$$ \begin{aligned}\frac{x+7}{x^2+5x-14}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x-2}\end{aligned} $$ | |
① | Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+7}$. $$ \begin{aligned} \frac{x+7}{x^2+5x-14} & =\frac{ 1 \cdot \color{blue}{ \left( x+7 \right) }}{ \left( x-2 \right) \cdot \color{blue}{ \left( x+7 \right) }} = \\[1ex] &= \frac{1}{x-2} \end{aligned} $$ |