Inequality: x+3 < x-1 - What Does It Mean?

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Homework Help Overview

The discussion revolves around the inequality involving absolute values: |x+3| < |x-1|. Participants are attempting to understand the implications of this inequality and the reasoning behind the assertion that x is closer to -3 than to 1.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the interpretation of the inequality and the reasoning that leads to the conclusion about the distances of x from -3 and 1. There are attempts to clarify the meaning of the inequality and its implications for the values of x.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the original poster's understanding of the inequality. Some guidance has been offered regarding the evaluation of the inequality across different intervals, but there is no explicit consensus on the interpretation yet.

Contextual Notes

Participants have noted the need for the original poster to provide more context or work to facilitate assistance, highlighting the forum's emphasis on showing thought processes before seeking help.

stat643
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| x+3 | < |x-1|

The solution states that this inequality implies x is closer to -3 therefore the solution is -1... why?
 
Last edited:
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Huh?

You need to state the exact question, and show some work before we can help you.
 
i don't know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." I am just trying to see why this is the case ... and why x is closer to -3?
 
Last edited:
stat643 said:
i don't know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." I am just trying to see why this is the case .


Well as i can see you are trying to solve the following inequality, right?
THen here it goes.

[tex]|x+3|<|x-1|[/tex]

we divide this in the following intervals

[tex]1)x\in(-\infty,-3)...2)x\in[-3,1),...and...3) x\in[1,\infty )[/tex]

Then evaluate separately the expression[tex]|x+3|<|x-1|[/tex] in each interval. Pay heed to the sign of |x+3| and |x-1|, separately in each interval as you evaluate it. THen at the end you will be able to find the interval for which that inequality holds.

Regards,

P.S. SHow what are your thoughts on the problem first, forum rules! lol...
 
stat643 said:
i don't know how to solve it, the sol is "because x is closer to -3 than it is to 1 Hence every x for which x < −1 satisfies the inequality." I am just trying to see why this is the case ... and why x is closer to -3?


|x+3|=|x-(-3)| is the distance between the number "x" and the number (-3).
|x-1| is the distance between the number "x" and the number 1.

Thus, the inequality |x+3| less than |x-1| has as its solutions all those x's whose distances to (-3) is less than their distances to 1.
 
thanks
 

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