# How to solve inequalities with more than one absolute value expression

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?

• I didn't check the whole thing, but I notice one mistake in Case 2—the inequality sign is pointing the wrong way. – Wildcard Oct 5 '15 at 7:38
• Indeed. I covered that in my edit. – Brion Ye Oct 6 '15 at 23:40
• The systematic approach you describe in step 2 is totally workable, by the way. – Wildcard Oct 7 '15 at 6:18

We know for real $y,$
$|y|=+y$ if $y\ge0$ else $|y|=-y$
Check for $x\ge3; -\dfrac52\le x<3; x< -\dfrac52$
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 $(-\infty, -8/3] \cup (-2,+\infty)$.