2(4y-2)+12=6(y-2)-(4y+4)

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Solution for 2(4y-2)+12=6(y-2)-(4y+4) equation: 



2(4y-2)+12=6(y-2)-(4y+4)

We move all terms to the left:

2(4y-2)+12-(6(y-2)-(4y+4))=0

We multiply parentheses

8y-(6(y-2)-(4y+4))-4+12=0



We calculate terms in parentheses: -(6(y-2)-(4y+4)), so:

6(y-2)-(4y+4)

We multiply parentheses

6y-(4y+4)-12

We get rid of parentheses

6y-4y-4-12

We add all the numbers together, and all the variables

2y-16

Back to the equation:

-(2y-16)



We add all the numbers together, and all the variables

8y-(2y-16)+8=0

We get rid of parentheses

8y-2y+16+8=0

We add all the numbers together, and all the variables

6y+24=0

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

6y=-24

y=-24/6

y=-4

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