$$ \begin{aligned}\frac{6x+18}{x^2-9}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{x-3}\end{aligned} $$ | |
① | Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$. $$ \begin{aligned} \frac{6x+18}{x^2-9} & =\frac{ 6 \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x-3 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &= \frac{6}{x-3} \end{aligned} $$ |