3. Three brigades, working together, must complete a certain job. It is known that the first and second brigades together can complete it 36 minutes faster than the third brigade. In the time it takes for the first and third brigades to complete the job together, the second brigade can complete only half of the job. In the time it takes for the second and third brigades to complete the job together, the first brigade can complete $\frac{2}{7}$ of the job. How long will it take for all three brigades to complete the job together?
Answer: 1 hour 20 minutes.

Solution.

Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be the productivity of the first, second, and third teams, respectively, that is, the part of the work that a team completes in 1 hour. Using the first condition of the problem, we form an equation. Since the productivity of the first and second teams working together is $(x+y)$, they will complete the entire work in $\frac{1}{x+y}$ hours. One third of the third team will complete the work in $\frac{1}{z}$ hours. We have: $\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z}$.

Similarly, using the other conditions of the problem, we obtain the system: $\left\{\begin{array}{l}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ \frac{1}{x+z}=\frac{1}{2 y} \\ \frac{1}{y+z}=\frac{2}{7 x}\end{array}\right.$.

$\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ x+z=2 y \\ 2 y+2 z=7 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ z=2 y-x \\ 2 y+2(2 y-x)=7 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ z=2 x \\ y=1.5 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+1.5 x}+\frac{3}{5}=\frac{1}{2 x} \\ z=2 x \\ y=1.5 x\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{1}{6} \\ y=\frac{1}{4} \\ z=\frac{1}{3}\end{array}\right.\right.\right.\right.\right.$

Thus, all three teams together complete the work in $\frac{1}{x+y+z}$ hours, that is, in 1 hour 20 minutes.