-(4z-3)+1=4+(4z+2)

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Solution for -(4z-3)+1=4+(4z+2) equation: 



-(4z-3)+1=4+(4z+2)

We move all terms to the left:

-(4z-3)+1-(4+(4z+2))=0

We get rid of parentheses

-4z-(4+(4z+2))+3+1=0



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

4+(4z+2)

determiningTheFunctionDomain
(4z+2)+4

We get rid of parentheses

4z+2+4

We add all the numbers together, and all the variables

4z+6

Back to the equation:

-(4z+6)



We add all the numbers together, and all the variables

-4z-(4z+6)+4=0

We get rid of parentheses

-4z-4z-6+4=0

We add all the numbers together, and all the variables

-8z-2=0

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

-8z=2

z=2/-8

z=-1/4

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