Question
Answer
$$\dfrac{2d^2-5d}{4d^2-20d+25}=\dfrac{d}{2d-5}$$
Explanation
| $$ \begin{aligned}\frac{2d^2-5d}{4d^2-20d+25}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{d}{2d-5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2d^2-5d}{4d^2-20d+25} $ to $ \dfrac{d}{2d-5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2d-5}$. $$ \begin{aligned} \frac{2d^2-5d}{4d^2-20d+25} & =\frac{ d \cdot \color{blue}{ \left( 2d-5 \right) }}{ \left( 2d-5 \right) \cdot \color{blue}{ \left( 2d-5 \right) }} = \\[1ex] &= \frac{d}{2d-5} \end{aligned} $$ |
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Simplify Rational Expressions