Answer
$$\frac{21-3x}{42x-6x^2}=\frac{1}{2x}$$
Explanation
| $$ \begin{aligned}\frac{21-3x}{42x-6x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{2x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{21-3x}{42x-6x^2} $ to $ \dfrac{1}{2x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{-3x+21}$. $$ \begin{aligned} \frac{21-3x}{42x-6x^2} & =\frac{ 1 \cdot \color{blue}{ \left( -3x+21 \right) }}{ 2x \cdot \color{blue}{ \left( -3x+21 \right) }} = \\[1ex] &= \frac{1}{2x} \end{aligned} $$ |
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Simplify Rational Expressions