-2(3y-4)-y=04(y-2)

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Solution for -2(3y-4)-y=04(y-2) equation: 



-2(3y-4)-y=04(y-2)

We move all terms to the left:

-2(3y-4)-y-(04(y-2))=0

We add all the numbers together, and all the variables

-1y-2(3y-4)-(04(y-2))=0

We multiply parentheses

-1y-6y-(04(y-2))+8=0



We calculate terms in parentheses: -(04(y-2)), so:

04(y-2)

We multiply parentheses

4y-8

Back to the equation:

-(4y-8)



We add all the numbers together, and all the variables

-7y-(4y-8)+8=0

We get rid of parentheses

-7y-4y+8+8=0

We add all the numbers together, and all the variables

-11y+16=0

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

-11y=-16

y=-16/-11

y=1+5/11

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