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8:55 AM
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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 ...

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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,

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…
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…
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…
1:38 PM
It seems that the post already went through reopen review: math.stackexchange.com/review/reopen/258431
 

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