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hohaddict
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Why can I flip the reciprical? 09 Jul 2019, 23:18
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Can someone please let me know when I can flip and when I cannot flip a fraction in an inequality?
Typically, this is "unsafe" since the variable's sign is not known. However, in this problem, it appears necessary.
If not, is there another way to solve it? I tried doing the wavy line method after solving for two sides independently and trying to overlap, but it didn't work....

Problem:
If 3 ≤ 6/(x+1) ≤ 6, find the range of x

My approach:
Because I don't know the sign of the expression (x+1), I resorted to solving two inequalities and seeing if there was an overlap:
3-6/(x+1)≤ 0
and
0≤6-6/(x+1)

The source (below) appears to reference the need to memorize some rule...but there must be a better way than this.

Thank you!

Source: crackverbal
gmat-inequalities/
Section: D) Reciprocal Inequalities

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Why can I flip the reciprical? 09 Jul 2019, 23:55
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WE need not reciprocate for this question.

Since it is given that 3 ≤ 6/(x+1) ≤ 6

x has to be between 0 and 1.
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billionaire
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Why can I flip the reciprical? 11 Jul 2019, 13:42
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hohaddict
Can someone please let me know when I can flip and when I cannot flip a fraction in an inequality?
Typically, this is "unsafe" since the variable's sign is not known. However, in this problem, it appears necessary.
If not, is there another way to solve it? I tried doing the wavy line method after solving for two sides independently and trying to overlap, but it didn't work....

Problem:
If 3 ≤ 6/(x+1) ≤ 6, find the range of x

My approach:
Because I don't know the sign of the expression (x+1), I resorted to solving two inequalities and seeing if there was an overlap:
3-6/(x+1)≤ 0
and
0≤6-6/(x+1)

The source (below) appears to reference the need to memorize some rule...but there must be a better way than this.

Thank you!

Source: crackverbal
gmat-inequalities/
Section: D) Reciprocal Inequalities
Show more


Hello hohaddict

I really dont understand what is confusing here for you.

Why complicate things ?

3= 6/x+1

Gives solution of x=1

6/x+1 = 6

Gives solution of x=0

Therefore, range of x for requested inequality is x= [0...1]



You are welcome :cool:
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DeeptiManyaExpert
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Why can I flip the reciprical? 17 Jul 2019, 02:33
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hohaddict: Inequalities can be really baffling for many in the beginning. So, do not worry. You are not alone. I liked @billionaire's reply. I might have used the same on the test. But, I am going to give you another way of approaching the problem, in case you didn't understand Billionaire's solution or are looking forward to an alternate approach.

If 3 ≤ 6/(x+1) ≤ 6, find the range of x.

Looking at the problem it is clear that the middle term i.e. 6/(x+1) should be between 3 and 6. So, (x + 1) could be either 1 or 2.

x + 1 = 1 => x = 0
x + 2 = 2 => x = 1

So, range is 0 to 1.

However, you may get a question in your mind if 0.5 could be the answer. Please go ahead and check it if you any such question pops up in your mind. You are using Plugging in method. You can take a liberty to check the numbers which are outside the range. Try plugging denominator of 6/(x+1) i.e. X+1 = 0.5 or 1.5 in the original inequality. See for yourself if it satisfies the inequality.

6/0.5 = 12 No
6/1.5 = 4 No

Hence proved.

Please give kudos if you liked the solution. :)


Deepti Singh
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Manya - The Princeton Review
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Why can I flip the reciprical? 17 Jul 2019, 14:55
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DeeptiManyaExpert
hohaddict: Inequalities can be really baffling for many in the beginning. So, do not worry. You are not alone. I liked @billionaire's reply. I might have used the same on the test. But, I am going to give you another way of approaching the problem, in case you didn't understand Billionaire's solution or are looking forward to an alternate approach.

If 3 ≤ 6/(x+1) ≤ 6, find the range of x.

Looking at the problem it is clear that the middle term i.e. 6/(x+1) should be between 3 and 6. So, (x + 1) could be either 1 or 2.

x + 1 = 1 => x = 0
x + 2 = 2 => x = 1

So, range is 0 to 1.

However, you may get a question in your mind if 0.5 could be the answer. Please go ahead and check it if you any such question pops up in your mind. You are using Plugging in method. You can take a liberty to check the numbers which are outside the range. Try plugging denominator of 6/(x+1) i.e. X+1 = 0.5 or 1.5 in the original inequality. See for yourself if it satisfies the inequality.

6/0.5 = 12 No
6/1.5 = 4 No

Hence proved.

Please give kudos if you liked the solution. :)


Deepti Singh
Master Trainer - GMAT
Manya - The Princeton Review
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Hello DeeptiManyaExpert

I am glad you liked my explanation :cool:

Here is another one ;)

x=0.5
or
x+1=1.5 does satisfy inequality :idea:

In fact x can be any number between 0 and 1, both of them included.

when x=0.5

3 ≤ 6/(x+1) ≤ 6 = 3 ≤ 4 ≤ 6 :idea:

Furthermore, x cannot be -0.5 and therefore cannot be x+1=0.5 :idea:
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Why can I flip the reciprical? 26 Jul 2019, 08:06
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hohaddict
Can someone please let me know when I can flip and when I cannot flip a fraction in an inequality?
Typically, this is "unsafe" since the variable's sign is not known. However, in this problem, it appears necessary.
If not, is there another way to solve it? I tried doing the wavy line method after solving for two sides independently and trying to overlap, but it didn't work....

Problem:
If 3 ≤ 6/(x+1) ≤ 6, find the range of x

My approach:
Because I don't know the sign of the expression (x+1), I resorted to solving two inequalities and seeing if there was an overlap:
Show more

Pause for a moment and let your logical brain take a step back. Are you sure you don't know the sign of the expression (x+1)? The reason we worry about that is because if it's negative, there are problems with multiplying by it. But in this case, 6/(x+1) is definitely positive, since it's between two positive numbers. And the numerator, 6, is positive. So, for the whole fraction to be positive, x+1 must be positive as well. We can safely multiply by it - everything is positive here!

3(x+1) ≤ 6 ≤ 6(x+1)

3x + 3 ≤ 6 ≤ 6x + 6
3x ≤ 3 ≤ 6x + 3
x ≤ 1 ≤ 2x + 1

You can split that into two inequalities: x ≤ 1, and 1 ≤ 2x + 1, which simplifies to 0 ≤ x. Therefore, 0≤x≤1.

But, that isn't the most efficient way to solve this thing - I just wanted to show that it's possible and gives the right answer! What you probably want to do instead, to be quicker, is to notice that everything is a multiple of 3. We're dividing 6 by something, and the result comes out to be somewhere between 3 and 6. So, we divided 6 by 2 (at the most) or 1 (at the least). That means x+1 is between 1 and 2, so x is between 0 and 1.

Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!