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For what range of values of x will the inequality [#permalink]
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For what range of values of 'x' will the inequality 15x  (2/x) > 1? A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5
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Re: Range for variable x in a given inequality [#permalink]
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Re: Range for variable x in a given inequality [#permalink]
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fluke wrote: It's not a good question because it is not formed correctly or because it is subpar or both?
What if the question read, "Which of the following represents the complete range of x over which 15x  (2/x) > 1"? Then; D would be the only choice, right? Right. fluke wrote: More important question is: How x<1/3, 1/3<x<0, 0<x<2/5 and x>2/5; get reduced to 1/3<x<0, x>2/5.
Is there any comprehensive material available for these types of inequalities and ranges apart from the shortcut method suggested by gurpreetsingh? This is actually quite simple. We want to solve \(\frac{(3x+1)(5x2)}{x}>0\): 1. Find zeros of the multiples (the values of x for which x, 3x+1 and 5x2 equal to zero): 1/3, 2/5 and 0; 2. Arrange them in ascending order to get 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\); 3. Check extreme value: if \(x\) is some very large number (so some number from the 4th range) then \(\frac{(3x+1)(5x2)}{x}\) will obviously be positive; 4. Trick: as in the 4th range expression is positive then in 3rd it'll be negative, in 2nd it'l be positive again and finally in 1st it'll be negative:  +  +, so expression is positive for 2nd range: \(\frac{1}{3}<x<0\), and for fourth range: \(x>\frac{2}{5}\). This works for all such kind of inequalities. Links in from previous post: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html xyplane71492.html?hilit=solving%20quadratic#p841486One more: datasuffinequalities109078.html
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Re: For what range of values of x will the inequality [#permalink]
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19 Jan 2012, 15:37
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bhandariavi wrote: Why this way doesn't work? When X is +ve 15 x^2  2 > x OR 15x^2x2 >0 OR (5x2) (3x+1) >0 ==> x>2/5, x>1/3 ====> X>2/5 When X is ve (everything stays same only inequality sign changes) (5x2) (3x+1) <0 ==> x<2/5, x<1/3 ====> X<  1/3 So E. The red parts are not correct/ \(x>0\) > \((3x+1)(5x2)>0\) > roots are 1/3 and 2/5 > ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<\frac{1}{3}\) and \(x>\frac{2}{5}\) > since we consider \(x>0\) range then the solution for this range is: \(x>\frac{2}{5}\); \(x<0\) > \((3x+1)(5x2)>0\) > the same roots: 1/3 and 2/5 > "<" sign indicates that the solution lies between the roots: \(\frac{1}{3}<x<\frac{2}{5}\) > since we consider \(x<0\) range then the solution for this range is: \(\frac{1}{3}<x<0\). Thus \(15x\frac{2}{x}>1\) holds true for \(\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\). See my 1st and 2nd posts for alternate approach and the links there for theory on this kind of questions. Hope it helps.
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Re: Range for variable x in a given inequality [#permalink]
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Bunuel wrote: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) I've been looking at this for 20 minutes, but if someone could explain this particular step a little more thoroughly I would appreciate it. I get the rest of the problem. I'm not understanding why there is an x in the denominator.



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ohsballer wrote: Bunuel wrote: \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) I've been looking at this for 20 minutes, but if someone could explain this particular step a little more thoroughly I would appreciate it. I get the rest of the problem. I'm not understanding why there is an x in the denominator. Basically he moved the 1 over to the LHS, then used x as the common denominator for the LHS



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Re: Range for variable x in a given inequality [#permalink]
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Stiv wrote: Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. If someone could explain why we cannot multiply the whole equation with x? 15x  (2/x) > 1 when multiplied with x we get 15x^2  x  2 > 0 I see that is is wrong, but don't understand why... Why is x in the denominator important? Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.So you cannot multiply both parts of inequality 15x  (2/x) > 1 by x as you don't know the sign of this unknown: if x>0 you should write 15x^2  x  2 > 0 BUT if x<0 you should write 15x^2  x  2 < 0 (flip the sign). Hope it helps.
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Re: Range for variable x in a given inequality [#permalink]
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Stiv wrote: Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. If someone could explain why we cannot multiply the whole equation with x? 15x  (2/x) > 1 when multiplied with x we get 15x^2  x  2 > 0 I see that is is wrong, but don't understand why... Why is x in the denominator important? You can not multiply by x because you don't know about the sign of x. Take this example, x+\(\frac{1}{x}\)>2. If we multiply across by x, we get x^2+1>2x or > (x+1)^2>0. As the square of any real number is always positive, thus the solution of the given inequality : for all real values of x. But, we can see that for x=5, the inequality doesn't hold good. So where is the mistake? It is in the fact that we multiplied it by x, without knowing the sign of x. Instead , if we multiply by x^2 throughout, we get x^3+x>2x^2 > or x(x^2+2x+1)>0 > x(x+1)^2>0.Thus, as (x+1)^2 is always positive, this boils down to x>0. Thus the ACTUAL solution of the given inequality is : for any value of x>0. So, basically, even if you WANT to multiply, do it by multiplying by x^2, or x^4. Basically, any power that makes the factor that you are multiplying across positive.
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Re: For what range of values of x will the inequality [#permalink]
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18 Jun 2013, 10:00
WholeLottaLove wrote: I understand the way to solve this problem, but I'm still not sure as to why d. is correct and not a. Is is because d is a more "complete" range of correct values?
Thanks! The question asks you for the range of values for which the inequality holds. So basically, you need to give all values of x for which the inequality holds. Since 1/3 < x< 0 is also a part of this solution, (D) includes all value. (A) is only a part of this solution.
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Re: For what range of values of x will the inequality [#permalink]
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25 Jul 2016, 06:01
fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 Responding to a pm: Quote: Could you please advice me when to consider zero when solving such question of ranges that hold the inequality true. Some question you and Bunuel consider zero on the number line to define the range points becuase it is very important to know your approach of  +  + . if I ignore zero as a point so I will get this approach wrong. I hope you got my issue idea.
When x is a factor of the inequality, you need to consider 0 as a transition point. The idea is no different from the other transition points. On the right of 0, x is positive and on the left of 0, x is negative. If the inequality is: x * (x + 3) > 0, the transition points are 0 and 3. If the inequality is: (x  2)(x + 3) > 0, the transition points are 2 and 3. Does this help?
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Re: Range for variable x in a given inequality [#permalink]
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17 Feb 2011, 02:58
fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question.
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Re: Range for variable x in a given inequality [#permalink]
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17 Feb 2011, 03:23
Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
Not a good question. It's not a good question because it is not formed correctly or because it is subpar or both? What if the question read, "Which of the following represents the complete range of x over which 15x  (2/x) > 1"? Then; D would be the only choice, right? More important question is: How x<1/3, 1/3<x<0, 0<x<2/5 and x>2/5; get reduced to 1/3<x<0, x>2/5. Is there any comprehensive material available for these types of inequalities and ranges apart from the shortcut method suggested by gurpreetsingh?
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Re: Range for variable x in a given inequality [#permalink]
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17 Feb 2011, 13:18
can someone plz explain me how u go from 15x^2x2 into (3x+1)(5x2) can u show me how u do it fast? thanks.
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Re: Range for variable x in a given inequality [#permalink]
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17 Feb 2011, 19:40
144144 wrote: can someone plz explain me how u go from 15x^2x2 into (3x+1)(5x2)
can u show me how u do it fast?
thanks. You factorize by splitting the middle term. Check out a discussion on this concept: http://gmatclub.com/forum/hardfactoringquestion109006.html#p870223
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Re: For what range of values of x will the inequality [#permalink]
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19 Jan 2012, 15:15
Why this way doesn't work? When X is +ve 15 x^2  2 > x OR 15x^2x2 >0 OR (5x2) (3x+1) >0 ==> x>2/5, x>1/3 ====> X>2/5 When X is ve (everything stays same only inequality sign changes) (5x2) (3x+1) <0 ==> x<2/5, x<1/3 ====> X<  1/3 So E.
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Re: Range for variable x in a given inequality [#permalink]
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14 Oct 2012, 07:24
Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. Could anyone please explain why 0 is a root? I thought that if 0 was in the denominator then the result would be undefined?



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Re: Range for variable x in a given inequality [#permalink]
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14 Oct 2012, 08:06
arnivorous wrote: Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. Could anyone please explain why 0 is a root? I thought that if 0 was in the denominator then the result would be undefined? As shown by Bunuel, the range for which the inequality works is 1/3 < x < 0 and x > 2/5. Notice that x cannot be 0 here. 0 is not a root here. Note that the term 'root' is generally used for equations (the value which when substituted for the unknown satisfies the equation). When working with inequalities, you deal with the range of values that work. The inequality is undefined at 0 but it changes sign around it. So you include it while working on finding the ranges where the expression is positive/negative. To understand how and why you do this, check out the inequalities trick link given by Bunuel above or my post: http://www.veritasprep.com/blog/2012/06 ... efactors/
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Re: Range for variable x in a given inequality [#permalink]
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06 Apr 2013, 02:46
Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. If someone could explain why we cannot multiply the whole equation with x? 15x  (2/x) > 1 when multiplied with x we get 15x^2  x  2 > 0 I see that is is wrong, but don't understand why... Why is x in the denominator important?
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Re: Range for variable x in a given inequality [#permalink]
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10 May 2013, 04:17
Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. Hi Banuel, When i factorsie 15x^2  X2 =0 i got only only 2 roots : 1/3 and 2/5. ( I cross multiplied the denominator x so it become "0" on the RHS.) So i got only 3 intervals. How did you got 3 roots , especially 0 as a root. please provide me the insight and tell me where did i went wrong. By the way thank you very much for your links to learn inequalities concepts. It's very useful. Thanks in advance. Regards
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Re: Range for variable x in a given inequality [#permalink]
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10 May 2013, 05:50
kabilank87 wrote: Bunuel wrote: fluke wrote: For what range of values of 'x' will the inequality 15x  (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. 1/3 < x < 0.4, x > 15/2 D. 1/3 < x < 0, x > 2/5 E. x < 1/3 and x > 2/5 \(15x\frac{2}{x}>1\) > \(\frac{15x^2x2}{x}>0\) > \(\frac{(3x+1)(5x2)}{x}>0\) >3 expressions change their signs at: 1/3, 0, 2/5, so 4 ranges: \(x<\frac{1}{3}\), \(\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) > in rightmost range expression is positive: so ++ > \(\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\). Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x\frac{2}{x}>1\) holds true than the answer could be D as well as A. P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero. Check for more about the approach used here: everythingislessthanzero108884.html?hilit=extreme#p868863, here: inequalitiestrick91482.html and here: xyplane71492.html?hilit=solving%20quadratic#p841486Not a good question. Hi Banuel, When i factorsie 15x^2  X2 =0 i got only only 2 roots : 1/3 and 2/5. ( I cross multiplied the denominator x so it become "0" on the RHS.) So i got only 3 intervals. How did you got 3 roots , especially 0 as a root. please provide me the insight and tell me where did i went wrong. By the way thank you very much for your links to learn inequalities concepts. It's very useful. Thanks in advance. Regards Check here: rangeforvariablexinagiveninequality109468.html#p1131289 and here: rangeforvariablexinagiveninequality109468.html#p1208802Hope it helps.
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Re: Range for variable x in a given inequality
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