13.107. Two brigades, working together, were supposed to repair a given section of a highway in 18 days. In reality, however, only the first brigade worked at first, and the second brigade, which has a higher labor productivity than the first, finished the repair of the road section. As a result, the repair of the given road section took 40 days, with the first brigade completing $2 / 3$ of the work during its working time. How many days would it take for each brigade to repair the given road section separately?
Solution.

Let $x$ be the work rate of the first team, and $y$ be the work rate of the second team. Working together, the teams will repair the section in $\frac{1}{x+y}$ days. According to the problem, $\frac{1}{x+y}=18$. The productivity of the first team is $\frac{1}{x}$, and the second team's productivity is $\frac{1}{y}$. According to the problem, $\frac{2}{3} \cdot \frac{1}{x}+\frac{1}{3} \cdot \frac{1}{y}=40$. Solving the system $\left\{\begin{array}{l}\frac{1}{x+y}=18, \\ \frac{2}{3} \cdot \frac{1}{x}+\frac{1}{3} \cdot \frac{1}{y}=40,\end{array}\right.$ we find: $\frac{1}{x}=45$ days; $\frac{1}{y}=30$ days or $\frac{1}{x}=24$ days, $\frac{1}{y}=72$ days. Considering the conditions of the problem, finally: $\frac{1}{x}=45$ days, $\frac{1}{y}=30$ days. Therefore, the first team will repair the entire section in 45 days, and the second team in 30 days.

Answer: 45 and 30 days.