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

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Solution for 7(z+3)-2(z-3)=2(z-2)+2(z-1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

7z-2z-(2(z-2)+2(z-1))+21+6=0



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

2(z-2)+2(z-1)

We multiply parentheses

2z+2z-4-2

We add all the numbers together, and all the variables

4z-6

Back to the equation:

-(4z-6)



We add all the numbers together, and all the variables

5z-(4z-6)+27=0

We get rid of parentheses

5z-4z+6+27=0

We add all the numbers together, and all the variables

z+33=0

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

z=-33

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