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drippydrop
Junior Member

USA
3 Posts

 Posted - 10/08/2007 :  21:46:52 3/(x-1)-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 (x-1).3/(x-1) - 4/x 13x/(x-1) - 4 xetc...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 drippydrop3/(x-1)-4/x 1Critical points: 0, 1Multiply each term by x(x-1).In (-infinity, 0),x(x-1)> 03x - 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).

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/(x-1)-4/x >=13/(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)] <= 0Now, 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) > 0One 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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