Solving $|x-2| + |x-5|=3$ 
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,
 A: Hint:
One characterization of "$x$ is between $a$ and $b$" is
$$
|x-a|+|x-b|=|a-b|
$$
A: 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 $2\leq x \leq 5$. 
