Why can I flip the reciprical?

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

Hello DeeptiManyaExpert

I am glad you liked my explanation

Here is another one

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

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

Furthermore, x cannot be -0.5 and therefore cannot be x+1=0.5

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.