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

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \color{blue}{5} $ by $ \left( x+2\right) $ $$ \color{blue}{5} \cdot \left( x+2\right) = 5x+10 $$

Subtract $ \dfrac{1}{5x+10} $ from $ \dfrac{2x}{-x+3} $ to get $ \dfrac{ \color{purple}{ 10x^2+21x-3 } }{ -5x^2+5x+30 }$.

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}{ 5x+10 }$ and the second by $\color{blue}{ -x+3 }$.

$$ \begin{aligned} \frac{2x}{-x+3} - \frac{1}{5x+10} & = \frac{ 2x \cdot \color{blue}{ \left( 5x+10 \right) }}{ \left( -x+3 \right) \cdot \color{blue}{ \left( 5x+10 \right) }} - \frac{ 1 \cdot \color{blue}{ \left( -x+3 \right) }}{ \left( 5x+10 \right) \cdot \color{blue}{ \left( -x+3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 10x^2+20x } }{ -5x^2-10x+15x+30 } - \frac{ \color{purple}{ -x+3 } }{ -5x^2-10x+15x+30 } = \\[1ex] &=\frac{ \color{purple}{ 10x^2+20x - \left( -x+3 \right) } }{ -5x^2+5x+30 }=\frac{ \color{purple}{ 10x^2+21x-3 } }{ -5x^2+5x+30 } \end{aligned} $$