Algebra Seminar Session 3 Study Guide: Absolute Value
Definition of Absolute Value
The absolute value of a number is the number's distance from zero on a number line.
Since we never talk about distance as being negative, the absolute value of a number is always positive. Think of it as the
number of jumps it would take to get to zero from a number.


It would take five jumps to get to zero from -5, so the absolute value of -5 is 5.
It would also take five jumps to get to zero from +5, so the absolute value of +5 is also 5.
We use bars (vertical lines) on either side of a number absolute to mean absolute value.
|-4| = the absolute value of -4 = 4
Pretty simple. If the number is negative, make it positive. If the number is already positive, leave it alone.
From: http://www.shmoop.com/basic-operations/absolute-value.html
Solving Equations with Absolute Value
Absolute-value equations always work this way: to be able to remove the absolute-value bars, you have to isolate the absolute value onto
one side, and then split the equation into the two possible cases.
Strategy 1 (the 2 possible cases for absolute value can be represented by taking the positive and negative case
for what the absolute value is equal to, as in strategy 1)
When solving equations that involve absolute values, consider that the quantity within the absolute value bars
could be either both the positive and negative version of what it is equal to. Let's look at some more exercises:
C200 Reserved
 Solve | x + 2 | = 7
If the absolute value of| x + 2 |= 7, then whatever quantity is within the absolute value sign could only
possibly be equal to 7 or -7 (since the only quantities that have an absolute of 7 are 7 and -7) To clear the
absolute-value bars, I must split the equation into its two possible two cases, one case for each sign:
x+2=7
x=5
or x + 2 = -7
or –9 = x
Then the solution is x = –9, 5.
Solve | 2x - 3 | -4 =3
First, I'll isolate the absolute-value part; that is, I'll get the absolute-value expression by itself on one side of the
"equals" sign, with everything else on the other side:

| 2x – 3 | – 4 = 3
| 2x – 3 | = 7
Now I'll clear the absolute-value bars by splitting the equation into its two cases, one for each sign:
2x – 3 = 7
2x = 10
x=5
or
or
or
2x – 3 = -7
2x = 4
x = –2
So the solution is x = –2, 5.
Strategy 2 (the 2 possible cases for absolute value can also be represented by taking the positive and negative
case for the absolute value itself, as in strategy 2)

Solve | x + 2 | = 7
To clear the absolute-value bars, I must split the equation into its two possible two cases, one case for each sign:
(x + 2) = 7 or –(x + 2) = 7
x+2=7
or
–x – 2 = 7
x=5
or
–9 = x
Then the solution is x = –9, 5.

Solve | 2x – 3 | – 4 = 3
First, I'll isolate the absolute-value part; that is, I'll get the absolute-value expression by itself on one side of the
"equals" sign, with everything else on the other side:
| 2x – 3 | – 4 = 3
| 2x – 3 | = 7
Now I'll clear the absolute-value bars by splitting the equation into its two cases, one for each sign:
(2x – 3) = 7
2x – 3 = 7
2x = 10
x=5
or
or
or
or
–(2x – 3) = 7
–2x + 3 = 7
–2x = 4
x = –2
So the solution is x = –2, 5.
From: http://www.purplemath.com/modules/solveabs.htm
Summary of Strategies
Strategy 1
(taking the positive and negative case for what the
absolute value is equal to)
1. Isolate the absolute value to one side of the
equation
| 2x – 3 | – 4 = 3
| 2x – 3 | = 7
Strategy 2
(taking the positive and negative case for the absolute
value itself)
1. Isolate the absolute value to one side of the
equation
| 2x – 3 | – 4 = 3
| 2x – 3 | = 7
2. taking the positive and negative case for what
the absolute value is equal to
3.
2x – 3 = 7
or 2x – 3 = -7
2x = 10
or
2x = 4
x=5
or x = –2
2. taking the positive and negative case for the
absolute value itself
3. Solve your 2 equations
4.
2x – 3 = 7
or 2x – 3 = -7
2x = 10
or
2x = 4
x=5
or x = –2
3. Solve your 2 equations
So the solution is x = –2, 5.
(2x – 3) = 7
2x – 3 = 7
2x = 10
x=5
(2x – 3) = 7
2x – 3 = 7
2x = 10
x=5
or
or
or
or
or
or
or
or
So the solution is x = –2, 5.
–(2x – 3) = 7
–2x + 3 = 7
–2x = 4
x = –2
–(2x – 3) = 7
–2x + 3 = 7
–2x = 4
x = –2