2z+6+4z=z-3+5z+12-z

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Solution for 2z+6+4z=z-3+5z+12-z equation: 



2z+6+4z=z-3+5z+12-z

We move all terms to the left:

2z+6+4z-(z-3+5z+12-z)=0

We add all the numbers together, and all the variables

2z+4z-(5z+9)+6=0

We add all the numbers together, and all the variables

6z-(5z+9)+6=0

We get rid of parentheses

6z-5z-9+6=0

We add all the numbers together, and all the variables

z-3=0

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

z=3

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