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