The case $x < - 3$ in the absolute value equation $|x + 3| + |x - 2| = 5$ In the absolute value equation $|x + 3| + |x - 2| = 5$, why do we replace $|x + 3|$ by $-x - 3$ rather than $3 - x$ when $-\infty < x < -3$?


$$|x+3|+|x-2|=5$$
  What is the result set?
$$\begin{array}{c|c|c|c}
    & \hphantom{xxx}-3 & \hphantom{xxx-2} & 2\hphantom{-xxx} \\\hline
x+3 & - & + & + \\\hline
x-2 & - & - & +
\end{array}$$
My problem is in the case $-\infty<x<3$.
  $-\infty<x<3$ $\implies$ $-x-3-x-2=5$ $\implies$ $-2x-1=5$ $\implies$ $-2x=6$ $\implies$ $x=-3$
  As a basic rule in absolute value we can change $|x+3|$ to $3-x$ when $x<0$ and we want to remove the absolute value.
  So that why we didn't solve it like:
  $3-x+2-x=-5$ $\implies$ $-2x=0$ $\implies$ $x=0$ and $x$ is inacceptable 

 A: Reminders about absolute value:
$$|x|:=\begin{cases} x&\text{when}~x\geq 0\\ -x&\text{when}~x<0\end{cases}$$
For your specific question, finding the values of $x$ such that $|x+3|+|x-2|=5$, it behooves us to break into cases so that we may remove the absolute value signs.
$|x+3|+|x-2| = \begin{cases} x+3+x-2&\text{when}~x\geq 2\\
x+3-x+2&\text{when}~-3<x<2\\
-x-3-x+2&\text{when}~x\leq -3\end{cases}$
Setting the expression equal to five gives us for the first case, $x\geq 2$ and $2x+1=5\Rightarrow 2x=4\Rightarrow x=2$.  Checking, $x=2$ does agree with $x\geq 2$ so it is a solution.  Indeed $|x+3|+|x-2|=|2+3|+|2-2|=|5|+|0|=5$
For the second case, we get $x+3-x+2$ which is equal to five always regardless the value of $x$, so all values in the second case will work.  For example, $x=0.1$, we have $|0.1+3|+|0.1-2|=|3.1|+|-1.9|=3.1+1.9=5$
For the third case, we get $x\leq -3$ and $-x-3-x+2=5\Rightarrow -2x-1=5\Rightarrow -2x=6\Rightarrow x=-3$.  Checking, $x=-3$ does agree with $x\leq -3$ so it is also a solution.
Putting all of this together, we have the full solution set is the set of all numbers $x$ such that $-3\leq x\leq 2$  (or using interval notation, $[-3,2]$)
In your work, you say something along the lines of "When $x<0$ we can change $|3+x|$ to $|3-x|$."  That is not true.  Take $x=-1$ for counterexample.  $|3+(-1)|=|2|=2$ whereas $|3-(-1)|=|3+1|=|4|=4$.  Refer again to the definition of absolute value given above.  It says that if whatever is in the absolute value as a whole is negative, then change its sign.  You do not change the sign of single pieces of what is in it.
$$|3+x|=\begin{cases}3+x&\text{when}~3+x\geq 0\\ -(3+x)&\text{when}~3+x<0\end{cases}$$
