Найдите координаты точек пересечения графиков функций y=x/(x-3) и y=(3x-4)/(2x).

\[y = \frac{x}{x - 3};\ \ y = \frac{3x - 4}{2x}\]

\[\frac{x^{\backslash 2x}}{x - 3} = \frac{3x - 4^{\backslash x - 3}}{2x};\ \ \ \ \ \ \ x \neq 3;\ \ \ x \neq 0\]

\[2x^{2} = 3x^{2} - 4x - 9x + 12\]

\[3x^{2} - 2x^{2} - 13x + 12 = 0\]

\[x^{2} - 13x + 12 = 0\]

\[x_{1} + x_{2} = 13;\ \ \ \ x_{1} \cdot x_{2} = 12\]

\[x_{1} = 12;\ \ \ x_{2} = 1.\]

\[y_{1} = \frac{12}{12 - 3} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{3};\]

\[y_{2} = \frac{1}{1 - 3} = - \frac{1}{2} = - 0,5.\]

\[Ответ:графики\ пересекаются\ в\ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ точках\ \left( 12;1\frac{1}{3} \right);(1;\ - 0,5).\]