Answer
$$35\cdot\frac{n^2}{14n-63}=\frac{35n^2}{14n-63}$$
Explanation
| $$ \begin{aligned}35 \cdot \frac{n^2}{14n-63}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{35n^2}{14n-63}\end{aligned} $$ | |
| ① | Multiply $35$ by $ \dfrac{n^2}{14n-63} $ to get $ \dfrac{ 35n^2 }{ 14n-63 } $. Step 1: Write $ 35 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 35 \cdot \frac{n^2}{14n-63} & \xlongequal{\text{Step 1}} \frac{35}{\color{red}{1}} \cdot \frac{n^2}{14n-63} \xlongequal{\text{Step 2}} \frac{ 35 \cdot n^2 }{ 1 \cdot \left( 14n-63 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 35n^2 }{ 14n-63 } \end{aligned} $$ |
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