Question
Answer
$$\dfrac{4y-8y^2}{10y-5}=-\dfrac{4y}{5}$$
Explanation
| $$ \begin{aligned}\frac{4y-8y^2}{10y-5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{4y}{5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4y-8y^2}{10y-5} $ to $ \dfrac{-4y}{5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2y-1}$. $$ \begin{aligned} \frac{4y-8y^2}{10y-5} & =\frac{ \left( -4y \right) \cdot \color{blue}{ \left( 2y-1 \right) }}{ 5 \cdot \color{blue}{ \left( 2y-1 \right) }} = \\[1ex] &= \frac{-4y}{5} \end{aligned} $$ |
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Simplify Rational Expressions