Solving modulus inequalities algebraically without the use of graphs The question is as follows: $|x+2|<|\frac12x-5|$
The book that I am using mentions that there are 2 useful properties that can be used when solving modulus inequalities, which are $|p|≤q ⇔ -q≤p≤q$ and $|p|≥q ⇔ p≤-q\;or\;p≥q$
However, this question involves 2 moduli which are on both sides. In another example, the book shows that you can use $|p|≥|q| ⇔ p^2≥q^2$
The part where I'm confused at is whether I should use $|p|≤q ⇔ -q≤p≤q$ or $|p|≥|q| ⇔ p^2≥q^2$ to solve this question and how to go forward from there.
 A: Whenever you encounter such dilemma, try both and see what problem you will encounter.
If you use the first method, in a single move, you get rid of one absolute value sign, there is still an absolute value left and there's more work to do. However for the purpose of exploration, you are still encouraged to try it.
If you do the second approach, you get rid of both absolute signs directly.
$$\left( \frac12 x - 5\right)^2-(x+2)^2>0$$
Now, you can use $a^2-b^2=(a-b)(a+b)$ to factorize and solve the problem.
A: $$|x+2|<|\frac12x-5|$$

When the relation is "simple", I like to replace the inequality symbol with an equals symbol. I should get zero, one or two points for a solution. If the inequality contains "=", then those points are solutions. Otherwise they are not.

*

*No points means that all x or no x will solve the inequality. Just test one point to see which it is.

*One or more points will break the number line into intervals. In each interval, all or none of the points will solve the inequality. Just test one point in each interval to see which it is.

In this case, we need to solve
$$|x+2| = |\frac12x-5|$$
I know of two ways to solve this.

*

*Break it into $x+2 = \frac12x-5$ and $x+2 = -(\frac12x-5)$ You get
$x \in \{2, -14\}$

*Square both sides. You get

\begin{align}
   x^2 + 4x + 4 &= \frac 14 x^2 - 5x + 25 \\
   4x^2 + 16x + 16 &= x^2 - 20x + 100 \\
   3x^2 + 36x - 84 &= 0 \\
   x^2 + 12x - 28 &= 0 \\
   (x-2)(x+14) &= 0 \\
   x &\in \{2, -14\}.
\end{align}
The numbers $2$ and $-14$ break the number line into two points and three intervals
$$(-\infty, -14) \cup \{-14\} \cup (-14, 2) \cup \{2\} \cup (2, \infty)$$
We will reject the points $x=-14$ and $x=2$ because they solve $|x+2|=|\frac12x-5|$.
So we test the interval $(-\infty, -14)$. Try to pick an "easy to use" number from this interval. I choose $x = -100$. Then $|x+2|<|\frac12x-5|$ becomes 98 < 45. This is FALSE, so we reject the whole interval.
Next we test the interval $(-14, 2)$. An easy point to test is $x=0$. Then $|x+2|<|\frac12x-5|$ becomes $2 < 5$ which is TRUE. So we accept thw whole interval $(-14, 2)$.
Finally we test the interval $(2, \infty)$. I pick $x=10$ for testing. $|x+2|<|\frac12x-5|$ becomes $12 < 0$ which is FALSE. So we reject the whole interval.
Hence the solution set is $(-14, 2)$.
