Originally Posted by
earboth 1. Use the definition of the absolute value:
$\displaystyle |x-3| = \left\{\begin{array}{lcr}x-3&if&x \geq 3\\-(x-3)&if&x < 3\end{array}\right.$
$\displaystyle |x-1| = \left\{\begin{array}{lcr}x-1&if&x \geq 1\\-(x-1)&if&x < 1\end{array}\right.$
2. Thus you get three different intervals. Now re-write the equation without absolute values:
$\displaystyle x < 1~:~\begin{aligned}-(x-3)-(-(x-1)) \leq 2\\-x+3+x-1\leq2\\2\leq2\end{aligned}$
$\displaystyle 1\leq x<3~:~\begin{aligned}-(x-3)-(x-1) \leq 2\\ -2x\leq 2\\x \geq 1\end{aligned}$
$\displaystyle x \geq 3~:~\begin{aligned}(x-3)-(x-1) \leq 2\\-3+1\leq2\\-2\leq2\end{aligned}$
Therefore the solution is: $\displaystyle x \in \mathbb{R}$