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LESSON
2 . 3
Name Class Date
Solving Absolute Value 2.3 Solving Absolute Value
Inequalities Inequalities
Essential Question: What are two ways to solve an absolute value inequality?
Resource
Common Core Math Standards Locker
The student is expected to:
Explore
COMMON A-CED.A.1 Visualizing the Solution Set of an Absolute
CORE Value Inequality
Create equations and inequalities in one variable and use them to solve You know that when solving an absolute value equation, it’s possible to get two solutions.
problems. Also A-REI.B.3, F-IF.C.7b Here, you will explore what happens when you solve absolute value inequalities.
Mathematical Practices A Determine whether each of the integers from -5 to 5 is a solution
COMMON Number Solution?
of the inequality x + 2 < 5. Write yes or no for each number
CORE MP.4 Modeling ⎜ ⎟ x = -5 no
in the table. If a number is a solution, plot it on the number line. x = -4 no
Language Objective x = -3 no
Match absolute value equations and inequalities with their graphs, x = -2 yes
explaining and justifying reasoning. - - - - -
5 4 3 2 1 0 1 2 3 4 5 x = -1 yes
x = 0 yes
x = 1 yes
ENGAGE x = 2 yes
x = 3 no
Essential Question: What are two x = 4 no
ways to solve an absolute value x = 5 no
inequality? y
an Determine whether each of the integers from -5 to 5 is a solution
B Number Solution?
Possible answer: You can solve an absolute value mp of the inequality x + 2 > 5. Write yes or no for each number
o ⎜ ⎟ x = -5 yes
C
g in the table. If a number is a solution, plot it on the number line.
n
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n 5 4 3 2 1 0 1 2 3 4 5 no
i
l
inequality symbol to determine the intervals on the f
f
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other. For an algebraic solution, isolate the absolute o x = 2 no
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value expression and rewrite the inequality as a no
compound inequality that doesn’t involve absolute x = 4 yes
value so that you can finish solving the inequality. x = 5 yes
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A2_MNLESE385894_U1M0
87 Lesson 2 . 3
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State the solutions of the equation x + 2 = 5 and relate them to the solutions you found
⎜ ⎟
for the inequalities in Steps A and B. EXPLORE
The solutions are -3 and 3. These are the only numbers that are not solutions of the
⎜ ⎟ ⎜ ⎟
inequalities x + 2 < 5 and x + 2 > 5. Visualizing the Solution Set of an
Absolute Value Inequality
If x is any real number and not just an integer, graph the solutions of x + 2 < 5 and
⎜ ⎟
x + 2 > 5 .
⎜ ⎟
Graph of all real solutions of x + 2 < 5: Graph of all real solutions of x + 2 > 5:
⎜ ⎟ ⎜ ⎟ QUESTIONING STRATEGIES
- - - - - - - - - - How would you characterize the solutions to
5 4 3 2 1 0 1 2 3 4 5 5 4 3 2 1 0 1 2 3 4 5
an absolute value inequality? Sample answer:
They lie either between two values or everywhere
Reflect else except between the two values, depending on
1. It’s possible to describe the solutions of x + 2 < 5 and x + 2 > 5 using inequalities that don’t
⎜ ⎟ ⎜ ⎟ the inequality.
involve absolute value. For instance, you can write the solutions of x + 2 < 5 as x > -3 and
⎜ ⎟
x < 3. Notice that the word and is used because x must be both greater than -3 and less than 3. Why do you write the solutions to the absolute
How would you write the solutions of x + 2 > 5? Explain.
⎜ ⎟
⎜ ⎟
Write the solutions of x + 2 > 5 as x < -3 or x > 3. Use the word or because x must be value inequality as a compound inequality
either less than -3 or greater than 3; it can’t be both. statement? Sample answer: because the solution
2. Describe the solutions of x + 2 ≤ 5 and x + 2 ≥ 5 using inequalities that don’t involve consists of the union or intersection of the solutions
⎜ ⎟ ⎜ ⎟
absolute value. of the two related linear inequalities.
⎜ ⎟
The solutions of x + 2 ≤ 5 are the values of x for which x ≥ -3 and x ≤ 3. The solutions
⎜ ⎟
of x + 2 ≥ 5 are the values of x for which x ≤ -3 or x ≥ 3.
Explain 1 Solving Absolute Value Inequalities Graphically EXPLAIN 1
( ) ( ) ( ) ( )
You can use a graph to solve an absolute value inequality of the form ƒ x > g x or ƒ x < g x ,
( ) ( )
where ƒ x is an absolute value function and g x is a constant function. Graph each function © Solving Absolute Value Inequalities
H
separately on the same coordinate plane and determine the intervals on the x-axis where one o
u
( ) ( ) gh Graphically
graph lies above or below the other. For ƒ x > g x , you want to find the x-values for which t
o
n
( ) ( ) ( ) ( ) M
the graph ƒ x is above the graph of g x . For ƒ x < g x , you want to find the x-values for
i
f
f
( ) ( ) l
which the graph of ƒ x is below the graph of g x .
i
n
H
ar
c QUESTIONING STRATEGIES
o
Example 1 Solve the inequality graphically. y ur
t
P
ub How do you interpret which points are
l
i
s
x + 3 + 1 > 4 h
⎜ ⎟ 2 i
n
g solutions of the inequality? Sample answer:
C
x o
( ) ( ) mp
The inequality is of the form ƒ x > g x , so determine the intervals on If f(x) > g(x), the graph of f(x) must be “above” the
- - - 0 an
( ) 6 4 2
the x-axis where the graph of ƒ x = x + 3 + 1 lies above the graph of
⎜ ⎟
( ) y
g x = 4. -2 graph of g(x). The solutions are the x-values in the
( ) ( ) interval along the x-axis where f(x) has y-values
The graph of ƒ x = x + 3 + 1 lies above the graph of g x = 4
⎜ ⎟ -4
to the left of x = -6 and to the right of x = 0, so the solution of greater than g(x). If f(x) < g(x), the graph of f(x)
x + 3 + 1 > 4 is x < -6 or x > 0.
⎜ ⎟ must be “below” the graph of g(x). The solutions are
the x-values in the interval along the x-axis where
Module 2 88 Lesson 3 f(x) has y-values less than g(x).
PROFESSIONAL DEVELOPMENT How do you know when the endpoints of
A2_MNLESE385894_U1M02L3 88 16/05/14 4:41 AM the solution interval on the x-axis are not
Math Background included in the solution? The original inequality
An absolute value inequality is often in the form |ax + b| < c or |ax + b| > c. is < or >.
If the inequality is in the form |ax + b| < c, then it can be rewritten as the
compound inequality –c < ax + b < c, and solved for x. The
-c - b c - b
_ _
solution will be of the form a < x and x < a . If the inequality is in the
form |ax + b| > c, it can be rewritten as the compound inequality c < ax + b or
c - b
_
ax + b < –c, and solved for x. The solution will be of the form a < x or
-c - b
x <_
. The solutions are easily adjusted for ≤ and ≥.
a
Solving Absolute Value Inequalities 88
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x - 2 - 3 < 1 y
B ⎜ ⎟ 4
( ) ( )
The inequality is of the form f x < g x , so determine the intervals 2
( ) below x
on the x-axis where the graph of f x = x - 2 - 3 lies
( ) ⎜ ⎟ 0
the graph of g x = 1. -
2 2 4 6
( ) below -2
The graph of f x = x - 2 - 3 lies the graph of
⎜ ⎟
( ) -4
g x = 1 between x = -2 and x = 6 , so the solution of
⎜ ⎟
x - 2 - 3 < 1 is x > -2 and x < 6 .
Reflect
3. Suppose the inequality in Part A is x + 3 + 1 ≥ 4 instead of x + 3 + 1 > 4. How does the
⎜ ⎟ ⎜ ⎟
solution change?
The solution now includes the endpoints of the interval: x ≤ -6 or x ≥ 0.
4. In Part B, what is another way to write the solution x > -2 and x < 6?
-2 < x < 6
( )
5. Discussion Suppose the graph of an absolute value function ƒ x lies entirely above the graph of
( ) ( ) ( )
the constant function g x . What is the solution of the inequality ƒ x > g x ? What is the solution
( ) ( )
of the inequality ƒ x < g x ?
( ) ( ) ( )
The solution of f x > g x is all real numbers, because every point on the graph of f x is
( ) ( ) ( )
above the corresponding point on the graph of g x . The solution of f x < g x is no real
( )
number, because no point on the graph of f x is below the corresponding point on the
( )
graph of g x .
Your Turn
6. Solve x + 1 - 4 ≤ -2 graphically.
⎜ ⎟
y ( ) ( )
an The inequality is of the form ƒ x ≤ g x , so determine the intervals on the x-axis
mp ( ) ( )
where the graph of ƒ x = x + 1 - 4 intersects or lies below the graph of g x = -2.
o ⎜ ⎟
C
g
n
i ( ) ( )
The graph of ƒ x = x + 1 - 4 intersects the graph of g x = -2 at x = -3 and x = 1
h ⎜ ⎟
s
i
l ( )
ub and lies below the graph of g x = -2 between those x-values, so the solution of
P
t x + 1 -4 ≤ -2 is x ≥ -3 and x ≤ 1.
ur ⎜ ⎟
o
c
ar
H
n
i
l
f
f
i
M
n
o
t
gh
u
o
H
©
Module 2 89 Lesson 3
COLLABORATIVE LEARNING
A2_MNLESE385894_U1M02L3.indd 89 19/03/14 1:49 PM
Small Group Activity
Have students work in groups to make a flowchart that explains how to solve each
of the four types of an absolute value inequality of the form |ax + b| c or
|ax – b| c, where the box represents the inequality symbol. For example:
|2x + 3| ≤ 6 |2x + 3| < 6 |2x + 3| > 6 |2x + 3| ≥ 6
Ask each student in a group to finish one branch of the flowchart. Then have the
group collate the branches to make the entire flowchart.
89 Lesson 2 . 3
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Explain 2 Solving Absolute Value Inequalities Algebraically
To solve an absolute value inequality algebraically, start by isolating the absolute value expression. EXPLAIN 2
When the absolute value expression is by itself on one side of the inequality, apply one of the
following rules to finish solving the inequality for the variable. Solving Absolute Value Inequalities
Solving Absolute Value Inequalities Algebraically Algebraically
1. If x > a where a is a positive number, then x < -a or x > a.
⎜ ⎟
2. If x < a where a is a positive number, then -a < x < a.
⎜ ⎟
Example 2 Solve the inequality algebraically. Graph the solution on a number line. QUESTIONING STRATEGIES
⎜ ⎟ ⎜ ⎟ When does the graph of the solution to an
4 - x + 15 > 21 x + 4 - 10 ≤ -2
inequality include the endpoints? when the
⎜ ⎟
4 - x > 6 ⎜ ⎟
x + 4 ≤ 8 original inequality is ≤ or ≥
4 - x < -6 or 4 - x > 6 x + 4 ≥ -8 and x + 4 ≤ 8 When does the graph of the solution include
- x < -10 or - x > 2 x ≥ -12 and x ≤ 4 the points between the endpoints found in the
x > 10 or x < -2 The solution is x ≥ -12 and x ≤ 4 , solution? When the original inequality is < or ≤,
The solution is x > 10 or x < -2. or ≤ x ≤ 4 . the compound inequality is an “and” statement, so
-12
- - - its graph includes the intersection of two graphs.
6 4 2 0 2 4 6 8 10 12 14
- - -
12 8 4 0 4 These graphs intersect between the endpoints.
Reflect INTEGRATE TECHNOLOGY
⎜ ⎟ ⎜ ⎟
7. In Part A, suppose the inequality were 4 - x + 15 > 14 instead of 4 - x + 15 > 21.
How would the solution change? Explain. A graphing calculator can be used to check the
⎜ ⎟
The first step in solving would be to subtract 15 from both sides and get 4 - x > -1.
solution graph for an inequality. For example,
At this point, the solving process can stop, because the absolute value of every number is
you would graph y = |4 – x| + 15 > 21 to check the
©
H
greater than -1. So, the solution is all real numbers. o
u solution for |4 – x| + 15 > 21. The graph will be a
gh
t
o
n
M broken horizontal line above the x-axis with
i
f
f
l
i
n
8. In Part B, suppose the inequality were x + 4 - 10 ≤ -11 instead of x + 4 - 10 ≤ -2. H
⎜ ⎟ ⎜ ⎟ endpoints at –2 and 10. You must interpret the graph
ar
How would the solution change? Explain. c
o
ur to be open at the endpoints. So, the solution graph is
The first step in solving would be to add 10 to both sides and get x + 4 ≤ -1. At this t
⎜ ⎟ P
ub
l x < –2 or x > 10.
i
s
point, the solving process can stop, because there are no real numbers whose absolute h
i
n
g
C
value is less than or equal to -1. So, there is no solution. o
mp
an
y AVOID COMMON ERRORS
Some students confuse when to use or and and when
rewriting an absolute value inequality. Remind
students that when the inequality is |ax + b| < c or
|ax + b| ≤ c, they should rewrite the inequality using
Module 2 90 Lesson 3 and. When the inequality is |ax + b| > c or
DIFFERENTIATE INSTRUCTION |ax + b| ≥ c, they should rewrite the inequality using
or. Emphasize the importance of checking some of
A2_MNLESE385894_U1M02L3.indd 90 15/05/14 5:17 PM the solutions in the original inequality to help avoid
Visual Cues
You may want students to use visual models to help them understand some simple this error.
inequalities as well as some real-world inequalities. For simple inequalities of the
form |x| < a or |x|> a, constructing a graph of possible solutions on a number
line as a first step may be helpful. For a real-world problem, involving allowable
tolerance, for example, drawing an unlabeled conjunction graph centered at the
middle value with the endpoints representing the minimum and maximum
tolerance values may be most helpful. This should help students visualize how to
construct an inequality based on the graph.
Solving Absolute Value Inequalities 90
