Multiply $3$ by $ \dfrac{x}{x+2} $ to get $ \dfrac{ 3x }{ x+2 } $.

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.

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

Add $ \dfrac{x}{x-1} $ and $ \dfrac{3x}{x+2} $ to get $ \dfrac{ \color{purple}{ 4x^2-x } }{ x^2+x-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}{ x+2 }$ and the second by $\color{blue}{ x-1 }$.

$$ \begin{aligned} \frac{x}{x-1} + \frac{3x}{x+2} & = \frac{ x \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x-1 \right) \cdot \color{blue}{ \left( x+2 \right) }} + \frac{ 3x \cdot \color{blue}{ \left( x-1 \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ x^2+2x } }{ x^2+2x-x-2 } + \frac{ \color{purple}{ 3x^2-3x } }{ x^2+2x-x-2 }=\frac{ \color{purple}{ 4x^2-x } }{ x^2+x-2 } \end{aligned} $$