By definition of absolute value, $|x|=a$ becomes $x=±a$, $a>0$
With $|x| < a$ however, why does this no longer happen as $x < ±a$ is no longer true, and that it instead becomes $-a < x < a$?
Also, I really need to understand how modulus function works, so could explanations be detailed? Thanks very much!
On
For real $x,y$, regarded as points on the number line, $$|x-y|=\text{the distance from $x$ to $y$ (or equivalently, from $y$ to $x$)}$$ As a special case, using $y=0$, $$|x| = \text{the distance from $x$ to $0$}$$
Now fix $a > 0$.
Using the distance interpretation of absolute value, \begin{align*} &|x| > a\\[3pt] \iff\;&\text{The distance from $x$ to $0$ is greater than $a$}\\[-1pt] &\text{(i.e., $x$ is more then $a$ away from the origin).}\\[3pt] \iff\;&x < -a\;\;\text{or}\;\;x > a\\[10pt] &\text{Analogously,}\\[4pt] &|x| < a\\[3pt] \iff\;&\text{The distance from $x$ to $0$ is less than $a$}\\[-1pt] &\text{(i.e., $x$ is less then $a$ away from the origin).}\\[3pt] \iff\;&-a < x < a\\[3pt] \end{align*}
Thus, the distance interpretation justifies the algebraic results.
On
I think the confusion arises due to the idea that for $a > 0$, $|x|=a$ "becomes" $x= \pm{a}$.
I think it will help to say, for $a > 0$, $|x|=a$ is true if and only if either $x= -a$ or $x = a$.
For $a>0$.
Informally, the inequality $|x|<a$ is true if and only if $x$ is closer to $0$ than the numbers whose modulus (absolute value) is $a$.
Better, the inequality $|x|<a$ is true if and only if $x$ is closer to $0$ than the solutions to $|x|=a$.
Finally, the inequality $|x|<a$ is true if and only if $-a<x<a$.
By definition: $$|x|=a, a>0 \Rightarrow \pm x=a \Rightarrow x= -a \ or \ x=a.$$ Similarly: $$|x|<a, a>0 \Rightarrow \pm x<a \Rightarrow -x<a \ and \ x<a \Rightarrow x>-a \ and \ x<a.$$ Also: $$|x|>a, a>0 \Rightarrow \pm x> a \Rightarrow -x>a \ or \ x>a \Rightarrow x<-a \ or \ x>a.$$ Alternatively, the inequalities can be squared to be solved by the method of intervals and to be explained why $and$ and $or$.