ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1422

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\[1)\ \left\{ \begin{matrix} \frac{x - y}{5} - \frac{x + y}{2} = 10\ \ \ \ \ | \bullet 10 \\ \frac{x}{5} + \frac{y}{2} = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \bullet 10 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2(x - y) - 5(x + y) = 100 \\ 2x + 2y = 100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 2x - 2y - 5x - 5y = 100 \\ 2x + 5y = 100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 3x - 7y = 100 \\ 2x + 5y = 100\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} - 3x = 100 + 7y \\ 2x = 100 - 5y\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = - \frac{100}{3} - \frac{7}{3}y \\ x = 50 - 2,5y\ \ \ \ \\ \end{matrix} \right.\ \]

\[- \frac{100}{3} - \frac{7}{3}y = 50 - 2,5y\ \ \ \ \ | \bullet 3\]

\[- 100 - 7y = 150 - 7,5y\]

\[0,5y = 250\]

\[y = 500;\]

\[x = 50 - 2,5 \bullet 500 =\]

\[= 50 - 1250 = - 1200.\]

\[Ответ:\ \ ( - 1200;\ 500).\]

\[2)\ \left\{ \begin{matrix} \frac{x + y}{2} + \frac{x - y}{3} = 6\ \ \ \ \ | \bullet 6\ \ \\ \frac{x + y}{4} - \frac{x - y}{3} = 0\ \ \ \ \ | \bullet 12 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3(x + y) + 2(x - y) = 36 \\ 3(x + y) - 4(x - y) = 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} 3x + 3y + 2x - 2y = 36 \\ 3x + 3y - 4x + 4y = 0\ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5x + y = 36\ \\ - x + 7y = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} y = 36 - 5x \\ 7y = x\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} y = 36 - 5x \\ y = \frac{x}{7}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[36 - 5x = \frac{x}{7}\ \ \ \ \ | \bullet 7\]

\[252 - 35x = x\]

\[36x = 252\]

\[x = 7;\]

\[y = 36 - 5 \bullet 7 = 36 - 35 = 1.\]

\[Ответ:\ \ (7;\ 1).\]