3x+3+4x=(7x-2)+(7x-2)+(2x+4)

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Solution for 3x+3+4x=(7x-2)+(7x-2)+(2x+4) equation: 



3x+3+4x=(7x-2)+(7x-2)+(2x+4)

We move all terms to the left:

3x+3+4x-((7x-2)+(7x-2)+(2x+4))=0

We add all the numbers together, and all the variables

7x-((7x-2)+(7x-2)+(2x+4))+3=0



We calculate terms in parentheses: -((7x-2)+(7x-2)+(2x+4)), so:

(7x-2)+(7x-2)+(2x+4)

We get rid of parentheses

7x+7x+2x-2-2+4

We add all the numbers together, and all the variables

16x

Back to the equation:

-(16x)



We add all the numbers together, and all the variables

-9x+3=0

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

-9x=-3

x=-3/-9

x=1/3

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