Add $ \dfrac{b-a}{4i} $ and $ \dfrac{b}{2i} $ to get $ \dfrac{ \color{purple}{ -2ai+6bi } }{ 8i^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}{ 2i }$ and the second by $\color{blue}{ 4i }$.

$$ \begin{aligned} \frac{b-a}{4i} + \frac{b}{2i} & = \frac{ \left( b-a \right) \cdot \color{blue}{ 2i }}{ 4i \cdot \color{blue}{ 2i }} + \frac{ b \cdot \color{blue}{ 4i }}{ 2i \cdot \color{blue}{ 4i }} = \\[1ex] &=\frac{ \color{purple}{ 2bi-2ai } }{ 8i^2 } + \frac{ \color{purple}{ 4bi } }{ 8i^2 }=\frac{ \color{purple}{ -2ai+6bi } }{ 8i^2 } \end{aligned} $$

Multiply $ \dfrac{2a}{2} $ by $ i $ to get $ \dfrac{ 2ai }{ 2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2a}{2} \cdot i & \xlongequal{\text{Step 1}} \frac{2a}{2} \cdot \frac{i}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2a \cdot i }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2ai }{ 2 } \end{aligned} $$

$$ 8i^2 = 8 \cdot (-1) = -8 $$

Multiply $ \dfrac{2a}{2} $ by $ i $ to get $ \dfrac{ 2ai }{ 2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2a}{2} \cdot i & \xlongequal{\text{Step 1}} \frac{2a}{2} \cdot \frac{i}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2a \cdot i }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2ai }{ 2 } \end{aligned} $$

Add $ \dfrac{-2ai+6bi}{-8} $ and $ \dfrac{2ai}{2} $ to get $ \dfrac{ \color{purple}{ 0 } }{ 0 }$.

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}{ 0 }$ and the second by $\color{blue}{ 0 }$.

$$ \begin{aligned} \frac{-2ai+6bi}{-8} + \frac{2ai}{2} & = \frac{ \left( -2ai+6bi \right) \cdot \color{blue}{ 0 }}{ \left( -8 \right) \cdot \color{blue}{ 0 }} + \frac{ 2ai \cdot \color{blue}{ 0 }}{ 2 \cdot \color{blue}{ 0 }} = \\[1ex] &=\frac{ \color{purple}{ 0ai0bi } }{ 0 } + \frac{ \color{purple}{ 0ai } }{ 0 }=\frac{ \color{purple}{ 0 } }{ 0 } \end{aligned} $$