Answer
$$\frac{9b^2-63b}{7b^3-49b^2}=\frac{9}{7b}$$
Explanation
| $$ \begin{aligned}\frac{9b^2-63b}{7b^3-49b^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{7b}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{9b^2-63b}{7b^3-49b^2} $ to $ \dfrac{9}{7b} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b^2-7b}$. $$ \begin{aligned} \frac{9b^2-63b}{7b^3-49b^2} & =\frac{ 9 \cdot \color{blue}{ \left( b^2-7b \right) }}{ 7b \cdot \color{blue}{ \left( b^2-7b \right) }} = \\[1ex] &= \frac{9}{7b} \end{aligned} $$ |
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Simplify Rational Expressions