-(3z-(7z+4))=4+(3z+2)

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Solution for -(3z-(7z+4))=4+(3z+2) equation: 



-(3z-(7z+4))=4+(3z+2)

We move all terms to the left:

-(3z-(7z+4))-(4+(3z+2))=0



We calculate terms in parentheses: -(3z-(7z+4)), so:

3z-(7z+4)

We get rid of parentheses

3z-7z-4

We add all the numbers together, and all the variables

-4z-4

Back to the equation:

-(-4z-4)





We calculate terms in parentheses: -(4+(3z+2)), so:

4+(3z+2)

determiningTheFunctionDomain
(3z+2)+4

We get rid of parentheses

3z+2+4

We add all the numbers together, and all the variables

3z+6

Back to the equation:

-(3z+6)



We get rid of parentheses

4z-3z+4-6=0

We add all the numbers together, and all the variables

z-2=0

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

z=2

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