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

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

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

We move all terms to the left:

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

We multiply parentheses

9z-2z-(2(z-2)+4(z-4))+27+8=0


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

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

We multiply parentheses

2z+4z-4-16

We add all the numbers together, and all the variables

6z-20

Back to the equation:

-(6z-20)


We add all the numbers together, and all the variables

7z-(6z-20)+35=0

We get rid of parentheses

7z-6z+20+35=0

We add all the numbers together, and all the variables

z+55=0

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

z=-55