Inequality with two absolute values

[paech]

Homework Statement

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

Homework Equations

[itex]|x-3| > |x + 1|[/itex]

The Attempt at a Solution

Splitting up the inequality into cases I get:

1. [itex] |x-3| > x + 1[/itex] and 2. [itex]|x-3| < -x - 1[/itex]


1. [itex] x-3 > x + 1[/itex] or [itex]x-3 < -x - 1[/itex]

2. [itex] x-3 < x + 1[/itex] or [itex]x-3 > -x - 1[/itex]

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.

 

[MarneMath]

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.

 

[SammyS]

Homework Statement

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

Homework Equations

[itex]|x-3| > |x + 1|[/itex]

The Attempt at a Solution

Splitting up the inequality into cases I get:

1. [itex] |x-3| > x + 1[/itex] and 2. [itex]|x-3| < -x - 1[/itex]

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. [itex] x-3 > x + 1[/itex] or [itex]x-3 < -x - 1[/itex]

2. [itex] x-3 < x + 1[/itex] or [itex]x-3 > -x - 1[/itex]

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.