Answer
$$(1-yi)\cdot(1-5i)-(6-yi)\cdot(1-5i)=25i-5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-yi)\cdot(1-5i)-(6-yi)\cdot(1-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1-5i-iy+5i^2y-(6-30i-iy+5i^2y) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}1-5i-iy+5i^2y-6+30i+iy-5i^2y \xlongequal{ } \\[1 em] & \xlongequal{ }1-5i -\cancel{iy}+ \cancel{5i^2y}-6+30i+ \cancel{iy} -\cancel{5i^2y} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}25i-5\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-iy}\right) $ by each term in $ \left( 1-5i\right) $. $$ \left( \color{blue}{1-iy}\right) \cdot \left( 1-5i\right) = 1-5i-iy+5i^2y $$Multiply each term of $ \left( \color{blue}{6-iy}\right) $ by each term in $ \left( 1-5i\right) $. $$ \left( \color{blue}{6-iy}\right) \cdot \left( 1-5i\right) = 6-30i-iy+5i^2y $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 6-30i-iy+5i^2y \right) = -6+30i+iy-5i^2y $$ |
| ③ | Combine like terms: $$ \color{blue}{1} \color{red}{-5i} \, \color{green}{ -\cancel{iy}} \,+ \, \color{blue}{ \cancel{5i^2y}} \, \color{blue}{-6} + \color{red}{30i} + \, \color{green}{ \cancel{iy}} \, \, \color{blue}{ -\cancel{5i^2y}} \, = \color{red}{25i} \color{blue}{-5} $$ |
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