ГДЗ по алгебре 7 класс Макарычев Задание 633

\[\boxed{\text{633\ (633).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2 \cdot (3 - 5c) + 1 = 4 \cdot (1 - c)\]

\[6 - 10c + 1 = 4 - 4c\]

\[- 10c + 4c = 4 - 7\]

\[- 6c = - 3\]

\[c = 0,5\]

\[Ответ:при\ c = 0,5.\]

\[\textbf{б)} - 3 \cdot (2x + 1) - 20 = 8x + 5\]

\[- 6x - 3 - 8x = 20 + 5\]

\[- 14x = 25 + 3\]

\[- 14x = 28\]

\[x = - 2\]

\[Ответ:при\ x = - 2.\]

\[\textbf{в)}\ 3 \cdot (5x + 7) = 61 - 10x\]

\[15x + 21 = 61 - 10x\]

\[15x + 10x = 61 - 21\]

\[25x = 40\]

\[x = 40\ :25 = \frac{40}{25} = \frac{8}{5} = 1\frac{3}{5}\]

\[x = 1,6\]

\[Ответ:\ \ при\ x = 1,6.\]

\[\textbf{г)}\ 8 - y = 2 \cdot (7 + y)\]

\[8 - y = 14 + 2y\]

\[- y - 2y = 14 - 8\]

\[- 3y = 6\]

\(y = - 2\)

\[Ответ:при\ y = - 2.\]