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

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Solution for 10(z+3)-3(z-3)=3(z-2)+3(z-2) equation: 



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

We move all terms to the left:

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

We multiply parentheses

10z-3z-(3(z-2)+3(z-2))+30+9=0



We calculate terms in parentheses: -(3(z-2)+3(z-2)), so:

3(z-2)+3(z-2)

We multiply parentheses

3z+3z-6-6

We add all the numbers together, and all the variables

6z-12

Back to the equation:

-(6z-12)



We add all the numbers together, and all the variables

7z-(6z-12)+39=0

We get rid of parentheses

7z-6z+12+39=0

We add all the numbers together, and all the variables

z+51=0

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

z=-51

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