Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 5. Решение уравнений Задание 12

\[\boxed{\text{5.\ 12}}\]

\[\textbf{а)}\ 9,8 - (7,8 - x) = 7,3\]

\[9,8 - 7,8 + x = 7,3\]

\[2 + x = 7,3\]

\[x = 7,3 - 2\]

\[x = 5,3.\]

\[\textbf{б)} - 9 + (c - 31) = - 6\]

\[- 9 + c - 31 = - 6\]

\[c - 40 = - 6\]

\[c = - 6 + 40\]

\[c = 34.\]

\[\textbf{в)}\ \frac{39}{44} - \left( \frac{5}{44} - x \right) = \frac{7}{11}\]

\[\frac{39}{44} - \frac{5}{44} + x = \frac{7}{11}\]

\[\frac{34}{44} - x = \frac{7}{11}\]

\[x = \frac{34}{44} - \frac{28}{44}\]

\[x = \frac{6}{44}\]

\[x = \frac{3}{22}.\]

\[\textbf{г)}\ (z + 5) - 21 = - 30\]

\[z + 5 - 21 = - 30\]

\[z - 16 = - 30\]

\[z = - 30 + 16\]

\[z = - 14.\]

\[\textbf{д)} - (20 - c) + 41,2 = - 23,8\]

\[- 20 + c + 41,2 = - 23,8\]

\[c + 21,2 = - 23,8\]

\[c = - 23,8 - 21,2\]

\[c = - 50.\]

\[\textbf{е)}\ \left( y + \frac{10}{15} \right) - \frac{4}{15} = 1,9\]

\[y + \frac{10}{15} - \frac{4}{15} = 1,9\]

\[y + \frac{6}{15} = 1,6\]

\[y + 0,4 = 1,6\]

\[y = 1,6 - 0,4\]

\[y = 1,2.\]