3z-4(1-3z)=2z-(z+1)

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Solution for 3z-4(1-3z)=2z-(z+1) equation: 



3z-4(1-3z)=2z-(z+1)

We move all terms to the left:

3z-4(1-3z)-(2z-(z+1))=0

We add all the numbers together, and all the variables

3z-4(-3z+1)-(2z-(z+1))=0

We multiply parentheses

3z+12z-(2z-(z+1))-4=0



We calculate terms in parentheses: -(2z-(z+1)), so:

2z-(z+1)

We get rid of parentheses

2z-z-1

We add all the numbers together, and all the variables

z-1

Back to the equation:

-(z-1)



We add all the numbers together, and all the variables

15z-(z-1)-4=0

We get rid of parentheses

15z-z+1-4=0

We add all the numbers together, and all the variables

14z-3=0

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

14z=3

z=3/14

z=3/14

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