Question
Answer
$$\dfrac{7b-35}{b^2+3b-40}=\dfrac{7}{b+8}$$
Explanation
| $$ \begin{aligned}\frac{7b-35}{b^2+3b-40}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{b+8}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{7b-35}{b^2+3b-40} $ to $ \dfrac{7}{b+8} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b-5}$. $$ \begin{aligned} \frac{7b-35}{b^2+3b-40} & =\frac{ 7 \cdot \color{blue}{ \left( b-5 \right) }}{ \left( b+8 \right) \cdot \color{blue}{ \left( b-5 \right) }} = \\[1ex] &= \frac{7}{b+8} \end{aligned} $$ |
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Simplify Rational Expressions