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

Solution for 8(z+2)-3(z-2)=2(z-4)+2(z-2) equation:

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

We move all terms to the left:

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

We multiply parentheses

8z-3z-(2(z-4)+2(z-2))+16+6=0


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

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

We multiply parentheses

2z+2z-8-4

We add all the numbers together, and all the variables

4z-12

Back to the equation:

-(4z-12)


We add all the numbers together, and all the variables

5z-(4z-12)+22=0

We get rid of parentheses

5z-4z+12+22=0

We add all the numbers together, and all the variables

z+34=0

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

z=-34