Question
Answer
$$\dfrac{2n-8}{n^2-16}=\dfrac{2}{n+4}$$
Explanation
| $$ \begin{aligned}\frac{2n-8}{n^2-16}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{n+4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2n-8}{n^2-16} $ to $ \dfrac{2}{n+4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{n-4}$. $$ \begin{aligned} \frac{2n-8}{n^2-16} & =\frac{ 2 \cdot \color{blue}{ \left( n-4 \right) }}{ \left( n+4 \right) \cdot \color{blue}{ \left( n-4 \right) }} = \\[1ex] &= \frac{2}{n+4} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions