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

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

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

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

Step 1: Write $ 3 $ 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 $\color{blue}{x}$.

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

Add $ \dfrac{-2x+8}{x} $ and $ \dfrac{-3x+3}{x} $ to get $ \dfrac{-5x+11}{x} $.

To add expressions with the same denominators, we add the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{-2x+8}{x} + \frac{-3x+3}{x} & = \frac{-2x+8}{\color{blue}{x}} + \frac{-3x+3}{\color{blue}{x}} =\frac{ -2x+8 + \left( -3x+3 \right) }{ \color{blue}{ x }} = \\[1ex] &= \frac{-5x+11}{x} \end{aligned} $$