Question 823811
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This is the correct complete solution:
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Pipe A can fill a tank in 6 hours,<pre>
So Pipe A's tank filling rate is 1 tank per 6 hr or {{{1_tank/6_hr}}} or {{{1/6}}}{{{tank/hr}}}</pre>
pipe B can fill it in 2 hours less time than it takes drainpipe C to empty the tank <I>(with pipes A and B closed)</i>.<pre> 
Suppose drainpipe C can drain 1 tank in x hours.  Then drainpipe C's "filling"
rate is a negative quantity -1 tank per x hours or {{{-1_tank/x_hr}}} or {{{-1/x}}}{{{tank/hr}}}
</pre>pipe B can fill it in 2 hours less time than it takes drainpipe C to empty the tank.<pre> 
So Pipe B's tank filling rate is 1 tank per x-2 hr or {{{1_tank/(x-2_hr)}}} or {{{1/(x-2)}}}{{{tank/hr}}}
</pre>With all three pipes open, it takes 3 hours and 20 minutes to fill the tank.<pre>
3 hours and 20 minutes is {{{3&1/3}}} or {{{10/3}}} hours.  

So A, B, and C's combined filling rate is 1 tank per {{{10/3}}} hr or {{{1_tank/((10/3)hr)}}} or {{{1/(10/3)}}}{{{tank/hr}}} or {{{3/10}}}{{{tank/hr}}}

So we have this equation:

{{{(matrix(4,1,Pipe,"A's",filling,rate))}}}{{{""+""}}}{{{(matrix(4,1,Pipe,"B's",filling,rate))}}}{{{""+""}}}{{{(matrix(8,1,Drain,Pipe,"A's","'filling'","rate,",which,is, negative))}}}{{{""=""}}}{{{(matrix(4,1,Their,combine,filling,rate))}}}

   {{{1/6}}}{{{""+""}}}{{{1/(x-2)}}}{{{""+""}}}{{{-1/x}}}{{{""=""}}}{{{3/10}}}

   {{{1/6}}}{{{""+""}}}{{{1/(x-2)}}}{{{""-""}}}{{{1/x}}}{{{""=""}}}{{{3/10}}}
Multiply through by LCD = 30x(x-2)

   5x(x-2) + 30x - 30(x-2) = 3·3x(x-2)
5x² - 10x + 30x - 30x + 60 = 9x(x-2)
            5x² - 10x + 60 = 9x² - 18x
                         0 = 4x² - 8x - 60

Divide through by 4      0 = x² - 2x - 15
                         0 = (x - 5)(x + 3)
                             x-5=0;   x+3=0
                               x=5;     x=-3 (ignore)

Aswer: 5 hours

Edwin</pre>