(2 3/9-1 7/9)+6 4/9-(4 2/9+8/9)= столбиком

Решите задачу:

$(2 \frac{3}{9} -1 \frac{7}{9} )+6 \frac{4}{9} -(4 \frac{2}{9} + \frac{8}{9} )=1 \frac{8}{9} \\ \\ 1) 2 \frac{3}{9} -1 \frac{7}{9} =1 \frac{12}{9} -1 \frac{7}{9} = \frac{5}{9} \\ \\ 2)4 \frac{2}{9} + \frac{8}{9} =4 \frac{10}{9} =5 \frac{1}{9} \\ \\ 3) \frac{5}{9} +6 \frac{4}{9} =6+1=7 \\ \\ 4) 6 \frac{9}{9} -5 \frac{1}{9} =1 \frac{8}{9}$