Solve for x
x\in \left(\frac{4}{3},2\right)
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3x-4>0 x-2<0
For the quotient to be negative, 3x-4 and x-2 have to be of the opposite signs. Consider the case when 3x-4 is positive and x-2 is negative.
x\in \left(\frac{4}{3},2\right)
The solution satisfying both inequalities is x\in \left(\frac{4}{3},2\right).
x-2>0 3x-4<0
Consider the case when x-2 is positive and 3x-4 is negative.
x\in \emptyset
This is false for any x.
x\in \left(\frac{4}{3},2\right)
The final solution is the union of the obtained solutions.
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