13.426 Two pumps were installed to fill a swimming pool with water. The first pump can fill the pool 8 hours faster than the second one. Initially, only the second pump was turned on for a time equal to twice the amount of time it would take to fill the pool if both pumps were working simultaneously. Then, the first pump was also turned on, and 1.5 hours after the first pump was turned on, the pool was filled with water. How many hours does it take for each pump, working separately, to fill the pool?
Solution. Let $x$ and $y$ be the pump efficiencies, and $z$ be the volume of work. Then $\frac{z}{x+y}$ is the time it takes to fill the pool when both pumps are working simultaneously. According to the conditions of the problem,
$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=8, \\ \frac{2 z}{x+y} \cdot y+1.5(x+y)=z\end{array} \Leftrightarrow\left\{\begin{array}{l}z(x-y)=8 x y, \\ \frac{3}{2}(x+y)=\frac{z(x-y)}{x+y}\end{array} \Rightarrow \frac{3}{2}(x+y)=\frac{8 x y}{x+y} \Leftrightarrow\right.\right.$

$\Leftrightarrow 3 x^{2}-10 x y+3 y^{2}=0 \Leftrightarrow\left[\begin{array}{l}x=3 y, \\ x=\frac{y}{3} .\end{array}\right.$

Since $x>y$, then $x=3 y$. Substituting this into the first equation of the system, we get

$\frac{z}{y}=12, \frac{z}{x}=4$.

Answer: 4 hours, 12 hours.