Question
Answer
$$\dfrac{b^2-3b-10}{10b^2+20b}=\dfrac{b-5}{10b}$$
Explanation
| $$ \begin{aligned}\frac{b^2-3b-10}{10b^2+20b}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{b-5}{10b}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{b^2-3b-10}{10b^2+20b} $ to $ \dfrac{b-5}{10b} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b+2}$. $$ \begin{aligned} \frac{b^2-3b-10}{10b^2+20b} & =\frac{ \left( b-5 \right) \cdot \color{blue}{ \left( b+2 \right) }}{ 10b \cdot \color{blue}{ \left( b+2 \right) }} = \\[1ex] &= \frac{b-5}{10b} \end{aligned} $$ |
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Simplify Rational Expressions