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Question

Answer

$$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=\dfrac{-24i+10}{13}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{1-2i}{2+i}+\frac{4-i}{3+2i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-i+\frac{10-11i}{13} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-24i+10}{13}\end{aligned} $$
①
Divide $ \, 1-2i \, $ by $ \, 2+i \, $ to get $\,\, -i $.
( view steps )Divide $ \, 4-i \, $ by $ \, 3+2i \, $ to get $\,\, \dfrac{10-11i}{13} $.
( view steps )
②

Add $-i$ and $ \dfrac{10-11i}{13} $ to get $ \dfrac{ \color{purple}{ -24i+10 } }{ 13 }$.

Step 1: Write $ -i $ 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 first fraction by $\color{blue}{ 13 }$.

$$ \begin{aligned} -i+ \frac{10-11i}{13} & \xlongequal{\text{Step 1}} \frac{-i}{\color{red}{1}} + \frac{10-11i}{13} = \frac{ \left( -i \right) \cdot \color{blue}{ 13 }}{ 1 \cdot \color{blue}{ 13 }} + \frac{ 10-11i }{ 13 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -13i } }{ 13 } + \frac{ \color{purple}{ 10-11i } }{ 13 }=\frac{ \color{purple}{ -24i+10 } }{ 13 } \end{aligned} $$
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