Question
Answer
$$\dfrac{-2j+6}{j}(-2j-2)=\dfrac{4j^2-8j-12}{j}$$
Explanation
| $$ \begin{aligned}\frac{-2j+6}{j}(-2j-2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4j^2-8j-12}{j}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{-2j+6}{j} $ by $ -2j-2 $ to get $ \dfrac{4j^2-8j-12}{j} $. Step 1: Write $ -2j-2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{-2j+6}{j} \cdot -2j-2 & \xlongequal{\text{Step 1}} \frac{-2j+6}{j} \cdot \frac{-2j-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( -2j+6 \right) \cdot \left( -2j-2 \right) }{ j \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4j^2+4j-12j-12 }{ j } = \frac{4j^2-8j-12}{j} \end{aligned} $$ |
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