2(y-4)=-3(3y+2)+8

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Solution for 2(y-4)=-3(3y+2)+8 equation: 



2(y-4)=-3(3y+2)+8

We move all terms to the left:

2(y-4)-(-3(3y+2)+8)=0

We multiply parentheses

2y-(-3(3y+2)+8)-8=0



We calculate terms in parentheses: -(-3(3y+2)+8), so:

-3(3y+2)+8

We multiply parentheses

-9y-6+8

We add all the numbers together, and all the variables

-9y+2

Back to the equation:

-(-9y+2)



We get rid of parentheses

2y+9y-2-8=0

We add all the numbers together, and all the variables

11y-10=0

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

11y=10

y=10/11

y=10/11

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