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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(4(z-1)+3(z-2)), so:

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

We multiply parentheses

4z+3z-4-6

We add all the numbers together, and all the variables

7z-10

Back to the equation:

-(7z-10)


We add all the numbers together, and all the variables

8z-(7z-10)+24=0

We get rid of parentheses

8z-7z+10+24=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