6(z+2)=4(z+1)-8

Solution for 6(z+2)=4(z+1)-8 equation:

6(z+2)=4(z+1)-8

We move all terms to the left:

6(z+2)-(4(z+1)-8)=0

We multiply parentheses

6z-(4(z+1)-8)+12=0


We calculate terms in parentheses: -(4(z+1)-8), so:

4(z+1)-8

We multiply parentheses

4z+4-8

We add all the numbers together, and all the variables

4z-4

Back to the equation:

-(4z-4)


We get rid of parentheses

6z-4z+4+12=0

We add all the numbers together, and all the variables

2z+16=0

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

2z=-16

z=-16/2

z=-8