4(z+12)=40+2(4+2z)

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Solution for 4(z+12)=40+2(4+2z) equation: 



4(z+12)=40+2(4+2z)

We move all terms to the left:

4(z+12)-(40+2(4+2z))=0

We add all the numbers together, and all the variables

4(z+12)-(40+2(2z+4))=0

We multiply parentheses

4z-(40+2(2z+4))+48=0



We calculate terms in parentheses: -(40+2(2z+4)), so:

40+2(2z+4)

determiningTheFunctionDomain
2(2z+4)+40

We multiply parentheses

4z+8+40

We add all the numbers together, and all the variables

4z+48

Back to the equation:

-(4z+48)



We get rid of parentheses

4z-4z-48+48=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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