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16 Oct 2017, 21:29
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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.
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.
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
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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
inequalities-tips-and-hints-175001.html#p1379270
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
inequalities-tips-and-hints-175001.html#p1379270
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!
