Subtract $ \dfrac{3}{2} $ from $ x $ to get $ \dfrac{ \color{purple}{ 2x-3 } }{ 2 }$.

Step 1: Write $ x $ 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 first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} x- \frac{3}{2} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} - \frac{3}{2} = \frac{ x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} - \frac{ 3 }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x } }{ 2 } - \frac{ \color{purple}{ 3 } }{ 2 }=\frac{ \color{purple}{ 2x-3 } }{ 2 } \end{aligned} $$

Divide $ \dfrac{2x-3}{2} $ by $ x-2 $ to get $ \dfrac{ 2x-3 }{ 2x-4 } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{2x-3}{2} }{x-2} & \xlongequal{\text{Step 1}} \frac{2x-3}{2} \cdot \frac{\color{blue}{1}}{\color{blue}{x-2}} \xlongequal{\text{Step 2}} \frac{ \left( 2x-3 \right) \cdot 1 }{ 2 \cdot \left( x-2 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x-3 }{ 2x-4 } \end{aligned} $$