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

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



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

We move all terms to the left:

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

We multiply parentheses

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



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

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

We multiply parentheses

2z+2z-4-4

We add all the numbers together, and all the variables

4z-8

Back to the equation:

-(4z-8)



We add all the numbers together, and all the variables

5z-(4z-8)+27=0

We get rid of parentheses

5z-4z+8+27=0

We add all the numbers together, and all the variables

z+35=0

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

z=-35

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