Multiply $5$ by $ \dfrac{a}{a-2} $ to get $ \dfrac{ 5a }{ a-2 } $.

Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 5 \cdot \frac{a}{a-2} & \xlongequal{\text{Step 1}} \frac{5}{\color{red}{1}} \cdot \frac{a}{a-2} \xlongequal{\text{Step 2}} \frac{ 5 \cdot a }{ 1 \cdot \left( a-2 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5a }{ a-2 } \end{aligned} $$

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

Step 1: Write $ 4 $ 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 $\color{blue}{a-2}$.

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

Add $ \dfrac{a+8}{a-2} $ and $ \dfrac{2}{a} $ to get $ \dfrac{ \color{purple}{ a^2+10a-4 } }{ a^2-2a }$.

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}{ a }$ and the second by $\color{blue}{ a-2 }$.

$$ \begin{aligned} \frac{a+8}{a-2} + \frac{2}{a} & = \frac{ \left( a+8 \right) \cdot \color{blue}{ a }}{ \left( a-2 \right) \cdot \color{blue}{ a }} + \frac{ 2 \cdot \color{blue}{ \left( a-2 \right) }}{ a \cdot \color{blue}{ \left( a-2 \right) }} = \\[1ex] &=\frac{ \color{purple}{ a^2+8a } }{ a^2-2a } + \frac{ \color{purple}{ 2a-4 } }{ a^2-2a }=\frac{ \color{purple}{ a^2+10a-4 } }{ a^2-2a } \end{aligned} $$