Question 194736
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You got off on the wrong foot from the git-go.  You said *[tex \LARGE r = \frac{40}{6}], but that can't be true because 40 is the one-way distance and 6 hours is the total time for both directions.  Not to mention that you then said *[tex \LARGE r - 5 = \frac{40}{6}] and both of those statements cannot be simultaneously true.


Let's begin anew.


Let <b><i>r</i></b> represent the speed of the boat in still water. Then <b><i>r</i> - 5</b> is the speed of the boat upriver (against the current) and <b><i>r</i> + 5</b> is the speed downriver.  Let <b><i>t</i></b> represent the amount of time it took to go upriver, then <b>6 - <i>t</i></b> is the time it took to go downriver.


For the upriver trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = 40 = (r - 5)t]


and for the downriver trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = 40 = (r + 5)(6 - t)]


Solving both equations for <b><i>t</i></b> we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{40}{r - 5}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = 6 - \frac{40}{r + 5} ]


Now that we have two things equal to <b><i>t</i></b>, we can set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{40}{r - 5} = 6 - \frac{40}{r + 5}]


LCD is *[tex \LARGE (r - 5)(r + 5)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{40(r + 5)}{(r - 5)(r + 5)} = \frac{\left(6(r - 5)(r + 5)\right)-\left(40(r -5)\right)}{(r - 5)(r + 5)}]


Expanding binomials, removing parentheses, collecting terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{40r + 200}{r^2 - 25} = \frac{6r^2 - 40r + 50}{r^2 - 50}]


Now, you can either multiply both sides by *[tex \LARGE r^2 - 50] or realize that if you have two equal fractions with identical denominators, then the numerators have to be equal as well to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 40r + 200 = 6r^2 - 40r + 50]


Put in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6r^2 - 80r - 150 = 0]


Fortunately, this little puppy factors rather neatly:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (6r + 10)(r - 15) = 0]


Exclude the negative root because we are reasonably certain that the boat wasn't going backwards as a negative rate through the water would suggest, and the rate in still water is 15 mph.


Check:


A 40 mile upriver trip would take 40 divided by (15 - 5) = 4 hours.  A 40 mile downriver trip would take 40 divided by (15 + 5) = 2 hours.  Total time: 6 hours, answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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