Question
Answer
$$(4+3yi)\cdot(3+4i)-(4+5yi)\cdot(3+4i)=-8i^2y-6iy$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4+3yi)\cdot(3+4i)-(4+5yi)\cdot(3+4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12+16i+9iy+12i^2y-(12+16i+15iy+20i^2y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}12+16i+9iy+12i^2y-12-16i-15iy-20i^2y \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{12}+ \cancel{16i}+9iy+12i^2y -\cancel{12} -\cancel{16i}-15iy-20i^2y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-8i^2y-6iy\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4+3iy}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{4+3iy}\right) \cdot \left( 3+4i\right) = 12+16i+9iy+12i^2y $$Multiply each term of $ \left( \color{blue}{4+5iy}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{4+5iy}\right) \cdot \left( 3+4i\right) = 12+16i+15iy+20i^2y $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 12+16i+15iy+20i^2y \right) = -12-16i-15iy-20i^2y $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{12}} \,+ \, \color{green}{ \cancel{16i}} \,+ \color{blue}{9iy} + \color{red}{12i^2y} \, \color{blue}{ -\cancel{12}} \, \, \color{green}{ -\cancel{16i}} \, \color{blue}{-15iy} \color{red}{-20i^2y} = \color{red}{-8i^2y} \color{blue}{-6iy} $$ |
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