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