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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

3z+2z-9-4

We add all the numbers together, and all the variables

5z-13

Back to the equation:

-(5z-13)


We add all the numbers together, and all the variables

6z-(5z-13)+12=0

We get rid of parentheses

6z-5z+13+12=0

We add all the numbers together, and all the variables

z+25=0

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

z=-25