9(z+2)-2(z-2)=3(z-1)+3(z-3)

Solution for 9(z+2)-2(z-2)=3(z-1)+3(z-3) equation:

9(z+2)-2(z-2)=3(z-1)+3(z-3)

We move all terms to the left:

9(z+2)-2(z-2)-(3(z-1)+3(z-3))=0

We multiply parentheses

9z-2z-(3(z-1)+3(z-3))+18+4=0


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

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

We multiply parentheses

3z+3z-3-9

We add all the numbers together, and all the variables

6z-12

Back to the equation:

-(6z-12)


We add all the numbers together, and all the variables

7z-(6z-12)+22=0

We get rid of parentheses

7z-6z+12+22=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