8/9(3i+27)=2/9(i+68)

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Solution for 8/9(3i+27)=2/9(i+68) equation: 



8/9(3i+27)=2/9(i+68)

We move all terms to the left:

8/9(3i+27)-(2/9(i+68))=0



Domain of the equation: 9(3i+27)!=0

i∈R





Domain of the equation: 9(i+68))!=0

i∈R



We calculate fractions


(72ii/(9(3i+27)*9(i+68)))+(-18i3/(9(3i+27)*9(i+68)))=0



We calculate terms in parentheses: +(72ii/(9(3i+27)*9(i+68))), so:

72ii/(9(3i+27)*9(i+68))

We multiply all the terms by the denominator


72ii

Back to the equation:

+(72ii)





We calculate terms in parentheses: +(-18i3/(9(3i+27)*9(i+68))), so:

-18i3/(9(3i+27)*9(i+68))

We multiply all the terms by the denominator


-18i3

We add all the numbers together, and all the variables

-18i^3

We do not support eipression: i^3

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