3(z+1)=11-2(z+13)

Solution for 3(z+1)=11-2(z+13) equation:

3(z+1)=11-2(z+13)

We move all terms to the left:

3(z+1)-(11-2(z+13))=0

We multiply parentheses

3z-(11-2(z+13))+3=0


We calculate terms in parentheses: -(11-2(z+13)), so:

11-2(z+13)

determiningTheFunctionDomain
-2(z+13)+11

We multiply parentheses

-2z-26+11

We add all the numbers together, and all the variables

-2z-15

Back to the equation:

-(-2z-15)


We get rid of parentheses

3z+2z+15+3=0

We add all the numbers together, and all the variables

5z+18=0

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

5z=-18

z=-18/5

z=-3+3/5