Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 5. Решение уравнений Задание 28

\[\boxed{\text{5.\ 28}}\]

\[\textbf{а)}\ 7,2 - (z - 6,1) = 6,3\]

\[7,2 - z + 6,1 = 6,3\]

\[- z + 13,3 = 6,3\]

\[- z = 6,3 - 13,3\]

\[- z = - 7\]

\[z = 7.\]

\[\textbf{б)} - 2,9 + (y - 5,3) = - 3,4\]

\[- 2,9 + y - 5,3 = - 3,4\]

\[y - 8,2 = - 3,4\]

\[y = - 3,4 + 8,2\]

\[y = 4,8.\]

\[\textbf{в)}\ 4,4 - (a - 5,6) = 100\]

\[4,4 - a + 5,6 = 100\]

\[10 - a = 100\]

\[a = 10 - 100\]

\[a = - 90.\]

\[\textbf{г)} - \frac{8}{9} - (n - 1) = \frac{7}{18}\]

\[- \frac{8}{9} - n + 1 = \frac{7}{18}\]

\[\frac{1}{9} + n = \frac{7}{18}\]

\[n = \frac{7}{18} - \frac{2}{18}\]

\[n = \frac{5}{18}.\]

\[\textbf{д)}\ 1\frac{5}{9} - \left( s + \frac{4}{9} \right) = \frac{2}{3}\]

\[1\frac{5}{9} - s - \frac{4}{9} = \frac{2}{3}\]

\[1\frac{1}{9} - s = \frac{4}{9}\]

\[s = \frac{10}{9} - \frac{4}{9}\]

\[s = \frac{5}{9}.\]

\[\textbf{е)} - 5\frac{4}{7} + \left( - \frac{5}{14} + z \right) = 3\frac{1}{7}\]

\[- 5\frac{8}{14} - \frac{5}{14} + z = 3\frac{2}{14}\]

\[z - 5\frac{13}{14} = 3\frac{2}{14}\]

\[z = 3\frac{2}{14} + 5\frac{13}{14}\]

\[z = 8\frac{15}{14}\]

\[z = 9\frac{1}{14}\text{.\ }\]