Question
Answer
$$\dfrac{49d^2-9}{49d^2+42d+9}=\dfrac{7d-3}{7d+3}$$
Explanation
| $$ \begin{aligned}\frac{49d^2-9}{49d^2+42d+9}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7d-3}{7d+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{49d^2-9}{49d^2+42d+9} $ to $ \dfrac{7d-3}{7d+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{7d+3}$. $$ \begin{aligned} \frac{49d^2-9}{49d^2+42d+9} & =\frac{ \left( 7d-3 \right) \cdot \color{blue}{ \left( 7d+3 \right) }}{ \left( 7d+3 \right) \cdot \color{blue}{ \left( 7d+3 \right) }} = \\[1ex] &= \frac{7d-3}{7d+3} \end{aligned} $$ |
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Simplify Rational Expressions