3=2+(2/z+2)

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Solution for 3=2+(2/z+2) equation: 



3=2+(2/z+2)

We move all terms to the left:

3-(2+(2/z+2))=0



Domain of the equation: z+2))!=0

z∈R



We multiply all the terms by the denominator


-(2+(2+3*z+2))=0



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

2+(2+3*z+2)

determiningTheFunctionDomain
(2+3*z+2)+2

We add all the numbers together, and all the variables

(3z+4)+2

We get rid of parentheses

3z+4+2

We add all the numbers together, and all the variables

3z+6

Back to the equation:

-(3z+6)



We get rid of parentheses

-3z-6=0

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

-3z=6

z=6/-3

z=-2

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