Interpretation of $|x-a|>c$ I am confused on what $|x-a|>c$ represents, where $x$ is a variable and $a$ and $c$ are constants. More precisely i want to know that when the modulus sign is removed how can we represent it? i.e. if it would have been$|x-a|<c$ i could have represented it as $-c<x-a<c$ but what about $|x-a|>c$
 A: The easiest way to think about statements of the form $|x-a|>c$ is to recall that
$$\text{$|a-b|$ represents the distance of $a$ from $b$}$$
assuming $a$ and $b$ denote real numbers.
Therefore, the statement $|x-a|>c$ literally means "the distance of $x$ from $a$ is more than $c$." Or, put more elegantly: $x$ is more than $c$ units away from $a$. What points are $c$ units away from $a$? Well, $a+c$ and $a-c$. So thinking of the statement this way immediately gives the solution: $x$ is either more than $a+c$ or less than $a-c$.
For a more difficult application of this idea, consider $|x^2-5x+6|>1$. You can use the above translation of $|a-b|$ to interpret this statement as saying "the distance of $x^2$ from $5x-6$ is more than 1," or, more elegantly, "$x^2$ is more than 1 unit away from $5x-6$." This might not sound very helpful, but it implies that either $x^2>5x-6+1$ or $x^2<5x-6-1$. You can then use other techniques (viz., factoring and sign analysis) to solve these quadratic inequalities.
And just to whet your appetite for wonders to come: because $|x-a|$ denotes the distance between $x$ and $a$, it is really your first example of a metric. In place of $|x-a|$ we could use the notation $d(x,a)$, and that is exactly the step we take when we generalize the notion of distance to other types of sets, beyond the set of real numbers. (Or, we may even impose other kinds of distance functions on the real numbers.)
As for your question "what about $|x-a|>c$" given your knowledge of $|x-a|<c$: well, note that $|x-a|$, no matter what it is, is either below $c$, equal to $c$, or above $c$. You already know that if it's below $c$, then $x-a$ must be between $-c$ and $c$. If $|x-a|$ equals $c$, then $x-a$ must be either $c$ or $-c$, the endpoints of the interval. So if $|x-a|$ is more than $c$, the only options are that $x-a$ is above $c$ or below $-c$. (I am of course assuming that $c$ is not negative.)
A: $$|x - a| \gt c  \qquad  \text {where}\space c \gt 0 $$
If $x - a \ge 0$   (This means the value is positive so the thing in modulus can be written directly)
$$\implies x\ge a $$
$$x -a \gt c $$
If $x - a \lt 0$   (This means the value is negative  so the thing in modulus in should be multiply -1 to remove the modulus . )
$$\implies x\lt a $$
$$-(x -a) \gt c $$
$$(x -a) \lt -c $$
A: 
For $|x-3|>2$ then $x > 5$ or $x < 1$
For $|x-a|>c$ then $x > a+c $ or $x < a - c$
A: An intuitive interpretation of $|x-a| >c$  to get you started would be the following: 
$\textbf{All}$ values of $x$ on the $x$-axis whose distance from $a$, a fixed point on the $x$-axis, is greater than $c$. 
