12(3-y)=10y+40

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Solution for 12(3-y)=10y+40 equation: 



12(3-y)=10y+40

We move all terms to the left:

12(3-y)-(10y+40)=0

We add all the numbers together, and all the variables

12(-1y+3)-(10y+40)=0

We multiply parentheses

-12y-(10y+40)+36=0

We get rid of parentheses

-12y-10y-40+36=0

We add all the numbers together, and all the variables

-22y-4=0

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

-22y=4

y=4/-22

y=-2/11

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