Multiply $ \dfrac{2}{5} $ by $ x+2 $ to get $ \dfrac{ 2x+4 }{ 5 } $.

Step 1: Write $ x+2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Subtract $1$ from $ \dfrac{2x+4}{5} $ to get $ \dfrac{ \color{purple}{ 2x-1 } }{ 5 }$.

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

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