If you're going through this entire online
Algebra I course,
then you already know that I love WolframAlpha.
Well, WolframAlpha allows you to create
widgets to
share its amazing computational power!
You can play with the widget below, before learning to solve absolute value sentences yourself.
It can also be useful in checking your work.
See that ‘WolframAlpha widget’ in the righthand column?
When you press ‘Submit’, a popup window opens;
Have fun playing with the widget!
If you want, copyandpaste absolute value sentences from this page.

WolframAlpha Widget 
Now, let's talk about the concepts involved in solving absolute value equations:
Recall first that normal mathematical conventions dictate
that
‘$\,x = k\ $’
represents an entire class of sentences,
including the members
‘$\,x = 2\ $’,
‘$\,x = 5.7\ $’,
and
‘$\,x = \frac{1}{3}\,$’.
The variable
$\,k\ $ changes from sentence to sentence,
but is constant within a given sentence.
Also recall that ‘$\,x=\pm k\ $’ is a shorthand for ‘$\,x = k\ \text{ or }\ x = k\ $’.
When you see a sentence of the form
$\,x = k\ $, here's what you should do:
Recall that ‘$\,\iff \,$’ is a symbol for ‘is equivalent to’.
The power of the sentencetransforming tool
‘$\,x = k \iff x = \pm k\ $’
goes far beyond solving simple sentences like
$\,x = 5\,$!
Since $\,x\,$ can be any real number,
you should think of
$\,x\,$
as merely representing
the stuff inside the absolute value symbols.
Thus, you could think of rewriting the tool as:
‘$\,\text{stuff} = k \iff \text{stuff} = \pm k\ $’
See how this idea is used in the following examples:
$2  3x = 7$  (original equation) 
$23x = \pm 7$  (check that $\,k\ge 0\,$; use the theorem) 
$23x = 7\ \text{ or }\ 23x = 7$  (expand the plus/minus) 
$3x = 5\ \text{ or }\ 3x = 9$  (subtract $\,2\,$ from both sides of both equations) 
$\displaystyle x = \frac{5}{3}\ \text{ or } x = 3$  (divide both sides of both equations by $\,3\,$) 
$5  23  4x = 7$  (original equation) 
$23  4x = 12$  (subtract $\,5\,$ from both sides) 
$3  4x = 6$  (divide both sides by $\,2\,$) 
$3  4x = \pm 6$  (check that $\,k\ge 0\,$; use the theorem) 
$3  4x = 6\ \text{ or }\ 3  4x = 6$  (expand the plus/minus) 
$4x = 3\ \text{ or }\ 4x = 9$  (subtract $\,3\,$ from both sides of both equations) 
$\displaystyle x = \frac{3}{4}\ \text{ or }\ x = \frac{9}{4}$  (divide both sides of both equations by $\,4\,$) 
Solve the given absolute value equation.
Write the result in the most conventional way.
For more advanced students, a graph is displayed.
For example, the equation $\,2  3x = 7\,$
is optionally accompanied by the
graph of $\,y = 2  3x\,$ (the left side of the equation, dashed green)
and the graph of
$\,y = 7\,$ (the right side of the equation, solid purple).
In this example, you are finding the values of $\,x\,$ where the green
graph intersects the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
