Answer
$$\frac{9n+72}{3n^2+25n+8}=\frac{9}{3n+1}$$
Explanation
| $$ \begin{aligned}\frac{9n+72}{3n^2+25n+8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{3n+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{9n+72}{3n^2+25n+8} $ to $ \dfrac{9}{3n+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{n+8}$. $$ \begin{aligned} \frac{9n+72}{3n^2+25n+8} & =\frac{ 9 \cdot \color{blue}{ \left( n+8 \right) }}{ \left( 3n+1 \right) \cdot \color{blue}{ \left( n+8 \right) }} = \\[1ex] &= \frac{9}{3n+1} \end{aligned} $$ |
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Simplify Rational Expressions