Gone fishing Debbie traveled by boat 5 miles upstream to fis
Gone fishing. 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?
Solution
Upstream DATA:
 distance = 5 miles; time = x + 20 ; rate = d/t = 5/(x+2) hrs.
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 Downstream DATA:
 distance = 5 miles ; time = x ; rate = d/t = 5/x
 ----------------------------
 Equation:
 rate upstream = b-4=5/(x+2)
 rate downstrea= b+4=5/x
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 Subtract 1st from 2nd to get:
 8 = 5/x - 5/(x+2)
 Solve for \"x\":
 8x(x+2) = 5(x+2) - 5x
 8x^2+16x = 10
 4x^2+8x-5 = 0
 4x^2+10x-2x-5 = 0
 2x(2x+5)-(2x+5) = 0
 (2x+5)(2x-1) = 0
 Positive solution:
 x = 1/2
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 Solve for \"b\" when b+4 = 5/x
 b+4 = 5/(1/2)
 b+4 = 10
 b = 6 mph (speed of the boat in still water)

