6(z+3)=2(2z+12)

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Solution for 6(z+3)=2(2z+12) equation: 



6(z+3)=2(2z+12)

We move all terms to the left:

6(z+3)-(2(2z+12))=0

We multiply parentheses

6z-(2(2z+12))+18=0



We calculate terms in parentheses: -(2(2z+12)), so:

2(2z+12)

We multiply parentheses

4z+24

Back to the equation:

-(4z+24)



We get rid of parentheses

6z-4z-24+18=0

We add all the numbers together, and all the variables

2z-6=0

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

2z=6

z=6/2

z=3

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