Question
Answer
$$\dfrac{4y^2-100}{y^2+4y-5}=\dfrac{4y-20}{y-1}$$
Explanation
| $$ \begin{aligned}\frac{4y^2-100}{y^2+4y-5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4y-20}{y-1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4y^2-100}{y^2+4y-5} $ to $ \dfrac{4y-20}{y-1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y+5}$. $$ \begin{aligned} \frac{4y^2-100}{y^2+4y-5} & =\frac{ \left( 4y-20 \right) \cdot \color{blue}{ \left( y+5 \right) }}{ \left( y-1 \right) \cdot \color{blue}{ \left( y+5 \right) }} = \\[1ex] &= \frac{4y-20}{y-1} \end{aligned} $$ |
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