$$5-5\cdot\dfrac{i}{1-3i}\cdot(1-i)=\dfrac{-20i+60}{10}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}5-5 \cdot \frac{i}{1-3i}\cdot(1-i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}5-5 \cdot \frac{-3+i}{10}\cdot(1-i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}5-\frac{5i-15}{10}\cdot(1-i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}5-\frac{-5i^2+20i-15}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}5-\frac{5+20i-15}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}5-\frac{20i-10}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-20i+60}{10}\end{aligned} $$ | |
| ① | Divide $ \, i \, $ by $ \, 1-3i \, $ to get $\,\, \dfrac{-3+i}{10} $. ( view steps ) |
| ② | Multiply $5$ by $ \dfrac{-3+i}{10} $ to get $ \dfrac{5i-15}{10} $. Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 5 \cdot \frac{-3+i}{10} & \xlongequal{\text{Step 1}} \frac{5}{\color{red}{1}} \cdot \frac{-3+i}{10} \xlongequal{\text{Step 2}} \frac{ 5 \cdot \left( -3+i \right) }{ 1 \cdot 10 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -15+5i }{ 10 } = \frac{5i-15}{10} \end{aligned} $$ |
| ③ | Multiply $ \dfrac{5i-15}{10} $ by $ 1-i $ to get $ \dfrac{-5i^2+20i-15}{10} $. Step 1: Write $ 1-i $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{5i-15}{10} \cdot 1-i & \xlongequal{\text{Step 1}} \frac{5i-15}{10} \cdot \frac{1-i}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 5i-15 \right) \cdot \left( 1-i \right) }{ 10 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 5i-5i^2-15+15i }{ 10 } = \frac{-5i^2+20i-15}{10} \end{aligned} $$ |
| ④ | $$ -5i^2 = -5 \cdot (-1) = 5 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{5} +20i \color{blue}{-15} = 20i \color{blue}{-10} $$ |
| ⑥ | Subtract $ \dfrac{20i-10}{10} $ from $ 5 $ to get $ \dfrac{ \color{purple}{ -20i+60 } }{ 10 }$. Step 1: Write $ 5 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |