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Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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22 Feb 2012, 19:59
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Re: [tricky one] Is x > 0? [#permalink]
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
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).
Please discuss.
Regards



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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25 Feb 2012, 10:52
saxenaashi wrote: Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
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).
Please discuss.
Regards It seems that you understand this method very well. Everything is correct except the red parts: (x3) = (2x 3) > x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient. Hope it's clear.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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25 Feb 2012, 11:07
Bunuel wrote: saxenaashi wrote: Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)
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).
Please discuss.
Regards It seems that you understand this method very well. Everything is correct except the red parts: (x3) = (2x 3) > x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient. Hope it's clear. Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation. Regards



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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25 Feb 2012, 11:15
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saxenaashi wrote: Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation. Regards A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). So in statement (2) since both parts of expression are nonnegative we can safely apply squaring. Hope it helps.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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27 Feb 2012, 23:09
Well, I have a little bit another solution for this problem Consider (1): if x>3 then x+3=4x3, so x=2, sufficient if x<3, then x3=4x3, which Leeds to answer x=0, which is wrong because we consider only x<3 Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficient In your logic, you gave a mistake, because you've missed the root. Posted from GMAT ToolKit



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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27 Feb 2012, 23:30
terance wrote: Well, I have a little bit another solution for this problem Consider (1): if x>3 then x+3=4x3, so x=2, sufficient if x<3, then x3=4x3, which Leeds to answer x=0, which is wrong because we consider only x<3 Sum up (1) is sufficient The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient So the answer is D either (1) is sufficient or (2) is sufficient In your logic, you gave a mistake, because you've missed the root. Posted from GMAT ToolKitWelcome to GMAT Club. Unfortunately your answer is not correct: OA for this question is A, not D (you can see it under the spoiler in the initial post). (2) x – 3 = 2x – 3 has two roots x=0 and x=2 (just substitute these values to see that they both satisfy the given equation), so you can not get the single numerical value of x, which makes this statement insufficient. You can refer to above solutions for two different approaches of how to get these roots for (2). Please ask if anything remains unclear.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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15 Apr 2012, 15:38
A simple solution for this is solve for x in each statement, but remember when you solve for x with abs value you must make the term positive and negative. in both cases you will find x = +# and zero. plug the numbers back into the equation and you will find zero will not fit the equation so x must be > 0
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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17 Apr 2012, 06:19
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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



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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17 Apr 2012, 08:19
Is this a correct way to approach 2) x – 3 = 2x – 3 ? Step 1: (x3) = (2x3) => x = 0 Substitute x = 0 in x – 3 = 2x – 3 => VALID
Step 2: (x3) = (2x3) => x = 2 Substitute x = 2 in x – 3 = 2x – 3 => VALID
So 2) x – 3 = 2x – 3 is NOT SUFFICIENT



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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17 Apr 2012, 08:41
kirncha wrote: Is this a correct way to approach 2) x – 3 = 2x – 3 ? Step 1: (x3) = (2x3) => x = 0 Substitute x = 0 in x – 3 = 2x – 3 => VALID
Step 2: (x3) = (2x3) => x = 2 Substitute x = 2 in x – 3 = 2x – 3 => VALID
So 2) x – 3 = 2x – 3 is NOT SUFFICIENT Yes but you can skip the substitution part, read my statement above



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Re: [tricky one] Is x > 0? [#permalink]
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18 Apr 2012, 07:46
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. Pretty slick answer. Saves a lot of time.



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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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09 May 2013, 13:38
Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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10 May 2013, 02:01
nikhil007 wrote: Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device If you substitute x=0 in x + 3 = 4x – 3 you'll get: LHS=x + 3=3 and RHS=4x – 3=3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation. When expanding x+3: When x<3, then x+3=(x+3), so in this case we'll have (x+3)=4x3 > x=0 > discard this value since 0 is not less than 3 (we consider the range when x<3). When x>=3, then x+3=x+3, so in this case we'll have x+3=4x3 > x=2 > this value of x is OK since 2>3. So, x + 3 = 4x  3 has only one root, x=2. Hope it's clear.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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11 May 2013, 06:07
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Bunuel wrote: nikhil007 wrote: Hi Bunuel This is how I solved the 2 statements, http://campl.us/fWdgqiT5EsKand I get 0 & 2 for both statements, can u tell me what wrong am I doing? Posted from my mobile device If you substitute x=0 in x + 3 = 4x – 3 you'll get: LHS=x + 3=3 and RHS=4x – 3=3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation. When expanding x+3: When x<3, then x+3=(x+3), so in this case we'll have (x+3)=4x3 > x=0 > discard this value since 0 is not less than 3 (we consider the range when x<3). When x>=3, then x+3=x+3, so in this case we'll have x+3=4x3 > x=2 > this value of x is OK since 2>3. So, x + 3 = 4x  3 has only one root, x=2. Hope it's clear. Thanks Bunuel, this is how I solve the Modulus questions, so as a rule i guess, I should check both values by putting them back in eq, I use to think that they both are the roots since I considered both scenarios, ie one with X+3 and one with (x+3). I guess I will have to add this additional step in my methodology. Thanks.
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Re: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 [#permalink]
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03 Jun 2013, 23:42
dvinoth86 wrote: Is x > 0? (1) x + 3 = 4x – 3 (2) x – 3 = 2x – 3 Liked the question? encourage by giving kudos I generally think using graphs are faster for simple linear equations but you should be fast at plotting them as soon as you see the equations. For statement 2, if you plot the modulus graphs , you will see that one root is at x=0 and another root is some +ve x, 2 solutions hence insufficient i.e f(x) = x3 , f(0) = 3(y intercept) and x intercept is @ x=3 second f(x) = 2x 3 , f(0) = 3( y intercept is same as previous graph or solution 1) and x intercept is 3/2.



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Re: [tricky one] Is x > 0? [#permalink]
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04 Jun 2013, 07:56
Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! 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.



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Re: [tricky one] Is x > 0? [#permalink]
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04 Jun 2013, 08:00
WholeLottaLove wrote: Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! 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. One can do this way too, but the way shown in my post is faster.
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Re: [tricky one] Is x > 0? [#permalink]
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04 Jun 2013, 08:26
Hmmm... When I solve for the pos. and neg. values of x+3=4x3 I get: I. x+3=4x3 ==> 3x=6 ==> x=2II. x+3=(4x3) ==> x+3=4x+3 ==> 5x=0 ==> x=0So here in my presumably incorrect simplification, I have x=2 and x=0 in which case we can't be sure if x>0 For many abs. value questions it seems that you have to find the positive and negative cases each equation. I get that in this case, x+3=4x3 means that 4x3 is positive but why for other, similar questions, is solving for the positive and negative cases necessary? Bunuel wrote: WholeLottaLove wrote: Hi! Why do we not find the positive and negative values for x+3=4x3? Is it because this isn't a <, >, >= problem? Thanks! 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. One can do this way too, but the way shown in my post is faster.




Re: [tricky one] Is x > 0?
[#permalink]
04 Jun 2013, 08:26



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