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.