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2((2y-1)+y)=5(y+3)
We move all terms to the left:
2((2y-1)+y)-(5(y+3))=0
We calculate terms in parentheses: -(5(y+3)), so:We get rid of parentheses
5(y+3)
We multiply parentheses
5y+15
Back to the equation:
-(5y+15)
2((2y-1)+y)-5y-15=0
We add all the numbers together, and all the variables
-5y+2((2y-1)+y)-15=0
We move all terms containing y to the left, all other terms to the right
-5y+2((2y-1)+y)=15
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