SOLVING ABSOLUTE VALUE EQUATIONS

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!

See that ‘WolframAlpha widget’ in the right-hand column?
It has a starting equation already in place, which happens to be an example from this page.
Go ahead and press the ‘Submit’ button—and behold the power of WolframAlpha!
You'll see a graph (for slightly more advanced users), the solutions,
and a number line visualization of the solutions.

Next, start playing!
Type in any absolute value equation or inequality;
I like to press ‘Enter’ instead of clicking the ‘Submit’ button.
(There is a vertical bar ‘ | ’ for absolute value somewhere on your keyboard—look around!)

You can even copy-and-paste from the randomly-generated exercises and worksheets.

|x| = 3       |x| < 3       |x| <= 3       |x| > 3       |x| >= 3

Feel free to move the widget around to keep it nearby while you're reading the lesson.
When you're done exploring, just close the widget window.
Have fun!

## WolframAlpha Widget

Now, let's talk about the concepts involved in solving absolute value equations:

THEOREM solving absolute value equations
Let $\,x\in\mathbb{R}\,$, and let $\,k\ge 0\,$.   Then, $$|x| = k\ \ \text{ is equivalent to }\ \ x = \pm k$$
TRANSLATING THE THEOREM

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:

• Check that $\,k\,$ is a nonnegative number (zero, or greater than zero).
• The symbol $\,|x|\,$ represents the distance between $\,x\,$ and $\,0\,$.
• Thus, you want the numbers $\,x\,$, whose distance from $\,0\,$ is $\,k\$.
• You can walk from $\,0\,$ in two directions: to the right, or to the left.
Walk to the right a distance $\,k\$, and you get to the number $\,k\$.
Walk to the left a distance $\,k\$, and you get to the number $\,-k\$.
• Thus,   $\,|x| = k\$   is equivalent to   $\,x = k\ \text{ or }\ x = -k\$, which goes by the shorthand   $\,x=\pm k\$.
• Equivalent sentences are completely interchangeable, and you can use whichever is easiest to work with.
In this case, you're getting rid of the troublesome absolute value in exchange for a less-troublesome ‘plus or minus’ sign.

Recall that   ‘$\,\iff \,$’   is a symbol for ‘is equivalent to’.

The power of the sentence-transforming 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:

EXAMPLE:
Solve: $|2 - 3x| = 7$
Solution:
Write a nice, clean list of equivalent sentences:
 $|2 - 3x| = 7$ (original equation) $2-3x = \pm 7$ (check that $\,k\ge 0\,$; use the theorem) $2-3x = 7\ \text{ or }\ 2-3x = -7$ (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 $\,-3\,$)
It's a good idea to check your solutions:

$|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$,     $|2 + 5| = 7$     Check!

$|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$,     $|2 - 9| = 7$     Check!
EXAMPLE:
Solve: $5 - 2|3 - 4x| = -7$
Solution:
To use the theorem, you must have the absolute value all by itself on one side of the equation.
Thus, your first job is to isolate the absolute value:
 $5 - 2|3 - 4x| = -7$ (original equation) $-2|3 - 4x| = -12$ (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 $\,-4\,$)
EXAMPLE:
Solve: $|3x + 1| = -5$
Solution:
The theorem can't be used here, since $\,k\,$ is negative.
In this case, you need to stop and think.
Can absolute value ever be negative?   NO!
No matter what number you substitute for $\,x\,$,
the left-hand side of the equation will always be a number that is greater than or equal to zero.
Therefore, this sentence has no solutions. It is always false.
Master the ideas from this section

When you're done practicing, move on to:
Solving Absolute Value Inequalities Involving ‘Less Than’

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: