$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} * \frac{y}{y - 2}$|*(y + 1)(y − 2)
$\begin{equation*} \begin{cases} 6(y - 2) + y(y + 1) = 6y &\\ y ≠ {-1;2} & \end{cases} \end{equation*}$
$6y - 12 + y^2 + y = 6y$
$y^2 + y - 12 = 0$
(y + 4)(y − 3) = 0
$\begin{equation*} \begin{cases} y_1 = -4, y_2 = 3 &\\ y ≠ {-1;2} & \end{cases} \end{equation*}$
Ответ:
$y_1 = -4$;
$y_2 = 3$.

$\frac{2}{y - 3} + \frac{6}{y + 3} = \frac{2}{y - 3} : \frac{6}{y + 3}$
$\frac{2}{y - 3} : \frac{6}{y + 3} = \frac{2}{y - 3} * \frac{y + 3}{3(y - 3)}$
$\frac{2}{y - 3} + \frac{6}{y + 3} = \frac{2}{y - 3} * \frac{y + 3}{3(y - 3)}$|*3(y − 3)(y + 3)
$\begin{equation*} \begin{cases} 2 * 3(y + 3) + 6 * 3(y - 3) = (y + 3)^2 &\\ y ≠ ±3 & \end{cases} \end{equation*}$
$6y + 18 + 18y - 54 = y^2 + 6y + 9$
$y^2 - 18y + 45 = 0$
(y − 3)(y − 15) = 0
$\begin{equation*} \begin{cases} y_1 = 3, y_2 = 15 &\\ y ≠ ±3 & \end{cases} \end{equation*}$
Ответ: y = 15

$\frac{y + 12}{y - 4} - \frac{y}{y + 4} = \frac{y + 12}{y - 4} * \frac{y}{y + 4}$|*(y − 4)(y + 4)
$\begin{equation*} \begin{cases} (y + 12)(y + 4) - y(y - 4) = y(y + 12) &\\ y ≠ ±4 & \end{cases} \end{equation*}$
$y^2 + 16y + 48 - y^2 + 4y = y^2 + 12y$
$y^2 - 8y - 48 = 0$
(y + 4)(y + 12) = 0
$\begin{equation*} \begin{cases} y_1 = -4, y_2 = 12 &\\ y ≠ ±4 & \end{cases} \end{equation*}$
Ответ: y = 12