Inequality with two absolute values

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The discussion focuses on solving the inequality |x-3| > |x + 1| by breaking it into cases. Participants note that the initial case breakdown may have errors, particularly in flipping the inequality signs. They emphasize the importance of understanding the conditions under which the absolute values change, specifically at x = -1. Graphing the two functions is suggested to visualize their intersection, which is expected to occur at a single point. The conversation highlights the need for careful case analysis in solving absolute value inequalities.
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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