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Martin Sleziak
Martin Sleziak
8:55 AM
1
Q: Adding $2$ absolute values together: $|x+2| + |x-3| =5.$

CharlieI came across a very basic absolute value question $|x+2| + |x-3| =5.$ Initially, I thought the answer was $x=-2$ and $x=3$ because I let each absolute values be either positive and negative and that's what you get. But the correct answer was an inequality; $-2\le x \le 3.$ Now, If you try ...

algebra-precalculus absolute-value
1
Q: Solving $|x-2| + |x-5|=3$

meg_1997 Possible Duplicate: How could we solve $x$, in $|x+1|-|1-x|=2$? How should I solve: $|x-2| + |x-5|=3$ Please suggest a way that I could use in other problems of this genre too Any help to solve this problem would be greatly appreciated. Thank you,

algebra-precalculus absolute-value
The two questions are basically the same, they are of the form $|x-a|+|x-b|=b-a$, Should they be closed as duplicates?
The older one is closed as a duplicate of a question of the type $|x-a|-|x-b|=a-b$. (I.e., difference instead of sum.| I am not sure that these two should be considered duplicates.)
 
 
2 hours later…
Martin Sleziak
Martin Sleziak
10:30 AM
I mentioned these two questions also in the main chatroom:
in Mathematics, 2 mins ago, by Martin Sleziak
Adding $2$ absolute values together: $|x+2| + |x-3| =5$ and Solving $|x-2| + |x-5|=3$. The two questions are basically the same, they are of the form $|x-a|+|x-b|=b-a$, Should they be closed as duplicates?
in Mathematics, 2 mins ago, by Martin Sleziak
The older one is closed as a duplicate of a question of the type $|x-a|-|x-b|=a-b$. (I.e., difference instead of sum.) I am not sure that these two should be considered duplicates. (This remark is directed to @robjohn, who closed the question using mod-dupe-hammer.)
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
11:49 AM
I even get some response starting here:
in Mathematics, 8 mins ago, by robjohn
@MartinSleziak They are both solved by $|x-y|+|y-z|=|x-z|\iff y$ is between $x$ and $z$. I figured that a difference can be written as a sum, but if enough people think otherwise, it can be reopened. Can you not vote to reopen because I closed it?
I voted to close on the post I posted below.
I voted to reopen on the older question, since I don't think they are duplicates. Although robjohn think otherwise
in Mathematics, 12 mins ago, by robjohn
@MartinSleziak They are both solved by $|x-y|+|y-z|=|x-z|\iff y$ is between $x$ and $z$. I figured that a difference can be written as a sum, but if enough people think otherwise, it can be reopened. Can you not vote to reopen because I closed it?
 
 
2 hours later…
Martin Sleziak
Martin Sleziak
1:38 PM
It seems that the post already went through reopen review: math.stackexchange.com/review/reopen/258431
 

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