Решите систему уравнений: y-3/x=-4; x+6/y=1.

\[x\left( \frac{3}{x} - 4 \right) + 6 = \frac{3}{x} - 4\]

\[3 - 4x + 6 = \frac{3}{x} - 4\]

\[4x + \frac{3}{x} - 4 - 9 = 0\]

\[4x + \frac{3}{x} - 13 = 0\ \ \ \ \ | \cdot x\]

\[4x² + 3 - 13x = 0\]

\[4x^{2} - 13x + 3 = 0\]

\[D = ( - 13)^{2} - 4 \cdot 4 \cdot 3 =\]

\[= 169 - 48 = 121\]

\[x_{1} = \frac{13 + \sqrt{121}}{2 \cdot 4} = \frac{13 - 11}{8} =\]

\[= \frac{24}{8} = 3\]

\[x_{2} = \frac{13 - \sqrt{121}}{2 \cdot 4} = \frac{13 - 11}{8} =\]

\[= \frac{2}{8} = \frac{1}{4}\]

\[x_{1} = 3 \Longrightarrow \text{\ \ \ \ \ \ \ \ \ \ }y_{1} = \frac{3}{3} - 4 =\]

\[= 1 - 4 = - 3.\]

\[x_{2} = \frac{1}{4}\ \Longrightarrow \text{\ \ \ \ \ \ \ \ }y_{2} = \frac{3}{\frac{1}{4}} - 4 =\]

\[= 12 - 4 = 8.\]

\[Ответ:(3;\ - 3),\ \left( \frac{1}{4};8 \right).\]