4(3y-3)+12=3(y-2)-(-8y+2)

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Solution for 4(3y-3)+12=3(y-2)-(-8y+2) equation: 



4(3y-3)+12=3(y-2)-(-8y+2)

We move all terms to the left:

4(3y-3)+12-(3(y-2)-(-8y+2))=0

We multiply parentheses

12y-(3(y-2)-(-8y+2))-12+12=0



We calculate terms in parentheses: -(3(y-2)-(-8y+2)), so:

3(y-2)-(-8y+2)

We multiply parentheses

3y-(-8y+2)-6

We get rid of parentheses

3y+8y-2-6

We add all the numbers together, and all the variables

11y-8

Back to the equation:

-(11y-8)



We add all the numbers together, and all the variables

12y-(11y-8)=0

We get rid of parentheses

12y-11y+8=0

We add all the numbers together, and all the variables

y+8=0

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

y=-8

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