Using two pumps

Hi, Ramel! This is a working together problem. The rate of the 8 hp pump is: $$\frac{1}{8}$$ The rate of the 3 hp pump is: 1/12 $$\frac{1}{12}$$ The time together will be 4 hours and the time that the 3 hp pump works alone would be $x$ hrs. We want to fill only 1 tank. The formula we now need then is: $$\left (\frac{1}{8} \times 4 \right ) + \left (\frac{1}{12} \times 4 \right ) + \left (\frac{1}{12} \times x \right ) = 1$$ we get... $$\frac{4}{8} + \frac{4}{12} + \frac{x}{12} = 1$$ Adding the fractions, we get: $$\frac{5}{6} + \frac{x}{12} = 1$$ Subtract $\frac{5}{6}$ to get: $$\frac{x}{12} = \frac{1}{6}$$ The value of $x$ then is $2$. It means that the 3 hp pump must at least work for additional $\color{blue}{2hours}$ to fill the tank. Alternate solutions are encouraged. Thanks!