Question
Answer
$$\dfrac{n^2+7n}{4n^2+28n}=\dfrac{1}{4}$$
Explanation
| $$ \begin{aligned}\frac{n^2+7n}{4n^2+28n}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{n^2+7n}{4n^2+28n} $ to $ \dfrac{1}{4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{n^2+7n}$. $$ \begin{aligned} \frac{n^2+7n}{4n^2+28n} & =\frac{ 1 \cdot \color{blue}{ \left( n^2+7n \right) }}{ 4 \cdot \color{blue}{ \left( n^2+7n \right) }} = \\[1ex] &= \frac{1}{4} \end{aligned} $$ |
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Simplify Rational Expressions