Add $ \dfrac{3x^2-3}{x^2} $ and $ 2x $ to get $ \dfrac{ \color{purple}{ 2x^3+3x^2-3 } }{ x^2 }$.

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

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

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

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

Step 1: Write $ 3 $ 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 second fraction by $\color{blue}{x^2}$.

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