3x+3x+3x+3x=(2x+5)+(2x+5)+(x+1)+(x+1)

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Solution for 3x+3x+3x+3x=(2x+5)+(2x+5)+(x+1)+(x+1) equation: 



3x+3x+3x+3x=(2x+5)+(2x+5)+(x+1)+(x+1)

We move all terms to the left:

3x+3x+3x+3x-((2x+5)+(2x+5)+(x+1)+(x+1))=0

We add all the numbers together, and all the variables

12x-((2x+5)+(2x+5)+(x+1)+(x+1))=0



We calculate terms in parentheses: -((2x+5)+(2x+5)+(x+1)+(x+1)), so:

(2x+5)+(2x+5)+(x+1)+(x+1)

We get rid of parentheses

2x+2x+x+x+5+5+1+1

We add all the numbers together, and all the variables

6x+12

Back to the equation:

-(6x+12)



We get rid of parentheses

12x-6x-12=0

We add all the numbers together, and all the variables

6x-12=0

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

6x=12

x=12/6

x=2

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