Question
Answer
$$\dfrac{5x^2-15x}{x^3-9x}=\dfrac{5}{x+3}$$
Explanation
| $$ \begin{aligned}\frac{5x^2-15x}{x^3-9x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{5x^2-15x}{x^3-9x} $ to $ \dfrac{5}{x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2-3x}$. $$ \begin{aligned} \frac{5x^2-15x}{x^3-9x} & =\frac{ 5 \cdot \color{blue}{ \left( x^2-3x \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( x^2-3x \right) }} = \\[1ex] &= \frac{5}{x+3} \end{aligned} $$ |
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Simplify Rational Expressions