Inequality with two absolute values

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SUMMARY

The discussion focuses on solving the inequality |x-3| > |x + 1|. Participants analyze the problem by breaking it into cases, leading to confusion regarding the correct manipulation of the inequality signs. Specifically, the second case was incorrectly handled, which resulted in misleading conclusions. The key takeaway is the importance of correctly applying the properties of absolute values and understanding their graphical representations to find the intersection points accurately.

PREREQUISITES
  • Understanding of absolute value inequalities
  • Familiarity with case analysis in algebra
  • Graphing techniques for visualizing inequalities
  • Basic knowledge of solving linear inequalities
NEXT STEPS
  • Study the properties of absolute values in inequalities
  • Learn how to graph absolute value functions and their intersections
  • Practice solving complex inequalities involving absolute values
  • Explore case analysis techniques in algebraic problem-solving
USEFUL FOR

Students studying algebra, particularly those tackling inequalities, educators teaching mathematical concepts, and anyone looking to improve their problem-solving skills in mathematics.

paech
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Homework Statement


Find all real values of x that satisfy the following inequality.

Homework Equations


|x-3| > |x + 1|



The Attempt at a Solution


Splitting up the inequality into cases I get:

1. |x-3| > x + 1 and 2. |x-3| < -x - 1


1. x-3 > x + 1 or x-3 < -x - 1

2. x-3 < x + 1 or x-3 > -x - 1

The solutions to these inequalities just don't make sense. I've done something wrong with splitting them up, but I'm not sure what.
 
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Well! That looks fine to me. I understand that one case gives you 'no real information' but the other one should tell you something about x = 1. my advice plot these two graphs and notice their relation. Secondly you can notice that one is a V shape and the other is an upside down V, and infer that they'll only intersect at 1 point.
 
paech said:

Homework Statement


Find all real values of x that satisfy the following inequality.

Homework Equations


|x-3| > |x + 1|

The Attempt at a Solution


Splitting up the inequality into cases I get:

1. |x-3| > x + 1 and 2. |x-3| < -x - 1
You should not have flipped the inequality sign around for item 2 above.

|x + 1| = x + 1 when x+1≥0, that is to say, when x ≥ -1 .

Similarly, |x + 1| = -x - 1 when x+1≤0, that is to say, when x ≤ -1 .

1. x-3 > x + 1 or x-3 < -x - 1

2. x-3 < x + 1 or x-3 > -x - 1

The solutions to these inequalities just don't make sense. I've done something wrong with splitting them up, but I'm not sure what.
 

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