$$ \begin{aligned}\frac{y^2-16}{y^2-2y-8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{y+4}{y+2}\end{aligned} $$ | |
① | Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y-4}$. $$ \begin{aligned} \frac{y^2-16}{y^2-2y-8} & =\frac{ \left( y+4 \right) \cdot \color{blue}{ \left( y-4 \right) }}{ \left( y+2 \right) \cdot \color{blue}{ \left( y-4 \right) }} = \\[1ex] &= \frac{y+4}{y+2} \end{aligned} $$ |