Question
Answer
$$\dfrac{x^2+5x+6}{x^2-4x}\cdot4\dfrac{x}{x+2}=\dfrac{4x+12}{x-4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+5x+6}{x^2-4x}\cdot4\frac{x}{x+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+5x+6}{x^2-4x}\frac{4x}{x+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4x^2+12x}{x^2-4x} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{4x+12}{x-4}\end{aligned} $$ | |
| ① | Multiply $4$ by $ \dfrac{x}{x+2} $ to get $ \dfrac{ 4x }{ x+2 } $. Step 1: Write $ 4 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 4 \cdot \frac{x}{x+2} & \xlongequal{\text{Step 1}} \frac{4}{\color{red}{1}} \cdot \frac{x}{x+2} \xlongequal{\text{Step 2}} \frac{ 4 \cdot x }{ 1 \cdot \left( x+2 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4x }{ x+2 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{x^2+5x+6}{x^2-4x} $ by $ \dfrac{4x}{x+2} $ to get $ \dfrac{ 4x^2+12x }{ x^2-4x } $. Step 1: Factor numerators and denominators. Step 2: Cancel common factors. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2+5x+6}{x^2-4x} \cdot \frac{4x}{x+2} & \xlongequal{\text{Step 1}} \frac{ \left( x+3 \right) \cdot \color{blue}{ \left( x+2 \right) } }{ x^2-4x } \cdot \frac{ 4x }{ 1 \cdot \color{blue}{ \left( x+2 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x+3 }{ x^2-4x } \cdot \frac{ 4x }{ 1 } \xlongequal{\text{Step 3}} \frac{ \left( x+3 \right) \cdot 4x }{ \left( x^2-4x \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 4x^2+12x }{ x^2-4x } \end{aligned} $$ |
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