7y+3(y-3)=6(y+1)-2

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Solution for 7y+3(y-3)=6(y+1)-2 equation: 



7y+3(y-3)=6(y+1)-2

We move all terms to the left:

7y+3(y-3)-(6(y+1)-2)=0

We multiply parentheses

7y+3y-(6(y+1)-2)-9=0



We calculate terms in parentheses: -(6(y+1)-2), so:

6(y+1)-2

We multiply parentheses

6y+6-2

We add all the numbers together, and all the variables

6y+4

Back to the equation:

-(6y+4)



We add all the numbers together, and all the variables

10y-(6y+4)-9=0

We get rid of parentheses

10y-6y-4-9=0

We add all the numbers together, and all the variables

4y-13=0

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

4y=13

y=13/4

y=3+1/4

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