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

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 first fraction by $\color{blue}{ x }$.

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

Subtract $4$ from $ \dfrac{2x^2+3}{x} $ to get $ \dfrac{ \color{purple}{ 2x^2-4x+3 } }{ x }$.

Step 1: Write $ 4 $ 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 $\color{blue}{x}$.

$$ \begin{aligned} \frac{2x^2+3}{x} -4 & \xlongequal{\text{Step 1}} \frac{2x^2+3}{x} - \frac{4}{\color{red}{1}} = \frac{ 2x^2+3 }{ x } - \frac{ 4 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2+3 } }{ x } - \frac{ \color{purple}{ 4x } }{ x }=\frac{ \color{purple}{ 2x^2-4x+3 } }{ x } \end{aligned} $$

Subtract $x$ from $ \dfrac{2x^2-4x+3}{x} $ to get $ \dfrac{ \color{purple}{ x^2-4x+3 } }{ x }$.

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 second fraction by $\color{blue}{x}$.

$$ \begin{aligned} \frac{2x^2-4x+3}{x} -x & \xlongequal{\text{Step 1}} \frac{2x^2-4x+3}{x} - \frac{x}{\color{red}{1}} = \frac{ 2x^2-4x+3 }{ x } - \frac{ x \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2-4x+3 } }{ x } - \frac{ \color{purple}{ x^2 } }{ x }=\frac{ \color{purple}{ x^2-4x+3 } }{ x } \end{aligned} $$

Subtract $ \dfrac{5}{x} $ from $ \dfrac{x^2-4x+3}{x} $ to get $ \dfrac{x^2-4x-2}{x} $.

To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{x^2-4x+3}{x} - \frac{5}{x} & = \frac{x^2-4x+3}{\color{blue}{x}} - \frac{5}{\color{blue}{x}} =\frac{ x^2-4x+3 - 5 }{ \color{blue}{ x }} = \\[1ex] &= \frac{x^2-4x-2}{x} \end{aligned} $$

Add $ \dfrac{x^2-4x-2}{x} $ and $ 2 $ to get $ \dfrac{ \color{purple}{ x^2-2x-2 } }{ x }$.

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

$$ \begin{aligned} \frac{x^2-4x-2}{x} +2 & \xlongequal{\text{Step 1}} \frac{x^2-4x-2}{x} + \frac{2}{\color{red}{1}} = \frac{ x^2-4x-2 }{ x } + \frac{ 2 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^2-4x-2 } }{ x } + \frac{ \color{purple}{ 2x } }{ x }=\frac{ \color{purple}{ x^2-2x-2 } }{ x } \end{aligned} $$