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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,

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    [Robjohn argues](http://chat.stackexchange.com/transcript/message/17423082#17423082) that the two questions are the same. I think that there is a slight difference between $|x-1|-|x+1|=2$ and $|x-1|+|x+1|=2$, so I voted to reopen.2014-08-30

2 Answers 2

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Well, there are a couple of ways.

Method 1. By cases.

One is to consider cases: note that $\begin{align*} |x-2|&=\left\{\begin{array}{ll} x-2 & \text{if }x\geq 2\\ 2-x &\text{if }x\lt 2 \end{array}\right.\\ |x-5|&=\left\{\begin{array}{ll} x-5 &\text{if }x\geq 5\\ 5-x &\text{if }x\lt 5 \end{array}\right. \end{align*}$ So, you consider what happens if $x\geq 5$, if $2\leq x\lt 5$, and if $x\lt 2$. In the first case, you have $x-2+x-5 = 3$ which is the same as $2x-7 = 3$, or $2x=10$, or $x=5$. Since this satisfies $x\geq 5$, that is one solution.

If $2\leq x \lt 5$, then you get $x-2 + 5-x = 3$ which is always true. So all numbers between $2$ and $5$ work (check and see this is true).

And if $x\lt 2$, you get $2-x + 5-x = 3$ which is the same as $7-2x = 3$, or $2x=4$; that is, $x=2$. But $x=2$ does not satisfy $x\lt 2$, so there are no solutions here.

So the solution is that $x$ satisfies the equation if and only if $2\leq x\leq 5$.

Method 2. The absolute value is a measure of distance. $|x-2|$ is how far $x$ is from $2$, and $|x-5|$ is how far $x$ is from $5$. you are trying to find all numbers whose distance from $2$ plus their distance from $5$ equal $3$. Note that no number greater than $5$ can work, because then their distance to $2$ is already greater than $3$. No number smaller than $2$ can work because their distance to $5$ is already greater than $3$. And any number between $2$ and $5$, inclusively, will work, because if $2\leq x \leq 5$, then adding the distance from $x$ to $2$ and from $x$ to $5$ will necessarily add up to $3$. So the answer is that $x$ satisfies the equation if and only if $2\leq x \leq 5$.

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    but now i got even that!2012-06-04
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Hint:

One characterization of "$x$ is between $a$ and $b$" is $ |x-a|+|x-b|=|a-b| $