Алгебра. Задание, решить уравнение.

Решите задачу:

$\frac{4y+24}{5y^2-45} + \frac{y+3}{5y^2-15y} = \frac{y-3}{y^2+3y} \\\\ \frac{4y+24}{5(y-3)(y+3)} + \frac{y+3}{5y(y-3)} = \frac{y-3}{y(y+3)} \\\\ \frac{(4y+24)y+(y+3)^2-5(y-3)^2}{5y(y-3)(y+3)} =0\\\\ \frac{4y^2+24y+y^2+6y+9-5y^2+30y-45}{5y(y-3)(y+3)} =0\\\\ \frac{60y-36}{5y(y-3)(y+3)} =0\; ,\; \; ODZ:\; y\ne 0,\; y\ne -3,\; y\ne 3\\\\60y=36\\\\10y=6\\\\y=\frac{6}{10}\\\\y=\frac{3}{5}$