5(2z+2)=6-2(z-2)

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Solution for 5(2z+2)=6-2(z-2) equation: 



5(2z+2)=6-2(z-2)

We move all terms to the left:

5(2z+2)-(6-2(z-2))=0

We multiply parentheses

10z-(6-2(z-2))+10=0



We calculate terms in parentheses: -(6-2(z-2)), so:

6-2(z-2)

determiningTheFunctionDomain
-2(z-2)+6

We multiply parentheses

-2z+4+6

We add all the numbers together, and all the variables

-2z+10

Back to the equation:

-(-2z+10)



We get rid of parentheses

10z+2z-10+10=0

We add all the numbers together, and all the variables

12z=0

z=0/12

z=0

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