Question
Answer
$$\dfrac{-3xy^4}{x^3y}\dfrac{7x^2z^3}{6y^3z}=-\dfrac{21x^3y^4z^3}{6x^3y^4z}$$
Explanation
| $$ \begin{aligned}\frac{-3xy^4}{x^3y}\frac{7x^2z^3}{6y^3z}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{21x^3y^4z^3}{6x^3y^4z}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{-3xy^4}{x^3y} $ by $ \dfrac{7x^2z^3}{6y^3z} $ to get $ \dfrac{ -21x^3y^4z^3 }{ 6x^3y^4z } $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{-3xy^4}{x^3y} \cdot \frac{7x^2z^3}{6y^3z} & \xlongequal{\text{Step 1}} \frac{ \left( -3xy^4 \right) \cdot 7x^2z^3 }{ x^3y \cdot 6y^3z } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ -21x^3y^4z^3 }{ 6x^3y^4z } \end{aligned} $$ |
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