Answer
$$\frac{x^2-9}{x^2-7x+12}=\frac{x+3}{x-4}$$
Explanation
| $$ \begin{aligned}\frac{x^2-9}{x^2-7x+12}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+3}{x-4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2-9}{x^2-7x+12} $ to $ \dfrac{x+3}{x-4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-3}$. $$ \begin{aligned} \frac{x^2-9}{x^2-7x+12} & =\frac{ \left( x+3 \right) \cdot \color{blue}{ \left( x-3 \right) }}{ \left( x-4 \right) \cdot \color{blue}{ \left( x-3 \right) }} = \\[1ex] &= \frac{x+3}{x-4} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions