| $$ \begin{aligned}\frac{x^2+2x-3}{x^2+12x+27}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-1}{x+9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+2x-3}{x^2+12x+27} $ to $ \dfrac{x-1}{x+9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$. $$ \begin{aligned} \frac{x^2+2x-3}{x^2+12x+27} & =\frac{ \left( x-1 \right) \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x+9 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &= \frac{x-1}{x+9} \end{aligned} $$ |