Question
Answer
$$\dfrac{3v^2-14v+15}{v^2-12v+27}=\dfrac{3v-5}{v-9}$$
Explanation
| $$ \begin{aligned}\frac{3v^2-14v+15}{v^2-12v+27}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3v-5}{v-9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{3v^2-14v+15}{v^2-12v+27} $ to $ \dfrac{3v-5}{v-9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{v-3}$. $$ \begin{aligned} \frac{3v^2-14v+15}{v^2-12v+27} & =\frac{ \left( 3v-5 \right) \cdot \color{blue}{ \left( v-3 \right) }}{ \left( v-9 \right) \cdot \color{blue}{ \left( v-3 \right) }} = \\[1ex] &= \frac{3v-5}{v-9} \end{aligned} $$ |
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Simplify Rational Expressions