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Solving Inequalities [#permalink]
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Updated on: 16 May 2013, 00:24
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Solving Inequalities I was going through the posts on inequalities and found that many good concepts are explained here, but still people do have trouble solving the question using these concept. In these posts, there were quadratic equations, curves, graphs and other mathematical stuff. With this post, I am trying to provide a simple method to solve such questions quickly. I won't be writing the concepts behind it. Remember this is the same OLD concept, it's just presented differently.Case 1: Multiplication for example: \((x1)(x2)(x3)(x7) \leq 0\) To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression  (+)1 ()2 (+)3 ()7 (+) Thus, inequality would hold true in the intervals: \(1 \leq x \leq 2\) \(3 \leq x \leq 7\), Note that intervals are inclusive of 3 & 7 Case 2: Division In case of division: \(\frac{(x1)(x2)}{(x3)(x7)} \leq 0\) Using the same approach as above; \(1 \leq x \leq 2\) \(3 < x < 7,\) \((x\neq3,\) \(7)\) Since (x3)(x7) is in denominator, its value can't be 0. Following things to be kept in mind while using above method:1. Cofficient of x should be positive: for ex  \((xa)(bx)>0\), can be written as \((xa)(xb)<0\) 2. Even powers: for ex  \((x9)^2(x+3) \geq 0\), \((x9)^2\) is always greater than 0, so, it should be only considered to check the equality (=0)3. Odd powers: \((xa)^3(xb)^5>0\), will be same as \((xa)(xb)>0\) 4. Cancelling the common terms: for ex  \(\frac {(x^2+x6)(x11)}{(x+3)} >0\), it can be simplified as (x2)(x11)>0 or, (+)2()11(+) thus \(x <2\) or \(x>11\), but since at x = 3 (in the original expression), we get undefined form, so, \(x
\neq 3\) A question for you:For what values of x, does the following inequality holds true? \((xa)(xb)...(xn)...(xz) \geq 0\), where {a, b, c,...} are integers. The expression has (xx), thus, it is always 0=0 for every value of x. Reference post: http://gmatclub.com/forum/inequalitiestrick91482.htmlPS: I hope you find this post useful, please provide feedback to improve the quality of the post. Thanks,
Originally posted by cyberjadugar on 19 Jun 2012, 03:34.
Last edited by cyberjadugar on 16 May 2013, 00:24, edited 1 time in total.



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Re: Solving Inequalities [#permalink]
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19 Jun 2012, 05:21
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This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.
Example for practice from OG13 PS229:
How many of the integers that satisfy the inequality \(\frac{{(x+2)(x+3)}}{{x2}}x\geq{0}\) are less than 5?
A 1 B 2 C 3 D 4 E 5



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Re: Solving Inequalities [#permalink]
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19 Jun 2012, 05:59
macjas wrote: This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.
Example for practice from OG13 PS229:
How many of the integers that satisfy the inequality \(\frac{{(x+2)(x+3)}}{{x2}}x\geq{0}\) are less than 5?
A 1 B 2 C 3 D 4 E 5 Hi, Check here: http://gmatclub.com/forum/howmanyoftheintegersthatsatisfytheinequalityx2x134194.html#p1094729You can get the range for x using the mentioned method. Regards,



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Re: Solving Inequalities [#permalink]
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19 Jun 2012, 06:09
cyberjadugar wrote: macjas wrote: This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.
Example for practice from OG13 PS229:
How many of the integers that satisfy the inequality \(\frac{{(x+2)(x+3)}}{{x2}}x\geq{0}\) are less than 5?
A 1 B 2 C 3 D 4 E 5 Hi, Check here: http://gmatclub.com/forum/howmanyoftheintegersthatsatisfytheinequalityx2x134194.html#p1094729You can get the range for x using the mentioned method. Regards, haha I know; I started that thread. I reposted this problem to link a real GMAT question to this technique to add value to this thread...



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Re: Solving Inequalities [#permalink]
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20 Jun 2012, 01:24
So that problem that you posted, let me give it a try:
A question for you: For what values of x, does the following inequality holds true? \((xa)(xb)...(xn)...(xz) \geq 0\), where {a, b, c,...} are integers.
Roots are: a,b,c,d...z The condition will hold true for these intervals: \(x\geq{z}\) \(x =x\) \({x}\leq{x}\leq{y}\) \({v}\leq{x\leq{w}\) \({s}\leq{x\leq{t}\) \({p}\leq{x\leq{t}\) \({m}\leq{x\leq{n}\) \({j}\leq{x\leq{k}\) \({g}\leq{x\leq{h}\) \({d}\leq{x\leq{e}\) \({a}\leq{x\leq{b}\)
Is this correct??



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Re: Solving Inequalities [#permalink]
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20 Jun 2012, 01:33
macjas wrote: So that problem that you posted, let me give it a try:
A question for you: For what values of x, does the following inequality holds true? \((xa)(xb)...(xn)...(xz) \geq 0\), where {a, b, c,...} are integers.
Roots are: a,b,c,d...z The condition will hold true for these intervals: \(x\geq{z}\) \(x =x\) \({x}\leq{x}\leq{y}\) \({v}\leq{x\leq{w}\) \({s}\leq{x\leq{t}\) \({p}\leq{x\leq{t}\) \({m}\leq{x\leq{n}\) \({j}\leq{x\leq{k}\) \({g}\leq{x\leq{h}\) \({d}\leq{x\leq{e}\) \({a}\leq{x\leq{b}\)
Is this correct?? Hi, You have identified everything, but give a close tought again. The answer is pretty much straight forward. Regards,



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Re: Solving Inequalities [#permalink]
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20 Jun 2012, 01:48
cyberjadugar wrote: macjas wrote: So that problem that you posted, let me give it a try:
A question for you: For what values of x, does the following inequality holds true? \((xa)(xb)...(xn)...(xz) \geq 0\), where {a, b, c,...} are integers.
Roots are: a,b,c,d...z The condition will hold true for these intervals: \(x\geq{z}\) \(x =x\) \({x}\leq{x}\leq{y}\) \({v}\leq{x\leq{w}\) \({s}\leq{x\leq{t}\) \({p}\leq{x\leq{t}\) \({m}\leq{x\leq{n}\) \({j}\leq{x\leq{k}\) \({g}\leq{x\leq{h}\) \({d}\leq{x\leq{e}\) \({a}\leq{x\leq{b}\)
Is this correct?? Hi, You have identified everything, but give a close tought again. The answer is pretty much straight forward. Regards, I have no idea... what am I missing here??



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Re: Solving Inequalities [#permalink]
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21 Jun 2012, 11:08
hey cj, looking forward to your Solving Set theory post...



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Re: Solving Inequalities [#permalink]
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25 Jun 2012, 05:05
nice one  and a great link .. inequalities kudos to cyberjadugar as well as gurpreet singh, veritas karishma... is the answer to macjas quest is  4 ( 2, 3, 3,4) ? kindly correct with explaination. and to the solution to cyberjadugar  is it as follows? x<=a = +ve a>=x>=b = ve b>=x>=n = +ve n>=x>=z = ve x>=z = +ve i may have done above wrong. pls do correct. and if the above is right , how the whole thing can be put together? like is it possible to write  z<=x<=a ot like z<=x<=a is +ve.



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Re: Solving Inequalities [#permalink]
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25 Jun 2012, 08:38
3,2,3,4 Therefore, there are 4 integers less than 5 that satisfy the inequality. Cheers, Der alte Fritz.
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Re: Solving Inequalities [#permalink]
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02 Nov 2012, 14:36
Edit: The 'x' before the inequality caused the confusion. I checked and that 'x' doe not exist. What about '0' and '4', '5' '6'...? 0<5, 4<5 ... macjas wrote: This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.
Example for practice from OG13 PS229:
How many of the integers that satisfy the inequality \(\frac{{(x+2)(x+3)}}{{x2}}x\geq{0}\) are less than 5?
A 1 B 2 C 3 D 4 E 5 OldFritz wrote: 3,2,3,4
Therefore, there are 4 integers less than 5 that satisfy the inequality.
Cheers, Der alte Fritz.



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Re: Solving Inequalities [#permalink]
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05 Nov 2012, 06:23
gpk wrote: Edit: The 'x' before the inequality caused the confusion. I checked and that 'x' doe not exist. What about '0' and '4', '5' '6'...? 0<5, 4<5 ... macjas wrote: This one is a really useful trick cyberjadugar. Kudos to you and gurpreetsingh.
Example for practice from OG13 PS229:
How many of the integers that satisfy the inequality \(\frac{{(x+2)(x+3)}}{{x2}}x\geq{0}\) are less than 5?
A 1 B 2 C 3 D 4 E 5 OldFritz wrote: 3,2,3,4
Therefore, there are 4 integers less than 5 that satisfy the inequality.
Cheers, Der alte Fritz. Yes, the extra 'x' is a typo. Of course, none of the options match if that x was still present in the inequality.
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Re: Solving Inequalities [#permalink]
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05 Nov 2012, 06:33
cyberjadugar wrote: 2. Even powers: for ex  \((x9)^2(x+3) \geq 0\), \((x9)^2\) is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: \((xa)^3(xb)^5>0\), will be same as \((xa)(xb)>0\)
Hello cyberjadugar, Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2).
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Re: Solving Inequalities [#permalink]
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20 Apr 2013, 20:16
closed271 wrote: cyberjadugar wrote: 2. Even powers: for ex  \((x9)^2(x+3) \geq 0\), \((x9)^2\) is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: \((xa)^3(xb)^5>0\), will be same as \((xa)(xb)>0\)
Hello cyberjadugar, Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2). Could some one please explain point 2 ??



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Re: Solving Inequalities [#permalink]
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15 May 2013, 23:55
shelrod007 wrote: closed271 wrote: cyberjadugar wrote: 2. Even powers: for ex  \((x9)^2(x+3) \geq 0\), \((x9)^2\) is always greater than 0, so, it should be only considered to check the equality (=0) 3. Odd powers: \((xa)^3(xb)^5>0\), will be same as \((xa)(xb)>0\)
Hello cyberjadugar, Could you illustrate (2) and (3) with examples. Though (3) is kind of clear, I do not understand what exactly you mean in (2). Could some one please explain point 2 ?? Hi, Sorry for the late reply. For even powers, such as \((x9)^2\), if you check for various values of x, for example, \(x = 1, (x9)^2 = 64(>0)\) \(x=1, (x9)^2 = 100(>0)\) \(x =10, (x9)^2=1(>0)\) but for \(x = 9, (x9)^2=0\) so, for every value of x, the even powers will always be greater or equal to 0, i.e. the sign of the expression doesn't change from positive to negative. Let me know if you need further clarification. Regards,



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Re: Solving Inequalities [#permalink]
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24 Jul 2013, 07:13
cyberjadugar wrote: Solving Inequalities I was going through the posts on inequalities and found that many good concepts are explained here, but still people do have trouble solving the question using these concept. In these posts, there were quadratic equations, curves, graphs and other mathematical stuff. With this post, I am trying to provide a simple method to solve such questions quickly. I won't be writing the concepts behind it. Remember this is the same OLD concept, it's just presented differently.Case 1: Multiplication for example: \((x1)(x2)(x3)(x7) \leq 0\) To check the intervals in which this inequality holds true, we need to pick only one value from the number line. Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression  (+)1 ()2 (+)3 ()7 (+) Thus, inequality would hold true in the intervals: \(1 \leq x \leq 2\) \(3 \leq x \leq 7\), Note that intervals are inclusive of 3 & 7 Case 2: Division In case of division: \(\frac{(x1)(x2)}{(x3)(x7)} \leq 0\) Using the same approach as above; \(1 \leq x \leq 2\) \(3 < x < 7,\) \((x\neq3,\) \(7)\) Since (x3)(x7) is in denominator, its value can't be 0. Following things to be kept in mind while using above method:1. Cofficient of x should be positive: for ex  \((xa)(bx)>0\), can be written as \((xa)(xb)<0\) 2. Even powers: for ex  \((x9)^2(x+3) \geq 0\), \((x9)^2\) is always greater than 0, so, it should be only considered to check the equality (=0)3. Odd powers: \((xa)^3(xb)^5>0\), will be same as \((xa)(xb)>0\) 4. Cancelling the common terms: for ex  \(\frac {(x^2+x6)(x11)}{(x+3)} >0\), it can be simplified as (x2)(x11)>0 or, (+)2()11(+) thus \(x <2\) or \(x>11\), but since at x = 3 (in the original expression), we get undefined form, so, \(x
\neq 3\) A question for you:For what values of x, does the following inequality holds true? \((xa)(xb)...(xn)...(xz) \geq 0\), where {a, b, c,...} are integers. The expression has (xx), thus, it is always 0=0 for every value of x. Reference post: http://gmatclub.com/forum/inequalitiestrick91482.htmlPS: I hope you find this post useful, please provide feedback to improve the quality of the post. Thanks, Hi, I have question with the below one, [list]for ex  \(\frac {(x^2+x6)(x11)}{(x+3)} >0\), it can be simplified as (x2)(x11)>0 or,  (+)2 ()11 (+) thus \(x <2\) or \(x>11\), but since at x = 3 (in the original expression), we get undefined form, so, [m]x using the plot of + , , how to identify x<2 or x>2 . I have a impression like if > is the used then it should be always x>2 & x>11 but the answer u have mentioned again confused me. Please help



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Re: Solving Inequalities [#permalink]
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24 Jul 2013, 07:17
Hi there, I wrote a post about how to choose the correct interval(s): tipsandtricksinequalities150873.html#p1211920. Hope it helps
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If 4<(7x)/3, which of the following must be true?
I. 5<x II. x+3>2 III. (x+5) is positive
A. II only B. III only C. I and II only D. II and III only E. I, II and III
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