Question
Answer
$$\dfrac{x}{x-5}\cdot2\dfrac{x}{x+4}=\dfrac{2x^2}{x^2-x-20}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x}{x-5}\cdot2\frac{x}{x+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x}{x-5}\frac{2x}{x+4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2x^2}{x^2-x-20}\end{aligned} $$ | |
| ① | Multiply $2$ by $ \dfrac{x}{x+4} $ to get $ \dfrac{ 2x }{ x+4 } $. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2 \cdot \frac{x}{x+4} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{x}{x+4} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 1 \cdot \left( x+4 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x }{ x+4 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{x}{x-5} $ by $ \dfrac{2x}{x+4} $ to get $ \dfrac{2x^2}{x^2-x-20} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{x}{x-5} \cdot \frac{2x}{x+4} & \xlongequal{\text{Step 1}} \frac{ x \cdot 2x }{ \left( x-5 \right) \cdot \left( x+4 \right) } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2x^2 }{ x^2+4x-5x-20 } = \frac{2x^2}{x^2-x-20} \end{aligned} $$ |
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