Factor out 2 on both numerator and denominator
[tex]\begin{gathered} \frac{2-2y}{2y-2} \\ =\frac{2(1-y)}{2(y-1)} \\ \\ \text{cancel out }2\text{ on both numerator and denominator} \\ =\frac{\cancel{2}(1-y)}{\cancel{2}(y-1)} \\ =\frac{(1-y)}{(y-1)} \\ \\ \text{factor out }-1\text{ on numerator},\text{ and rearrange to cancel out common binomial} \\ =\frac{(1-y)}{(y-1)} \\ =\frac{-1(-1+y)}{(y-1)} \\ =\frac{-1(y-1)}{(y-1)} \\ =\frac{-1\cancel{(y-1)}}{\cancel{(y-1)}} \\ =-1 \\ \\ \text{Therefore,} \\ \frac{2-2y}{2y-2}=-1 \end{gathered}[/tex]Part 2:
Since the given expression is in fraction, we cannot let the denominator equal to zero. Find values of y that makes the denominator by zero
[tex]\begin{gathered} \text{Denominator: }2y-2 \\ \\ \text{Equate to zero} \\ 2y-2=0 \\ 2y-2+2=0+2 \\ 2y\cancel{-2+2}=2 \\ \frac{2y}{2}=\frac{2}{2} \\ y=1 \\ \\ \text{If }y=1,\text{ the denominator }2y-2\text{ becomes zero therefore}, \\ y\neq1 \end{gathered}[/tex]