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

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

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

We move all terms to the left:

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

We multiply parentheses

12z-4z-(3(z-4)+4(z-1))+24+12=0


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

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

We multiply parentheses

3z+4z-12-4

We add all the numbers together, and all the variables

7z-16

Back to the equation:

-(7z-16)


We add all the numbers together, and all the variables

8z-(7z-16)+36=0

We get rid of parentheses

8z-7z+16+36=0

We add all the numbers together, and all the variables

z+52=0

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

z=-52