Subtract $ \dfrac{5}{y^2} $ from $ y^2-4y $ to get $ \dfrac{ \color{purple}{ y^4-4y^3-5 } }{ y^2 }$.

Step 1: Write $ y^2-4y $ 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 $\color{blue}{ y^2 }$.

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

Add $ \dfrac{y^4-4y^3-5}{y^2} $ and $ 5y $ to get $ \dfrac{ \color{purple}{ y^4+y^3-5 } }{ y^2 }$.

Step 1: Write $ 5y $ 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 $\color{blue}{y^2}$.

$$ \begin{aligned} \frac{y^4-4y^3-5}{y^2} +5y & \xlongequal{\text{Step 1}} \frac{y^4-4y^3-5}{y^2} + \frac{5y}{\color{red}{1}} = \frac{ y^4-4y^3-5 }{ y^2 } + \frac{ 5y \cdot \color{blue}{ y^2 }}{ 1 \cdot \color{blue}{ y^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ y^4-4y^3-5 } }{ y^2 } + \frac{ \color{purple}{ 5y^3 } }{ y^2 }=\frac{ \color{purple}{ y^4+y^3-5 } }{ y^2 } \end{aligned} $$

Add $ \dfrac{y^4+y^3-5}{y^2} $ and $ 4 $ to get $ \dfrac{ \color{purple}{ y^4+y^3+4y^2-5 } }{ y^2 }$.

Step 1: Write $ 4 $ 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 $\color{blue}{y^2}$.

$$ \begin{aligned} \frac{y^4+y^3-5}{y^2} +4 & \xlongequal{\text{Step 1}} \frac{y^4+y^3-5}{y^2} + \frac{4}{\color{red}{1}} = \frac{ y^4+y^3-5 }{ y^2 } + \frac{ 4 \cdot \color{blue}{ y^2 }}{ 1 \cdot \color{blue}{ y^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ y^4+y^3-5 } }{ y^2 } + \frac{ \color{purple}{ 4y^2 } }{ y^2 }=\frac{ \color{purple}{ y^4+y^3+4y^2-5 } }{ y^2 } \end{aligned} $$