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Notes 6.6: Solving
Absolute Value
Inequalities
I. Review of the Steps to Solve a Compound Inequality
A. Conjunction
● Example:
● This is a conjunction because the two inequality statements are
joined by the word “and”.
● You must solve each part of the inequality.
● The graph of the solution of the conjunction is the intersection of the
two inequalities. Both conditions of the inequalities must be met.
● In other words, the solution is wherever the two inequalities
overlap.
● If the solution does not overlap, there is no solution.
2 3 2 and 5 10x x
● Example:
● This is a disjunction because the two inequality statements are joined by the word “or”.
● You must solve each part of the inequality.
● The graph of the solution of the disjunction is the unionof the two inequalities. Only one condition of the inequality must be met.
● In other words, the solution will include each of the graphed lines. The graphs can go in opposite directions or towards each other, thus overlapping.
● If the inequalities do overlap, the solution is all reals.
3 15 or -2 +1 0 x x
B. Disjunction
Definition:Definition:
means and x a x a x a
-a a0 -a a0
means or x a x a x a
I. Keys to Solving Absolute Value Inequalities
x a x a
GreatOR Less ThAND
OR ANDx a x a
x a x a
Step 1: ISOLATE the absolute value expressions.Step 2: SPLIT the inequality using the keys below.
Conjunction, IntersectionDisjunction, Union
23 2
3x GreatOR
OR
OR
23 2
3x
23 2
3x
32
333 2x
2
23
3 33x
2 9 6x 2 9 6x
2 3x 2 15x
3
2x
15
2x
II. ExamplesEx 1: Solve the inequality and
graph the solution set on a number line.
5 2 4x 5 2 4x
2 1x 2 9x
1
2x
9
2x
Less ThAND
AND
AND
5 2 4x Ex 2: Solve the inequality
and graph the solution set on the number line
“AND” is conjunction, or intersection
Ex 3: Solve the inequality and graph the solutions.
|x| + 14 ≥ 19
|x| ≥ 5
x ≤ –5 OR x ≥ 5
Isolate the absolute value expression first. Since 14 is added to |x|,
subtract 14 from both sides to undo the addition.
Write as a compound inequality. The
solution set is {x: x ≤ –5 OR x ≥ 5}.
–10 –8 –6 –4 –2 0 2 4 6 8 10
5 units 5 units
– 14 –14
|x| + 14 ≥ 19
Your Turn… Solve | 2x + 3 | < 6.
2 3 6 2 3 6
2 3 2 9
3 9
2 2
AND
AND
AND
x x
x x
x x
Make two cases.
Solve the two cases
independently.
“AND” is conjunction, or intersection
Your Turn… Solve | 2x – 3 | > 5.
2 3 5 2 3 5
2 8 2 2
4 1
OR
OR
OR
x x
x x
x x
Make two cases.
Solve the two cases
independently.
III. “No Solution” and “All Real
Numbers”
• Lets look at what happens on the previous
two problems when we switch AND to OR
and OR to AND
Ex 1: Solve | 2x + 3 | < 6.
2 3 6 2 3 6
2 3 2 9
3 9
2 2
x x
x OR x
x x
R
R
O
O
Make two cases.
Solve the two cases
independently.
“OR” Means Union, so anything in either of the areas is a solution
“All Real Numbers”
Ex 2: Solve | 2x – 3 | > 5.
2 3 5 2 3 5
2 8 2 2
4 1
ANDx x
x AND x
x AND x
Make two cases.
Solve the two cases
independently.
“AND” Means where they intersect. Since
they do not intersect, the solution is:
No Solutions
You Try Ex 3:
Solve the inequality.
|x| – 9 ≥ –11
|x| – 9 ≥ –11 +9 ≥ +9
|x| ≥ –2
Add 9 to both sides.
Absolute-value expressions are always nonnegative.
Therefore, the statement is true for all real numbers.
The solution set is all real numbers.
You Try Ex 4:
Solve the inequality.
4|x – 3.5| ≤ –8
4|x – 3.5| ≤ –8
4 4|x – 3.5| ≤ –2
Absolute-value expressions are
always nonnegative. Therefore, the
statement is false for all values of x.
Divide both sides by 4.
The inequality has no solutions. The solution set is ø.