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Confused with flipping inequality signs.

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Confused with flipping inequality signs. [#permalink]

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New post 16 Oct 2017, 21:29
This is a question from Manhattan Book, and I am confused with one particular step in solving this problem:

If 4/x < -(1/3), what is the possible range of values for x?

We don't know if x is positive or negative, so we have to assume both scenarios.

x can't be positive though, since it will give us: 4 < -(x/3), which is not possible.

When x is negative, we go through the following steps:

1. Multiply both sides by 3, we will get:
12/x < -1

2. We have to multiply both sides by x to get rid of denominator. Here is the confusing part, since in this scenario x is negative, do we have to flip the inequality sign after the multiplication? The Rule says that we have to flip the sign when we multiply/divide both sides by a negative number, and since in this scenario x is a negative number, we have to get following inequality:

12 > x

So the range of x will be (-infinity ; 0) and (12 ; +infinity)

But, the in the Manhattan book, during step 2, they don't flip the sign when they multiply both sides by x, when assuming that x is negative. Why? is this some kind of exception?

They got the answer with explanation that I have attached to this post.
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Re: Confused with flipping inequality signs. [#permalink]

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New post 16 Oct 2017, 22:05
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Hi g1zmo,

1. Multiply both sides by 3, we will get:
12/x < -1

2. Multiply by x and flip the sign:

12 > -x

Drag x to the left side:
12 + x > 0

x > -12

So, x > -12, and, since x < 0, the range is (-12;0)

Check out this post also
https://gmatclub.com/forum/inequalities ... l#p1379270
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Re: Confused with flipping inequality signs.   [#permalink] 16 Oct 2017, 22:05
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