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