we are given
[tex] \frac{y^2}{y-3}*\frac{y^2-y-6}{y^2+1y} [/tex]
we know that denominators can not be zero
so, restrictions would be values of y where denominator is not zero
[tex] y-3\neq 0 [/tex]
[tex] y\neq 3 [/tex]
[tex] y^2+1y \neq 0 [/tex]
[tex] y(y+1) \neq 0 [/tex]
[tex] y \neq 0 [/tex]
[tex] y \neq -1 [/tex]
so, restrictions are
[tex] y \neq 0 [/tex]
[tex] y \neq -1 [/tex]
[tex] y\neq 3 [/tex]
[tex] \frac{y^2}{y-3}*\frac{y^2-y-6}{y^2+1y} [/tex]
now, we can factor numerators and denominator
and then we can simplify it
[tex] \frac{y^2}{y-3}*\frac{y^2-y-6}{y^2+1y}=\frac{y^2}{y-3}*\frac{(y-3)(y+2)}{y(y+1)} [/tex]
now, we can cancel it
[tex] \frac{y^2}{y-3}*\frac{y^2-y-6}{y^2+1y}=\frac{y(y+2)}{y+1} [/tex]
[tex] \frac{y^2}{y-3}*\frac{y^2-y-6}{y^2+1y}=\frac{y^2+2y}{y+1} [/tex]
so,
option-B..........Answer
