Add $ \dfrac{x^2-9}{x^2} $ and $ 3x $ to get $ \dfrac{ \color{purple}{ 3x^3+x^2-9 } }{ x^2 }$.

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

$$ \begin{aligned} \frac{x^2-9}{x^2} +3x & \xlongequal{\text{Step 1}} \frac{x^2-9}{x^2} + \frac{3x}{\color{red}{1}} = \frac{ x^2-9 }{ x^2 } + \frac{ 3x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^2-9 } }{ x^2 } + \frac{ \color{purple}{ 3x^3 } }{ x^2 }=\frac{ \color{purple}{ 3x^3+x^2-9 } }{ x^2 } \end{aligned} $$

Subtract $18$ from $ \dfrac{3x^3+x^2-9}{x^2} $ to get $ \dfrac{ \color{purple}{ 3x^3-17x^2-9 } }{ x^2 }$.

Step 1: Write $ 18 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{3x^3+x^2-9}{x^2} -18 & \xlongequal{\text{Step 1}} \frac{3x^3+x^2-9}{x^2} - \frac{18}{\color{red}{1}} = \frac{ 3x^3+x^2-9 }{ x^2 } - \frac{ 18 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^3+x^2-9 } }{ x^2 } - \frac{ \color{purple}{ 18x^2 } }{ x^2 }=\frac{ \color{purple}{ 3x^3-17x^2-9 } }{ x^2 } \end{aligned} $$