Coordinate Algebra
Absolute Value Inequalities Notes
Date: ______
COMPLETED
Absolute Value Inequalities have multiple solutions and the answers can be represented both algebraically and
graphically. There are two types of absolute value inequalities: conjunctions and disjunctions.
Conjunctions are when the inequality is less than (or less than or equal to).
Algebraically
We can translate absolute value inequalities into the following:
|ax + b| < '
|ax + b| ≤ c
Less
Th”and”
AND
-c < ax + b < c
-c ≤ ax + b ≤ c
Step 1: Isolate the absolute value (if necessary)
Step 2: Copy the basic inequality without the absolute value brackets.
Step 3: Write the additional inequality sign
Step 4: Negate the c value
This creates a compound conjunction.
Step 5: Solve & Graph
Ex1) |x + 1| < 3
x+1<3
<x+1<3
-3 < x + 1 < 3
-1
-1 -1
-4 < x < 2
-5 -4 -3 -2 -1
Step 1
Step 2
Step 3
Subtract 1 to all three sections
Solution
0
1
2
3
< : open circle
≤ : closed circle
4
For conjunctions we generally shade in-between the two points.
Remember: You must isolate the absolute value BEFORE you set up the compound inequality.
Ex2) |x − 2| < 12
Ex3) |x − 6| − 2 ≤ 5
+2 +2
|0 − 1| ≤ 2
-12 < x – 2 < 12
+2
+2 +2
-7 ≤ x – 6 ≤ 7
+6
+6 +6
-10 < x < 14
-1 ≤ x ≤ 13
-16 -12 -8 -4
0
4
8
12 16 20
-2
0
2 4
6
8 10 12 14 16
Ex4) |2x − 5| < 11
Ex5) |−x − 5| ≤ 7
-11 < 2x – 5 < 11
+5
+5 +5
-7 ≤ -x – 5 ≤ 7
+5
+5 +5
-6 < 2x < 16
-2 ≤ -x ≤ 12
-6 < 2x < 16
2
2 2
-2 ≤ -x ≤ 12
-1 -1 -1
-3 < x < 8
2 ≥ x ≥ -12
-6 -4
-2
0
2
4
6
8
10 12
Ex6) |2x − 10| − 4 ≤ −2
+4
+4
|23 − 45| ≤ 2
0
OR
8 ≤ 2x ≤ 12
-60 < 10x < 40
8 ≤ 2x ≤ 12
2 2
2
-60 < 10x < 40
10
10 10
4≤x≤6
-6 < x < 4
4
0
2
4
4
6
8
Ex7) |10x + 10| + 8 < 58
-8
-8
|453 + 45| < 50
-50 < 10x + 10 < 50
-10
-10 -10
2 3
-12 ≤ x≤
≤2
-14 -12 -10 -8 -6 -4 -2
-2 ≤ 2x – 10 ≤ 2
+10 +10 +10
1
When multiplying/
dividing by -1 you
must switch the
sign
5
6
7 8
9
-10 -8 -6 -4 -2
0
2
Disjunctions are when the inequality is greater than (or greater than or equal to). This is different from
conjunctions because either of the solutions can be true to satisfy the inequalities.
Algebraically
We can translate absolute value inequalities into the following:
|ax + b| > 9
ax + b < -c OR ax + b > c
Great”or
”
|ax + b| ≥ c
ax + b ≤ -c OR ax + b ≥ c
Step 1:
Step 2:
Step 3:
Step 4:
Isolate the Absolute Value if necessary
Copy the basic inequality without the absolute value brackets.
For the 2nd inequality, switch the inequality sign and negate the c value
Solve & Graph
Ex1) |x + 3| ≥ 9
x+3≥9
-3 -3
x + 3 ≤ -9
-3 -3
OR
x ≥ 6 OR
x ≤ -12
> : open circle
≥ : closed circle
-12 -10 -8 -6 -4 -2
0
2
4
6
For disjunctions we generally shade out from the two points.
Remember: You must isolate the absolute value BEFORE you set up the two inequalities.
Ex2) |x + 9| > 1
x+9>1
-9 -9
OR
x > -8 OR
Ex3) |3x − 3| ≥ 18
x + 9 < -1
-9 -9
3x – 3 ≥ 18
+3 +3
OR
3x – 3 ≤ -18
+3
+3
x < -10
3x ≥ 21
OR
3x ≤ -15
3x ≥ 21
3
3
OR
3x ≤ -15
3
3
x ≥ 7 OR
-6 -4
-15 -14 -13 -12 -11 -10 -9 -8 -7 -6
Ex4) |−x − 2| > 2
-x – 2 > 2
+2 +2
-x > 4
OR
OR
-x > 4
-1 -1
-2
0
x ≤ -5
2
4
6
-x – 2 < -2
+2 +2
3x – 6 ≥ 3
+6 +6
-x < 0
-x < 0
-1 -1
OR
1
2
3
3x – 6 ≤ -3
+6 +6
3x ≥ 9
3x ≤ 3
3x ≥ 9
3
3
3x ≤ 3
3
3
x ≤1
x ≥ 3 OR
0
10 12
Ex5) |3x − 6| − 2 ≥ 1
+2
+2
|3. − 6| ≥ 3
x < -4 OR x > 0
-5 -4 -3 -2 -1
8
4
-3 -2 -1
0
1
2
3
4
5
6
Ex6) |x − 5| + 4 ≥ 10
-4 -4
Ex7) |−4x + 4| − 1 > 31
+1
+1
|, − -| ≥ .
x–5≥6
+5 +5
OR
|−/, + /| > 01
x – 5 ≤ -6
+5 +5
x ≥ 11 OR x ≤ -1
-4x + 4 > 32
-4 -4
OR
-4x + 4 < -32
-4 -4
-4x > 28
-4
-4
OR
-4x < -36
-4
-4
x < -7 OR
-6 -4
-2
0
2
4
6
8
10 12
-8
-6 -4
-2
0
x>9
2
4
6
8 10