12(z-1)-(z-3)=13(z+3)+12

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Solution for 12(z-1)-(z-3)=13(z+3)+12 equation: 



12(z-1)-(z-3)=13(z+3)+12

We move all terms to the left:

12(z-1)-(z-3)-(13(z+3)+12)=0

We multiply parentheses

12z-(z-3)-(13(z+3)+12)-12=0

We get rid of parentheses

12z-z-(13(z+3)+12)+3-12=0



We calculate terms in parentheses: -(13(z+3)+12), so:

13(z+3)+12

We multiply parentheses

13z+39+12

We add all the numbers together, and all the variables

13z+51

Back to the equation:

-(13z+51)



We add all the numbers together, and all the variables

11z-(13z+51)-9=0

We get rid of parentheses

11z-13z-51-9=0

We add all the numbers together, and all the variables

-2z-60=0

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

-2z=60

z=60/-2

z=-30

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