Ну, не знаю, удовлетворит ли мое решение уровень 5-9 класса, но предложу:)
Пусть первоначальное кол-во жидкости таково:
x л - I, у л - II, z л - III.
После переливания из первого во второй получим:
![x- \frac{1}{3}x= \frac{2}{3} x](https://tex.z-dn.net/?f=x-+%5Cfrac%7B1%7D%7B3%7Dx%3D+%5Cfrac%7B2%7D%7B3%7D++x)
л - осталось в I
![(y+ \frac{1}{3}x)](https://tex.z-dn.net/?f=%28y%2B+%5Cfrac%7B1%7D%7B3%7Dx%29)
л стало во II
После переливания из второго в третий получим:
![(y+ \frac{1}{3}x)- \frac{1}{4} (y+ \frac{1}{3}x)= (\frac{1}{4}x+ \frac{3}{4}y)](https://tex.z-dn.net/?f=%28y%2B+%5Cfrac%7B1%7D%7B3%7Dx%29-+%5Cfrac%7B1%7D%7B4%7D+%28y%2B+%5Cfrac%7B1%7D%7B3%7Dx%29%3D+%28%5Cfrac%7B1%7D%7B4%7Dx%2B+%5Cfrac%7B3%7D%7B4%7Dy%29)
л - осталось во II
![z+ \frac{1}{4} (y+ \frac{1}{3}x)=( \frac{1}{12} x+ \frac{1}{4}y+z)](https://tex.z-dn.net/?f=z%2B+%5Cfrac%7B1%7D%7B4%7D+%28y%2B+%5Cfrac%7B1%7D%7B3%7Dx%29%3D%28+%5Cfrac%7B1%7D%7B12%7D+x%2B+%5Cfrac%7B1%7D%7B4%7Dy%2Bz%29+)
л - стало в III.
Наконец, после переливания из III в I получим:
![\frac{1}{12} x+ \frac{1}{4}y+z- \frac{1}{10}( \frac{1}{12} x+ \frac{1}{4}y+z)= (\frac{9}{120}x+ \frac{9}{40}y+ \frac{9}{10}z)](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B12%7D+x%2B+%5Cfrac%7B1%7D%7B4%7Dy%2Bz-+%5Cfrac%7B1%7D%7B10%7D%28+%5Cfrac%7B1%7D%7B12%7D+x%2B+%5Cfrac%7B1%7D%7B4%7Dy%2Bz%29%3D+%28%5Cfrac%7B9%7D%7B120%7Dx%2B+%5Cfrac%7B9%7D%7B40%7Dy%2B+%5Cfrac%7B9%7D%7B10%7Dz%29)
л - осталось в III
![\frac{2}{3}x+ \frac{1}{10}(z+ \frac{1}{4}(y+ \frac{1}{3}x))= (\frac{81}{120}x+ \frac{1}{40}y + \frac{1}{10}z)](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B3%7Dx%2B+%5Cfrac%7B1%7D%7B10%7D%28z%2B+%5Cfrac%7B1%7D%7B4%7D%28y%2B+%5Cfrac%7B1%7D%7B3%7Dx%29%29%3D+%28%5Cfrac%7B81%7D%7B120%7Dx%2B+%5Cfrac%7B1%7D%7B40%7Dy++%2B+%5Cfrac%7B1%7D%7B10%7Dz%29)
л - стало в I.
По условию, во всех сосудах стало по 9 л жидкости.
Решаем систему уравнений:
![\begin{cases} \frac{81}{120}x+ \frac{1}{40}y + \frac{1}{10}z=9 \\\frac{1}{4}x+ \frac{3}{4}y=9 \\ \frac{9}{120}x+ \frac{9}{40}y+ \frac{9}{10}z=9 \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} 81x+3y+12z=1080 \\ x+3y=36 \\ x+3y+12z=120 \end{cases} \ \textless \ =\ \textgreater \](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+%5Cfrac%7B81%7D%7B120%7Dx%2B+%5Cfrac%7B1%7D%7B40%7Dy+%2B+%5Cfrac%7B1%7D%7B10%7Dz%3D9+%5C%5C%5Cfrac%7B1%7D%7B4%7Dx%2B+%5Cfrac%7B3%7D%7B4%7Dy%3D9+%5C%5C+%5Cfrac%7B9%7D%7B120%7Dx%2B+%5Cfrac%7B9%7D%7B40%7Dy%2B+%5Cfrac%7B9%7D%7B10%7Dz%3D9+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C+++%5Cbegin%7Bcases%7D+81x%2B3y%2B12z%3D1080+%5C%5C+x%2B3y%3D36+%5C%5C+x%2B3y%2B12z%3D120+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C+)
![\begin{cases} 81x+120-x=1080 \\ x+3y=36 \\ 36+12z=120 \end{cases} \ \textless \ =\ \textgreater \ \begin{cases} x=12 \\ y=8 \\ z=7 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+81x%2B120-x%3D1080+%5C%5C+x%2B3y%3D36+%5C%5C+36%2B12z%3D120+%5Cend%7Bcases%7D+%5C+%5Ctextless+%5C+%3D%5C+%5Ctextgreater+%5C++%5Cbegin%7Bcases%7D+x%3D12+%5C%5C+y%3D8+%5C%5C+z%3D7+%5Cend%7Bcases%7D+)
Итак, первоначально было:
12 л - в I сосуде, 12 л - во II сосуде, 8 л - в I сосуде, 7 л - в III сосуде.
Ответ: 12 л, 8 л, 7 л.