Question
Answer
$$\dfrac{18x^3-8x}{12x^2-8x}=\dfrac{3x+2}{2}$$
Explanation
| $$ \begin{aligned}\frac{18x^3-8x}{12x^2-8x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3x+2}{2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{18x^3-8x}{12x^2-8x} $ to $ \dfrac{3x+2}{2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{6x^2-4x}$. $$ \begin{aligned} \frac{18x^3-8x}{12x^2-8x} & =\frac{ \left( 3x+2 \right) \cdot \color{blue}{ \left( 6x^2-4x \right) }}{ 2 \cdot \color{blue}{ \left( 6x^2-4x \right) }} = \\[1ex] &= \frac{3x+2}{2} \end{aligned} $$ |
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Simplify Rational Expressions