Question
Answer
$$\dfrac{g-6}{g^2-8g+12}=\dfrac{1}{g-2}$$
Explanation
| $$ \begin{aligned}\frac{g-6}{g^2-8g+12}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{g-2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{g-6}{g^2-8g+12} $ to $ \dfrac{1}{g-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{g-6}$. $$ \begin{aligned} \frac{g-6}{g^2-8g+12} & =\frac{ 1 \cdot \color{blue}{ \left( g-6 \right) }}{ \left( g-2 \right) \cdot \color{blue}{ \left( g-6 \right) }} = \\[1ex] &= \frac{1}{g-2} \end{aligned} $$ |
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Simplify Rational Expressions