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10(z+1)-4(z-4)=3(z-1)+2(z-1)
We move all terms to the left:
10(z+1)-4(z-4)-(3(z-1)+2(z-1))=0
We multiply parentheses
10z-4z-(3(z-1)+2(z-1))+10+16=0
We calculate terms in parentheses: -(3(z-1)+2(z-1)), so:We add all the numbers together, and all the variables
3(z-1)+2(z-1)
We multiply parentheses
3z+2z-3-2
We add all the numbers together, and all the variables
5z-5
Back to the equation:
-(5z-5)
6z-(5z-5)+26=0
We get rid of parentheses
6z-5z+5+26=0
We add all the numbers together, and all the variables
z+31=0
We move all terms containing z to the left, all other terms to the right
z=-31
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