Answer
$$-9\cdot\frac{x}{x-x^2}=-\frac{9x}{-x^2+x}$$
Explanation
| $$ \begin{aligned}-9 \cdot \frac{x}{x-x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{9x}{-x^2+x}\end{aligned} $$ | |
| ① | Multiply $9$ by $ \dfrac{x}{x-x^2} $ to get $ \dfrac{9x}{-x^2+x} $. Step 1: Write $ 9 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 9 \cdot \frac{x}{x-x^2} & \xlongequal{\text{Step 1}} \frac{9}{\color{red}{1}} \cdot \frac{x}{x-x^2} \xlongequal{\text{Step 2}} \frac{ 9 \cdot x }{ 1 \cdot \left( x-x^2 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x }{ x-x^2 } = \frac{9x}{-x^2+x} \end{aligned} $$ |
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Simplify Rational Expressions