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

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

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

We move all terms to the left:

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

We multiply parentheses

11z-4z-(3(z-3)+3(z-2))+33+4=0


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

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

We multiply parentheses

3z+3z-9-6

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)+37=0

We get rid of parentheses

7z-6z+15+37=0

We add all the numbers together, and all the variables

z+52=0

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

z=-52