z/5(z=1)=2(3-2z)+17

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Solution for z/5(z=1)=2(3-2z)+17 equation:



z/5(z=1)=2(3-2z)+17
We move all terms to the left:
z/5(z-(1))=0
Domain of the equation: 5(z-1)!=0
z∈R
We multiply all the terms by the denominator
z=0

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