1/9(18j-27)=2/3(15j-12)

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Solution for 1/9(18j-27)=2/3(15j-12) equation: 



1/9(18j-27)=2/3(15j-12)

We move all terms to the left:

1/9(18j-27)-(2/3(15j-12))=0



Domain of the equation: 9(18j-27)!=0

j∈R





Domain of the equation: 3(15j-12))!=0

j∈R



We calculate fractions


(3j1/(9(18j-27)*3(15j-12)))+(-18j1/(9(18j-27)*3(15j-12)))=0



We calculate terms in parentheses: +(3j1/(9(18j-27)*3(15j-12))), so:

3j1/(9(18j-27)*3(15j-12))

We multiply all the terms by the denominator


3j1

We add all the numbers together, and all the variables

3j

Back to the equation:

+(3j)





We calculate terms in parentheses: +(-18j1/(9(18j-27)*3(15j-12))), so:

-18j1/(9(18j-27)*3(15j-12))

We multiply all the terms by the denominator


-18j1

We add all the numbers together, and all the variables

-18j

Back to the equation:

+(-18j)



We get rid of parentheses

3j-18j=0

We add all the numbers together, and all the variables

-15j=0

j=0/-15

j=0

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