# Thread: solving an inequation...having difficulties

1. ## solving an inequation...having difficulties

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...

2. Originally Posted by solarscott
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...
$\frac{2x + 1}{2x - 1} = \frac{(2x - 1) + 2}{2x - 1} = 1 + \frac{2}{2x-1}$ so clearly the expression is never equal to 1. So your approaches to the problem are doomed.

However, if you draw a graph of $y = 1 + \frac{2}{2x-1}$ it's clear that the expression is greater than 1 for x > 1/2.

3. Originally Posted by solarscott
Hi All,

Just trying to solve:

(2x+1)/(2x-1)>=1...Not sure what to do now...by looking at the graph I think x>=1/2...but like i said, this is undefined...
This is the strangest thing...

But, any way, sometimes intuition proves to be invaluable.

Hint:

If both the numerator and the denominator are greater than zero, then the given quotient will be greater than one.

If both the numerator and the denominator are less than 0, then the given quotient will be less than one.

If both the numerator and the denominator are zero, then the given quotient is undefined.

You were on to something when you were thinking about x=1/2. Look at that again...

4. Thanks for the help guys. I appreciate it!

To Mr. Fantastic....this is a question for a maths test (well, its a test to test my maths knowledge to be a maths teacher), and Im sure the answer is something more than looking at the graph (which I have done, and see the answer) I just need to show it mathematically.

To VonNemo19....I sort of understand what you are saying...but not sure how it helps me...

I'm wondering if I should take a limit from the right hand side? ie
lim x-->1/2+ of (2x+1)/(2x-1)

to show that it is greater than 1?

5. Personally, I'd just keep things simple.

Asymptotes exist at y= 1 and x = 1/2
Hyperbola exists in Quadrant 1 and 4, relative to the asymptotes. Hence, for the graph to be greater 1 it must exist in Quadrant 1. In other words, x > 1/2.

6. Originally Posted by solarscott
Thanks for the help guys. I appreciate it!
I just need to show it mathematically.
In my post, I gave you the information necessary "to show it mathematically". You must keep in mind that logic plays a central role in mathematics. When I say:

$f(x)>1\Longleftrightarrow2x+1>{2}$, I have shown mathematically (math is just a language bounded by the rules of logic) that this must be so.

7. Here is how I would go about it

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

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

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

Now set your denominator to 0 and you get .5. Now test plug in a value below and above .5 and see where it makes the inequality true. You see that it is when x greater than or equal to .5. Hope that helps

8. Thanks everyone for the help! I may be on here a lot in the next week as my math teacher seems to have thrown in many 'twists' in his questions. haha. Thanks