Solve for x
x\in \left(\frac{2}{3},23\right)
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\frac{23-x}{3x-2}>0
Add 2 and 21 to get 23.
23-x<0 3x-2<0
For the quotient to be positive, 23-x and 3x-2 have to be both negative or both positive. Consider the case when 23-x and 3x-2 are both negative.
x\in \emptyset
This is false for any x.
3x-2>0 23-x>0
Consider the case when 23-x and 3x-2 are both positive.
x\in \left(\frac{2}{3},23\right)
The solution satisfying both inequalities is x\in \left(\frac{2}{3},23\right).
x\in \left(\frac{2}{3},23\right)
The final solution is the union of the obtained solutions.
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