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

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

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ x+3 }$.

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

Divide $ \dfrac{x^2+2x}{x+3} $ by $ x+2 $ to get $ \dfrac{ x }{ x+3 } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{x^2+2x}{x+3} }{x+2} & \xlongequal{\text{Step 1}} \frac{x^2+2x}{x+3} \cdot \frac{\color{blue}{1}}{\color{blue}{x+2}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x \cdot \color{blue}{ \left( x+2 \right) } }{ x+3 } \cdot \frac{ 1 }{ 1 \cdot \color{blue}{ \left( x+2 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x }{ x+3 } \cdot \frac{ 1 }{ 1 } \xlongequal{\text{Step 4}} \frac{ x \cdot 1 }{ \left( x+3 \right) \cdot 1 } \xlongequal{\text{Step 5}} \frac{ x }{ x+3 } \end{aligned} $$