Hello,
Multiplying by 1 means multiply by $\displaystyle \frac aa$
While you're here :
$\displaystyle \frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}$, do this :
$\displaystyle =\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}$
$\displaystyle =\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}$
$\displaystyle =\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}$
$\displaystyle =\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}$
but we used the multiplication by 1 lol!