2y-3+2(3y+2)=-6(y+1)

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Solution for 2y-3+2(3y+2)=-6(y+1) equation: 



2y-3+2(3y+2)=-6(y+1)

We move all terms to the left:

2y-3+2(3y+2)-(-6(y+1))=0

We multiply parentheses

2y+6y-(-6(y+1))+4-3=0



We calculate terms in parentheses: -(-6(y+1)), so:

-6(y+1)

We multiply parentheses

-6y-6

Back to the equation:

-(-6y-6)



We add all the numbers together, and all the variables

8y-(-6y-6)+1=0

We get rid of parentheses

8y+6y+6+1=0

We add all the numbers together, and all the variables

14y+7=0

We move all terms containing y to the left, all other terms to the right

14y=-7

y=-7/14

y=-1/2

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