14(z+3)-2(z-1)=5(z-1)+6(z-3)

Solution for 14(z+3)-2(z-1)=5(z-1)+6(z-3) equation:

14(z+3)-2(z-1)=5(z-1)+6(z-3)

We move all terms to the left:

14(z+3)-2(z-1)-(5(z-1)+6(z-3))=0

We multiply parentheses

14z-2z-(5(z-1)+6(z-3))+42+2=0


We calculate terms in parentheses: -(5(z-1)+6(z-3)), so:

5(z-1)+6(z-3)

We multiply parentheses

5z+6z-5-18

We add all the numbers together, and all the variables

11z-23

Back to the equation:

-(11z-23)


We add all the numbers together, and all the variables

12z-(11z-23)+44=0

We get rid of parentheses

12z-11z+23+44=0

We add all the numbers together, and all the variables

z+67=0

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

z=-67