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

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Solution for 12(z+1)-4(z-2)=3(z-3)+4(z-4) equation: 



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

We move all terms to the left:

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

We multiply parentheses

12z-4z-(3(z-3)+4(z-4))+12+8=0



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

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

We multiply parentheses

3z+4z-9-16

We add all the numbers together, and all the variables

7z-25

Back to the equation:

-(7z-25)



We add all the numbers together, and all the variables

8z-(7z-25)+20=0

We get rid of parentheses

8z-7z+25+20=0

We add all the numbers together, and all the variables

z+45=0

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

z=-45

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