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Divide the real number line (as nothing is given regarding the domain, & complex domain need not be considered) into $5$ parts based on $4$ points:

(i) $x\lt -1:$ $$-(x+1) +x -3(x-1)+2(x-2) = x+2\implies -x -2 = x+2\implies x = -2$$ The cumulative equality is satisfied at point $x=-2$. Only the point $x=-2$ provides solution in this interval.
As indicated by @Green.H comment to his answer, would check value by substitution.
$1-2+3(3)-2(4) = 0\implies 0=0$, hence proved.

(ii) $-1\le x\lt 0$: $$(x+1) +x -3(x-1)+2(x-2) = x+2\implies x= x+2\implies 0 = 2$$ Could not understand its significance, may be no value satisfies this interval.

(iii) $0\le x \lt 1$: $$(x+1)-x -3(x-1)+2(x-2) = x+2\implies -x=x+2\implies x = -1$$ The cumulative equality is satisfied at point $x=-1$. But, $x=-1$ is outside the value range, so no value satisfies this interval as solution.

(iv) $1\le x \lt 2$: $$(x+1)-x +3(x-1) +2(x-2) = x+2\implies 5x -6 = x+2\implies x=2$$ The cumulative equality is satisfied at point $x=2$.. But, $x=2$ is outside the range, so again no value satisfies this interval

(v) $x\gt 2$: $$(x+1)-x+3(x-1) -2(x-2) = x+2\implies x+2 = x+2\implies 0=0 $$ So, it is a tautology, & hence all values in this interval satisfy it.

The solution set is given by : $x \in -2 \cup [2, \infty )$.
Need vetting, in particular the case (ii).

Martin Sleziak
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jiten
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4 Answers4

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$$|x+1|-|x|+3|x-1|-2|x-2|= x+2$$

\begin{array}{|c|c|c|c|c|c|c|} \hline \text{interval}&|x+1|&-|x|&3|x-1|&-2|x-2|&\text{sum}&\text{solution}\\ &&&&&&\text{set}\\ \hline (-\infty, -1] & \color{red}{-x-1} & x & -3x+3 & 2x-4 & -x-2 & \{-2\}\\ \hline (-1, 0] & x+1 & \color{red}{x} & -3x+3 & 2x-4 & x & \emptyset\\ \hline (0,1] & x+1 & -x & \color{red}{-3x+3} & 2x-4 & -x & \emptyset\\ \hline (1,2] & x+1 & -x & 3x-3 & \color{red}{2x-4} & 5x-6 & \{2\}\\ \hline (2,\infty) & x+1 & -x & 3x-3 & -2x+4 & x+2 & (2,\infty)\\ \hline \end{array}

ADDENDUM.

I thought that the table might be easier to read if I transposed it.

\begin{array}{r|ccccc} \text{interval} & (-\infty,-1] & (-1,0] & (0, 1] & (1,2] & (2,\infty) \\ \hline |x+1| & -x-1 & x+1 & x+1 & x+1 & x+1 \\ -|x| & x & x & -x & -x & -x \\ 3|x-1| & -3x+3 &-3x+3 & -3x+3 & 3x-3 & 3x-3 \\ -2|x-2| & 2x-4 & 2x-4 & 2x-4 & 2x-4 & -2x+4 \\ \hline \text{sum} & -x-2 & x & -x & 5x-6 & x + 2 \\ \hline \text{soln set} & \{-2\} & \emptyset & \emptyset & \{2\} & (2, \infty) \end{array}

Steven Alexis Gregory
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  • I know of $($ or $)$ to represent the 'open-ended' or exclusion of that part (left, or right, respectively) of interval; and $[$ or $]$ to represent the inclusion of that part (left, or right, respectively) of interval. But, may be am wrong, as you have taken the reverse meaning associations. It is causing confusion to me, as have taken the reverse approach to all prevalent & logical; say as at : https://math.stackexchange.com/a/506127/424260. I am stating 'logical' as lower end is stated to be part of interval, in approaching from left-to-right. This also is cause for difference in case (iv). – jiten Jun 02 '18 at 16:16
  • Also, as per your solution : for the interval $(-\infty, -1]$ have for point $x=-1$, the value for $|x+1|=0$, which is not less than zero. So, the term $|x+1|$ should not be multiplied by $-1$. This shows that your logic is wrong. – jiten Jun 03 '18 at 04:02
  • But, for absolute value terms the criteria to multiply by $-1$ is when the term is $\lt 0$ rather than $\le 0$. This part confuses, as being $=0$ is never a criteria to multiply by $-1$. In your last comment, it should be: "If $x\lt -1$ then $x+1\lt 0$. Hence $|x+1|=-(x+1)$." If you could prove by some way that for $|x-a|=b$, the cause for multiplying $(x-a)$ by $-1$ can be also $x-a=0$, apart from the usual condition of $x-a\lt 0$, then can understand your rule of $x-a\le 0$ being the criteria. – jiten Jun 03 '18 at 21:32
  • For the third time, it also works for $\le$. Forget the symbols and try it out with some actual numbers. – Steven Alexis Gregory Jun 03 '18 at 22:18
  • Even if it works, the theory is of the absolute value quantity to be $\lt 0$ for change of sign. So, would you offer alternate theory. I am in pedagogy as well as preparing for research, & the environment at both places is of zero tolerance. For me, every thing should substantiate; so emphasizing on theory. – jiten Jun 03 '18 at 22:42
  • Nonsense. You are restricting yourself unnecessarily. What is the value of $-0$? – Steven Alexis Gregory Jun 04 '18 at 01:54
  • It is the same as $0$, but how does it help is not clear. – jiten Jun 04 '18 at 02:25
  • It is the difference between $<$ and $\le$. – Steven Alexis Gregory Jun 04 '18 at 02:37
  • I will try to work through many problems, and bug you if get a different solution (as theoretically am not clear still, & the only option is to see practically). – jiten Jun 04 '18 at 02:42
1

(ii)

Due to the contradiction, there does not exist solutions for $-1\leq x<0.$

(iii)

Since $ x=-1$ does not fall into $0 \leq x <1$, it cannot be the solution.

(iv)

As $ x=2$ does not fall into $1 \leq x <2$, it cannot be the solution either.

(v)

The condition must be $x\geq2$.

Green.H
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  • I hope there is a value $x=-2$ satisfying the case (i). – jiten Jun 02 '18 at 10:42
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    As you indicated in OP, the solutions must be $x \in -2 \cup [2,\infty)$, which includes $-2$. Also you can easily check that $-2$ works just by substituting it in the original equation. – Green.H Jun 02 '18 at 10:47
  • Do you feel there is some error left in the edited OP. – jiten Jun 02 '18 at 10:52
0

In Case (ii) you have reached a contradiction, so there cannot be any solutions when $-1\leqslant x\leqslant 0$.

You also need to ignore the solution $x=-1$, because you get this in the case that $0\leqslant x\leqslant 1$, and $-1$ is not in this interval.

A. Goodier
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0

Acttually we get $$x=-2$$ or $$x\geq 2$$

Dr. Sonnhard Graubner
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