3x+3x+(4x+1)=3x+(3x+7)+(3x+8)

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Solution for 3x+3x+(4x+1)=3x+(3x+7)+(3x+8) equation: 



3x+3x+(4x+1)=3x+(3x+7)+(3x+8)

We move all terms to the left:

3x+3x+(4x+1)-(3x+(3x+7)+(3x+8))=0

We add all the numbers together, and all the variables

6x+(4x+1)-(3x+(3x+7)+(3x+8))=0

We get rid of parentheses

6x+4x-(3x+(3x+7)+(3x+8))+1=0



We calculate terms in parentheses: -(3x+(3x+7)+(3x+8)), so:

3x+(3x+7)+(3x+8)

We get rid of parentheses

3x+3x+3x+7+8

We add all the numbers together, and all the variables

9x+15

Back to the equation:

-(9x+15)



We add all the numbers together, and all the variables

10x-(9x+15)+1=0

We get rid of parentheses

10x-9x-15+1=0

We add all the numbers together, and all the variables

x-14=0

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

x=14

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