Chapter 3 — Graphing Fundamentals
Section 3.2
Solving and Graphing Compound
Inequalities of a Single Variable
TERMINOLOGY
3.2
Prerequisite Terms:
Absolute Value Inequality
Your definition
New Terms to Learn:
Compound Inequality
Your definition
Formal definition
Example
READING ASSIGNMENT
3.2
Sections 8.3 and 8.5
READING AND SELF-DISCOVERY QUESTIONS
1. What are the conjunctions that are used presenting the solution set to a compound inequality?
2. How can you identify a compound inequality?
118
3.2
3. What are two ways of presenting a compound inequality?
KEY CONCEPTS
3.2
A compound inequality is two or more inequalities linked together with a conjunction (and, or). For
example, [x < 4 or x > 5] as well as [x > 4 and x < 5]. Such inequalities occur in situations where conditions
are put on the variables.
Limitation/Caution: In the and situation both inequalities must be true, whereas in the or situation
only one needs to be true (including when both are true).
A three-part inequality has two inequality signs in the same direction, for example, 2 < 3 < 5 which is the
compound inequality [2 < 3 and 3 < 5].
Limitation/Caution: The inequality signs need to be in the same direction and the compound inequality
2 < 3 and 4 < 5 cannot be written as a three-part inequality.
TECHNIQUE
3.2
CHANGING AN ABSOLUTE VALUE INEQUALITY INTO A COMPOUND INEQUALITY
Absolute value inequalities occur in scientific measurements which are not exact. A measured length, x,
is off by no more than 2 cm either way from the exact length, R, that is to say, |R cm – x cm| < 2 cm or
|R – x| < 2
Limitation/Caution: Absolute value inequalities with more than one absolute value are difficult to
understand and to solve.
|x| < 2 implies –2 < x < 2
–2
0
2
|x| > 2 implies x < –2 or x > 2
–2
0
2
Example 1:
Example 2:
Change into a compound inequality:
Change into a compound inequality:
|y – 2| < 3
|3 – 2x| > 5
–3 < y – 2 < 3
3 – 2x < –5 or 3 – 2x > 5
119
Chapter 3 — Graphing Fundamentals
METHODOLOGY
3.2
SOLVING COMPOUND INEQUALITIES
The methodology highlights, in its steps, how to change compound linear inequalities into simple linear
inequalities and then solve these linear inequalities and then present the solution set symbolically and
graphically including the appropriate notation for boundaries of the solution set.
Limitation/Caution: The methodology does not work for non-linear compound inequalities.
Example 1
Steps
1
≤ 2(3 x − 1) < 5 x
3
Set up the
problem
Separate
Compound
Inequality
If you are presented
with a compound
inequality, you must
solve each inequality
separately and then
move to step 4
Solve the
inequalities
Use the Methodology
for Solving
Inequalities in One
Variable.
1
1
( x − 3) < (4 x + 2) ≤ 10
2
3
Problem 1:
Problem 2:
1
1
( x − 3) < (4 x + 2)
2
3
1
(4 x + 2) ≤ 10
3
1
1
1
1
( x) − (3) < (4 x) + (2)
2
2
3
3
 x 3
 4x 2 
6 −  < 6 + 
2 2
 3 3
3x − 9 < 8 x + 4
−9 < 5x + 4
− 13 < 5 x
13
− <x
5
1
1
(4 x) + (2) ≤ 10
3
3
 4x 2 
3  +  ≤ 3  10
 3 3
4 x + 2 ≤ 30
4 x + 2 ≤ 30
4 x ≤ 28
Example 2
2
Solve for x:
Discussion
Example 2 Example 1
1
1
1
( x − 3) < (4 x + 2) ≤ 10
2
3
Example 1
Solve for x:
Example 2
Example 1
3
120
x≤7
Discussion
3
Solve the
inequalities
Use the Methodology
for Solving
Inequalities in One
Variable.
4
Present the
compound
solution set
Both solutions
must be true. If
appropriate, present
as a three-part
inequality.
5
Describe the
solution set
Identify the end of
the line segments,
determine if open or
closed, and identify
direction of the
solution sets
6
Present the
compound
solution set
graphically
Determine the end
point(s), open or
closed and the
interval(s)
Example 2
Steps
Example 1
13
< x and x ≤ 7
5
13
− <x≤7
5
Point is −
13
is open and point is 7 is closed.
5
Example 2 Example 1
Example 2
Example 1
Example 2
−
–
13
2
0
7
121
2
3
3 2

− < Chapter
≤ 10 (yes)
3 — Graphing
Fundamentals
2 3
Steps
7
Discussion
Test and
Validate
solution set
Pick two or more points, some in the solution 1
set
some
(4and
2) ≤ not
10 in solution set (note:
 7+
pick easy calculation points if possible)
3
11
28
Example
(28) ≤ 10,
≤ 10 (yes) 
3
3
We choose 0, 7 (in the solution set) and –10, 20 (not in the solution set)
1
1
( x − 3) < (4 x + 2) ≤ 10
2
3
1
1
(0 − 3) < (4  0 + 2) ≤ 10
2
3
3 2
− < ≤ 10 (yes) 
2 3
1
1
(− 10 − 3) < (4(− 10) + 2) ≤ 10
2
3
13
38
− <−
≤ 10 (no) 
2
3
1
(4  7 + 2) ≤ 10
3
1
28
(28) ≤ 10,
≤ 10 (yes) 
3
3
1
1
(19 − 3) < (4  19 + 2) ≤ 10
2
3
1
1
(16) < (78) ≤ 10
2
3
8 < 26 ≤ 10 (no) 
Example 2
1
1
(− 10 − 3) < (4(− 10) + 2) ≤ 10
2
3
13
38
− <−
≤ 10 (no) 
2
3
1
1
(19 − 3) < (4  19 + 2) ≤ 10
2
3
1
1
(16) < (78) ≤ 10
2
3
8 < 26 ≤ 10 (no) 
TIP FOR SUCCESS
3.2
You can solve a compound inequality by applying the properties to the whole inequality rather than
splitting it into two parts if there are constants on both ends.
Example:
7 < 4 x − 5 < 11
7 + 5 < 4 x − 5 + 5 < 11 + 5
12 < 4 x < 16
3< x < 4
122
Validate with x = 0, 6 (not in the solution set) and x = 3.5 (in the solution set)
7 < 4 (0) – 5 < 11
7 < –5 < 11 (no)
7 < 4(6) – 5 < 11
7 < 4 (3.5) – 5 < 11
7 < 24 – 5 < 11
7 < 14 – 5 < 11
7 < 19 < 11 (no)
7 < 9 < 11 (yes)
CRITICAL THINKING QUESTIONS
3.2
1. What three things do you have to know to be able to represent a solution set graphically and in interval
notation?
2. How do you represent an end point not in the solution set symbolically, graphically, and in interval
notation?
3. How is the methodology for solving a compound inequality different from the methodology for solving
a simple inequality?
4. What is the most common error that occurs when multiplying by the multiplicative inverse of a negative
coefficient?
5. Conceptually, how is the meaning of and involved in finding the common solution set of a compound
inequality?
6. What are the issues you can identify for why solving linear inequalities are difficult for some students?
123
Chapter 3 — Graphing Fundamentals
DEMONSTRATE YOUR UNDERSTANDING
Solve each of the following inequalities.
1
1. 2 < ( x − 2) < 5 x
3
2. − 2 < 2 x − 5 ≤ 7
3. 7 ≥ − 3 x + 5 ≥ − 3
4. − 3( x − 1) ≤
124
2
3
(5 − x) <
3
4
3.2
5.
2x − 5 > 7
6. On the number line, represent the set of numbers for which x >
9
or x < − 2 .
2
IDENTIFY AND CORRECT THE ERRORS
3.2
In the second column, identify the error you find in each of the following worked solutions and describe
the error made. Solve the problem correctly in the third column.
Problem
1. Solve: 2 ≤ 2 − x < 4
Worked Solution
(What is wrong here?)
Describe Error
Correct Process
Wrong process: Failure to
change the direction of the
inequality when multiplying
by a negative number.
2 ≤ 2− x < 4
2 − 2 ≤ −x < 4 − 2
0 ≤ −x < 2
0 ≤ x < −2
2. Solve: –7 < x – 4 < 5
Worked Solution
(What is wrong here?)
Wrong process:
Misinterpreting the “and”
conjunction.
−7 < x − 4 < 5
− 7 < x − 4 and x − 4 < 5
− 3 < x and x < 9
So, x < 9
125
Chapter 3 — Graphing Fundamentals
Problem
3. Solve: − 2 x < 5( x − 3) < 7
Worked Solution
(What is wrong here?)
Describe Error
Wrong process: Failure to
change the direction of the
inequality when multiplying
by a negative number.
− 2 x < 5( x − 3) < 7
− 2 x < 5( x − 3) and 5( x − 3) < 7
− 2 x < 5 x − 15 and 5 x − 15 < 7
− 7 x < − 15 and 5 x < 22
15
22
x<
and x <
7
5
15
x<
7
4. Solve: |2x – 3| > 4
Worked Solution
(What is wrong here?)
Wrong process:
Misinterpreting an absolute
value inequality with an
“and” rather than an “or”.
2x − 3 > 4
( 2 x − 3) < − 4
and (2 x − 3) > 4
2 x < − 1 and 2 x > 7
x>−
1
7
and x >
2
2
So, x >
7
2
5. Solve: –8 < –2x – 4 < 2
Worked Solution
(What is wrong here?)
−8 < −2 x − 4 < 2
− 8 < − 2 x − 4 and − 2 x − 4 < 2
− 4 < − 2 x and − 2 x > 6
2 > x and x < − 3
2 > x < −3
126
Wrong process: Making
an incorrect three-part
inequality for the solution
set.
Correct Process