Multiply $3$ by $ \dfrac{x}{x^2-49} $ to get $ \dfrac{ 3x }{ x^2-49 } $.

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.

$$ \begin{aligned} 3 \cdot \frac{x}{x^2-49} & \xlongequal{\text{Step 1}} \frac{3}{\color{red}{1}} \cdot \frac{x}{x^2-49} \xlongequal{\text{Step 2}} \frac{ 3 \cdot x }{ 1 \cdot \left( x^2-49 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3x }{ x^2-49 } \end{aligned} $$

Subtract $ \dfrac{5}{x+7} $ from $ \dfrac{3x}{x^2-49} $ to get $ \dfrac{ \color{purple}{ -2x+35 } }{ x^2-49 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{x-7}$.

$$ \begin{aligned} \frac{3x}{x^2-49} - \frac{5}{x+7} & = \frac{ 3x }{ x^2-49 } - \frac{ 5 \cdot \color{blue}{ \left( x-7 \right) }}{ \left( x+7 \right) \cdot \color{blue}{ \left( x-7 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3x } }{ x^2-49 } - \frac{ \color{purple}{ 5x-35 } }{ x^2-49 }=\frac{ \color{purple}{ 3x - \left( 5x-35 \right) } }{ x^2-49 } = \\[1ex] &=\frac{ \color{purple}{ -2x+35 } }{ x^2-49 } \end{aligned} $$