Q.The solution set of the following inequality $|x-1| \geq |x-3|$ is:

Definition. $|x|=x$ if $x\ge0$, and $|x|=-x$ if $x<0$ -- the distance of $x$ from $0$ on the number line. Consequently $|x|\ge0$ always, and $|x|=|-x|$.

Solving equations. $|u|=r$ (with $r\ge0$) splits into $u=r$ or $u=-r$; if $r<0$ there is no solution, since an absolute value can never be negative. $|u|=|v|$ splits into $u=v$ or $u=-v$.

Solving inequalities -- the two master rules:

proved by splitting into the cases $x\ge0$ and $x<0$. Shifted forms: $|x-a|\le r\iff x\in[a-r,a+r]$; $|x-a|\ge r\iff x\in(-\infty,a-r]\cup[a+r,\infty)$.

Algebraic identities. $|xy|=|x||y|$; $\left|\dfrac xy\right|=\dfrac{|x|}{|y|}$ ($y\ne0$); the triangle inequality $|x+y|\le|x|+|y|$; and if $|y+x|=|x-y|$ then $xy=0$. …