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

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



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

We move all terms to the left:

9(z+2)-2(z-3)-(2(z-4)+4(z-2))=0

We multiply parentheses

9z-2z-(2(z-4)+4(z-2))+18+6=0



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

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

We multiply parentheses

2z+4z-8-8

We add all the numbers together, and all the variables

6z-16

Back to the equation:

-(6z-16)



We add all the numbers together, and all the variables

7z-(6z-16)+24=0

We get rid of parentheses

7z-6z+16+24=0

We add all the numbers together, and all the variables

z+40=0

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

z=-40

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