Two brigades, working together, can do some work in $12$ days 
Two brigades, working together, can do some work in $12$ days. The
  brigade that works slowly can do $\dfrac{75}{7}\%$ of all work in 6
  days less than those required by the other brigade to do $\dfrac{3}{7}$ of all work. How many days can each brigade do the work on its own?

I haven't solved this kind of problems for a while so I would be very grateful if you could help me. As I see, we should model a system of equations that expresses the relationships given in the text of the problem. If $x$ and $y$ are the days required by the brigades ($x,y>0$), I think $\dfrac{12}{x}+\dfrac{12}{y}=1$.

75/7% is 75/700= 3/28.  Let x be the number of days it takes the first (slow) brigade to complete the job, y the number of days it takes the second brigade. Assuming a constant work rate, the second brigade will complete 3/7 of the job in (3/7)y.   The first (slow) brigade "can do 75/7% of all work in 6 days less than the other brigade can complete 3/7 of the job" so 75/7% x, which is (3/28)x, the time it would take the slow brigade to do 75/7% of the job, is equal to (3/7)y.  (3/28)x= (2/7)y- 6.  Yes, since x and y the time it would take each of them to complete the job, 1/x and 1/y are the rates at which they work.  Working together there rate is 1/x+ 1/y job per day.  Since, working together, they can finish the job in 12 days, there rate is 1/12 job per day:  1/x+ 1/y= 1/12.
Solve the two equations 1/x+ 1/y= 1/12 and  (3/28)x= (2/7)y- 6 for x and y.
