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Question

Answer

$$\dfrac{3}{6}-2x=\dfrac{-4x+1}{2}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{3}{6}-2x& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{ 3 : \color{orangered}{ 3 } }{ 6 : \color{orangered}{ 3 }} - 2x \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{1}{2}-2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-4x+1}{2}\end{aligned} $$
①
Divide both the top and bottom numbers by $ \color{orangered}{ 3 } $.
②

Subtract $2x$ from $ \dfrac{1}{2} $ to get $ \dfrac{ \color{purple}{ -4x+1 } }{ 2 }$.

Step 1: Write $ 2x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{2}$.

$$ \begin{aligned} \frac{1}{2} -2x & \xlongequal{\text{Step 1}} \frac{1}{2} - \frac{2x}{\color{red}{1}} = \frac{ 1 }{ 2 } - \frac{ 2x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 1 } }{ 2 } - \frac{ \color{purple}{ 4x } }{ 2 }=\frac{ \color{purple}{ -4x+1 } }{ 2 } \end{aligned} $$
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