12y-(y-3)=6y+8-(y+2)

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Solution for 12y-(y-3)=6y+8-(y+2) equation: 



12y-(y-3)=6y+8-(y+2)

We move all terms to the left:

12y-(y-3)-(6y+8-(y+2))=0

We get rid of parentheses

12y-y-(6y+8-(y+2))+3=0



We calculate terms in parentheses: -(6y+8-(y+2)), so:

6y+8-(y+2)

determiningTheFunctionDomain
6y-(y+2)+8

We get rid of parentheses

6y-y-2+8

We add all the numbers together, and all the variables

5y+6

Back to the equation:

-(5y+6)



We add all the numbers together, and all the variables

11y-(5y+6)+3=0

We get rid of parentheses

11y-5y-6+3=0

We add all the numbers together, and all the variables

6y-3=0

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

6y=3

y=3/6

y=1/2

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