Use the distributive property.

Add $ \dfrac{1}{x+8} $ and $ \dfrac{x}{x-1} $ to get $ \dfrac{ \color{purple}{ x^2+9x-1 } }{ x^2+7x-8 }$.

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}{ x-1 }$ and the second by $\color{blue}{ x+8 }$.

$$ \begin{aligned} \frac{1}{x+8} + \frac{x}{x-1} & = \frac{ 1 \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+8 \right) \cdot \color{blue}{ \left( x-1 \right) }} + \frac{ x \cdot \color{blue}{ \left( x+8 \right) }}{ \left( x-1 \right) \cdot \color{blue}{ \left( x+8 \right) }} = \\[1ex] &=\frac{ \color{purple}{ x-1 } }{ x^2-x+8x-8 } + \frac{ \color{purple}{ x^2+8x } }{ x^2-x+8x-8 }=\frac{ \color{purple}{ x^2+9x-1 } }{ x^2+7x-8 } \end{aligned} $$

Multiply $ \dfrac{x^2+9x-1}{x^2+7x-8} $ by $ 3 $ to get $ \dfrac{ 3x^2+27x-3 }{ x^2+7x-8 } $.

Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2+9x-1}{x^2+7x-8} \cdot 3 & \xlongequal{\text{Step 1}} \frac{x^2+9x-1}{x^2+7x-8} \cdot \frac{3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2+9x-1 \right) \cdot 3 }{ \left( x^2+7x-8 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x^2+27x-3 }{ x^2+7x-8 } \end{aligned} $$