13.213. Usually, two mechanisms are involved in performing a certain task simultaneously. The productivity of these mechanisms is not the same, and when working together, they complete the task in 30 hours. However, this time the joint operation of the two mechanisms lasted only 6 hours, after which the first mechanism was stopped and the second mechanism completed the remaining part of the task in 40 hours. How long would it take for each mechanism to complete the same task, working separately at its inherent productivity?
## Solution.

Let the entire volume of work be equal to one unit, the first mechanism completes the task in $x$ hours, and the second in $y$ hours. Then the productivity of the first mechanism is $\frac{1}{x}$, and the second is $-\frac{1}{y}$, the combined productivity when working together is $\frac{1}{x}+\frac{1}{y}$. Thus, we obtain the system of equations

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{1}{30} \\
1-6\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{40}{y}
\end{array} \Rightarrow x=75 \text { hours, } y=50\right. \text { hours }
$$

Answer: 75 and 50 hours.