ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1091

\[\boxed{\text{1091.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ \left\{ \begin{matrix} 3 \cdot (x - 5) - 1 = 6 - 2x \\ 3 \cdot (x - y) - 7y = - 4\ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3x - 15 - 1 = 6 - 2x \\ 3x - 3y - 7y = - 4\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5x = 22\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 10y = - 4 - 3x \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 4,4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 10y = - 4 - 3 \cdot 4,4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 4,4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = - \frac{17,2}{- 10} = 1,72 \\ \end{matrix} \right.\ \]

\[Ответ:(4,4;\ \ 1,72).\]

\[\textbf{б)}\ \left\{ \begin{matrix} 6 \cdot (x + y) - y = - 1\ \ \ \ \ \ \ \\ 7 \cdot (y + 4) - (y + 2) = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 6x + 6y - y = - 1\ \ \ \ \ \\ 7y + 28 - y - 2 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 6x = - 1 - 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6y = - 26 \rightarrow y = - \frac{26}{6} = - \frac{13}{3} = - 4\frac{1}{3} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = - 4\frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x = \frac{- 1 - 5 \cdot \left( - \frac{13}{3} \right)}{6} \\ \end{matrix} \right.\ \]

\[y = - 4\frac{1}{3}\]

\[x = \frac{62}{18} = \frac{31}{9} = 3\frac{4}{9}\]

\[Ответ:\left( 3\frac{4}{9};\ - 4\frac{1}{3} \right).\]