2(3z+1)-(z-6)=12

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Solution for 2(3z+1)-(z-6)=12 equation: 



2(3z+1)-(z-6)=12

We move all terms to the left:

2(3z+1)-(z-6)-(12)=0

We multiply parentheses

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

We get rid of parentheses

6z-z+6+2-12=0

We add all the numbers together, and all the variables

5z-4=0

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

5z=4

z=4/5

z=4/5

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