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1/3(y+2)=1/5(3y+2)
We move all terms to the left:
1/3(y+2)-(1/5(3y+2))=0
Domain of the equation: 3(y+2)!=0
y∈R
Domain of the equation: 5(3y+2))!=0We calculate fractions
y∈R
(5y3/(3(y+2)*5(3y+2)))+(-3yy/(3(y+2)*5(3y+2)))=0
We calculate terms in parentheses: +(5y3/(3(y+2)*5(3y+2))), so:
5y3/(3(y+2)*5(3y+2))
We multiply all the terms by the denominator
5y3
We add all the numbers together, and all the variables
5y^3
We do not support eypression: y^3
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