12(z+4)-3(z-1)=4(z-3)+4(z-1)

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Solution for 12(z+4)-3(z-1)=4(z-3)+4(z-1) equation: 



12(z+4)-3(z-1)=4(z-3)+4(z-1)

We move all terms to the left:

12(z+4)-3(z-1)-(4(z-3)+4(z-1))=0

We multiply parentheses

12z-3z-(4(z-3)+4(z-1))+48+3=0



We calculate terms in parentheses: -(4(z-3)+4(z-1)), so:

4(z-3)+4(z-1)

We multiply parentheses

4z+4z-12-4

We add all the numbers together, and all the variables

8z-16

Back to the equation:

-(8z-16)



We add all the numbers together, and all the variables

9z-(8z-16)+51=0

We get rid of parentheses

9z-8z+16+51=0

We add all the numbers together, and all the variables

z+67=0

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

z=-67

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