$\frac{3y + 9}{3y - 1} + \frac{2y - 13}{2y + 5} = 2$ |*(3y − 1)(2y + 5)
$\begin{equation*} \begin{cases} (3y + 9)(2y + 5) + (2y - 13)(3y - 1) = 2(3y - 1)(2y + 5) &\\ y ≠ {-2,5;\frac{1}{3}} & \end{cases} \end{equation*}$
$6y^2 + 33y + 45 + 6y^2 - 41y + 13 = 2(6y^2 + 13y - 5)$
−34y = −67
y = 2
$\begin{equation*} \begin{cases} y = 2 &\\ y ≠ {-2,5;\frac{1}{3}} & \end{cases} \end{equation*}$
Ответ: y = 2

$\frac{5y + 13}{5y + 4} - \frac{4 - 6y}{3y - 1} = 3$ |*(5y + 4)(3y − 1)
$\begin{equation*} \begin{cases} (5y + 13)(3y - 1) + (6y - 4)(5y + 4) = 3(5y + 4)(3y - 1) &\\ y ≠ {-\frac{4}{5};\frac{1}{3}} & \end{cases} \end{equation*}$
$15y^2 + 34y - 13 + 30y^2 + 4y - 16 = 3(15y^2 + 7y - 4)$
$15y^2 + 34y - 13 + 30y^2 + 4y - 16 = 45y^2 + 21y - 12$
$17y = 17$
y = 1
$\begin{equation*} \begin{cases} y = 1 &\\ y ≠ {-\frac{4}{5};\frac{1}{3}} & \end{cases} \end{equation*}$
Ответ: y = 1

$\frac{y + 1}{y - 5} + \frac{10}{y + 5} = \frac{y + 1}{y - 5} * \frac{10}{y + 5}$ |*(y − 5)(y + 5)
$\begin{equation*} \begin{cases} (y + 1)(y + 5) + 10(y - 5) = 10(y + 1) &\\ y ≠ ±5 & \end{cases} \end{equation*}$
$y^2 + 6y + 5 + 10y - 50 = 10y + 10$
$y^2 + 6y - 55 = 0$
(y + 11)(y − 5) = 0
$y_1 = -11$
$y_2 = 5$
$\begin{equation*} \begin{cases} y_1 = -11, y_2 = 5 &\\ y ≠ ±5 & \end{cases} \end{equation*}$
Ответ: y = −11

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