Question
Answer
$$\dfrac{4}{10}x\cdot\dfrac{6}{2}=\dfrac{6x}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{10}x\cdot\frac{6}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{ 4 : \color{orangered}{ 2 } }{ 10 : \color{orangered}{ 2 }} \cdot x \cdot \frac{ 6 : \color{orangered}{ 2 } }{ 2 : \color{orangered}{ 2 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2}{5}x\cdot\frac{3}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2x}{5}\cdot3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{6x}{5}\end{aligned} $$ | |
| ① | Divide both the top and bottom numbers by $ \color{orangered}{ 2 } $. |
| ② | Divide both the top and bottom numbers by $ \color{orangered}{ 2 } $. |
| ③ | Multiply $ \dfrac{2}{5} $ by $ x $ to get $ \dfrac{ 2x }{ 5 } $. Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2}{5} \cdot x & \xlongequal{\text{Step 1}} \frac{2}{5} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x }{ 5 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 2x }{ 5 } \end{aligned} $$ |
| ④ | Remove 1 from denominator. |
| ⑤ | Multiply $ \dfrac{2x}{5} $ by $ 3 $ to get $ \dfrac{ 6x }{ 5 } $. Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2x}{5} \cdot 3 & \xlongequal{\text{Step 1}} \frac{2x}{5} \cdot \frac{3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2x \cdot 3 }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 6x }{ 5 } \end{aligned} $$ |
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