32=12+(4z-1)

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Solution for 32=12+(4z-1) equation: 



32=12+(4z-1)

We move all terms to the left:

32-(12+(4z-1))=0



We calculate terms in parentheses: -(12+(4z-1)), so:

12+(4z-1)

determiningTheFunctionDomain
(4z-1)+12

We get rid of parentheses

4z-1+12

We add all the numbers together, and all the variables

4z+11

Back to the equation:

-(4z+11)



We get rid of parentheses

-4z-11+32=0

We add all the numbers together, and all the variables

-4z+21=0

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

-4z=-21

z=-21/-4

z=5+1/4

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