Multiply $5$ by $ \dfrac{x}{x-3} $ to get $ \dfrac{ 5x }{ x-3 } $.

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.

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

Subtract $ \dfrac{1}{x} $ from $ \dfrac{5x}{x-3} $ to get $ \dfrac{ \color{purple}{ 5x^2-x+3 } }{ x^2-3x }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x }$ and the second by $\color{blue}{ x-3 }$.

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