document.write( "Question 251528: Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4-mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water? \n" ); document.write( "
Algebra.Com's Answer #183249 by stanbon(75887)\"\" \"About 
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Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4-mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?
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\n" ); document.write( "Upstream DATA:
\n" ); document.write( "distance = 5 miles ; rate = b-4 mph ; time = d/r = 5/(b-4) hrs.
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\n" ); document.write( "Downstream DATA:
\n" ); document.write( "distance = 5 miles ; rate = b+4 mph ; time = 5/(b+4) hrs.
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\n" ); document.write( "Equation:
\n" ); document.write( "up - down = (1/5) hr
\n" ); document.write( "5/(b-4) - 5/(b+4) = 1/5
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\n" ); document.write( "25(b+4) - 25(b-4) = (b^2-16)
\n" ); document.write( "100+100 = b^2-16
\n" ); document.write( "b^2-216 = 0
\n" ); document.write( "Positive solution:
\n" ); document.write( "b = sqrt(216)
\n" ); document.write( "b = 14.7 mph (speed of the boat in quiet water)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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