9(z+2)-2(z-2)=2(z-3)+4(z-2)

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Solution for 9(z+2)-2(z-2)=2(z-3)+4(z-2) equation:



9(z+2)-2(z-2)=2(z-3)+4(z-2)
We move all terms to the left:
9(z+2)-2(z-2)-(2(z-3)+4(z-2))=0
We multiply parentheses
9z-2z-(2(z-3)+4(z-2))+18+4=0
We calculate terms in parentheses: -(2(z-3)+4(z-2)), so:
2(z-3)+4(z-2)
We multiply parentheses
2z+4z-6-8
We add all the numbers together, and all the variables
6z-14
Back to the equation:
-(6z-14)
We add all the numbers together, and all the variables
7z-(6z-14)+22=0
We get rid of parentheses
7z-6z+14+22=0
We add all the numbers together, and all the variables
z+36=0
We move all terms containing z to the left, all other terms to the right
z=-36

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