6y+2(y-3)=3(y+1)-2

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Solution for 6y+2(y-3)=3(y+1)-2 equation: 



6y+2(y-3)=3(y+1)-2

We move all terms to the left:

6y+2(y-3)-(3(y+1)-2)=0

We multiply parentheses

6y+2y-(3(y+1)-2)-6=0



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

3(y+1)-2

We multiply parentheses

3y+3-2

We add all the numbers together, and all the variables

3y+1

Back to the equation:

-(3y+1)



We add all the numbers together, and all the variables

8y-(3y+1)-6=0

We get rid of parentheses

8y-3y-1-6=0

We add all the numbers together, and all the variables

5y-7=0

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

5y=7

y=7/5

y=1+2/5

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