Решите уравнение (3y+2)/(4y^2+y)+(y-3)/(16y^2-1)=3/(4y-1).

\[\frac{3y + 2}{4y^{2} + y} + \frac{y - 3}{16y^{2} - 1} = \frac{3}{4y - 1}\]

\[\frac{3y + 2^{\backslash 4y - 1}}{y(4y + 1)} + \frac{y - 3^{\backslash y}}{(4y - 1)(4y + 1)} - \frac{3^{\backslash 4y^{2} + y}}{4y - 1} = 0\]

\[ОДЗ:\ \ y \neq 0;\ \ y = \pm \frac{1}{4}.\]

\[12y^{2} + 8y - 3y - 2 + y^{2} - 3y - 12y^{2} - 3y = 0\]

\[y^{2} - y - 2 = 0\]

\[y_{1} + y_{2} = 1;\ \ \ \ y_{1} \cdot y_{2} = - 2\]

\[y_{1} = 2;\ \ \ y_{2} = - 1.\]

\[Ответ:y = - 1;\ \ y = 2.\]