Answer
$$\frac{n-6}{n^2-13n+42}=\frac{1}{n-7}$$
Explanation
| $$ \begin{aligned}\frac{n-6}{n^2-13n+42}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{n-7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{n-6}{n^2-13n+42} $ to $ \dfrac{1}{n-7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{n-6}$. $$ \begin{aligned} \frac{n-6}{n^2-13n+42} & =\frac{ 1 \cdot \color{blue}{ \left( n-6 \right) }}{ \left( n-7 \right) \cdot \color{blue}{ \left( n-6 \right) }} = \\[1ex] &= \frac{1}{n-7} \end{aligned} $$ |
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Simplify Rational Expressions