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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(4(z+1)-2(z+1)), so:

4(z+1)-2(z+1)

We multiply parentheses

4z-2z+4-2

We add all the numbers together, and all the variables

2z+2

Back to the equation:

-(2z+2)


We get rid of parentheses

3z-2z-2+9=0

We add all the numbers together, and all the variables

z+7=0

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

z=-7