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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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



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

3(y+2)-3

We multiply parentheses

3y+6-3

We add all the numbers together, and all the variables

3y+3

Back to the equation:

-(3y+3)



We get rid of parentheses

3y-3y-3+3=0

We add all the numbers together, and all the variables

=0

y=0/1

y=0

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