1/3(i+4)-6=2/3(5-i)

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Solution for 1/3(i+4)-6=2/3(5-i) equation: 



1/3(i+4)-6=2/3(5-i)

We move all terms to the left:

1/3(i+4)-6-(2/3(5-i))=0



Domain of the equation: 3(i+4)!=0

i∈R





Domain of the equation: 3(5-i))!=0

i∈R



We add all the numbers together, and all the variables

1/3(i+4)-(2/3(-1i+5))-6=0

We calculate fractions


(3i(-)/(3(i+4)*3(-1i+5)))+(-6ii/(3(i+4)*3(-1i+5)))-6=0



We calculate terms in parentheses: +(3i(-)/(3(i+4)*3(-1i+5))), so:

3i(-)/(3(i+4)*3(-1i+5))

We add all the numbers together, and all the variables

3i0/(3(i+4)*3(-1i+5))

We multiply all the terms by the denominator


3i0

We add all the numbers together, and all the variables

3i

Back to the equation:

+(3i)





We calculate terms in parentheses: +(-6ii/(3(i+4)*3(-1i+5))), so:

-6ii/(3(i+4)*3(-1i+5))

We multiply all the terms by the denominator


-6ii

Back to the equation:

+(-6ii)



We get rid of parentheses

3i-6ii-6=0

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

3i-6ii=6

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