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