Question
Answer
$$(6-yi)\cdot(3-4i)-(6-yi)\cdot(3+5i)=9i^2y-54i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6-yi)\cdot(3-4i)-(6-yi)\cdot(3+5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}18-24i-3iy+4i^2y-(18+30i-3iy-5i^2y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}18-24i-3iy+4i^2y-18-30i+3iy+5i^2y \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{18}-24i -\cancel{3iy}+4i^2y -\cancel{18}-30i+ \cancel{3iy}+5i^2y \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}9i^2y-54i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{6-iy}\right) $ by each term in $ \left( 3-4i\right) $. $$ \left( \color{blue}{6-iy}\right) \cdot \left( 3-4i\right) = 18-24i-3iy+4i^2y $$Multiply each term of $ \left( \color{blue}{6-iy}\right) $ by each term in $ \left( 3+5i\right) $. $$ \left( \color{blue}{6-iy}\right) \cdot \left( 3+5i\right) = 18+30i-3iy-5i^2y $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 18+30i-3iy-5i^2y \right) = -18-30i+3iy+5i^2y $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{18}} \, \color{green}{-24i} \, \color{orange}{ -\cancel{3iy}} \,+ \color{red}{4i^2y} \, \color{blue}{ -\cancel{18}} \, \color{green}{-30i} + \, \color{orange}{ \cancel{3iy}} \,+ \color{red}{5i^2y} = \color{red}{9i^2y} \color{green}{-54i} $$ |
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