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

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

Add $ \dfrac{3t}{-x+2} $ and $ \dfrac{5}{x-2} $ to get $ \dfrac{ \color{purple}{ 3t-5 } }{ -x+2 }$.

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

$$ \begin{aligned} \frac{3t}{-x+2} + \frac{5}{x-2} & = \frac{ 3t }{ -x+2 } + \frac{ 5 \cdot \color{blue}{ \left( -1 \right) }}{ \left( x-2 \right) \cdot \color{blue}{ \left( -1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3t } }{ -x+2 } + \frac{ \color{purple}{ -5 } }{ -x+2 }=\frac{ \color{purple}{ 3t-5 } }{ -x+2 } \end{aligned} $$