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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(4(z-3)+3(z-2)), so:

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

We multiply parentheses

4z+3z-12-6

We add all the numbers together, and all the variables

7z-18

Back to the equation:

-(7z-18)


We add all the numbers together, and all the variables

8z-(7z-18)+16=0

We get rid of parentheses

8z-7z+18+16=0

We add all the numbers together, and all the variables

z+34=0

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

z=-34