document.write( "Question 175008: 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 #130104 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( "uptime - downtime = 1/3 hr
\n" ); document.write( "5/(b-4) - 5/(b+4) = 1/3
\n" ); document.write( "5(b+4) - 5(b-4) = (1/3)(b^2-16)
\n" ); document.write( "40 = (1/3)(b^2-16)
\n" ); document.write( "b^2 - 16 = 120
\n" ); document.write( "b^2 = 136
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\n" ); document.write( "Positive solution:
\n" ); document.write( "b = 11.66 mph (boat speed in quiet water)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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