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