Subtract $ \dfrac{1}{2} $ from $ \dfrac{4}{x} $ to get $ \dfrac{ \color{purple}{ -x+8 } }{ 2x }$.

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 }$ and the second by $\color{blue}{ x }$.

$$ \begin{aligned} \frac{4}{x} - \frac{1}{2} & = \frac{ 4 \cdot \color{blue}{ 2 }}{ x \cdot \color{blue}{ 2 }} - \frac{ 1 \cdot \color{blue}{ x }}{ 2 \cdot \color{blue}{ x }} = \\[1ex] &=\frac{ \color{purple}{ 8 } }{ 2x } - \frac{ \color{purple}{ x } }{ 2x }=\frac{ \color{purple}{ -x+8 } }{ 2x } \end{aligned} $$

Divide $ \dfrac{-x+8}{2x} $ by $ x-8 $ to get $ \dfrac{ -1 }{ 2x } $.

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

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{-x+8}{2x} }{x-8} & \xlongequal{\text{Step 1}} \frac{-x+8}{2x} \cdot \frac{\color{blue}{1}}{\color{blue}{x-8}} \xlongequal{\text{Step 2}} \frac{ \left( -1 \right) \cdot \color{blue}{ \left( x-8 \right) } }{ 2x } \cdot \frac{ 1 }{ 1 \cdot \color{blue}{ \left( x-8 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -1 }{ 2x } \cdot \frac{ 1 }{ 1 } \xlongequal{\text{Step 4}} \frac{ \left( -1 \right) \cdot 1 }{ 2x \cdot 1 } \xlongequal{\text{Step 5}} \frac{ -1 }{ 2x } \end{aligned} $$