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Let the inequality equation be,
$||x|-5|<|x-6|$, the solution set for this I found to be $(-∞, 5.5) \cup (5.5, ∞)$
I got four different equations and two of them were no solutions, would this be a correct way to present the solution set? That is, should I have ended up with 4 equations? I'm very positive the algebra is correct but feel free to correct me if I am wrong.
By inspecting $||x|-5|<|x-6|$, we see there are five intervals of interest: \begin{align*}&x > 6\\6 > &x > 5\\ 5 > &x > 0\\ 0 > &x > -5\\ -5 > &x\end{align*} Before working with these, we can see that the inequality clearly holds for $x=-5, x= 0, x = 5$, but it does not hold for $x = 6$.
For $x > 6$, $||x|-5|<|x-6| \Leftrightarrow x - 5 < x - 6$, which is impossible. For $-5 < x < 5$, $||x|-5|<|x-6| \Leftrightarrow 5 - |x| < 6 - x$, which is clearly true. For $x < -5$, $||x|-5|<|x-6| \Leftrightarrow |x|-5 < 6 - x \Leftrightarrow |x|+x < 11$, which is clearly true as $|x|+x = 0$ for negative $x$.
Therefore we have established that $(-\infty, 5]$ is a subset of the solution set, and $[6, \infty)$ is a subset of the non-solution set. All that remains is the case where $5<x<6$.
If $5<x<6$, then $||x|-5|<|x-6| \Leftrightarrow x-5 < 6 - x \Leftrightarrow x+x < 11$. This holds when $x<5.5$. Hence we also know that $(5, 5.5)$ is part of the solution set, while $[5.5, 6)$ is not.
Therefore the set of solutions is $(-\infty, 5.5)$
To verify that this is truly all of the solutions, note that $\mathbb{R} = (–\infty, 5.5)\cup[5.5, \infty)$ where the first set has been shown to be consistent with the equation, and the second set has been shown to be inconsistent.
You can go step by step and use that (for $a>0$):
$$|y|<a\to -a<y<a \\|y|>a \to y<-a \text { or } y>a$$
So,
$$||x|-5|<|x-6|\to -|x-6|<|x|-5<|x-6|\to 5-|x-6|<|x|<5+|x-6|$$
$1.$ For the first case:
$$|x|<5+|x-6|\to -(5+|x-6|)<x<5+|x-6|$$
$1.1$ $$-(5+|x-6|)<x\to |x-6|>-x-5$$
so we have two cases: $$x-6<-(-x-5)\to-6<5 \to x\in \Bbb R$$
$$x-6>-x-5\to x>0.5$$
combining those two cases we get $x\in \Bbb R \quad (1)$
$1.2$
$$x<5+|x-6|\to |x-6|>x-5$$
again we have two cases:
$$x-6<-(x-5)\to x<5.5$$ $$x-6>x-5\to \text{ no solution}$$
combining those two cases we get $x< 5.5 \quad (2)$
Finally, from $(1)$ and $(2)$ we get $x<5.5$.
I will let the second case for you.
