Question
Answer
$$\dfrac{12a^3-3a}{12a^3+6a^2}=\dfrac{2a-1}{2a}$$
Explanation
| $$ \begin{aligned}\frac{12a^3-3a}{12a^3+6a^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2a-1}{2a}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{12a^3-3a}{12a^3+6a^2} $ to $ \dfrac{2a-1}{2a} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{6a^2+3a}$. $$ \begin{aligned} \frac{12a^3-3a}{12a^3+6a^2} & =\frac{ \left( 2a-1 \right) \cdot \color{blue}{ \left( 6a^2+3a \right) }}{ 2a \cdot \color{blue}{ \left( 6a^2+3a \right) }} = \\[1ex] &= \frac{2a-1}{2a} \end{aligned} $$ |
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Simplify Rational Expressions