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There are two parts to this question.
1. I'm seeing the correct method to solve these types of inequalities as something to do with "transition points". I don't quite understand this method. How do we find the specific inequalities, and how do we work with the boundaries once we set them?

  1. I personally thought of a more systematic approach: Simplify one absolute value expression (convert to positive/negative cases), then, for each of those cases, split the second expression to get a total of 4 cases. Then, I string together the inequalities that I get after solving them all, via union or intersection as the inequality sign marks.

Edit: I've noticed the mistake in my process. In the Case 2 line, I forgot to switch the inequality sign. Still, I'm wondering why this isn't a viable strategy for solving these types of inequalities. Why is this method inferior to the method described in (1)?

Example problem: |x-3|+|2x+5| > 6
|x-3| > 6-|2x+5|
Case 1: x-3 > 6-|2x+5|
|2x+5| > 9-x
Case 1a: 2x+5 > 9-x, 3x > 4, x > 4/3
Case 1b: 2x+5 < x-9, x < -14
Case 2: x-3 > |2x+5|-6
|2x+5| < x+3
Case 2a: 2x+5 < x+3, x < -2,
Case 2b: 2x+5 < -x-3, 3x < -8, x > -8/3

String together the 4 resulting inequalities:
((x < 4/3)U(x < -14))U((x < -2)U(x > -8/3))
So my answer would be x < -2 U x > -8/3
This is wrong. The answer would actually be x < -8/3 U x > -2.

What did I do wrong?

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  • $\begingroup$ I didn't check the whole thing, but I notice one mistake in Case 2—the inequality sign is pointing the wrong way. $\endgroup$
    – Wildcard
    Oct 5, 2015 at 7:38
  • $\begingroup$ Indeed. I covered that in my edit. $\endgroup$
    – Brion Ye
    Oct 6, 2015 at 23:40
  • $\begingroup$ The systematic approach you describe in step 2 is totally workable, by the way. $\endgroup$
    – Wildcard
    Oct 7, 2015 at 6:18

2 Answers 2

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We need to rewrite the inequation without using the absolute value. Let's recall the definition of the absolute value: $$ \lvert x \rvert = \begin{cases} -x & \text{if $x < 0$,} \\ x & \text{if $x \geq 0$.} \end{cases} $$ The transition points are the solutions of the equations $x-3 = 0$ and $2x+5 = 0$. It can be covenient to use a table like this one.

$$ \begin{array}{r|cccccccc} x & -\infty & & -5/2 & & 3 & & +\infty\\ \text{sign of $x-3$} & & - & & - & & + & \\ \lvert x-3 \rvert & & -x+3 & & -x+3 & & x-3 & \\ \text{sign of $2x+5$} & & - & & + & & + & \\ \lvert 2x+5 \rvert & & -2x-5 & & 2x+5 & & 2x+5 & \\ \lvert x-3 \rvert + \lvert 2x+5 \rvert > 6 & & -3x-2 > 6 & & x+8 > 6 & & 3x+2 > 6 & \\ \text{Solutions} & & x<-8/3& & x>-2 & & x > 4/3 & & \\ \end{array} $$ The set of solutions of the inequation is $(-\infty, -8/3] \cup (-2,+\infty)$.

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HINT:

We know for real $y,$

$|y|=+y$ if $y\ge0$ else $|y|=-y$

Check for $x\ge3; -\dfrac52\le x<3; x< -\dfrac52$

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