Question
Answer
$$\dfrac{u^2+4u-21}{27-3u^2}=\dfrac{u+7}{-3u-9}$$
Explanation
| $$ \begin{aligned}\frac{u^2+4u-21}{27-3u^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{u+7}{-3u-9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{u^2+4u-21}{27-3u^2} $ to $ \dfrac{u+7}{-3u-9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{u-3}$. $$ \begin{aligned} \frac{u^2+4u-21}{27-3u^2} & =\frac{ \left( u+7 \right) \cdot \color{blue}{ \left( u-3 \right) }}{ \left( -3u-9 \right) \cdot \color{blue}{ \left( u-3 \right) }} = \\[1ex] &= \frac{u+7}{-3u-9} \end{aligned} $$ |
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Simplify Rational Expressions