9(z-3)+6=3(z+3)+6

Solution for 9(z-3)+6=3(z+3)+6 equation:

9(z-3)+6=3(z+3)+6

We move all terms to the left:

9(z-3)+6-(3(z+3)+6)=0

We multiply parentheses

9z-(3(z+3)+6)-27+6=0


We calculate terms in parentheses: -(3(z+3)+6), so:

3(z+3)+6

We multiply parentheses

3z+9+6

We add all the numbers together, and all the variables

3z+15

Back to the equation:

-(3z+15)


We add all the numbers together, and all the variables

9z-(3z+15)-21=0

We get rid of parentheses

9z-3z-15-21=0

We add all the numbers together, and all the variables

6z-36=0

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

6z=36

z=36/6

z=6