-2(4y-3)-8y+6=4(y-2-2)

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Solution for -2(4y-3)-8y+6=4(y-2-2) equation: 



-2(4y-3)-8y+6=4(y-2-2)

We move all terms to the left:

-2(4y-3)-8y+6-(4(y-2-2))=0

We add all the numbers together, and all the variables

-2(4y-3)-8y-(4(y-4))+6=0

We add all the numbers together, and all the variables

-8y-2(4y-3)-(4(y-4))+6=0

We multiply parentheses

-8y-8y-(4(y-4))+6+6=0



We calculate terms in parentheses: -(4(y-4)), so:

4(y-4)

We multiply parentheses

4y-16

Back to the equation:

-(4y-16)



We add all the numbers together, and all the variables

-16y-(4y-16)+12=0

We get rid of parentheses

-16y-4y+16+12=0

We add all the numbers together, and all the variables

-20y+28=0

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

-20y=-28

y=-28/-20

y=1+2/5

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