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

Solution for 7(z+3)-2(z-3)=2(z-4)+2(z-1) equation:

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

We move all terms to the left:

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

We multiply parentheses

7z-2z-(2(z-4)+2(z-1))+21+6=0


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

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

We multiply parentheses

2z+2z-8-2

We add all the numbers together, and all the variables

4z-10

Back to the equation:

-(4z-10)


We add all the numbers together, and all the variables

5z-(4z-10)+27=0

We get rid of parentheses

5z-4z+10+27=0

We add all the numbers together, and all the variables

z+37=0

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

z=-37