Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{20+30j}{5-\frac{1}{3}j}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{20+30j}{5-\frac{j}{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{20+30j}{\frac{-j+15}{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{90j+60}{-j+15}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{3} $ by $ j $ to get $ \dfrac{ j }{ 3 } $. Step 1: Write $ j $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{3} \cdot j & \xlongequal{\text{Step 1}} \frac{1}{3} \cdot \frac{j}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot j }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ j }{ 3 } \end{aligned} $$ |
| ② | Subtract $ \dfrac{j}{3} $ from $ 5 $ to get $ \dfrac{ \color{purple}{ -j+15 } }{ 3 }$. 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. |
| ③ | Divide $20+30j$ by $ \dfrac{-j+15}{3} $ to get $ \dfrac{90j+60}{-j+15} $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Write $ 20+30j $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{20+30j}{ \frac{\color{blue}{-j+15}}{\color{blue}{3}} } & \xlongequal{\text{Step 1}} 20+30j \cdot \frac{\color{blue}{3}}{\color{blue}{-j+15}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{20+30j}{\color{red}{1}} \cdot \frac{3}{-j+15} \xlongequal{\text{Step 3}} \frac{ \left( 20+30j \right) \cdot 3 }{ 1 \cdot \left( -j+15 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 60+90j }{ -j+15 } = \frac{90j+60}{-j+15} \end{aligned} $$ |