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

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

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

We move all terms to the left:

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

We multiply parentheses

9z-3z-(3(z-2)+2(z-1))+27+9=0


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

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

We multiply parentheses

3z+2z-6-2

We add all the numbers together, and all the variables

5z-8

Back to the equation:

-(5z-8)


We add all the numbers together, and all the variables

6z-(5z-8)+36=0

We get rid of parentheses

6z-5z+8+36=0

We add all the numbers together, and all the variables

z+44=0

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

z=-44