3(4x-4)+3(x-4)=(3x-1)+(3x-1)+(2x)

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Solution for 3(4x-4)+3(x-4)=(3x-1)+(3x-1)+(2x) equation: 



3(4x-4)+3(x-4)=(3x-1)+(3x-1)+(2x)

We move all terms to the left:

3(4x-4)+3(x-4)-((3x-1)+(3x-1)+(2x))=0

We multiply parentheses

12x+3x-((3x-1)+(3x-1)+2x)-12-12=0



We calculate terms in parentheses: -((3x-1)+(3x-1)+2x), so:

(3x-1)+(3x-1)+2x

We add all the numbers together, and all the variables

2x+(3x-1)+(3x-1)

We get rid of parentheses

2x+3x+3x-1-1

We add all the numbers together, and all the variables

8x-2

Back to the equation:

-(8x-2)



We add all the numbers together, and all the variables

15x-(8x-2)-24=0

We get rid of parentheses

15x-8x+2-24=0

We add all the numbers together, and all the variables

7x-22=0

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

7x=22

x=22/7

x=3+1/7

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