4(3y-4)-12=3(2y-4)-18

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Solution for 4(3y-4)-12=3(2y-4)-18 equation: 



4(3y-4)-12=3(2y-4)-18

We move all terms to the left:

4(3y-4)-12-(3(2y-4)-18)=0

We multiply parentheses

12y-(3(2y-4)-18)-16-12=0



We calculate terms in parentheses: -(3(2y-4)-18), so:

3(2y-4)-18

We multiply parentheses

6y-12-18

We add all the numbers together, and all the variables

6y-30

Back to the equation:

-(6y-30)



We add all the numbers together, and all the variables

12y-(6y-30)-28=0

We get rid of parentheses

12y-6y+30-28=0

We add all the numbers together, and all the variables

6y+2=0

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

6y=-2

y=-2/6

y=-1/3

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