Multiply each term of $ \left( \color{blue}{x+2}\right) $ by each term in $ \left( x+3\right) $.

$$ \left( \color{blue}{x+2}\right) \cdot \left( x+3\right) = x^2+3x+2x+6 $$

Multiply each term of $ \left( \color{blue}{x+2}\right) $ by each term in $ \left( x+4\right) $.

$$ \left( \color{blue}{x+2}\right) \cdot \left( x+4\right) = x^2+4x+2x+8 $$

Combine like terms:

$$ x^2+ \color{blue}{3x} + \color{blue}{2x} +6 = x^2+ \color{blue}{5x} +6 $$

Combine like terms:

$$ x^2+ \color{blue}{4x} + \color{blue}{2x} +8 = x^2+ \color{blue}{6x} +8 $$

Subtract $ \dfrac{x-6}{x^2+6x+8} $ from $ \dfrac{2x}{x^2+5x+6} $ to get $ \dfrac{ \color{purple}{ x^2+11x+18 } }{ x^3+9x^2+26x+24 }$.

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+4 }$ and the second by $\color{blue}{ x+3 }$.

$$ \begin{aligned} \frac{2x}{x^2+5x+6} - \frac{x-6}{x^2+6x+8} & = \frac{ 2x \cdot \color{blue}{ \left( x+4 \right) }}{ \left( x^2+5x+6 \right) \cdot \color{blue}{ \left( x+4 \right) }} - \frac{ \left( x-6 \right) \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x^2+6x+8 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 2x^2+8x } }{ x^3+4x^2+5x^2+20x+6x+24 } - \frac{ \color{purple}{ x^2+3x-6x-18 } }{ x^3+4x^2+5x^2+20x+6x+24 } = \\[1ex] &=\frac{ \color{purple}{ 2x^2+8x - \left( x^2+3x-6x-18 \right) } }{ x^3+9x^2+26x+24 }=\frac{ \color{purple}{ x^2+11x+18 } }{ x^3+9x^2+26x+24 } \end{aligned} $$

Simplify $ \dfrac{x^2+11x+18}{x^3+9x^2+26x+24} $ to $ \dfrac{x+9}{x^2+7x+12} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$.

$$ \begin{aligned} \frac{x^2+11x+18}{x^3+9x^2+26x+24} & =\frac{ \left( x+9 \right) \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x^2+7x+12 \right) \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] &= \frac{x+9}{x^2+7x+12} \end{aligned} $$