\[\boxed{\text{474\ (474).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ x\frac{км}{ч} - скорость\ \]
\[мотоциклиста,\ который\]
\[\ приехал\ раньше.\]
\[\text{y\ }\frac{км}{ч} - скорость\ второго\ \]
\[мотоциклиста.\ \]
\[30\ мин = 0,5\ часа.\]
\[Тогда:\ \ 0,5x + 0,5y = 50.\ \ \]
\[Время,\ которое\ на\ весь\]
\[\ путь\ затратит\ \]
\[первый:\ \frac{50}{x}\ часов.\ А\ второй\ \]
\[потратит\ \frac{50}{y}\ часов.\ 25\ мин =\]
\[= \frac{25}{60} = \frac{5}{12}\ ч.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} 0,5x + 0,5y = 50 \\ \frac{50}{x} + \frac{25}{60} = \frac{50}{y}\text{\ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x + y = 100 \\ \frac{2}{x} + \frac{1}{60} = \frac{2}{y}\text{\ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 100 - y\ \ \ \ \ \ \ \ \ \\ \frac{2}{100 - y} + \frac{1}{60} = \frac{2}{y} \\ \end{matrix} \right.\ \Longrightarrow\]
\[y^{2} - 340y + 12000 = 0\]
\[y_{1} + y_{2} = 340;\ \ y_{1} \cdot y_{2} = 12\ 000\]
\[y_{1} = 40;\ \ \ y_{2} = 300.\]
\[\left\{ \begin{matrix} y_{1} = 40 \\ x_{1} = 60 \\ \end{matrix} \right.\ \ \ \ или\ \ \]
\[\left\{ \begin{matrix} y_{2} = 300\ \ \\ x_{2} = - 200 \\ \end{matrix} \right.\ \Longrightarrow но\ x > 0 \Longrightarrow\]
\[\Longrightarrow не\ подходит.\]
\[Ответ:40\ \frac{км}{ч}\ и\ \ 60\frac{км}{ч}.\]
\[\boxed{\text{474.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 3xy + 14 = 0 \\ 3x^{2} + 2xy - 24 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2x^{2} - 6xy + 28 = 0 \\ 9x^{2} + 6xy - 72 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 11x^{2} = 44\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3xy + 14 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3xy + 14 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ \ }или\left\{ \begin{matrix} x_{2} = - 2 \\ y_{2} = - 3. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} 2x^{2} - 6y = xy\ \ \ \ \ \ \\ 3x^{2} - 8y = 0,5xy \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 6x^{2} - 18y = 3xy\ \ \ \ \\ - 6x^{2} + 16y = - xy \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} - 2y = 2xy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - 8y = x^{2} - 3y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2y(x + 1) = 0 \\ 2x² = 5y\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\left\{ \begin{matrix} y = 0 \\ x = 0 \\ \end{matrix} \right.\ ,\]
\[2)\ \left\{ \begin{matrix} x + 1 = 0 \\ y = \frac{2x^{2}}{5}\text{\ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = - 1\ \\ y = 0,4. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (2;3);( - 2;\ - 3);\ \ \]
\[\textbf{б)}\ (0;0);( - 1;0,4).\]