ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1092

\[\boxed{\text{1092.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 5y + 8 \cdot (x - 3y) = 7x - 12 \\ 9x + 3 \cdot (x - 9y) = 11y + 46 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5y + 8x - 24y = 7x - 12\ \ \\ 9x + 3x - 27y = 11y + 46 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 19y + x = - 12 \rightarrow x = 19y - 12 \\ 12x - 38y = 46\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 19y - 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ 12 \cdot (19y - 12) - 38y = 46\ \ \ (1) \\ \end{matrix} \right.\ \]

\[(1)228y - 144 - 38y = 46\]

\[190y = 190\]

\[y = 1\]

\[(2)\ \ x = 19 \cdot 1 - 12\]

\[x = 7\]

\[Ответ:(7;1).\]

\[\textbf{б)}\ \left\{ \begin{matrix} - 2 \cdot (a - b) + 16 = 3 \cdot (b + 7) \\ 6a - (a - 5) = - 8 - (b + 1)\text{\ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2a + 2b + 16 = 3b + 21 \\ 6a - a + 5 = - 8 - b - 1\ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2a - b = 5\ \ \\ 5a + b = - 14 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b = - 2a - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ 5a - 2a - 5 = - 14\ \ \ (1) \\ \end{matrix} \right.\ \]

\[(1)\ \ 3x = - 9\]

\[a = - 3\]

\[(2)\ \ b = - 2 \cdot ( - 3) - 5\]

\[b = 1\]

\[Ответ:( - 3;1).\]