Originally Posted by

**Krahl** your solution $\displaystyle 0 \leq 0 $ means that the equation is valid for all x.

notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get $\displaystyle -2\leq2$ or $\displaystyle 2\leq2$ because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

The way i would solve these is to solve the equations;

1. $\displaystyle -(x-3)-(x-1)\leq2$

2. $\displaystyle (x-3)-(x-1)\leq2$

3. $\displaystyle -(x-3)+(x-1)\leq2$

4. $\displaystyle (x-3)+(x-1)\leq2$

and take the union of the solutions. because if x solves any one of these then it solves the original inequality.