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3-(5y+2(y-1)+4)=5-(2(y-3)-3(y-2)
We move all terms to the left:
3-(5y+2(y-1)+4)-(5-(2(y-3)-3(y-2))=0
We calculate terms in parentheses: -(5y+2(y-1)+4), so:We get rid of parentheses
5y+2(y-1)+4
We multiply parentheses
5y+2y-2+4
We add all the numbers together, and all the variables
7y+2
Back to the equation:
-(7y+2)
-7y-(5-(2(y-3)-3(y-2))+3-2=0
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