Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?
I always have an awful problem getting my head around questions like this so I made the question easier to see how it would work.
I said:
Let the tank be 36L and let the times to drain be 4 hours and 6 hours respectively
So tank 1 will flow at 36/4 = 9L/hour
and tank 2 will flow at 36/6 = 6L/hour
if they are both draining then the water will flow at (9+6) L/hour = 15L/hour
so
\(\frac{15L}{1hour}=\frac{15L*\frac{36}{15}}{1hour*\frac{36}{15}}=\frac{36L}{2.4hours}\)
It will take 2.4 hours to drain the tank if both pipes are emptying it.
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Ok now I can do the same thing for your more difficult problem.
Let the tank be V litres
Pipe1 will take t hours to drain it so that is \(\frac{V}{t}\;\;litres/hour\)
Pipe2 will take (t+2) hours to drain it so that is \(\frac{V}{t+2}\;\;litres/hour\)
So the 2 pipes together will drain
\(\frac{V}{t}+\frac{V}{t+2}\;\;litres/hour\\ =V[\frac{1}{t}+\frac{1}{t+2}]\;\;litres/hour\\ =V[\frac{(t+2)+t}{t(t+2)}]\;\;litres/hour\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\;\;litres}{1hour}\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\left[\frac{t^2+2t}{2t+2}\right]\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\ =\frac{V\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\\)
So it will take
\(\frac{t^2+2t}{2t+2}\;\;\text{hours to drain the tank}\\~\\ now\\ \frac{t^2+2t}{2t+2}=6\\ t^2+2t=6(2t+2)\\ t^2+2t=12t+12\\ t^2-10t-12=0\\ t=\frac{10\pm\sqrt{100+48}}{2}\\ t=\frac{10\pm\sqrt{148}}{2}\\ t=\frac{10\pm2\sqrt{37}}{2}\\ t=5\pm\sqrt{37}\\ \text{t cannot be negative so }\\t=5+\sqrt{37}\approx 11.08276253\approx 11hours\;\;4\;minutes \;and\;58\;seconds\\\)
Which is near enough to 11 hours and 5 minutes
So the pipes will take 11 hours and 5 minutes and 13 hours and 5 min respectively to empty the tank on their own.
Since they take 6 hours together that sounds reasonable 
+118727 