Question
Answer
$$\dfrac{3x^2+25x-18}{3x^2-23x+14}=\dfrac{x+9}{x-7}$$
Explanation
| $$ \begin{aligned}\frac{3x^2+25x-18}{3x^2-23x+14}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+9}{x-7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{3x^2+25x-18}{3x^2-23x+14} $ to $ \dfrac{x+9}{x-7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{3x-2}$. $$ \begin{aligned} \frac{3x^2+25x-18}{3x^2-23x+14} & =\frac{ \left( x+9 \right) \cdot \color{blue}{ \left( 3x-2 \right) }}{ \left( x-7 \right) \cdot \color{blue}{ \left( 3x-2 \right) }} = \\[1ex] &= \frac{x+9}{x-7} \end{aligned} $$ |
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Simplify Rational Expressions