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drippydrop
Junior Member
USA
3 Posts |
 Posted - 10/08/2007 : 21:46:52
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3/(x-1)-4/x  1 |
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uberclay
Advanced Member
Canada
159 Posts |
Posted - 10/08/2007 : 23:04:49
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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
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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
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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
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| Thanks Tarragon. I forgot about stating restrictions. |
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tarragon
Advanced Member
Philippines
356 Posts |
Posted - 10/11/2007 : 03:15:11
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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]. 
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Edited by - tarragon on 10/11/2007 03:18:03 |
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