# Write the solution set of the inequation

Question:

Write the solution set of the inequation |x − 1| ≥ |x − 3|.

Solution:

We have:

$|x-1| \geq|x-3|$

$\Rightarrow|x-1|-|x-3| \geq 0$

The LHS of the inequation has two seperate modulus. Equating these to zero, we obtain $x=1,3$ as critical points.

These points divide the real line in three regions, i.e $(-\infty, 1],[1,3],[3, \infty)$.

CASE 1: When $-\infty$\therefore|x-1|-|x-3| \geq 0\Rightarrow-(x-1)-[-(x-3)] \geq 0\Rightarrow-x+1+x-3 \geq 0\Rightarrow-2 \geq 0$But this is not possible. CASE 2 : When$1 \leq x \leq 3$, then$|x-1|=x-1$and$|x-3|=-(x-3)\therefore|x-1|-|x-3| \geq 0\Rightarrow x-1+x-3 \geq 0\Rightarrow \Rightarrow 2 x \geq 4\Rightarrow x \geq 2\Rightarrow x \in[2, \infty)$CASE 3 : When$3 \leq x<\infty$, then$|x-1|=x-1$and$|x-3|=x-3\therefore|x-1|-|x-3| \geq 0\Rightarrow x-1-x+3 \geq 0\Rightarrow 2 \geq 0$This is true. Hence, the solution to the given inequality comes from case$s 2$and 3 .$[2, \infty) \mathrm{U}[3, \infty)=[2, \infty)\therefore x \in[2, \infty)\$