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Re: What is the value of x? [#permalink]
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05 Jun 2013, 11:23
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WholeLottaLove wrote: Ok...I see that. But isn't the whole point to test cases where 63x is positive and negative? if X=2 then 63x is neither positive or negative, it's just zero. Yes, the main point is to understand where is +ve/ve but a function can also equal 0! Just consider the simpliest case possible \(x=0\) \(x>0\) is positive so \(x=0\) but 0 is not in the interval \(x<0\) is negative so \(x=0\) but once more 0 is not in the interval This seems to have no solution, but the solution is obvious: \(x=0\) So just add an = under \(x>0\) for example => \(x\geq{0}\) and everything will make sense again. I don't wanna confuse you, just add an = under one of the <,> signs in order to include all the possible values
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Re: What is the value of x? [#permalink]
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05 Jun 2013, 11:26
OK. Fair enough. Thanks! Zarrolou wrote: WholeLottaLove wrote: Ok...I see that. But isn't the whole point to test cases where 63x is positive and negative? if X=2 then 63x is neither positive or negative, it's just zero. Yes, the main point is to understand where is +ve/ve but a function can also equal 0! Just consider the simpliest case possible \(x=0\) \(x>0\) is positive so \(x=0\) but 0 is not in the interval \(x<0\) is negative so \(x=0\) but once more 0 is not in the interval This seems to have no solution, but the solution is obvious: \(x=0\) So just add an = under \(x>0\) for example => \(x\geq{0}\) and everything will make sense again. I don't wanna confuse you, just add an = under one of the <,> signs in order to include all the possible values



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Re: What is the value of x? [#permalink]
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06 Jun 2013, 00:51



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Re: Absolute Value [#permalink]
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10 Jun 2013, 12:53
Why did you not test positive and negative cases for 63x=x2 like you did for 5x+3=2x+9? Also, in 5x+3=2x+9 what is the significance of 9/2. I understand that it makes 2x+9 greater than/equal to zero, but why is that important? Bunuel wrote: Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps.



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Re: Absolute Value [#permalink]
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10 Jun 2013, 12:58
WholeLottaLove wrote: Why did you not test positive and negative cases for 63x=x2 like you did for 5x+3=2x+9? Also, in 5x+3=2x+9 what is the significance of 9/2. I understand that it makes 2x+9 greater than/equal to zero, but why is that important? Bunuel wrote: Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps. Because for this particular question we CAN solve with another approach shown in my solution. As for your other question check here: whatisthevalueofx95012.html#p1232524Similar question: whatisthevalueofx95012.htmlHope it helps.
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Re: Absolute Value [#permalink]
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10 Jun 2013, 13:27
I think I am misunderstanding what you are saying. Let me rephrase my question.
Why do you solve I. and II. differently? I guess the fact that they are done differently is throwing me off.
Thanks!
For 63x=x2, x2 must be greater than or equal to zero because it is equal to an absolute value so it must be positive. for x2 to be greater than or equal to zero, x must be greater than or equal to two.
When x is greater than or equal to two, 63x is negative, thus 63x = (63x) ==> 6+3x=x2 ==> 2x=4 ==> x=2
In this case, the result falls within the range of "x must be greater than or equal to two."
HOWEVER
5x+3=2x+9, 2x+9 must be greater than or equal to zero because it is equal to an absolute value. for 2x+9 to be greater than or equal to zero, x must be greater than or equal to 9/2.
When x is greater than or equal to 9/2 then 5x+3 is negative, thus 5x+3 = (5x+3) = 5x3 ==> 5x3=2x+9 ==> x=7/12
Here are my issues.
Firstly, it seems for the first problem you tested only the negative result (i.e. 6+3x) and I thought for these types of problem you are supposed to test positive and negative cases.
Secondly, for 5x+3=2x+9, if we do indeed test negative cases, we look for what would make 2x+9 greater than or equal to zero because it needs to be as it equals an absolute value. However, if we're looking at just numbers



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Re: Absolute Value [#permalink]
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10 Jun 2013, 13:30



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Re: Absolute Value [#permalink]
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10 Jun 2013, 17:19
Thanks. Let's say I solve I.) by doing the following: 63x=x2 When x≤2 then 63x is positive, thus: 63x=x2 x=2 When x>2 then 63x is negative, thus: 6+3x=x2 x=2 For the second case, x2 yet we found that it is negative only when x>2. Why wouldn't that fail? What am I missing here? Thanks! Bunuel wrote: WholeLottaLove wrote: I think I am misunderstanding what you are saying. Let me rephrase my question.
Why do you solve I. and II. differently? I guess the fact that they are done differently is throwing me off.
Thanks. Because for (1) I can use another approach, which is easier. This approach does not work for (2), so I use another for that one.



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Re: Absolute Value [#permalink]
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11 Jun 2013, 05:37
WholeLottaLove wrote: Thanks. Let's say I solve I.) by doing the following: 63x=x2 When x≤2 then 63x is positive, thus: 63x=x2 x=2 When x>2 then 63x is negative, thus: 6+3x=x2 x=2 For the second case, x2 yet we found that it is negative only when x>2. Why wouldn't that fail? What am I missing here? Thanks! Bunuel wrote: WholeLottaLove wrote: I think I am misunderstanding what you are saying. Let me rephrase my question.
Why do you solve I. and II. differently? I guess the fact that they are done differently is throwing me off.
Thanks. Because for (1) I can use another approach, which is easier. This approach does not work for (2), so I use another for that one. For the second case we consider the range x>2. x=2 it out of this range.
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Re: Absolute Value [#permalink]
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11 Jun 2013, 07:56
Exactly. If x=2 and is out of range then wouldn't that mean it fails and thus 1.) is insufficient?



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Re: Absolute Value [#permalink]
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Re: Absolute Value [#permalink]
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11 Jun 2013, 12:45
Exactly my point. If one fails doesn't that mean the answer is insufficient? Isn't that the reason II.) isn't sufficient because one answer is out of the range? Bunuel wrote: WholeLottaLove wrote: Exactly. If x=2 and is out of range then wouldn't that mean it fails and thus 1.) is insufficient? When x≤2 > x=2. Solution belongs to the range, thus OK.When x>2 > x=2. Solution does not belongs to the range, thus not OK.Solution from the first range is OK > x=2.



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Re: Absolute Value [#permalink]
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Re: Absolute Value [#permalink]
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11 Jun 2013, 13:41
Haha! I have been going over the basics (again) for a while now.
Tell me, does this way work:
63x=x2
For positive cases: 63x=x2 x=2 63(2)=22 0=0
For negative cases: 6+3x=x2 x=2 6+3(2)=22 0=0
5x+3=2x+9
For positive cases: 5x+3=2x+9 x=2 5(2)+3=2(2)+= 13=13
For negative cases: 5x3=2x+9 x=7/12 5(7/12)+3=2(7/12)+9 35/12 DOES NOT EQUAL 14/12+9
Therefore, both cases of x work for I. but only one case of x works for II.
I think I am greatly over complicating things. Thank you for your putting up with my thick head!



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Re: Absolute Value [#permalink]
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11 Jun 2013, 13:49
WholeLottaLove wrote: Haha! I have been going over the basics (again) for a while now.
Tell me, does this way work:
63x=x2
For positive cases: 63x=x2 x=2 63(2)=22 0=0
For negative cases: 6+3x=x2 x=2 6+3(2)=22 0=0
5x+3=2x+9
For positive cases: 5x+3=2x+9 x=2 5(2)+3=2(2)+= 13=13
For negative cases: 5x3=2x+9 x=7/12 5(7/12)+3=2(7/12)+9 35/12 DOES NOT EQUAL 14/12+9
Therefore, both cases of x work for I. but only one case of x works for II.
I think I am greatly over complicating things. Thank you for your putting up with my thick head! 5x3=2x+9 > x=12/7 not 7/12. If you plug x=12/7 into 5x+3=2x+9 it will hold true. Thus 5x+3=2x+9 has two solutions x=2 and x=12/7.
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Re: Absolute Value [#permalink]
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11 Jun 2013, 14:37
Thank you very much. At least my mistake was poor division/not logic! Bunuel wrote: WholeLottaLove wrote: Haha! I have been going over the basics (again) for a while now.
Tell me, does this way work:
63x=x2
For positive cases: 63x=x2 x=2 63(2)=22 0=0
For negative cases: 6+3x=x2 x=2 6+3(2)=22 0=0
5x+3=2x+9
For positive cases: 5x+3=2x+9 x=2 5(2)+3=2(2)+= 13=13
For negative cases: 5x3=2x+9 x=7/12 5(7/12)+3=2(7/12)+9 35/12 DOES NOT EQUAL 14/12+9
Therefore, both cases of x work for I. but only one case of x works for II.
I think I am greatly over complicating things. Thank you for your putting up with my thick head! 5x3=2x+9 > x=12/7 not 7/12. If you plug x=12/7 into 5x+3=2x+9 it will hold true. Thus 5x+3=2x+9 has two solutions x=2 and x=12/7.



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Re: Absolute Value [#permalink]
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19 Dec 2013, 05:36
Bunuel wrote: Hussain15 wrote: OA is "A".
My query::
6  3x = x  2
As a module is always positive, hence we can say that x2>0 If x2>0 then it implies x>2. But our answer is showing that x=2
What am I missing here?? Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)). Stepbystep solution: (1) \(6  3x = x  2\) > as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) > \(x2\geq{0}\) > \(x\geq{2}\). For \(x\geq{2}\), \(6  3x=6+3x\), so \(6  3x = x  2\) becomes \(6+3x=x2\) > \(x=2\). Sufficient. (2) \(5x + 3 = 2x + 9\). (note that by the same logic as above \(x\geq\frac{9}{2}=4.5\)) One check point \(x=\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero). \(x\leq{\frac{3}{5}}\) > \(5x3=2x+9\) > \(x=\frac{12}{7}\)  valid solution as \(\frac{9}{2}<\frac{12}{7}<\frac{3}{5}\); \(x>{\frac{3}{5}}\) > \(5x+3=2x+9\) > \(x=2\)  valid solution as \(\frac{9}{2}<\frac{3}{5}<2\); Two valid solutions. Not sufficient. Answer: A. Hope it helps. Hi Bunuel, Why did you take two checkpoints on Statement 2, and only 1 checkpoint in Statement 1? Just to expand a little bit on this. OK, from my understanding what you did in Statement 1 is a shortcut for not having to find the two cases. In statement 2, I supposed you went by the traditional approach where you need to test both cases, Is this correct? Now how do you decide when to use the shortcut in 1? I mean is there anyway you can tell whether the modulus will have only 1 solution or something? Cheers J



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Re: What is the value of x? [#permalink]
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11 Feb 2014, 13:09
I think the question here is, how do we know when we can just use the approach shown by Bunuel in statement 1, and when do we need to do both cases. I could also say in the second statement that since 5x+3>=0. then 2x+9>=0, x>=9/2. And then find that x=2 and it satisfies for this. How do I know whether I need to check the other solution too. Namely, when 5x+3<0?
I'm guessing it has something to do with the fact that on statement 1 we had 63x=x2, note that here the value of the range given as x>=2. will ALWAYS be positive therefore we only need to solve when 63x = 0, and x is in fact 2.
In the second we had x.=9.2, so it could be either positive or negative.
Is this the reason?
Thanks Cheers J



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Re: Absolute Value [#permalink]
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08 Apr 2014, 05:17
Bunuel wrote: WholeLottaLove wrote: I think I am misunderstanding what you are saying. Let me rephrase my question.
Why do you solve I. and II. differently? I guess the fact that they are done differently is throwing me off.
Thanks. Because for (1) I can use another approach, which is easier. This approach does not work for (2), so I use another for that one. Bunuel could you please explain what this 'other approach' used in statement 1? I don't quite get how can you know beforehand you will only get 1 valid solution for the modulus. Many thanks Cheers! J



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Re: What is the value of x? [#permalink]
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08 Apr 2014, 06:49
Hussain15 wrote: What is the value of x?
(1) 6  3x = x  2
(2) 5x + 3 = 2x + 9 Statement I is sufficient: 6  3x = x  2 then x = 2 6  3x = x + 2 then x = 2 Both ways we are getting x = 2 hence statement is sufficient Statement II is insufficient: 5x + 3 = 2x + 9 then x = 2 5x + 3 = 2x  9 then x will not be equal to 2 Getting two different answers hence answer is A
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