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Solve the inequality $$||y|-1|>2$$

From my trial and error, the answer is $y<-3$ or $y>3$. But I don't know any technical solution or working for this. Hope someone can provide a detailed solution. Thanks in advance.

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6 Answers 6

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Note that if $a>0$ then $\vert x\vert>a$ is equivalent to the statement that $x>a$ or $x<-a$. Therefore, $$\begin{align} &\vert\vert y\vert-1\vert >2 \\ \implies &\vert y\vert-1 > 2\,\,\,\, \mathrm{or}\,\,\,\,\vert y\vert-1 < -2\\ \implies &\vert y\vert > 3\,\,\,\, \mathrm{or}\,\,\,\,\vert y\vert < -1\\ \implies &y>3\,\,\,\, \mathrm{or}\,\,\,\, y<-3 \end{align}$$ Noting that $\vert y \vert < -1$ has no solutions.

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we will consider several cases: 1 $$y\geq 0$$ then we have $$|y-1|>2$$ 1.1 $$y\geq 1$$ then we get $$y>3$$ 1.2 $$y<1$$ then we get $$-y+1>2$$ or $$y<-1$$ 2. if $$y<0$$ then we get $$|-y-1|>2$$ or $$|y+1|>2$$ 2.1 $$y\geq -1$$ then we get $$y>1$$ 2.2 $$y<-1$$ then we get $$-y-1>2$$ or $$y<-3$$

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  • $\begingroup$ Its really big. Lol $\endgroup$
    – Mathxx
    Commented Jun 10, 2017 at 10:55
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    $\begingroup$ it was to big for you? $\endgroup$ Commented Jun 10, 2017 at 10:57
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If $a>0$, $|x|>a$ if and only if $x>a$ or $x<-a$; and $|x|<a$ if and only if $-a<x<a$.

So $||y|-1|>2$ if and only if $|y|<-1$ or $|y|>3$.

Note that it is impossible that $|y|<-1$. So the given inequality is equivalent to $|y|>3$. This gives the solution $y<-3$ or $y>3$.

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$$||y|-1|>2\implies$$

either $|y|-1>2\iff|y|>2+1\implies$ either $y>?$ or $y<?$

or $|y-1|<-2\iff|y|<-2+1<0$ which is untenable

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Graphical method: Draw the graph of $y=||x|-1|$:

enter image description here

Hence $||x|-1|>2 \Rightarrow x\in(-\infty,-3)\cup(3,+\infty).$

Additionally: If $||x|-1|>0.5$, then $x\in(-\infty,-1.5)\cup(-0.5,0.5)\cup(1.5,+\infty).$

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Not the quickest or easiet solution but I like it because it really shows off the link between polynomials and absolute value.

As both sides are positive you can square the equation to get $y^2-2|y|+1>4$ then rearrange it to get $y^2-3>2|y|$ then both sides are still positive and you can square it to recieve $$y^4-6y^2+9>4y^2$$$$y^4-10y^2+9>0$$$$(y^2-1)(y^2-9)>0$$$$(y-1)(y+1)(y-3)(y+3)>0$$$$x<-3,-1<x<1,x>3$$ only $x<-3$ and $x>3$ are true for the absolute value.

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