abs((2x-1)/(x-1)) = 3. I'm trying to figure out what to do and how to finish it since the absolute value is around both (2x+1) and (x-1).
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Originally Posted by AlphaRock abs((2x-1)/(x-1)) = 3. I'm trying to figure out what to do and how to finish it since the absolute value is around both (2x+1) and (x-1). Hi AlphaRock, Let's consider case 1. Check this solution in your original absolute value equation. Now consider case 2. Check this solution in your original absolute value equation.
Originally Posted by AlphaRock abs((2x-1)/(x-1)) = 3. I'm trying to figure out what to do and how to finish it since the absolute value is around both (2x+1) and (x-1). Here is a second way. Solve:
You should, however, note that what you give is NOT an inequality, it is an equation.
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