How to solve this multiple-absolute-value equation using three-region number line?

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I can solve it with combination of giving each absolute value a negative sign and leaving it as it is. There are four combinations. The method using regions in number line only requires three combinations instead of four. But it fails. So please help me to solve it using region-in-a number line method. (The answer should be {-6, -2/3})

  • $\begingroup$ There are four combinations Write those down, and you'll find that one of them is empty, which leaves only three to consider. $\endgroup$ – dxiv Jun 8 '18 at 21:04
  • $\begingroup$ how to know the one that is empty in quick way/at glance (without having to solve the equation)? even the -6 and -2/3 don't fit in the conditioned regions. $\endgroup$ – jonforall Jun 8 '18 at 21:28
  • $\begingroup$ $2x+4=0$ at $x=-2\,$, and $3-x=0$ at $x=3\,$, so the three intervals are $(-\infty, -2)\,$, $[-2,3]\,$, and $(3,\infty)\,$. On the first one, for example, $2x+4 \lt 0$ and $3-x \gt 0$ so the equation becomes $-(2x+4)-(3-x)=-1 \iff x = -6 \in (-\infty, -2)\,$. Repeat for the other two intervals. $\endgroup$ – dxiv Jun 8 '18 at 21:31
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    $\begingroup$ Now I understand.... thanks so much..... $\endgroup$ – jonforall Jun 8 '18 at 23:06
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    $\begingroup$ 'how to know the one [of four] that is empty in quick way/at glance' There are two critical points of $\operatorname{abs}$ function in the equation, and two points on a line (the real line, in this case) divide the line into three parts. Just find the points, sort them by value and you'll get three intervals. $\endgroup$ – CiaPan Jun 25 '18 at 11:49

Case 1: Let $x<-2$ therefore $$|2x+4|-|3-x|=-2x-4-(3-x)=-1$$which yields to valid answer $x=-6$

Case 2: Let $-2\le x\le3$ therefore $$|2x+4|-|3-x|=2x+4-(3-x)=-1$$which yields to invalid answer $x=-\dfrac{2}{3}$

Case 3: Let $x>3$ therefore $$|2x+4|-|3-x|=2x+4+(3-x)=-1$$which yields to invalid answer $x=-8$


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