-41+3y+7=2(3y-5)-3

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Solution for -41+3y+7=2(3y-5)-3 equation: 



-41+3y+7=2(3y-5)-3

We move all terms to the left:

-41+3y+7-(2(3y-5)-3)=0

We add all the numbers together, and all the variables

3y-(2(3y-5)-3)-34=0



We calculate terms in parentheses: -(2(3y-5)-3), so:

2(3y-5)-3

We multiply parentheses

6y-10-3

We add all the numbers together, and all the variables

6y-13

Back to the equation:

-(6y-13)



We get rid of parentheses

3y-6y+13-34=0

We add all the numbers together, and all the variables

-3y-21=0

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

-3y=21

y=21/-3

y=-7

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