Решите уравнение: (5y-2)/(2y+1)=(3y+2)/(y+3).

\[\frac{5y - 2}{2y + 1} = \frac{3y + 2}{y + 3}\]

\[\frac{5y - 2}{2y + 1} - \frac{3y + 2}{y + 3} = 0\]

\[ОДЗ:\ \ \ y \neq - \frac{1}{2}\]

\[\ \ \ \ \ \ \ \ \ \ \ \ y \neq - 3\]

\[\frac{5y² + 13y - 6 - 6y^{2} - 7y - 2}{(2y + 1)(y + 3)} = 0\]

\[\frac{- y^{2} + 6y - 8}{(2y + )(y + 3)} = 0\]

\[- y^{2} + 6y - 8 = 0\ \ \ \ \ | \cdot ( - 1)\ \]

\[y^{2} - 6y + 8 = 0\]

\[y_{1} + y_{2} = 6\]

\[y_{1} \cdot y_{2} = 8 \Longrightarrow y_{1} = 4;\ \ y_{2} = 2\]

\[Ответ:\ \ y = 4\ \ \ \ и\ \ \ y = 2.\]