Subtract $2$ from $ \dfrac{8x-1}{x+2} $ to get $ \dfrac{ \color{purple}{ 6x-5 } }{ x+2 }$.

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 $\color{blue}{x+2}$.

$$ \begin{aligned} \frac{8x-1}{x+2} -2 & \xlongequal{\text{Step 1}} \frac{8x-1}{x+2} - \frac{2}{\color{red}{1}} = \frac{ 8x-1 }{ x+2 } - \frac{ 2 \cdot \color{blue}{ \left( x+2 \right) }}{ 1 \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 8x-1 } }{ x+2 } - \frac{ \color{purple}{ 2x+4 } }{ x+2 }=\frac{ \color{purple}{ 8x-1 - \left( 2x+4 \right) } }{ x+2 } = \\[1ex] &=\frac{ \color{purple}{ 6x-5 } }{ x+2 } \end{aligned} $$