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# Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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29 Dec 2015, 23:12
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<x-3<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| [#permalink]

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18 Aug 2016, 05:20
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

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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Joined: 02 Sep 2009
Posts: 44599
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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18 Aug 2016, 07:31
MarkusKarl wrote:
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

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 non-absolute 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=37

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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22 Dec 2016, 19:55
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

Hope it helps.

Thanks Bunuel - this was such a quick & efficient way of solving this problem.
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Joined: 07 Mar 2017
Posts: 3
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 Dec 2017, 10:30
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

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, |x-3|=|2x-3|, 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
Math Expert
Joined: 02 Sep 2009
Posts: 44599
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 Dec 2017, 10:34
bapoon1991 wrote:
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

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, |x-3|=|2x-3|, 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 $$|x-3|=|x-3/2|$$ but it's $$|x-3|=2*|x-3/2|$$. So, the distance from x to 3 is TWICE the distance from x to 3/2.
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Joined: 07 Mar 2017
Posts: 3
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 Dec 2017, 10:41
Bunuel wrote:
bapoon1991 wrote:
Bunuel wrote:
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, 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) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.

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, |x-3|=|2x-3|, 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 $$|x-3|=|x-3/2|$$ but it's $$|x-3|=2*|x-3/2|$$. So, the distance from x to 3 is TWICE the distance from x to 3/2.

Thanks a lot for your super fast response! After your explanation, everything is clear now. So thanks again!

Lastly, I have my GMAT in the next 8 days, and I am facing a bit of an issue solving 700+ DS questions. Hence, it will be great if you can share with me any link, PDF which has a group of 700+ DS questions which I can practice before my Exam.

Thanks,
Dev
Math Expert
Joined: 02 Sep 2009
Posts: 44599
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 Dec 2017, 10:44
bapoon1991 wrote:
Thanks a lot for your super fast response! After your explanation, everything is clear now. So thanks again!

Lastly, I have my GMAT in the next 8 days, and I am facing a bit of an issue solving 700+ DS questions. Hence, it will be great if you can share with me any link, PDF which has a group of 700+ DS questions which I can practice before my Exam.

Thanks,
Dev

Check our Questions' Bank: https://gmatclub.com/forum/search.php?view=search_tags You can find many 700 level and sort them by category.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|   [#permalink] 12 Dec 2017, 10:44

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