Subtract $ \dfrac{y}{5} $ from $ \dfrac{x}{3} $ to get $ \dfrac{ \color{purple}{ 5x-3y } }{ 15 }$.

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}{ 5 }$ and the second by $\color{blue}{ 3 }$.

$$ \begin{aligned} \frac{x}{3} - \frac{y}{5} & = \frac{ x \cdot \color{blue}{ 5 }}{ 3 \cdot \color{blue}{ 5 }} - \frac{ y \cdot \color{blue}{ 3 }}{ 5 \cdot \color{blue}{ 3 }} = \\[1ex] &=\frac{ \color{purple}{ 5x } }{ 15 } - \frac{ \color{purple}{ 3y } }{ 15 }=\frac{ \color{purple}{ 5x-3y } }{ 15 } \end{aligned} $$

Subtract $3x$ from $ \dfrac{5x-3y}{15} $ to get $ \dfrac{ \color{purple}{ -40x-3y } }{ 15 }$.

Step 1: Write $ 3x $ 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 $\color{blue}{15}$.

$$ \begin{aligned} \frac{5x-3y}{15} -3x & \xlongequal{\text{Step 1}} \frac{5x-3y}{15} - \frac{3x}{\color{red}{1}} = \frac{ 5x-3y }{ 15 } - \frac{ 3x \cdot \color{blue}{ 15 }}{ 1 \cdot \color{blue}{ 15 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 5x-3y } }{ 15 } - \frac{ \color{purple}{ 45x } }{ 15 }=\frac{ \color{purple}{ -40x-3y } }{ 15 } \end{aligned} $$

Add $ \dfrac{-40x-3y}{15} $ and $ 3y $ to get $ \dfrac{ \color{purple}{ -40x+42y } }{ 15 }$.

Step 1: Write $ 3y $ 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 $\color{blue}{15}$.

$$ \begin{aligned} \frac{-40x-3y}{15} +3y & \xlongequal{\text{Step 1}} \frac{-40x-3y}{15} + \frac{3y}{\color{red}{1}} = \frac{ -40x-3y }{ 15 } + \frac{ 3y \cdot \color{blue}{ 15 }}{ 1 \cdot \color{blue}{ 15 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -40x-3y } }{ 15 } + \frac{ \color{purple}{ 45y } }{ 15 }=\frac{ \color{purple}{ -40x+42y } }{ 15 } \end{aligned} $$