How do I evaluate $|x-1|+|x-2|-|x-3|<5$? $|x-1|+|x-2|-|x-3|<5$
How would I solve this inequality? I am trying to find all the values of $x$ that satisfy this inequality. I am not sure how I should go about doing this.
 A: Consider several cases:


*

*If $x<1$, your inequality becomes $1-x+2-x-3+x<5$.

*If $x\in[1,2)$, your inequality becomes $x-1+2-x-3+x<5$.

*If $x\in[2,3)$, your inequality becomes $x-1+x-2-3+x<5$.

*If $x\geqslant3$, …

A: Compare with Jose's answer. The graph is continuous and piecewise linear, and is allowed to change slope (have corners) only when $x=1,2,3$

A: When $x>3$, the equation behaves like $x-1+x-2-x+3=x$, which is less than 5 when $x<5$, when $x<1$, the equation behaves like $-(x-1)-(x-2)+(x-3)=-x$, which is less than x when $x>-5$. It is still possible however that the equation is greater than 5 between 1 and 3, to check if this is the case, consider each of the following cases, $1\ge x\ge2$ and $2\ge x\ge3$. In the first case, the equation becomes $(x-1)-(x-2)+(x-3)=x-2$ which is less than 5 on the interval 1 to 2. In the second case, the equation becomes $(x-1)+(x-2)+(x-3)=3x-6$ which is less than 5 on the interval 2 to 3, therefore, the equation $|x-1|+|x-2|-|x-3|$ is less than 5 for $-5<x<5$. 
