Add $b$ and $ \dfrac{1}{b-4} $ to get $ \dfrac{ \color{purple}{ b^2-4b+1 } }{ b-4 }$.

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

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ b-4 }$.

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

Subtract $3$ from $ \dfrac{b^2-4b+1}{b-4} $ to get $ \dfrac{ \color{purple}{ b^2-7b+13 } }{ b-4 }$.

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 $\color{blue}{b-4}$.

$$ \begin{aligned} \frac{b^2-4b+1}{b-4} -3 & \xlongequal{\text{Step 1}} \frac{b^2-4b+1}{b-4} - \frac{3}{\color{red}{1}} = \frac{ b^2-4b+1 }{ b-4 } - \frac{ 3 \cdot \color{blue}{ \left( b-4 \right) }}{ 1 \cdot \color{blue}{ \left( b-4 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ b^2-4b+1 } }{ b-4 } - \frac{ \color{purple}{ 3b-12 } }{ b-4 }=\frac{ \color{purple}{ b^2-4b+1 - \left( 3b-12 \right) } }{ b-4 } = \\[1ex] &=\frac{ \color{purple}{ b^2-7b+13 } }{ b-4 } \end{aligned} $$