2(y-4)=8y+1-2(-3y-3)

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Solution for 2(y-4)=8y+1-2(-3y-3) equation: 



2(y-4)=8y+1-2(-3y-3)

We move all terms to the left:

2(y-4)-(8y+1-2(-3y-3))=0

We multiply parentheses

2y-(8y+1-2(-3y-3))-8=0



We calculate terms in parentheses: -(8y+1-2(-3y-3)), so:

8y+1-2(-3y-3)

determiningTheFunctionDomain
8y-2(-3y-3)+1

We multiply parentheses

8y+6y+6+1

We add all the numbers together, and all the variables

14y+7

Back to the equation:

-(14y+7)



We get rid of parentheses

2y-14y-7-8=0

We add all the numbers together, and all the variables

-12y-15=0

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

-12y=15

y=15/-12

y=-1+1/4

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