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 September 11th, 2009, 04:39 AM #1 Newbie   Joined: Sep 2009 Posts: 4 Thanks: 0 solving (2x+1)/(2x-1)>=1 Hi All, Just trying to solve: (2x+1)/(2x-1)>=1 I try: 2x+1>=2x-1 0>=-2 (not the answer I want) I tried: (2x+1)(2x-1)/(2x-1)(2x-1)>=1 4x^2-1>=4x^2-4x+1 -2>=-4x x>=1/2 (can't be this, it is undefined) I also tried: (2x+1)(2x+1)/(2x-1)(2x+1)>=1 4x^2+4x+1>=4x^2-1 4x+1>=-1 4x>=-2 x>=-1/2 (but this answer doesnt work with the inequality...0>=1??) Not sure what to do now...by looking at the graph I think x>=1/2...but like i said, this is undefined...
 September 11th, 2009, 03:58 PM #2 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 670 Re: solving (2x+1)/(2x-1)>=1 You have to consider 2 cases separately x>1/2 and x<1/2. When multiplying by 2x-1 you will get 1 > -1 in the first case (true) and 1 < -1 in the second case (false). So the answer you are expecting x > 1/2 is correct.
 September 12th, 2009, 04:01 PM #3 Newbie   Joined: Sep 2009 Posts: 4 Thanks: 0 Re: solving (2x+1)/(2x-1)>=1 is that really the answer, even though it is undefined at x=1/2
September 13th, 2009, 03:24 AM   #4
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Quote:
 Originally Posted by solarscott Just trying to solve: (2x+1)/(2x-1)>=1 I try: 2x+1>=2x-1
When solving rational inequalities, you can not multiply through by the denominator!

Why? Because you do not know what the sign on the denominator is (it's a variable expression!), so you can't know whether or not to flip the inequality sign!

Instead, move everything over to one side of the inequality:

$\frac{2x\, +\, 1}{2x\, -\, 1}\, -\, 1\, \geq\, 0$

Convert to a common denominator:

$\frac{2x\, +\, 1}{2x\, -\, 1}\, -\, \frac{2x\, -\, 1}{2x\, -\, 1}\, \geq\, 0$

$\frac{(2x\, +\, 1)\, -\, (2x\, -\, 1)}{2x\, -\, 1}\, \geq\, 0$

$\frac{2}{2x\, -\, 1}\, \geq\, 0$

Now find the x-values which make for zeroes of the numerator and undefined values for the denominator, use these x-values to split up the number line, and then find the intervals on which the rational expression is positive....

 September 13th, 2009, 06:28 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,372 Thanks: 2008 Clearly, 2/(2x - 1) ? 0 ? (2x - 1)/2 > 0 (since equality isn't possible in the leftmost inequality), i.e., x - 1/2 > 0, i.e., x > 1/2.

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