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

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



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

We move all terms to the left:

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

We multiply parentheses

7y+3y-(3(y+1)-1)-9=0



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

3(y+1)-1

We multiply parentheses

3y+3-1

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

10y-(3y+2)-9=0

We get rid of parentheses

10y-3y-2-9=0

We add all the numbers together, and all the variables

7y-11=0

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

7y=11

y=11/7

y=1+4/7

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