9(z+2)-2(z-2)=2(z-2)+4(z-2)

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Solution for 9(z+2)-2(z-2)=2(z-2)+4(z-2) equation: 



9(z+2)-2(z-2)=2(z-2)+4(z-2)

We move all terms to the left:

9(z+2)-2(z-2)-(2(z-2)+4(z-2))=0

We multiply parentheses

9z-2z-(2(z-2)+4(z-2))+18+4=0



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

2(z-2)+4(z-2)

We multiply parentheses

2z+4z-4-8

We add all the numbers together, and all the variables

6z-12

Back to the equation:

-(6z-12)



We add all the numbers together, and all the variables

7z-(6z-12)+22=0

We get rid of parentheses

7z-6z+12+22=0

We add all the numbers together, and all the variables

z+34=0

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

z=-34

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