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

Solution for 9(z+3)-2(z+2)=3(z-3)+3(z-4) equation:

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

We move all terms to the left:

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

We multiply parentheses

9z-2z-(3(z-3)+3(z-4))+27-4=0


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

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

We multiply parentheses

3z+3z-9-12

We add all the numbers together, and all the variables

6z-21

Back to the equation:

-(6z-21)


We add all the numbers together, and all the variables

7z-(6z-21)+23=0

We get rid of parentheses

7z-6z+21+23=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