Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 4. Действия с рациональными числами Задание 235

\[\boxed{\text{4.\ 235}}\]

\[\textbf{а)} - 4 + x = 8,7\]

\[x = 8,7 - ( - 4) = 8,7 + 4\]

\[x = 12,7.\]

\[Проверка:\]

\[- 4 + 12,7 = 8,7\]

\[8,7 = 8,7.\]

\[\textbf{б)}\ 9,3 + x = - 8\]

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

\[x = - 17,3.\]

\[Проверка:\]

\[9,3 + ( - 17,3) = - 8\]

\[- 8 = - 8.\]

\[\textbf{в)}\ 7 - y = 2,4\]

\[y = 7 - 2,4\]

\[y = 4,6.\]

\[Проверка:\]

\[7 - 4,6 = 2,4\]

\[2,4 = 2,4.\]

\[\textbf{г)}\ 6 - y = - 3\frac{5}{7}\]

\[y = 6 - \left( - 3\frac{5}{7} \right) = 6 + 3\frac{5}{7}\]

\[y = 9\frac{5}{7}.\]

\[Проверка:\]

\[6 - 9\frac{5}{7} = - 3\frac{5}{7}\]

\[- 3\frac{5}{7} = - 3\frac{5}{7}.\]

\[\textbf{д)}\ c + \frac{5}{14} = - \frac{3}{7}\]

\[c = - \frac{3}{7} - \frac{5}{14} = - \frac{6}{14} - \frac{5}{14}\]

\[c = - \frac{11}{14}.\]

\[Проверка:\]

\[- \frac{11}{14} + \frac{5}{14} = - \frac{3}{7}\]

\[- \frac{6}{14} = - \frac{3}{7}\]

\[- \frac{3}{7} = - \frac{3}{7}.\]

\[\textbf{е)}\ c + 1,2 = - 1\frac{2}{5}\]

\[c = - 1,4 - 1,2\]

\[c = - 2,6.\]

\[Проверка:\]

\[- 2,6 + 1,2 = - 1\frac{2}{5}\]

\[- 1,4 = - 1,4.\]