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