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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

2z+3z-8-12

We add all the numbers together, and all the variables

5z-20

Back to the equation:

-(5z-20)


We add all the numbers together, and all the variables

6z-(5z-20)+34=0

We get rid of parentheses

6z-5z+20+34=0

We add all the numbers together, and all the variables

z+54=0

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

z=-54