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2(-3+2(z-1)=2/3(6z-15)
We move all terms to the left:
2(-3+2(z-1)-(2/3(6z-15))=0
Domain of the equation: 3(6z-15))!=0We multiply all the terms by the denominator
z∈R
2(-3+2(z-1)-(2=0
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