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11(z+2)-4(z-3)=3(z-4)+3(z-1)
We move all terms to the left:
11(z+2)-4(z-3)-(3(z-4)+3(z-1))=0
We multiply parentheses
11z-4z-(3(z-4)+3(z-1))+22+12=0
We calculate terms in parentheses: -(3(z-4)+3(z-1)), so:We add all the numbers together, and all the variables
3(z-4)+3(z-1)
We multiply parentheses
3z+3z-12-3
We add all the numbers together, and all the variables
6z-15
Back to the equation:
-(6z-15)
7z-(6z-15)+34=0
We get rid of parentheses
7z-6z+15+34=0
We add all the numbers together, and all the variables
z+49=0
We move all terms containing z to the left, all other terms to the right
z=-49
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