6(2+3z)=14-(2-18z)

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Solution for 6(2+3z)=14-(2-18z) equation: 



6(2+3z)=14-(2-18z)

We move all terms to the left:

6(2+3z)-(14-(2-18z))=0

We add all the numbers together, and all the variables

6(3z+2)-(14-(-18z+2))=0

We multiply parentheses

18z-(14-(-18z+2))+12=0



We calculate terms in parentheses: -(14-(-18z+2)), so:

14-(-18z+2)

determiningTheFunctionDomain
-(-18z+2)+14

We get rid of parentheses

18z-2+14

We add all the numbers together, and all the variables

18z+12

Back to the equation:

-(18z+12)



We get rid of parentheses

18z-18z-12+12=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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