3=(z+1)+11-2(z+13)

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Solution for 3=(z+1)+11-2(z+13) equation: 



3=(z+1)+11-2(z+13)

We move all terms to the left:

3-((z+1)+11-2(z+13))=0



We calculate terms in parentheses: -((z+1)+11-2(z+13)), so:

(z+1)+11-2(z+13)

determiningTheFunctionDomain
(z+1)-2(z+13)+11

We multiply parentheses

(z+1)-2z-26+11

We get rid of parentheses

z-2z+1-26+11

We add all the numbers together, and all the variables

-1z-14

Back to the equation:

-(-1z-14)



We get rid of parentheses

1z+14+3=0

We add all the numbers together, and all the variables

z+17=0

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

z=-17

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