Question
Answer
$$\dfrac{x^2+7x}{4x^2+28x}=\dfrac{1}{4}$$
Explanation
| $$ \begin{aligned}\frac{x^2+7x}{4x^2+28x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+7x}{4x^2+28x} $ to $ \dfrac{1}{4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2+7x}$. $$ \begin{aligned} \frac{x^2+7x}{4x^2+28x} & =\frac{ 1 \cdot \color{blue}{ \left( x^2+7x \right) }}{ 4 \cdot \color{blue}{ \left( x^2+7x \right) }} = \\[1ex] &= \frac{1}{4} \end{aligned} $$ |
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Simplify Rational Expressions