13.296. Two excavator operators must complete a certain job. After the first one worked for 15 hours, the second one starts and finishes the job in 10 hours. If, working separately, the first one completed $1 / 6$ of the job, and the second one completed $1 / 4$ of the job, it would take an additional 7 hours of their combined work to finish the job. How many hours would it take each excavator operator to complete the job individually?
Solution.

Let's take the entire volume of work as 1. Let $x, y$ be the work efficiency of the first and second excavator operators, respectively. According to the conditions, we have the system $\left\{\begin{array}{l}15 x+10 y=1, \\ 7(x+y)=1-\left(\frac{1}{6}+\frac{1}{4}\right),\end{array}\right.$ from which $x=\frac{1}{30}, y=\frac{1}{20}$.

Then the first excavator operator will complete the entire work in $\frac{1}{x}=30$ (hours), and $x$

the second $-3 \mathrm{a} \frac{1}{=}=20$ (hours).

$y$

Answer: in 20 and 30 hours.