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