Answer
$$\frac{4y^2-4y}{4y^2-12y+8}=\frac{y}{y-2}$$
Explanation
| $$ \begin{aligned}\frac{4y^2-4y}{4y^2-12y+8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{y}{y-2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4y^2-4y}{4y^2-12y+8} $ to $ \dfrac{y}{y-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{4y-4}$. $$ \begin{aligned} \frac{4y^2-4y}{4y^2-12y+8} & =\frac{ y \cdot \color{blue}{ \left( 4y-4 \right) }}{ \left( y-2 \right) \cdot \color{blue}{ \left( 4y-4 \right) }} = \\[1ex] &= \frac{y}{y-2} \end{aligned} $$ |
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Simplify Rational Expressions