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

Question

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).

Answer 1

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

\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}

Answer 2

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.

Answer 3

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

Answer 4

(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 $x\geq2$.