-23-(3y+1)=3(y-2)+3y

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Solution for -23-(3y+1)=3(y-2)+3y equation: 



-23-(3y+1)=3(y-2)+3y

We move all terms to the left:

-23-(3y+1)-(3(y-2)+3y)=0

We get rid of parentheses

-3y-(3(y-2)+3y)-1-23=0



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

3(y-2)+3y

We add all the numbers together, and all the variables

3y+3(y-2)

We multiply parentheses

3y+3y-6

We add all the numbers together, and all the variables

6y-6

Back to the equation:

-(6y-6)



We add all the numbers together, and all the variables

-3y-(6y-6)-24=0

We get rid of parentheses

-3y-6y+6-24=0

We add all the numbers together, and all the variables

-9y-18=0

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

-9y=18

y=18/-9

y=-2

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