3(5z-7)+2(9z-11)=4(8z-28)-111

Solution for 3(5z-7)+2(9z-11)=4(8z-28)-111 equation:

3(5z-7)+2(9z-11)=4(8z-28)-111

We move all terms to the left:

3(5z-7)+2(9z-11)-(4(8z-28)-111)=0

We multiply parentheses

15z+18z-(4(8z-28)-111)-21-22=0


We calculate terms in parentheses: -(4(8z-28)-111), so:

4(8z-28)-111

We multiply parentheses

32z-112-111

We add all the numbers together, and all the variables

32z-223

Back to the equation:

-(32z-223)


We add all the numbers together, and all the variables

33z-(32z-223)-43=0

We get rid of parentheses

33z-32z+223-43=0

We add all the numbers together, and all the variables

z+180=0

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

z=-180