Question about $|x+6|\leq 10$ Why does $|x+6|\leq 10$ in interval notation equal $[-16,4]$ and not $(-\infty,-16]\cup[4,\infty)$?
Also, can you please explain to me the difference between $\cup$ and $\cap$?
 A: \begin{align}
|x+6|\leq 10&\iff -10\leq x+6\leq 10\tag{by defn.}\\[0.5em]
&\iff -16\leq x\leq 4\tag{subtract $6$ throughout}\\[0.5em]
&\iff x\in[-16,4].
\end{align}
A: Think of $\cup$ as meaning "or" as in $A \cup B$ means all the elements in $A$ or $B$ or both. $\cap$ is like "and" as in $A \cap B$ means all the elements in both $A$ and $B$. For example, let
$A = \{ 1,2,3,4,5\}$ and $B = \{ 4,5,6,7 \}$.
Then $A \cup B = \{1,2,3,4,5,6,7\}$ and $A \cap B = \{ 4,5\}$.
As far as the inequality goes:
$$|x+6| \leq 10$$
$$-10 \leq x+6 \leq 10$$
$$-16 \leq x \leq 4$$
which means $x\in[ -16,4]$
A: It is clearest with the geometrical interpretation of the absolute value: $\lvert x+6\rvert$ is the distance from the point $x$ to $-6$. The inequation means this distance is no more than $10$, in one direction or the other. In numbers, this means 
$$-10\le x-(-6)\le 10\iff -6-10\le x\le -6+10.$$
A: PLEASE do not confuse the rules for the inequalities $|\text{stuff}| \geq a$ and $|\text{stuff}| \leq a$.
For $|\text{stuff}| \leq a$, this means: $$-a \leq \text{stuff} \leq a.$$ 
For $|\text{stuff}| \geq a$, this means:
$$\text{stuff} \geq a \text{  or } \text{stuff} \leq -a.$$
So, here our "stuff" is $x + 6$, and $a$ is $10$, and we have the first case, so that means we have:
$|x + 6| \leq 10$ means $-10 \leq x + 6 \leq 10$.
Finally, to "solve" for $x$, subtract $6$ from all sides to get:
$-10 - 6 \leq x + 6 - 6 \leq 10 - 6$
which gives us:
$-16 \leq x \leq 4$
and in interval notation, this stands for the interval $[-16, 4]$.
