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

Solution for 11(z+3)-2(z-4)=4(z-1)+4(z-2) equation:

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

We move all terms to the left:

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

We multiply parentheses

11z-2z-(4(z-1)+4(z-2))+33+8=0


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

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

We multiply parentheses

4z+4z-4-8

We add all the numbers together, and all the variables

8z-12

Back to the equation:

-(8z-12)


We add all the numbers together, and all the variables

9z-(8z-12)+41=0

We get rid of parentheses

9z-8z+12+41=0

We add all the numbers together, and all the variables

z+53=0

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

z=-53