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1/2(2y-4)=2/3(9y+3)
We move all terms to the left:
1/2(2y-4)-(2/3(9y+3))=0
Domain of the equation: 2(2y-4)!=0
y∈R
Domain of the equation: 3(9y+3))!=0We calculate fractions
y∈R
(3y9/(2(2y-4)*3(9y+3)))+(-4y2/(2(2y-4)*3(9y+3)))=0
We calculate terms in parentheses: +(3y9/(2(2y-4)*3(9y+3))), so:
3y9/(2(2y-4)*3(9y+3))
We multiply all the terms by the denominator
3y9
We add all the numbers together, and all the variables
3y^9
We do not support eypression: y^9
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