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 Inequalities
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drippydrop
Junior Member

USA
3 Posts

Posted - 10/08/2007 :  21:46:52  Show Profile
3/(x-1)-4/x 1
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/08/2007 :  23:04:49  Show Profile
Since you haven't indicated where you are having a problem, I'll assume it's from the get go.

Start by eliminating the denominators by multiplying through, first by x and then by (x-1).

3/(x-1) - 4/x 1

3x/(x-1) - 4 x

etc...

Once you've eliminated the denominators, rearrange it so all terms are on the same side. You'll be left with a quadratic inequality, which you can factor. Then solve for the interval/s for which the original statement is true.

Edited by - uberclay on 10/08/2007 23:11:11
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tarragon
Advanced Member

Philippines
356 Posts

Posted - 10/09/2007 :  12:04:51  Show Profile
drippydrop...

In addition to what uberclay said, you have to take note that x = 1 and x = 0 must not be included in your solution set.
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sahsjing
Advanced Member

USA
2399 Posts

Posted - 10/09/2007 :  17:31:47  Show Profile
quote:
Originally posted by drippydrop

3/(x-1)-4/x 1


Critical points: 0, 1
Multiply each term by x(x-1).
In (-infinity, 0),
x(x-1)> 0
3x - 4(x-1) x(x-1)
Then you get a general inequality to solve.

In the same way, try in (0, 1) and (1, infinity).
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/09/2007 :  19:47:01  Show Profile
Thanks Tarragon. I forgot about stating restrictions.
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tarragon
Advanced Member

Philippines
356 Posts

Posted - 10/11/2007 :  03:15:11  Show Profile
To Uberclay... you're welcome.

Hi... I tried solving the problem again.

The problem is a very tricky one and the solution to the inequality is not easy to find. Here’s how I have done.

3/(x-1)-4/x >=1

3/(x-1)-4/x – 1 >= 0

(Get the LCD) [3x – 4(x-1) –x(x-1)] / [x(x-1)] >= 0

[-x^2 + 4] / [x(x-1)] >= 0

(Factoring; multiplying both sides of inequality by -1)

[(x – 2)(x+2)] / [x(x-1)] <= 0

Now, we need to have values of x such that the rational expression given is at most 0. To satisfy this condition, we need to analyze two cases:

(1) (x – 2)(x+2) >= 0 AND x(x-1) < 0 OR

(2) (x – 2)(x+2) <= 0 AND x(x-1) > 0

One can verify that the solutions to these two conditions are as follows: (note that we first consider each of them separately... by the way, "inf" means infinity)

(1) [-2, 2] AND (-inf, 0) or (1,+inf) OR

(2) (0,1) AND (-inf, -2] or [2, +inf)

By graphing each solution separately, the solution (2) is actually not possible because there are no intersections. But by looking at solution (1), the “real” solution is [-2, 0) OR (1, 2]. By combining the results of my two solutions, the solution to the inequality problem is:


[-2, 0) OR (1, 2].



Edited by - tarragon on 10/11/2007 03:18:03
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