4(3y-4)+8(6-y)=12

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Solution for 4(3y-4)+8(6-y)=12 equation: 



4(3y-4)+8(6-y)=12

We move all terms to the left:

4(3y-4)+8(6-y)-(12)=0

We add all the numbers together, and all the variables

4(3y-4)+8(-1y+6)-12=0

We multiply parentheses

12y-8y-16+48-12=0

We add all the numbers together, and all the variables

4y+20=0

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

4y=-20

y=-20/4

y=-5

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