2At%5B2%5D+\"
\n" ); document.write( "Also given:
\n" ); document.write( "(3) \"+t%5B1%5D+%2B+t%5B2%5D+=+3+\"
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\n" ); document.write( "From (1) and (2):
\n" ); document.write( "(2) \"+t%5B2%5D+=+5t%5B1%5D+\"
\n" ); document.write( "and
\n" ); document.write( "(3) \"+t%5B1%5D+%2B+5t%7B1%5D+=+3+\"
\n" ); document.write( "(3) \"++6t%5B1%5D+=+3+\"
\n" ); document.write( "(3) \"+t%5B1%5D+=+.5+\" hr
\n" ); document.write( "and
\n" ); document.write( "(3) \"+1%2F2+%2B+t%5B2%5D+=+3+\"
\n" ); document.write( "(3) \"+t%5B2%5D+=+2.5+\" hrs
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\n" ); document.write( "He can spend 1/2 hr going downstream
\n" ); document.write( "and
\n" ); document.write( "(1) \"+d+=+5t%5B1%5D+\"
\n" ); document.write( "(1) \"+d+=+5%2A%281%2F2%29+\"
\n" ); document.write( "(1) \"+d+=+2.5+\" mi
\n" ); document.write( "He can travel 2.5 mi downstream
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\n" ); document.write( "check:
\n" ); document.write( "(2) \"+d+=+1%2At%5B2%5D+\"
\n" ); document.write( "(2) \"+d+=+1%2A2.5+\"
\n" ); document.write( "(2) \"+d+=+2.5+\" mi
\n" ); document.write( "OK\r
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