\[\boxed{\text{441\ (441).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy - y^{2} = 11 \\ x - 2y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} + 4y + 1 + 2y^{2} + y - y^{2} = 11 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y^{2} + 5y - 10 = 0 \\ \end{matrix} \right.\ \]
\[5y^{2} + 5y - 10 = 0\ \ \ \ |\ :5\]
\[y^{2} + y - 2 = 0\]
\[y_{1} + y_{2} = - 1;\ \ \ y_{2} \cdot y_{2} = - 2\]
\[y_{1} = 1;\ \ \ y_{2} = - 2.\]
\[1)\ y_{1} = 1;\ \ x_{1} = 3;\]
\[2)\ y_{2} = - 2;\ \ x_{2} = - 3.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy - 3y = 9 \\ 3x + 2y = - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 1,5x^{2} - 0,5x + 4,5x + 1,5 - 9 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \\ - 0,5x^{2} + 4x - 7,5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[0,5x^{2} - 4x + 7,5 = 0\ \ \ \ \ \ | \cdot 2\]
\[x^{2} - 8x + 15 = 0\]
\[D_{1} = 16 - 15 = 1\]
\[x_{1} = 4 + 1 = 5;\ \ \ \]
\[x_{2} = 4 - 1 = 3.\]
\[1)\ x_{1} = 5;\ \ y_{1} = - 8;\]
\[2)\ x_{2} = 3;\ \ \ y_{2} = - 5.\]
\[Ответ:а)\ \ (3;1);( - 3;\ - 2);\ \ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (5;\ - 8);\ \ (3; - 5).\]
\(\boxed{\text{441.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[Пусть\ раствор\ содержал\ \text{x\ }\]
\[грамм\ воды.\ \]
\[(x + 150)\ г - воды\ в\ новом\ \]
\[растворе.\]
\[\frac{50}{x} - была\ концентрация\ соли;\]
\[\frac{50}{x + 150} - концентрация\ соли\ \]
\[в\ новом\ растворе.\]
\[Известно,\ что\ концентрация\]
\[\ уменьшилась\ на\ 7,5\% = 0,075.\]
\[Составим\ уравнение:\]
\[\ \frac{50}{x + 150} = \frac{50}{x} - 0,075\]
\[0,075x^{2} + 11,25x - 7500 = 0\]
\[D = {11,25}^{2} + 4 \cdot 0,075 \cdot 7500 =\]
\[= 2376,5625\]
\[x_{1,2} = \frac{- 11,25 \pm 48,75}{0,15}.\]
\[Так\ как\ x > 0 \Longrightarrow x = 250.\]
\[Концентрация:\ \ \]
\[\frac{50}{250} = 0,2.\]
\[Ответ:250\ грамм;\ \ 20\%.\]