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.
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.