3z-12=3-(1+z)

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Solution for 3z-12=3-(1+z) equation: 



3z-12=3-(1+z)

We move all terms to the left:

3z-12-(3-(1+z))=0

We add all the numbers together, and all the variables

3z-(3-(z+1))-12=0



We calculate terms in parentheses: -(3-(z+1)), so:

3-(z+1)

determiningTheFunctionDomain
-(z+1)+3

We get rid of parentheses

-z-1+3

We add all the numbers together, and all the variables

-1z+2

Back to the equation:

-(-1z+2)



We get rid of parentheses

3z+1z-2-12=0

We add all the numbers together, and all the variables

4z-14=0

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

4z=14

z=14/4

z=3+1/2

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