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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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



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

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

We get rid of parentheses

x+3x-4-2

We add all the numbers together, and all the variables

4x-6

Back to the equation:

-(4x-6)



We get rid of parentheses

2x-4x+6+6=0

We add all the numbers together, and all the variables

-2x+12=0

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

-2x=-12

x=-12/-2

x=+6

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