2(3y+1)+8=7+3(y-3)

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Solution for 2(3y+1)+8=7+3(y-3) equation: 



2(3y+1)+8=7+3(y-3)

We move all terms to the left:

2(3y+1)+8-(7+3(y-3))=0

We multiply parentheses

6y-(7+3(y-3))+2+8=0



We calculate terms in parentheses: -(7+3(y-3)), so:

7+3(y-3)

determiningTheFunctionDomain
3(y-3)+7

We multiply parentheses

3y-9+7

We add all the numbers together, and all the variables

3y-2

Back to the equation:

-(3y-2)



We add all the numbers together, and all the variables

6y-(3y-2)+10=0

We get rid of parentheses

6y-3y+2+10=0

We add all the numbers together, and all the variables

3y+12=0

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

3y=-12

y=-12/3

y=-4

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