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

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

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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(3(z-4)+3(z-4))+22+12=0


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

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

We multiply parentheses

3z+3z-12-12

We add all the numbers together, and all the variables

6z-24

Back to the equation:

-(6z-24)


We add all the numbers together, and all the variables

7z-(6z-24)+34=0

We get rid of parentheses

7z-6z+24+34=0

We add all the numbers together, and all the variables

z+58=0

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

z=-58