Divide $ \, 5-i \, $ by $ \, 1+i \, $ to get $\,\, 2-3i $.
( view steps )

Subtract $2-3i$ from $ \dfrac{6}{3-2i} $ to get $ \dfrac{ \color{purple}{ -6i^2+13i } }{ -2i+3 }$.

Step 1: Write $ 2-3i $ 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 $\color{blue}{-2i+3}$.

$$ \begin{aligned} \frac{6}{3-2i} -2-3i & \xlongequal{\text{Step 1}} \frac{6}{3-2i} - \frac{2-3i}{\color{red}{1}} = \frac{ 6 }{ 3-2i } - \frac{ \left( 2-3i \right) \cdot \color{blue}{ \left( -2i+3 \right) }}{ 1 \cdot \color{blue}{ \left( -2i+3 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 6 } }{ 3-2i } - \frac{ \color{purple}{ -4i+6+6i^2-9i } }{ 3-2i }=\frac{ \color{purple}{ 6 - \left( -4i+6+6i^2-9i \right) } }{ -2i+3 } = \\[1ex] &=\frac{ \color{purple}{ -6i^2+13i } }{ -2i+3 } \end{aligned} $$

$$ -6i^2 = -6 \cdot (-1) = 6 $$

Divide $ \, 6+13i \, $ by $ \, 3-2i \, $ to get $\,\, \dfrac{-8+51i}{13} $.
( view steps )