Question
Answer
$$\dfrac{6x^2-19x+3}{4x^2-36}=\dfrac{6x-1}{4x+12}$$
Explanation
| $$ \begin{aligned}\frac{6x^2-19x+3}{4x^2-36}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6x-1}{4x+12}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{6x^2-19x+3}{4x^2-36} $ to $ \dfrac{6x-1}{4x+12} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-3}$. $$ \begin{aligned} \frac{6x^2-19x+3}{4x^2-36} & =\frac{ \left( 6x-1 \right) \cdot \color{blue}{ \left( x-3 \right) }}{ \left( 4x+12 \right) \cdot \color{blue}{ \left( x-3 \right) }} = \\[1ex] &= \frac{6x-1}{4x+12} \end{aligned} $$ |
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Simplify Rational Expressions