document.write( "Question 47781: Gone fishing. Debbie traveled 5 miles by boat upstream to fish in her favorite spot. Because of the 4mph current, it took her 20 minutes longer to get there than return. How fast will her boat go in still water?\r
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Algebra.Com's Answer #31591 by stanbon(75887)\"\" \"About 
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Debbie traveled 5 miles by boat upstream to fish in her favorite spot. Because of the 4mph current, it took her 20 minutes longer to get there than return. How fast will her boat go in still water?\r
\n" ); document.write( "\n" ); document.write( "Upstream DATA:
\n" ); document.write( "distance=5 mi ; rate=boat speed -5 mph ; time= d/r = 5/(b-5)
\n" ); document.write( "Downstream DATA:
\n" ); document.write( "distance = 5 mi ; rate = bs+5 mph ; time = 5/(b+5)\r
\n" ); document.write( "\n" ); document.write( "EQUATION:
\n" ); document.write( "upstream time - downstream time = (1/3) hr
\n" ); document.write( "5/(b-5) - 5/(b+5) = 1/3
\n" ); document.write( "[5b+25-(5b-25)]/(b^2-25) = 1/3
\n" ); document.write( "50(3) = b^2-25
\n" ); document.write( "b^2=175
\n" ); document.write( "b=13.23 mph\r
\n" ); document.write( "\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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