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

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

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

We move all terms to the left:

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

We multiply parentheses

12z-4z-(3(z-4)+4(z-4))+48+8=0


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

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

We multiply parentheses

3z+4z-12-16

We add all the numbers together, and all the variables

7z-28

Back to the equation:

-(7z-28)


We add all the numbers together, and all the variables

8z-(7z-28)+56=0

We get rid of parentheses

8z-7z+28+56=0

We add all the numbers together, and all the variables

z+84=0

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

z=-84