Найдите корни уравнения: (3y-3)/(3y-2)+(6+2y)/(3y+2)=2.

\[\frac{3y - 3}{3y - 2} + \frac{6 + 2y}{3y + 2} = 2\]

\[ОДЗ:\ \ 3y - 2 \neq 0;y \neq \frac{2}{3}\]

\[\ \ \ \ \ \ 3y + 2 \neq 0;\ \ y \neq - \frac{2}{3}\]

\[15y^{2} + 11y - 18 = 18y^{2} - 8\]

\[3y^{2} - 11y + 10 = 0\]

\[D = b^{2} - 4ac =\]

\[= 121 - 4 \cdot 3 \cdot 10 =\]

\[= 121 - 120 = 1\]

\[y_{1} = \frac{11 + 1}{6} = \frac{12}{6} = 2\]

\[y_{2} = \frac{11 - 1}{6} = \frac{10}{6} = \frac{5}{3} = 1\frac{2}{3}\]

\[Ответ:y = 2\ \ и\ \ \ y = 1\frac{2}{3}.\]