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

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Solution for 9(z+4)-3(z-4)=3(z-3)+2(z-1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

9z-3z-(3(z-3)+2(z-1))+36+12=0



We calculate terms in parentheses: -(3(z-3)+2(z-1)), so:

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

We multiply parentheses

3z+2z-9-2

We add all the numbers together, and all the variables

5z-11

Back to the equation:

-(5z-11)



We add all the numbers together, and all the variables

6z-(5z-11)+48=0

We get rid of parentheses

6z-5z+11+48=0

We add all the numbers together, and all the variables

z+59=0

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

z=-59

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