Question
Answer
$$\dfrac{b^2+9b+18}{b^2+b-30}=\dfrac{b+3}{b-5}$$
Explanation
| $$ \begin{aligned}\frac{b^2+9b+18}{b^2+b-30}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{b+3}{b-5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{b^2+9b+18}{b^2+b-30} $ to $ \dfrac{b+3}{b-5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b+6}$. $$ \begin{aligned} \frac{b^2+9b+18}{b^2+b-30} & =\frac{ \left( b+3 \right) \cdot \color{blue}{ \left( b+6 \right) }}{ \left( b-5 \right) \cdot \color{blue}{ \left( b+6 \right) }} = \\[1ex] &= \frac{b+3}{b-5} \end{aligned} $$ |
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Simplify Rational Expressions