3(2z+2)-3=6(z-1)+9

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Solution for 3(2z+2)-3=6(z-1)+9 equation: 



3(2z+2)-3=6(z-1)+9

We move all terms to the left:

3(2z+2)-3-(6(z-1)+9)=0

We multiply parentheses

6z-(6(z-1)+9)+6-3=0



We calculate terms in parentheses: -(6(z-1)+9), so:

6(z-1)+9

We multiply parentheses

6z-6+9

We add all the numbers together, and all the variables

6z+3

Back to the equation:

-(6z+3)



We add all the numbers together, and all the variables

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

We get rid of parentheses

6z-6z-3+3=0

We add all the numbers together, and all the variables

=0

z=0/1

z=0

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