Question
Answer
$$\dfrac{4n^2-12n+9}{2n^2+15n-27}=\dfrac{2n-3}{n+9}$$
Explanation
| $$ \begin{aligned}\frac{4n^2-12n+9}{2n^2+15n-27}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2n-3}{n+9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4n^2-12n+9}{2n^2+15n-27} $ to $ \dfrac{2n-3}{n+9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2n-3}$. $$ \begin{aligned} \frac{4n^2-12n+9}{2n^2+15n-27} & =\frac{ \left( 2n-3 \right) \cdot \color{blue}{ \left( 2n-3 \right) }}{ \left( n+9 \right) \cdot \color{blue}{ \left( 2n-3 \right) }} = \\[1ex] &= \frac{2n-3}{n+9} \end{aligned} $$ |
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Simplify Rational Expressions