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