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6(1+1/2u=2(u+1)
We move all terms to the left:
6(1+1/2u-(2(u+1))=0
Domain of the equation: 2u-(2(u+1))!=0We multiply all the terms by the denominator
u∈R
6(1+1=0
We add all the numbers together, and all the variables
=0
u=0/1
u=0
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