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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- Sep 16th 2009, 10:33 PMAlphaRockHow to solve this inequality?
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). - Sep 17th 2009, 04:28 AMmasters
Hi AlphaRock,

$\displaystyle \left|\frac{2x-1}{x-1}\right|=3$

__Let's consider case 1.__

$\displaystyle \frac{2x-1}{x-1}=3$

$\displaystyle 2x-1=3(x-1)$

$\displaystyle 2x-1=3x-3$

$\displaystyle \boxed{x=2}$

Check this solution in your original absolute value equation.

__Now consider case 2.__

$\displaystyle \frac{2x-1}{x-1}=-3$

$\displaystyle 2x-1=-3(x-1)$

$\displaystyle 2x-1=-3x+3$

$\displaystyle \boxed{x=\frac{4}{5}}$

Check this solution in your original absolute value equation.

- Sep 17th 2009, 06:31 AMPlato
- Sep 17th 2009, 07:50 AMHallsofIvy
You should, however, note that what you give is NOT an inequality, it is an equation.