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

Solution for 8(z+4)-2(z-4)=3(z-1)+2(z-4) equation:

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

We move all terms to the left:

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

We multiply parentheses

8z-2z-(3(z-1)+2(z-4))+32+8=0


We calculate terms in parentheses: -(3(z-1)+2(z-4)), so:

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

We multiply parentheses

3z+2z-3-8

We add all the numbers together, and all the variables

5z-11

Back to the equation:

-(5z-11)


We add all the numbers together, and all the variables

6z-(5z-11)+40=0

We get rid of parentheses

6z-5z+11+40=0

We add all the numbers together, and all the variables

z+51=0

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

z=-51