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

Solution for 11(z+2)-3(z-2)=3(z-1)+4(z-1) equation:

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

We move all terms to the left:

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

We multiply parentheses

11z-3z-(3(z-1)+4(z-1))+22+6=0


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

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

We multiply parentheses

3z+4z-3-4

We add all the numbers together, and all the variables

7z-7

Back to the equation:

-(7z-7)


We add all the numbers together, and all the variables

8z-(7z-7)+28=0

We get rid of parentheses

8z-7z+7+28=0

We add all the numbers together, and all the variables

z+35=0

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

z=-35