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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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



We calculate terms in parentheses: -(3(2x-4)+2(-1x+2)), so:

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

We multiply parentheses

6x-2x-12+4

We add all the numbers together, and all the variables

4x-8

Back to the equation:

-(4x-8)



We add all the numbers together, and all the variables

9x-(4x-8)-12=0

We get rid of parentheses

9x-4x+8-12=0

We add all the numbers together, and all the variables

5x-4=0

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

5x=4

x=4/5

x=4/5

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