Animated Solution for Mathematics - Sets and Relations: Let $A=\{x:|x^2-10|\le 6\}$ and $B=\{x:|x-2|>1\}$. Then

First, we tackle set $A$, defined by the constraint $|x^2-10|\le 6$. In the language of mathematics, the absolute value $|u|\le a$ is a statement about distance, implying that the value of $u$ must reside within a distance of $a$ from the origin.

Thus, $|x^2-10|\le 6$ translates to the compound inequality:

By adding $10$ to every part of this inequality, we isolate the squared term:

We are looking for all real numbers $x$ whose square is trapped between $4$ and $16$. Taking the square root, we recall that $\sqrt{x^2}=|x|$, which yields:

This reveals the two symmetric regions of set $A$: $x\in [-4,-2]\cup [2,4]$. Imagine these as two islands on the number line, one in the negative territory and one in the positive.

Next, we turn our attention to set $B$, defined by $|x-2|>1$. Here, the distance from the point $2$ must be strictly greater than $1$.

This inequality splits the number line into two infinite rays:

1. $x-2>1\Rightarrow x>3$ 2. $x-2<-1\Rightarrow x<1$

Thus, our set $B$ is the union of two vast regions: $(-\infty ,1)\cup (3,\infty )$.

Now, we reach the climax of our journey: finding $B-A$. This is the set of all points that belong to $B$ but are strictly excluded from $A$, mathematically represented as $B\cap A^c$.

Think of it as taking the two infinite rays of $B$ and "punching holes" in them wherever $A$ exists.

Let us look at the first part of $B$, which is $(-\infty ,1)$. We must remove the portion of $A$ that overlaps with this. Since $A$ contains the interval $[-4,-2]$, which sits entirely within $(-\infty ,1)$, removing this interval splits the region into:

Now, let us look at the second part of $B$, which is $(3,\infty )$. We must remove the portion of $A$ that overlaps here. $A$ contains the interval $[2,4]$, which overlaps with $(3,\infty )$ specifically on the interval $(3,4]$.

When we remove this "chunk" from $(3,\infty )$, we are left with everything greater than $4$, which is $(4,\infty )$.

When we combine these fragments, we obtain the final set:

This result corresponds to the solution set for the given problem. By visualizing the number line and systematically carving out the forbidden regions, we have successfully navigated the logic of sets.