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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

2z+(z-4)-4

We get rid of parentheses

2z+z-4-4

We add all the numbers together, and all the variables

3z-8

Back to the equation:

-(3z-8)


We add all the numbers together, and all the variables

7z-(3z-8)+35=0

We get rid of parentheses

7z-3z+8+35=0

We add all the numbers together, and all the variables

4z+43=0

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

4z=-43

z=-43/4

z=-10+3/4