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For absolute value equalities, say $|x-a|=b$, my approach is :
There are two cases, based on which the sign of the absolute value is taken.
Case 1: If $x-a\ge b$, then $x-a\ge 0$; so for given interval $x\ge a+b$, $x = a+b$.
Case 2: If $x-a\lt b$ then $x-a\lt 0$; so for given interval $x\lt a+b$, $x = a-b$.

But, have two posts on mse that contradict.
One is an answer to my post, and other is here.
For the first example, have reverse inclusion, i.e. left end is excluded; while the right end is included. This has yielded an extra solution point. However, substitution of the extra solution $x=2$ yields a correct value on substitution back, as follows:
$|x+1|-|x|+3|x-1|-2|x-2|= x+2$
$3-2+3= 4$
$4=4$, which is correct.
It is logical too, as in the next interval all of the points on real axis for which $x\ge 2$ are valid.
But, still it is a logical error as per me.


The other post, has error (as per me) in the second case where both left and right end are included.


Thanks to @SiongThyeGoh for pointing out errors in my solution construction.
Have not specified (i) $b\ge 0$,
(ii) Should specify for the two cases instead as :
Case 1: If $x-a\ge 0$, then $x-a=b\implies x = a+b$,
Case 2: If $x-a\lt 0$, then $x-a=-b\implies x = a-b$,

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For $|x-a|=b$ to have a solution, we need $b$ to be nonnegative and

$$|x-a|=b$$

the distance of $x$ from $a$ is $b$.

Hence, $$x=a \pm b.$$

Note that if $b<0$ then there is no solution.

Altenatively:

  • Consider $x \ge a$ and $x < a$.

  • If $x \ge a$, then $|x-a|=b$ becomes $x-a=b$ and we have $x=a+b$ which is at least as big as $a$ if $b \ge 0$.

  • If $x < a$, then $|x-a|=b$ becomes $a-x=b$ and we have $x=a-b$ which is a solution if $b>0$.

Remark for your working:

  • You have to mention that there is no solution if $b<0$.
  • Case $1$, If $x-a \ge b$, since $b \ge 0$, we then have $x-a \ge 0$, and hence $x-a=b$ and $x=a+b$.
  • Case $2$, if $x-a < b$, then $x-a=-b$, hence $x=a-b$. While $x<a+b$ is a true statement, it is not as clear to me how do you conclude $x=a-b$.(it is a true statement but your phrasing makes me uncertain whether you get the concept).

Comment for the other two posts:

  • I am guessing your confusion comes from when to include the end point, it doesn't matter as long as you consider every single number in the domain. If really uncertain, we can even consider those bondary points separately on their own.
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  • $\begingroup$ Thanks, but no indication as to which approach is correct: mine, or any of the other two. $\endgroup$ – jiten Jun 3 '18 at 4:45
  • $\begingroup$ I am looking at those posts and I can't figure out what do you mean by contradiction yet. $\endgroup$ – Siong Thye Goh Jun 3 '18 at 4:46
  • $\begingroup$ I am having two intervals based on the boundary point $x= a+b$ : (i) $x\lt a+b$, (ii) $x\ge a+b$. ( (i) refers to Case 2 of your answer, while (ii) refers to Case 1). Till now, $x$ denotes value on real number line. The next part is the final equality achieved, which for (i) $x=a-b$, while for (ii) $x=a+b$. Although, am not sure how to interpret the two values of $x$ obtained, should I say that as for (i) the interval was $x\lt a+b$, so the value of $x=a-b$ is correct; and for (ii) $x\ge a+b$, the result $x=a+b$ is also valid. $\endgroup$ – jiten Jun 3 '18 at 5:22
  • $\begingroup$ Sorry for botched up solution approach. I have edited the OP. $\endgroup$ – jiten Jun 3 '18 at 6:19
  • $\begingroup$ Kindly look at my post for an advanced question on absolute value equalities from the book : Elementary Mathematics, by - Dorofeev, Potapov, Rozov. My approach leads to a different solution than the book. The post is at: math.stackexchange.com/q/2807159/424260 $\endgroup$ – jiten Jun 4 '18 at 0:29

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