To find the distance between any two real numbers,
take the greater number (the
one lying farthest to the right on a number line)
and subtract the lesser number (the one lying farthest to the left).
That is, if
[beautiful math coming... please be patient]
$x \ge y$, then the distance between them is just the difference, $\,x  y\,$.
For example, the distance between
[beautiful math coming... please be patient]
$\,3\,$ and $\,5\,$ is: $\,5  (3) = 8\,$
But what if you goof up the order of subtraction?
In this case, you get a number with the right size (distance from zero), but the wrong sign: $\,3  5 = 8\,$
To get from $\,8\,$ to the answer you really want, just take the absolute value!
Thus, absolute value provides a convenient way to talk about the distance between any two real numbers:
That is, to find the distance between any two real numbers:
Notice that $\,x  y\,$ and $\,y  x\,$ are opposites, since $\,x  y = (y  x)\,$.
Opposites have the same distance from zero, hence have the same absolute value.
(Remember: absolute value gives distance from zero!)
Thus, $\,x  y\, = y  x\,$. So we repeat—you can subtract in whatever order you want!
By recognizing an absolute value expression as representing a distance between
two numbers,
many simple absolute value sentences can be solved ‘by inspection’—just
look at them, think, and write down the solution set!
Here are some examples:
Solve: $x  1 < 2$
Solution:
View $\,x  1\,$ as the distance between $\,x\,$ and $\,1\,$.
Here, we want all numbers $\,x\,$ whose distance from $\,1\,$ is less than $\,2\,$:
$$
\overbrace{\strut x1}^{\text{want numbers $x$ whose distance from $1\ \ldots \ $ }}
\overbrace{\strut \ \ < 2\ \ }^{\text{is less than two}}
$$
interval notation for solution set: $\,(1,3)$
sentence form of solution: $\,1 < x < 3\,$
Solve: $x + 3 \ge 1$
Solution:
First, rewrite $\,x+3\,$ so there is a difference inside the absolute value:
$\,x + 3\, = x  (3)\,$
Thus, $\,x+3\,$ represents the distance between $\,x\,$ and $\,3\,$.
Here, we want all numbers $\,x\,$ whose distance from $\,3\,$ is greater than or equal to $\,1\,$:
$$
\begin{gather}
x+3 \ge 1\cr
x  (3) \ge 1\cr\cr
\overbrace{\strut x(3)}^{\text{want numbers $x$ whose distance from $3\ \ldots \ $ }}
\overbrace{\strut \ \ \ge 1\ \ }^{\text{is greater than or equal to $\,1$}}
\end{gather}
$$
interval notation for solution set: $\,(\infty,4] \cup [2,\infty)$
sentence form of solution: $\,x \le 4 \ \ \text{ or }\ \ x \ge 2\,$
This final example is important in Calculus.
When you study the precise definition of a
limit (which is the central concept in Calculus),
then you'll see this type of absolute value sentence.
Solve: $\,0 < x  2 < \delta\,$
Solution:
The symbol ‘$\,\delta\,$’ is the lowercase Greek letter delta.
It is used in this context to denote a small positive number.
The sentence ‘$\,0 < x  2 < \delta\,$’ is a compound inequality—it uses more than one inequality symbol.
This type of compound inequality is a shorthand for a mathematical ‘and’ sentence:
$$
0 < x  2 < \delta\qquad \qquad \qquad \qquad \qquad \text{ is equivalent to }\qquad \qquad \qquad \qquad \qquad
0 < x  2\ \ \text{ and }\ \ x  2 < \delta
$$
Recall that the only time an ‘and’ sentence is true is when both subsentences are true.
So, here's what you should think:
$$
\overbrace{\strut x  2}^{\text{we want numbers $x$ whose distance from $2\ \ldots$ } }
\overbrace{\strut \ \ > 0\ \ }^{\text{is greater than $0$} }
\ \ \ \
\overbrace{\strut \text{ and }}^{\text{ and }}\ \ \
\overbrace{\strut x  2}^{\text{ whose distance from $2\ \ldots$ } }
\overbrace{\strut < \delta}^{\text{is less than delta}}
$$
That is, the solutions to ‘$\,0 < x  2 < \delta\,$’ are all numbers that are within $\,\delta\,$ units of $\,2\,$, but not equal to $\,2\,$.
More precisely, study the number lines below:
The resulting solution set is often called a ‘punctured neighborhood of $\,2\,$’:
interval notation for solution set: $\,(2\delta,2) \cup (2,2+\delta)$
sentence form of solution: $\,2\delta < x < 2 \ \ \text{ or }\ \ 2 < x < 2+\delta\,$
equivalent sentence form of solution: $\,2\delta < x < 2+\delta \ \ \text{ and }\ \ x\ne 2\,$
For fun, zip up to WolframAlpha and type in any of these:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
