7 1\12-(k-1 1\3)=4 17\36 помогите решить плиз

Решение:
[tex]7 \frac{1}{12} - (k - 1 \frac{1}{3} ) = 4 \frac{17}{36} [/tex]
[tex]k - 1 \frac{1}{3} = 7 \frac{1}{12} - 4 \frac{17}{36} [/tex]
[tex]k - 1 \frac{1}{3} = 7 \frac{3}{36} - 4 \frac{17}{36} [/tex]
[tex]k - 1 \frac{1}{3} = 6 \frac{39}{36} - 4 \frac{17}{36} [/tex]
[tex]k - 1 \frac{1}{3} = 2 \frac{22}{36} [/tex]
[tex]k - 1 \frac{1}{3} = 2 \frac{11}{18} [/tex]
[tex]k = 1 \frac{1}{3} + 2 \frac{11}{18} [/tex]
[tex]k = 1 \frac{6}{18} + 2 \frac{11}{18} [/tex]
[tex]k = 3 \frac{17}{18} [/tex]
Ответ: [tex]3 \frac{17}{18} [/tex]
Проверка:

[tex]7 \frac{3}{36} - (3 \frac{34}{36} - 1 \frac{12}{36} ) = 4 \frac{17}{36} [/tex]