Correct. Using the equation for work, $\displaystyle{r}=\frac{{W}}{{t}}\usepackage{cmbright}\Huge\$ , you find that the rate of the slow pipe is $\displaystyle\frac{{1}}{{p}}\usepackage{cmbright}\Huge\$ (because it fills 1 pool in p hours), and the rate of the fast pipe is $\displaystyle\frac{{2}}{{p}}\usepackage{cmbright}\Huge\$ (because it can fill 2 pools in the time it takes the slow pipe to fill 1). Since you know that it takes both pipes 2 hours to fill 1 pool, you can write $\displaystyle\frac{{1}}{{p}}+\frac{{2}}{{p}}=\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ . Solving for p yields p=6, so it would take the slow pipe 6 hours.