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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

2z+2z-4-6

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)+9=0

We get rid of parentheses

5z-4z+10+9=0

We add all the numbers together, and all the variables

z+19=0

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

z=-19