3*(2x+1)+12*(3x-2)=6*(3x-1)+3*(x-4)+11

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Solution for 3*(2x+1)+12*(3x-2)=6*(3x-1)+3*(x-4)+11 equation: 



3(2x+1)+12(3x-2)=6(3x-1)+3(x-4)+11

We move all terms to the left:

3(2x+1)+12(3x-2)-(6(3x-1)+3(x-4)+11)=0

We multiply parentheses

6x+36x-(6(3x-1)+3(x-4)+11)+3-24=0



We calculate terms in parentheses: -(6(3x-1)+3(x-4)+11), so:

6(3x-1)+3(x-4)+11

We multiply parentheses

18x+3x-6-12+11

We add all the numbers together, and all the variables

21x-7

Back to the equation:

-(21x-7)



We add all the numbers together, and all the variables

42x-(21x-7)-21=0

We get rid of parentheses

42x-21x+7-21=0

We add all the numbers together, and all the variables

21x-14=0

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

21x=14

x=14/21

x=2/3

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