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

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

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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(3(z-4)+3(z-1))+11+8=0


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

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

We multiply parentheses

3z+3z-12-3

We add all the numbers together, and all the variables

6z-15

Back to the equation:

-(6z-15)


We add all the numbers together, and all the variables

7z-(6z-15)+19=0

We get rid of parentheses

7z-6z+15+19=0

We add all the numbers together, and all the variables

z+34=0

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

z=-34