СРОЧНО ПОМОГИТЕ РЕШИТЬ ПРИМЕР ПО АЛГЕБРЕ, 8 КЛАСС, ПОЖАЛУЙСТА ЗАРАНЕЕ БЛАГОДАРНА

$1)\frac{y+1}{3y-y^2}+\frac{y+5}{y^2-9}=\frac{y+1}{y(3-y)}+\frac{y+5}{(y-3)(y+3)}=\\\\=\frac{y+1}{y(3-y)}-\frac{y+5}{(3-y)(y+3)}=\frac{(y+1)*(y+3)-y*(y+5)}{y(3-y)(y+3)}=\\\\=\frac{y^2+y+3y+3-y^2-5y}{y(3-y)(y+3)}=\frac{3-y}{y(3-y)(y+3)}=\frac{1}{y(y+3)}\\\\$

\frac{1}{y(y+3)}\\\\\frac{1}{1,2(1,2+3)}=\frac{1}{1,2*4,2}=\frac{1}{5,04}=\frac{100}{504}=\frac{25}{126}" alt="2) y=1,2=>\frac{1}{y(y+3)}\\\\\frac{1}{1,2(1,2+3)}=\frac{1}{1,2*4,2}=\frac{1}{5,04}=\frac{100}{504}=\frac{25}{126}" align="absmiddle" class="latex-formula">