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

Solution for 9(z+4)-2(z-1)=3(z-4)+3(z-1) equation:

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

We move all terms to the left:

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

We multiply parentheses

9z-2z-(3(z-4)+3(z-1))+36+2=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)+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