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

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

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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(3(z-1)+3(z-4))+22+16=0


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

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

We multiply parentheses

3z+3z-3-12

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)+38=0

We get rid of parentheses

7z-6z+15+38=0

We add all the numbers together, and all the variables

z+53=0

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

z=-53