The document discusses absolute value and inequalities involving absolute value. It explains that absolute value provides the distance from zero, regardless of the number's sign. Absolute value inequalities can be expressed as intervals, with solutions falling between the interval boundaries or including the boundaries. For an inequality like x < a, the solution is the interval (-a, a); for x > a it is the union of the intervals (-∞, -a) and (a, ∞). Examples are provided to demonstrate solving absolute value inequalities algebraically.

2. 1 block 1 block
You walk directly east from your house one block. How far
from your house are you?
You walk directly west from your house one block. How far
from your house are you?
It didn't matter which direction you walked, you were still 1 block
from your house.
This is like absolute value. It is the distance from zero. It doesn't
matter whether we are in the positive direction or the negative
direction, we just care about how far away we are.
−4 = 4 4 units away from 0 4 units away from 0
4 =4
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

3. x =6 What we are after here are values of x such
that they are 6 away from 0.
x = 6 or x = −6 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
6 and -6 are both 6 units away from 0
x + 4 = 10
The "stuff" inside the absolute value signs could = 10 (the
positive direction) or the "stuff" inside the absolute value signs
could = -10 (the negative direction)
x + 4 = 10 x + 4 = −10 x = 6 or x = −14
Let's check it: 6 + 4 = 10 − 14 + 4 = 10
10 = 10 − 10 = 10

4. Let's look at absolute value with an inequality. x <5
This is asking, "For what numbers is the distance from 0 less than 5
units?"
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Everything
inbetween these
−5 < x < 5 lines is less than 5 (−5, 5)
units away from 0
Inequality notation Interval notation
So if we have x < a it is equivalent to −a < x < a
This means x is greater than -a and x is less than a
(or x is inbetween -a and a)

5. Based on what we just observed, the "stuff"
3x − 1 ≤ 5 inside the absolute value signs is inbetween
-5 and 5 or equal to either end since the
inequality sign has "or equal to".
To solve this we get x isolated in the middle.
− 5 ≤ 3x − 1 ≤ 5 Whatever steps we do to get it alone, we do
+1 +1 +1 to each end. We keep in our minds the fact
-4 6 that if we multiply or divide by a negative, we
must turn the signs the other way.
− 4 ≤ 3x ≤ 6
3 3 3
4 So x is inbetween or equal to  4 
− ≤x≤2 - 4/3 and 2 − 3 , 2
3  
Let's graph the solution: [ ]
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

6. What if the inequality is greater than? x >5
This is asking, "When is our distance from 0 more than 5 units away?"
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Everything outside these Everything outside these
lines is more than 5 lines is more than 5
units away from 0 We'll have to units away from 0
express this with two
difference pieces
x < −5 OR x>5
In interval notation: (−∞,−5) or (5, ∞)
x
So if we have x > a it is equivalent to < −a or x > a

7. 1− 2x − 3 ≥ 3 We must first isolate the absolute value.
+3 +3 This means if there are other things on the
6 left hand side of the inequality that are
outside of the absolute value signs, we
must get rid of them first.
From what we saw previously, the "stuff"
1− 2x ≥ 6 inside the absolute value is either less than
or equal to -6 or greater than or equal to 6
Isolate x, remembering that if
1 − 2 x ≤ −6 or 1 − 2 x ≥ 6 you multiply or divide by a
negative you must turn the sign.
− 2 x ≤ −7 − 2x ≥ 5 We are dividing by a negative
-2 -2 -2 -2 so turn the signs!
7 5 ] [
x≥ or x≤−
2 2 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8