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

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

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

We move all terms to the left:

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

We multiply parentheses

9z-3z-(2(z-1)+3(z-1))+9+12=0


We calculate terms in parentheses: -(2(z-1)+3(z-1)), so:

2(z-1)+3(z-1)

We multiply parentheses

2z+3z-2-3

We add all the numbers together, and all the variables

5z-5

Back to the equation:

-(5z-5)


We add all the numbers together, and all the variables

6z-(5z-5)+21=0

We get rid of parentheses

6z-5z+5+21=0

We add all the numbers together, and all the variables

z+26=0

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

z=-26