Question
Answer
$$\dfrac{5x-45}{x^2-9x}=\dfrac{5}{x}$$
Explanation
| $$ \begin{aligned}\frac{5x-45}{x^2-9x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{5x-45}{x^2-9x} $ to $ \dfrac{5}{x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-9}$. $$ \begin{aligned} \frac{5x-45}{x^2-9x} & =\frac{ 5 \cdot \color{blue}{ \left( x-9 \right) }}{ x \cdot \color{blue}{ \left( x-9 \right) }} = \\[1ex] &= \frac{5}{x} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions