9(1i+2)=3(1i-2)

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Solution for 9(1i+2)=3(1i-2) equation: 



9(1i+2)=3(1i-2)

We move all terms to the left:

9(1i+2)-(3(1i-2))=0

We add all the numbers together, and all the variables

9(i+2)-(3(i-2))=0

We multiply parentheses

9i-(3(i-2))+18=0



We calculate terms in parentheses: -(3(i-2)), so:

3(i-2)

We multiply parentheses

3i-6

Back to the equation:

-(3i-6)



We get rid of parentheses

9i-3i+6+18=0

We add all the numbers together, and all the variables

6i+24=0

We move all terms containing i to the left, all other terms to the right

6i=-24

i=-24/6

i=-4

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