One pipe can fill a tank in 36 minutes, a second can fill it in 9 minutes, and a third can fill it in 12 minutes. If the tank is empty, how long will the three pipes, operating together, take to fill it?

This is a classic work problem. These are solved by finding out how much of the work being done can get done in a single unit of time. In our case minutes.

If pipe 1 takes 36 minutes to fill the tank, then it can get 1/36 of the job done in a single minute.

If pipe 2 takes 9 minutes to fill the tank, then it can get 1/9 of the job done in a single minute.

If pipe 3 takes 12 minutes to fill the tank, then it can get 1/12 of the job done in a single minute.

The sum of all of these together will take x minutes. Our equation then is this:

[tex]\frac{1}{36} +\frac{1}{9} +\frac{1}{12} =\frac{1}{x}[/tex]

We need to find the LCD of those numbers and multiply through by that LCD to get rid of the fractions, cuz who likes fractions?! The LCD is 36x; each denominator, including the x, goes into 36x evenly and that's the lowest number that each goes into evenly. Multiplying through by the LCD:

[tex](36x)\frac{1}{36}+(36x)\frac{1}{9}+(36x)\frac{1}{12}=(36x)\frac{1}{x}[/tex]

Reducing gives us the equation:

x + 4x + 3x = 36 and

8x = 36 so

x = 4.5 minutes or 4 1/2 minutes