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1/22/2020 12:12:56 PM [ARCHIVED THREAD] - Would Anyone Like to Help With Some Simple Math?
 Member   Joined Nov 2008 Posts 628 EE Offline OH, USA Posted: 9/17/2009 4:52:30 PM EST Ok, this is really stupid, but we can't figure this one out. We need to solve this inequality. 3/(x-1) - 4/x >= 1 So x =/= 1 or 0 We bring the 1 over so we have 3/(x-1) - 4/x - 1 >= 0 Then get a common denominator and simplify (x^2 -2x +4)/(x(x-1)) >= 0 What now? That numerator doesn't factor Member   Joined Mar 2006 Posts 771 EE Offline MA, USA  Posted: 9/17/2009 4:59:15 PM EST [Last Edit: 9/17/2009 5:04:28 PM EST by wheel] The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1): 3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1) 3x -4(x-1) >= x(x-1) 3x -4x + 4 >= x^2 - x 4 >= x^2 x <= 2 Gun control is not the answer. Gun control is the question. The answer is NO.
 Member   Joined Nov 2008 Posts 629 EE Offline OH, USA  Posted: 9/17/2009 5:06:32 PM EST Originally Posted By wheel: The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1): 3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1) 3x -4(x-1) >= x(x-1) 3x -4x + 4 >= x^2 - x 4 >= x^2 x <= 2 Maybe, I dunno. Our Professor made a whole big deal about not multiplying by the denominator today. Can't quite remember why though.
 Member   Joined Mar 2006 Posts 772 EE Offline MA, USA  Posted: 9/17/2009 5:10:24 PM EST [Last Edit: 9/17/2009 5:15:00 PM EST by wheel] Uhh..... it works. Try plugging in 2 for x, then try 1.9 for instance. ETA: actually I goofed on the last line: x^2 <= 4 -2 >= x <= 2 Gun control is not the answer. Gun control is the question. The answer is NO.
 ((CH3)2Hg)   Joined Sep 2005 Posts 3534 EE Offline OH, USA  Posted: 9/17/2009 5:20:06 PM EST [Last Edit: 9/17/2009 5:26:56 PM EST by Ryan1021] Edit again: This is what I got -2<=x<0 1
 Active Duty Jarhead   Joined Oct 2004 Posts 8378 EE Offline USA  Posted: 9/17/2009 5:42:27 PM EST [Last Edit: 9/17/2009 5:43:37 PM EST by nicholsmf] Simple? 1 + X = 2 is simple 3/(x-1) - 4/x >= 1 So x =/= 1 or 0 is not!
 Member   Joined Sep 2008 Posts 2135 EE Offline IN, USA  Posted: 9/17/2009 5:43:58 PM EST [Last Edit: 9/17/2009 5:45:42 PM EST by msprooch] Originally Posted By Feier: Originally Posted By wheel: The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1): 3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1) 3x -4(x-1) >= x(x-1) 3x -4x + 4 >= x^2 - x 4 >= x^2 x <= 2 Maybe, I dunno. Our Professor made a whole big deal about not multiplying by the denominator today. Can't quite remember why though. Hes an idiot. EDIT the only time you have to worry about multipling or dividing on an inequallity is when you are x or / by a negative number. Obama the answer to WTF?
 Prefiero vivir durmiendo   Joined Dec 2005 Posts 3752 EE Offline OH, USA  Posted: 9/17/2009 5:46:55 PM EST Originally Posted By nicholsmf: Simple? 1 + X = 2 is simple 3/(x-1) - 4/x >= 1 So x =/= 1 or 0 is not! compared to a triple integral, its is simple Does "vaporized" mean nothing happened to it?
 Brought to you by Carl's Jr.®   Joined May 2004 Posts 4654 EE Offline TX, USA  Posted: 9/17/2009 5:51:39 PM EST Originally Posted By nicholsmf: Simple? 1 + X = 2 is simple 3/(x-1) - 4/x >= 1 So x =/= 1 or 0 is not! Not dividing by 0 IS simple. "During the second 100 days, we will design, build and open a library dedicated to my first 100 days." -Barack Obama, May 9 2009
 Sarah Palin 2012   Joined May 2008 Posts 2264 EE Offline FL, USA  Posted: 9/17/2009 6:46:18 PM EST [Last Edit: 9/17/2009 6:56:57 PM EST by Mudfish] 1) Bring the 1 over: 3/(x-1) - 4/x - 1 >= 0. 2) Multiply by common denominator: 3x - 4(x-1) / x(x-1) - x(x-1) / x(x-1) >= 0. 3) Combine: 3x - 4x + 4 -x^2 + x / x(x-1) >= 0. 4) Combine: 4 - x^2 / x(x-1) >= 0. 5) Factor numerator: (2+x) (2-x) / x(x-1) >= 0. For simplicity, we will call the left side of this inequality f(x). 6) Note: f(x) can change signs only when there is a zero in the numerator or denominator. i.e. x = -2, 0, 1, 2. So we must check the intervals on both sides of these critical points. 7) Check interval x < -2: (plug in x = -3), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0. 8) Check interval -2 < x < 0 (plug in x = -1), f(x) > 0, so this IS part of the solution set for f(x) >= 0. 9) Check interval 0 < x < 1 (plug in x = 1/2), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0. 10) Check interval 1 < x < 2 (plug in x = 3/2), f(x) > 0, so this IS part of the solution set for f(x) >= 0. 11) Check interval x > 2 (plug in x = 3), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0. So our solution set is: { -2 <= x < 0, 1 < x <= 2 }. (Note that f(x) is undefined for x = 0 and x = 1). Death before dishonor."Ever notice how you come across somebody once in awhile that you shouldn't have fucked with? That's me."  Joined Apr 2007 Posts 61 EE Offline TN, USA  Posted: 9/17/2009 7:28:41 PM EST [Last Edit: 9/17/2009 7:31:23 PM EST by Meche_03] I believe the MathFu in this post is weak........... your teacher is correct as far as I know, you can not multiply by the denominator when working with INEQUALITIES You have to get each section to have the same denominator though .. x(x-1) first subtract one from both sides then multiply each section by a unit of one be it x/x, x-1/x-1, x(x-1)/x(x-1) 3x 4(x-1) x(x-1) >= 0 ––––- - ––––––––- - ––––––- x(x-1) x(x-1) x(x-1) then combine all over one denominator 3x - 4(x-1) - x(x-1) >= 0 ––––––––––––––––––––––––- x(x-1) simplifiy 3x - 4x +4 - x^2 + x >= 0 ––––––––––––––––––––––- x(x-1) simplify -x^2 + 4 >= 0 ––––––––––- x(x-1) factor (x-2)(-x-2) >= 0 –––––––––––––– x(x-1) solve each factor equal to zero, this gives your points where the line crosses zero and changes signs x= 2, -2, 0, 1 plot matrix for sign of each factor when between each point, then at the bottom of matrix multiply signs, ie -*-=+, -*+=-, +*+=+ -2 0 1 2 x - - + + + -x-2 + - - - - x-2 - - - - + x-1 - - - + + ––––––––––––––––––––––––––––––––- - + - + - only the region with positives are greater than zero as required above with the equation >= 0 thus -2 <= x < 0 and 1 < x <= 2 , x can not be = to 0 or 1 since they are the cross points found in the denominator [-2,0),(1,2] I think F* math equations on forums, the format gets screwed up damn, seconds to late Mudfish's MathFu is strong   Joined Feb 2009 Posts 210 EE Offline TN, USA  Posted: 9/17/2009 7:56:54 PM EST Originally Posted By nicholsmf: Simple? 1 + X = 2 is simple 3/(x-1) - 4/x >= 1 So x =/= 1 or 0 is not! It's damn simple compared to Triple Integrals, Complex Partial Differential Equations, and Fourier Series. [ARCHIVED THREAD] - Would Anyone Like to Help With Some Simple Math? Top Top