(2x-2)+(x+1)+x+(x+1)=(2x-9)+(x+1)+(x+8)+8

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Solution for (2x-2)+(x+1)+x+(x+1)=(2x-9)+(x+1)+(x+8)+8 equation: 



(2x-2)+(x+1)+x+(x+1)=(2x-9)+(x+1)+(x+8)+8

We move all terms to the left:

(2x-2)+(x+1)+x+(x+1)-((2x-9)+(x+1)+(x+8)+8)=0

We add all the numbers together, and all the variables

x+(2x-2)+(x+1)+(x+1)-((2x-9)+(x+1)+(x+8)+8)=0

We get rid of parentheses

x+2x+x+x-((2x-9)+(x+1)+(x+8)+8)-2+1+1=0



We calculate terms in parentheses: -((2x-9)+(x+1)+(x+8)+8), so:

(2x-9)+(x+1)+(x+8)+8

We get rid of parentheses

2x+x+x-9+1+8+8

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

5x-(4x+8)=0

We get rid of parentheses

5x-4x-8=0

We add all the numbers together, and all the variables

x-8=0

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

x=8

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