Author 
Topic 

drippydrop
Junior Member
USA
3 Posts 
Posted  10/08/2007 : 21:46:52

3/(x1)4/x 1 


uberclay
Advanced Member
Canada
159 Posts 
Posted  10/08/2007 : 23:04:49

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 (x1).
3/(x1)  4/x 1
3x/(x1)  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 


tarragon
Advanced Member
Philippines
356 Posts 
Posted  10/09/2007 : 12:04:51

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. 


sahsjing
Advanced Member
USA
2399 Posts 
Posted  10/09/2007 : 17:31:47

quote: Originally posted by drippydrop
3/(x1)4/x 1
Critical points: 0, 1 Multiply each term by x(x1). In (infinity, 0), x(x1)> 0 3x  4(x1) x(x1) Then you get a general inequality to solve.
In the same way, try in (0, 1) and (1, infinity). 


uberclay
Advanced Member
Canada
159 Posts 
Posted  10/09/2007 : 19:47:01

Thanks Tarragon. I forgot about stating restrictions. 


tarragon
Advanced Member
Philippines
356 Posts 
Posted  10/11/2007 : 03:15:11

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/(x1)4/x >=1
3/(x1)4/x – 1 >= 0
(Get the LCD) [3x – 4(x1) –x(x1)] / [x(x1)] >= 0
[x^2 + 4] / [x(x1)] >= 0
(Factoring; multiplying both sides of inequality by 1)
[(x – 2)(x+2)] / [x(x1)] <= 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(x1) < 0 OR
(2) (x – 2)(x+2) <= 0 AND x(x1) > 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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