Ok our equation is now : |x+3|+|x-2|= 5

And examining the following 4 cases we have:

1) $\displaystyle x+3\geq 0\Longrightarrow x=2$ which is accepted

2)$\displaystyle x+3\geq 0\wedge x-2<0\Longrightarrow 5=5$

3)$\displaystyle x+3<0\wedge x-2\geq 0\Longrightarrow -5=5$

4)$\displaystyle x+3<0\wedge x-2<0\Longrightarrow x = -3$ and since x+3<0 => x<-3 ,for x=-3 we have -3<-3

Hence the solutions of the equation are:

x=2 or 5=5 or -5=5 or -3<-3 .And since $\displaystyle -5\neq 5,\neg -3<-3$ we have :

x=2 or 5=5.

And the question is how do we get rid of 5=5