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

Solution for 9(z+1)-2(z-2)=4(z-4)+2(z-3) equation:

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

4z+2z-16-6

We add all the numbers together, and all the variables

6z-22

Back to the equation:

-(6z-22)


We add all the numbers together, and all the variables

7z-(6z-22)+13=0

We get rid of parentheses

7z-6z+22+13=0

We add all the numbers together, and all the variables

z+35=0

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

z=-35