Решение:
\[(3\frac{1}{7} - n) + 1\frac{4}{7} = 3\frac{5}{7} + \frac{2}{7}\]
\[(3\frac{1}{7} - n) + 1\frac{4}{7} = 4\frac{7}{7}\]
\[(3\frac{1}{7} - n) + 1\frac{4}{7} = 4\frac{0}{1}\]
\[3\frac{1}{7} - n = 4\frac{0}{1} - 1\frac{4}{7}\]
\[3\frac{1}{7} - n = 2\frac{3}{7}\]
\[-n = 2\frac{3}{7} - 3\frac{1}{7}\]
\[-n = - \frac{5}{7}\]
\[n = \frac{5}{7}\]
Проверка:
\[(3\frac{1}{7} - \frac{5}{7}) + 1\frac{4}{7} = 3\frac{5}{7} + \frac{2}{7}\]
\[2\frac{3}{7} + 1\frac{4}{7} = 3\frac{7}{7}\]
\[4 = 4\]
Ответ: \(\frac{5}{7}\)