3/2(j+12)=1/2(2j-6)

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Solution for 3/2(j+12)=1/2(2j-6) equation: 



3/2(j+12)=1/2(2j-6)

We move all terms to the left:

3/2(j+12)-(1/2(2j-6))=0



Domain of the equation: 2(j+12)!=0

j∈R





Domain of the equation: 2(2j-6))!=0

j∈R



We calculate fractions


(6j2/(2(j+12)*2(2j-6)))+(-2jj/(2(j+12)*2(2j-6)))=0



We calculate terms in parentheses: +(6j2/(2(j+12)*2(2j-6))), so:

6j2/(2(j+12)*2(2j-6))

We multiply all the terms by the denominator


6j2

We add all the numbers together, and all the variables

6j^2

Back to the equation:

+(6j^2)





We calculate terms in parentheses: +(-2jj/(2(j+12)*2(2j-6))), so:

-2jj/(2(j+12)*2(2j-6))

We multiply all the terms by the denominator


-2jj

Back to the equation:

+(-2jj)



We get rid of parentheses

6j^2-2jj=0

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