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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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16 Apr 2014, 06:23
a. x+3 = 4x 3 x+3 = x + 3 for x + 3 >= 0 > x > 3. Solving for x, x + 3 = 4x 3 > 6 = 3x > x = 2 ( solution accepted since x > 3)  x + 3 = (x +3 ) for x + 3 < 0 > x < 3. Solving for x, x  3 = 4x 3 5x = 0 > x = 0 ( solution discarded as x is not < 3) A is sufficient. b. x 3 = 2x 3 x 3 = x 3 for x3 >= 0 > x >= 3 x 3 = (x 3) for x3 < 0 > x < 3 2x 3  = 2x 3 for 2x 3 >= 0 > x >=3/2 2x 3  = (2x 3) for 2x 3 < 0 > x < 3/2 So we have 3 ranges 1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution) Let me know if I have solved the question correctly. I know the process is lengthy. One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 May 2015, 10:56
vikram4689 wrote: dvinoth86 wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos Remember: When you have  on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have  on only one side you MUST VERIFY the answer by putting them back in eqn(1) x + 3 = 4x – 3 a) x + 3 = 4x – 3 => x=2 .. VALID b) (x + 3) = 4x – 3 => x=0 .. INVALID Hence Sufficient (2) x – 3 = 2x – 3 => x = 0 or 6 .. INVALID Hence InSufficient Your statement 1 has two sceneries. Then how does it become sufficient?
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 May 2015, 22:57
ranaazad wrote: vikram4689 wrote: dvinoth86 wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos Remember: When you have  on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have  on only one side you MUST VERIFY the answer by putting them back in eqn(1) x + 3 = 4x – 3 a) x + 3 = 4x – 3 => x=2 .. VALID b) (x + 3) = 4x – 3 => x=0 .. INVALID Hence Sufficient (2) x – 3 = 2x – 3 => x = 0 or 6 .. INVALID Hence InSufficient Your statement 1 has two sceneries. Then how does it become sufficient? Dear ranaazadThe solution quoted by you above has some tacit steps. I'll list them here: St. 1 says: x + 3 = 4x – 3 . . . (1) Case 1: x + 3 > = 0That is, x > = 3 In this case, x+3 = (x+3) So, Equation 1 becomes: x + 3 = 4x  3 => 6 = 3x => x = 2 Does this value of x satisfy the condition of Case 1, that x > = 3? Yes, it does. So, x = 2 is a valid value of x. Let's now consider Case 2: x + 3 < 0That is, x < 3 In this case, x+3 = (x+3) So, Equation 1 becomes: (x + 3) = 4x  3 => x  3 = 4x  3 => 0 = 5x => x = 0 Does this value of x satisfy the condition of Case 2, that x < 3? No, it doesn't. So, x = 0 is an INVALID value of x. Thus, we got only 1 valid value of X from Statement 1 (x = 2) and so, Statement 1 is sufficient to say that x > 0. Hope this answered your doubt! Japinder
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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16 May 2015, 21:37
Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. I got 0,2 in both statements. Reading your responses below, I would need to substitute to get A as the answer. However, if I input( x>=3/4 ) such as x=1 in statement 1, I get 1+3=4(RHS)4(1)3=1(LHS), any reason?



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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18 May 2015, 03:49
hak15 wrote: Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. I got 0,2 in both statements. Reading your responses below, I would need to substitute to get A as the answer. However, if I input( x>=3/4 ) such as x=1 in statement 1, I get 1+3=4(RHS)4(1)3=1(LHS), any reason? First of all, x=0 is not a solution of x+3=4x−3. It has only one solution x=2. Next, the question asks whether x>0 and from (1) we get that \(x\geq{\frac{3}{4}}\), so x must be greater than 0. So, the statement answers the question. \(x\geq{\frac{3}{4}}\) does NOT mean that any value greater than or equal to 3/4 will satisfy the equation, it means that the value(s) that satisfy it must be greater than or equal to 3/4. Hope it's clear.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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29 Dec 2015, 22:12
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution Is x > 0? (1) x + 3 < 4 (2) x  3 < 4 In case of inequality questions, it is important to note that conditions are sufficient if the range of the question includes the range of the conditions. There is 1 variable (x) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance that D is the answer. In case of the condition 1), we can obtain 4<x+3<4. This yields 7<x<1. Since the range of the question does not include the range of the condition, the condition 1) is not sufficient. In case of the condition 2), we can obtain 4<x3<4. This also yields, 1<x<7. Since the range of the question does not include the range of the condition, the condition 2) is not sufficient. Using both the condition 1) and 2), since 1<x<1, the range of the question does not include the range of the conditions. Therefore, the conditions are not sufficient and the correct answer is E. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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18 Aug 2016, 04:20
Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel! Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about? Best wishes
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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18 Aug 2016, 06:31
MarkusKarl wrote: Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel! Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about? Best wishes When there are absolute values on both sides you can safely square. When there is an absolute value on one side and nonabsolute value on another it becomes trickier  you should plug back the solutions to make sure that they are valid. You can search for similar questions here: search.php?search_id=tag&tag_id=37Hope it helps.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 Dec 2017, 09:30
Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, x3=2x3, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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12 Dec 2017, 09:34
bapoon1991 wrote: Bunuel wrote: Is x > 0?
(1) \(x+3=4x3\) > LHS is an absolute value, which is always nonnegative (\(some \ expression\geq{0}\)), so RHS must also be nonnegative > \(4x3\geq{0}\) > \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) \(x3=2x3\). Square both sides: \((x3)^2=(2x3)^2\) > \((2x3)^2(x3)^2=0\). Apply \(a^2b^2=(ab)(a+b)\), rather than squaring: > \(x(3x6)=0\) > \(x=0\) or \(x=2\). Not sufficient.
Answer: A.
Hope it helps. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, x3=2x3, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev You would have been right if it were \(x3=x3/2\) but it's \(x3=2*x3/2\). So, the distance from x to 3 is TWICE the distance from x to 3/2.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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15 Jul 2018, 05:37
Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution )[Did not understand, Why ve sign is here as the x range is positive]3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.
What am i missing here and how to solve these questions effectively Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). SOLUTION:We have two transition points for \(x3=2x3\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\) Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When \(x<\frac{3}{2}\), then \(x3\) is negative and \(2x3\) is negative too, thus \(x3=(x3)\) and \(2x3=(2x3)\). Therefore for this range \(x3=2x3\) transforms to \((x3)=(2x3)\) > \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)). 2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x3\) is negative and \(2x3\) is positive, thus \(x3=(x3)\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \((x3)=2x3\) > \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)). 3. When \(3<x\), then \(x3\) is positive and \(2x3\) is positive too, thus \(x3=x3\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \(x3=2x3\) > \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)). Thus \(x3=2x3\) has two solutions \(x=0\) and \(x=2\). Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/isx01x3 ... l#p1048512Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x  3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x  3, it will result in negative number, however if i plug in 3 in x  3, it will result in 0 or nonnegative, and therefore x  3 can be negative or nonnegative. In your post x  3 must be negative, how do we eliminate the nonnegative one ? Or do we simply ignore nonnegative ?



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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15 Jul 2018, 06:06
tixan wrote: Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...
1) x >=3 ==> x  3 = 2x 3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> (x3) = 2x 3 ====> x + 3 = 2x 3 ===> x =2 ( accepted solution )[Did not understand, Why ve sign is here as the x range is positive]3) x < 3/2 ===> (x3) = (2x 3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.
What am i missing here and how to solve these questions effectively Absolute value properties:When \(x \leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\); When \(x \geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). SOLUTION:We have two transition points for \(x3=2x3\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\) Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When \(x<\frac{3}{2}\), then \(x3\) is negative and \(2x3\) is negative too, thus \(x3=(x3)\) and \(2x3=(2x3)\). Therefore for this range \(x3=2x3\) transforms to \((x3)=(2x3)\) > \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)). 2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x3\) is negative and \(2x3\) is positive, thus \(x3=(x3)\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \((x3)=2x3\) > \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)). 3. When \(3<x\), then \(x3\) is positive and \(2x3\) is positive too, thus \(x3=x3\) and \(2x3=2x3\). Therefore for this range \(x3=2x3\) transforms to \(x3=2x3\) > \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)). Thus \(x3=2x3\) has two solutions \(x=0\) and \(x=2\). Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/isx01x3 ... l#p1048512Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x  3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x  3, it will result in negative number, however if i plug in 3 in x  3, it will result in 0 or nonnegative, and therefore x  3 can be negative or nonnegative. In your post x  3 must be negative, how do we eliminate the nonnegative one ? Or do we simply ignore nonnegative ? Yes, \({x}\leq{3}\), gives \(x  3 \leq 0\) but \(x3=(x3)\) is still true. That's because a = a when a <= 0. Meaning that if a is negative or 0, then a = a. For example, if say a = 1, then a = a = (1) = 1 and if say a = 0, then a = a = 0 = 0. Does this make sense?
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 00:32
dvinoth86 wrote: Is x > 0?
(1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
(1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 09:13
Mo2men wrote: dvinoth86 wrote: Is x > 0?
(1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
(1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x3 and 2x3) do NOT have to be nonnegative. One solution for Statement 2 is x=0, with the result that x3 = 3 and 2x3 = 3.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 12:08
GMATGuruNY wrote: Mo2men wrote: (1) x + 3 = 4x – 3RHS must be nonnegative....therefore \(4x3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive Sufficient Dear GMATGuruNY Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ? Thanks in advance Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x3 and 2x3) do NOT have to be nonnegative. One solution for Statement 2 is x=0, with the result that x3 = 3 and 2x3 = 3. Thanks GMATGuruNY. In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign?



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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03 Aug 2018, 22:07
Mo2men wrote: Thanks GMATGuruNY. In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign? Since the question stem asks whether x>0  and an equation with absolute value on both sides can have both a positive and a nonpositive solution  I recommend that you solve the equation, either by opening up the modulus or by squaring both sides.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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28 Nov 2018, 05:06
Bunuel does it mean that in all questions based on absolute values, one solution of it would be 0? I'm clear with Stat1 but in stat 2, I used the formula by squaring both sides of absolute figures to find the inequality range. Here is what I did. x−3=2x−3 Square both sides: (x−3)^2=(2x−3)^2 x^26x+9=4x^212x+9 3x^2=6x Therefore, x=2. So I chose option D. Can you help me to find what is the error? I'm not getting it. All I did was use the formula.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3
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28 Nov 2018, 05:38
topper97 wrote: Bunuel does it mean that in all questions based on absolute values, one solution of it would be 0? I'm clear with Stat1 but in stat 2, I used the formula by squaring both sides of absolute figures to find the inequality range. Here is what I did. x−3=2x−3 Square both sides: (x−3)^2=(2x−3)^2 x^26x+9=4x^212x+9 3x^2=6x Therefore, x=2. So I chose option D. Can you help me to find what is the error? I'm not getting it. All I did was use the formula. 1. Of course 0 is NOT a solution of all absolute value questions. Why would it be? 2. You solved 3x^2 = 6x incorrectly. You cannot reduce 3x^2 = 6x by x because x can be 0 and we cannot divide by 0. By doing so you loose a root, namely x = 0. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 &nbs
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