Subtract $ \dfrac{r+8}{r^2-5r+4} $ from $ \dfrac{r}{r-4} $ to get $ \dfrac{ \color{purple}{ r^2-2r-8 } }{ r^2-5r+4 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ r-1 }$.

$$ \begin{aligned} \frac{r}{r-4} - \frac{r+8}{r^2-5r+4} & = \frac{ r \cdot \color{blue}{ \left( r-1 \right) }}{ \left( r-4 \right) \cdot \color{blue}{ \left( r-1 \right) }} - \frac{ r+8 }{ r^2-5r+4 } = \\[1ex] &=\frac{ \color{purple}{ r^2-r } }{ r^2-r-4r+4 } - \frac{ \color{purple}{ r+8 } }{ r^2-r-4r+4 }=\frac{ \color{purple}{ r^2-r - \left( r+8 \right) } }{ r^2-5r+4 } = \\[1ex] &=\frac{ \color{purple}{ r^2-2r-8 } }{ r^2-5r+4 } \end{aligned} $$

Simplify $ \dfrac{r^2-2r-8}{r^2-5r+4} $ to $ \dfrac{r+2}{r-1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{r-4}$.

$$ \begin{aligned} \frac{r^2-2r-8}{r^2-5r+4} & =\frac{ \left( r+2 \right) \cdot \color{blue}{ \left( r-4 \right) }}{ \left( r-1 \right) \cdot \color{blue}{ \left( r-4 \right) }} = \\[1ex] &= \frac{r+2}{r-1} \end{aligned} $$