4z+6=2(z-6)+z(12+)

Solution for 4z+6=2(z-6)+z(12+) equation:

4z+6=2(z-6)+z(12+)

We move all terms to the left:

4z+6-(2(z-6)+z(12+))=0

We add all the numbers together, and all the variables

4z-(2(z-6)+z0)+6=0


We calculate terms in parentheses: -(2(z-6)+z0), so:

2(z-6)+z0

We add all the numbers together, and all the variables

z+2(z-6)

We multiply parentheses

z+2z-12

We add all the numbers together, and all the variables

3z-12

Back to the equation:

-(3z-12)


We get rid of parentheses

4z-3z+12+6=0

We add all the numbers together, and all the variables

z+18=0

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

z=-18