Страница 128 номер 460, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} \frac{1}{x} - \frac{1}{y} = \frac{1}{12} \\ 2x - y = 2 \end{cases}\ \Rightarrow \begin{cases} 12y - 12x = xy \\ y = 2x - 2 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} 12(2x - 2) - 12x = x(2x - 2) \\ y = 2x - 2 \end{cases}\)
\(12(2x - 2) - 12x = x(2x - 2)\)
\(24x - 24 - 12x = 2x^{2} - 2x\)
\(2x^{2} - 14x + 24 = 0\)
\(x^{2} - 7x + 12 = 0\)
\((x - 3)(x - 4) = 0\)
\(x_{1} = 3;\ y_{1} = 2 \cdot 3 - 2 = 4\)
\(x_{2} = 4;\ y_{2} = 2 \cdot 4 - 2 = 6\)

2) \(\begin{cases} \frac{4}{x} + \frac{3}{y} = 1 \\ x + 5y = 3 \end{cases}\ \Rightarrow \begin{cases} 4y + 3x = xy \\ x = 3 - 5y \end{cases}\)
\(4y + 3x = xy\)
\(4y + 3(3 - 5y) = y(3 - 5y)\)
\(4y + 9 - 15y = 3y - 5y^{2}\)
\(5y^{2} - 14y + 9 = 0\)
\(\dfrac{D}{4} = 49 - 5 \cdot 9 = 49 - 45 = 4\)
\(y_{1} = \dfrac{7 - 2}{5} = 1;\ x_{1} = 3 - 5 = - 2\)
\(y_{2} = \dfrac{7 + 2}{5} = \dfrac{9}{5};\ x_{2} = 3 - 5 \cdot \dfrac{9}{5} = - 6\)

Ответ: 1) \((3;\ 4);\ (4;\ 6)\); 2) \(( - 2;\ 1);\left( - 6;\frac{9}{5} \right)\).