Add $2$ and $ \dfrac{1}{x} $ to get $ \dfrac{ \color{purple}{ 2x+1 } }{ 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 first fraction by $\color{blue}{ x }$.

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

Divide $ \dfrac{2x+1}{x} $ by $ \dfrac{x}{y} $ to get $ \dfrac{ 2xy+y }{ x^2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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