Animated Solution for Mathematics - Sets and Relations: If $A=\{x\in R:|x|<2\}$ and $B=\{x\in R:|x-2|\ge 3\}$ then :

The real number line represents an infinite road stretching from $-\infty$ to $+\infty$. Our objective is to define two sets, $A$ and $B$, and determine the result of the set difference $B-A$.

The Neighborhood

Set $A$ is defined by the inequality $|x|<2$. Geometrically, $|x|$ represents the distance of a point $x$ from the origin $0$.

The condition $|x|<2$ identifies all points whose distance from the origin is strictly less than $2$. This creates a symmetric open interval centered at $0$.

The Repulsion

Set $B$ is defined by the inequality $|x-2|\ge 3$. This implies that the distance of $x$ from the point $2$ must be at least $3$.

This condition splits into two distinct linear inequalities: 1. $x-2\ge 3\Rightarrow x\ge 5$ 2. $x-2\le -3\Rightarrow x\le -1$

Thus, the set $B$ consists of two rays:

The operation $B-A$ requires us to take the set $B$ and remove any elements that are also contained within set $A$. We must identify the overlap between $B$ and $A=(-2,2)$.

The ray $(-\infty ,-1]$ overlaps with $A$ in the interval $(-2,-1]$. Removing this overlap from the ray $(-\infty ,-1]$ leaves us with the interval $(-\infty ,-2]$.

The second part of $B$, which is $[5,\infty )$, shares no common elements with $A$. Therefore, this portion remains entirely unchanged.

By combining the remaining segments, we arrive at the final set:

This result represents all real numbers except those in the open interval $(-2,5)$. We can express this elegantly as:

The final result is $(-\infty, -2] \cup