# Best strategy to solve absolut value inequality

Is there any best strategy to go with when solving inequalities involving absolute values?

Up until now I found three different methods, which work more or less for every example I have tried so far. I'm wondering though if these methods have any limitations to when they can be used and which I should generally go for.

Let's demonstrate those methods with the following example:

$$|x-1| \le |x+3|$$

1) Cases

We basically use the definition of the absolute value here.

|x+3|=\left\{ \begin{align} x+3 & \text{ , if }x\geq -3 \\ -(x+3) & \text{ , if }x <-3 \end{align} \right\}

|x-1|=\left\{ \begin{align} x-1 & \text{ , if }x\geq 1 \\ -(x-1) & \text{ , if }x < 1 \end{align} \right\}

Now we must distinguish the following cases:

For $x \lt -3$:

$$-(x-1) \le -(x+3)$$ $$\iff x+3 \le x-1 \iff 1 \le -3 \implies \text{ no solutions for } x \lt -3$$

For $-3 \le x \lt 1$:

$$-(x-1) \le (x+3)$$ $$\iff -2 \le 2x \iff x \ge -1$$

For $x \ge 1$:

$$x-1 \le x+3$$ $$\iff -1 \le 3 \implies \text{ all } x \ge 1 \text { are solutions }$$

So putting all cases together yields to the result:

$$x \ge -1$$

$$\\$$ 2) Squaring

We square each sides which doesn't change the inequality.

$$|x-1| \le |x+3| \iff (x-1)^2 \le (x+3)^2$$ $$\iff x^2 - 2x + 1 \le x^2 +6x +9 \iff -8 \le 8x \iff x \ge -1$$ $$\\$$ 3) "Less / Greater Than rules"

The rules I used here are:

• (i): $|a| \le b \Leftrightarrow -b \le a \le b$ (which can be written also as $-b \le a$ AND $a \le b$)
• (ii): $b \le |a| \Leftrightarrow b \le a$ OR $a \le -b$

Threat the right absolut value of the original inequality like b and proceed with (i):

$$x-1 \le |x+3| \text{ AND } -|x+3| \le x-1 \iff -(x-1) \le |x+3|$$

For the first inequality we can use (ii) to get the following results:

$$x-1 \le x+3 \iff -1 \le 3 \text{ (true statement) }$$ $$x+3 \le -(x-1) \iff 2x \le -2 \iff x \le -1$$

Now for the second inequality we use (ii) too:

$$-(x-1) \le x+3 \iff -2 \le 2x \iff x \ge -1$$ $$x+3 \le x-1 \iff 3 \le -1 \text{ (false statement) }$$

As we see from the rules, we now got a (a OR b) AND (c OR d) conjunction, so our result is

$$\text{ ( } true \text{ OR } x \le -1 \text{ ) AND ( } x \ge -1 \text{ OR } false \text{ ) }$$

Which reduces to:

$$true \text{ AND } x \ge -1 \text{ and gives again x \ge -1 }$$ $$\\$$

While the method with separating the cases should work every time (I guess?) it's sometimes confusing and not the fastest. Squaring can be quick (like here) but only works sometimes, it can often give you a difficult polynomial for which you need a calculator to plot it / get its roots. And for the "Less / Greater Than Rules", I'm still wondering when they can be applied and when not.

• Can I offer a more handy-wavy solution? Your inequality in words says that the distance between $x$ and $1$ is less than or equal to the distance between $x$ and $-3$. It's quite easy to imagine the solution $x \ge -1$ when you look at a number line. – dannum Jul 2 '17 at 19:35
• Thanks, but I just took this example to illustrate the methods. The inequality could be arbitrarily more complex and I wanted to know which to apply then. – philmcole Jul 2 '17 at 20:02
• Your 3rd method has currently a small flaw. When you apply $(ii)$ you implicitly assume by removing the $|\dots|$ that $x+3\geq 0$. This puts a restriction on the conclusion and you need to consider also the other option. If you do it correctly you will find the same result as before. As far as a best strategy is concerned, it is more what type of approach you prefer. – Ronald Blaak Jul 2 '17 at 20:08
• Yeah, I think I've made an error in my 3rd method. Which of the $|..|$ do you mean? But I don't really understand why I can't just apply the two laws like that. Would you mind editing my post and correcting it? – philmcole Jul 2 '17 at 20:30
• Sorry my mistake, you wanted to apply the logical method. Anyway, I corrected the last step $(true ~\text{OR}~ x \leq -1 ) = true$. – Ronald Blaak Jul 3 '17 at 15:14

Your greater than less than rules didn't really follow the form $-b \le a \le b$

To apply them properly you have

I) $-x- 3 \le x-1 \le x+3$ AND $x+3 \ge 0$

or

II) $x+3 \le x-1 \le -x -3$ AND $x+3 < 0$.

For I) $x - 1 \le x+3$ so redundant so we get

$-x-3 \le x-1 \implies -2 \le 2x \implies x \ge -1$ and $x+3 \ge 0\implies x \ge -3$ which is redundantly unnecessary.

So $x \ge -1$

For II) $x + 3 \le x-1$ is inconsinstant so this is impossible.

So $x \ge -1$.

I'd recommend this when applicable.

The only real distinction between this and "cases" is that the inconsistencies are ruled out at once.

I don't like "squares" as it leads to harder equations and potential extraneous extra incorrect solutions.

• Thanks a lot! The result make sense now! But can you explain why I need to make a distinction between $x+3 \ge 0$ and $\lt 0$ right at the start? In my original post I treat it like some b and forget about it until the next step, when I apply rule (ii) to it. Why is this not possible? – philmcole Jul 4 '17 at 7:43
• You could do it your way. I just thought it was too many steps and I could follow it at first. Your way also allows for a logical error. You had (true OR b) AND (false OR c). That means c AND maybe b/maybe not b. which means c. – fleablood Jul 4 '17 at 15:35