$||x-2|-3|>1$. I have made some cases but still the complete values don't come,plus I don't have any idea of how to sketch the graph for the lefy hand side of the inequality.
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$\begingroup$$$\left| \left| x-2 \right| -3 \right| > 1$$ is the same as $$ \left| x-2 \right| < 2 \text{ or } \left| x-2 \right| >4$$ which further reduces to $$0<x<4 \text{ or } x<-2 \text{ or } x>6$$ I think this will be enough to sketch the graph.
$\endgroup$ $\begingroup$Since there are no graphical method provided above, here I outline how to solve the solutions graphically.
The main technique to draw a graph involving absolute value, is to "flip" the part of curve below x-axis upwards. For instance, in this question, consider $y=x-2$. For $x<2$, $y<0$. So for the graph $y=|x-2|$, flip the $x<2$ part upwards and we obtain ($1$):
For the "$-3$" part, shift the whole $y=|x-2|$ graph downward for $3$ units.
Repeat step (1) to construct $y=||x-2|-3|$.
Draw a line $y=1$. The required domain should be the parts of curve above the line.
$\endgroup$ 3 $\begingroup$The inequality $||x-2|-3|>1$ has as an implication:
$$ |x-2|-3>1\text{ or }|x-2|-3<-1 \implies |x-2|>4\text{ or }|x-2|<2$$
This in turn implies:
$$ x-2>4\text{ or }x-2<-4\text{ or }-2<x-2<2 \implies x>6\text{ or }x<-2\text{ or }0<x<4$$
This means that $x \in (-\infty, -2)\cup(0,4)\cup(6,\infty)$.
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