A Conceptual Approach to
S
itting in the back of Ms. Corey’s sixthgrade mathematics class, I enjoyed seeing
students enthusiastically demonstrate
their understanding of absolute value. On
the giant number line on the classroom
floor, they counted the steps that they needed to
take to get back to zero. The old definition of absolute value of a number as its distance from zero—
learned by students and teachers of the previous
generation—has long ago been replaced with this
algebraic statement: |x| = x if x > 0 or – x if x < 0.
The absolute value learning objective in high school
mathematics requires students to solve far more
complex absolute value equations and inequalities.
Using number lines sweeps away
the mystery of working with
absolute values and empowers
students to make connections
between procedures and concepts.
However, I cannot remember students attacking
the task with enthusiasm or having any understanding beyond “make the inside positive.”
The desire to rekindle in students the conceptual
understanding and mathematical success with absolute value seen in Ms. Corey’s classroom prompted
consideration of a different approach to the topic:
combining the number-line distance definition with
the idea of using transformations of a “parent” function. This approach connects the visual representa-
tion of the number line and the verbal context of
distance from a fixed point to the abstract symbolic
form, thus preventing the process from becoming a
series of steps to be memorized and followed blindly.
What has been the result of this more conceptual
approach? We have witnessed an increased sense
of competence and confidence in students: They
can make sense of and correctly solve absolute
value problems. In classroom after classroom where
this approach has been introduced, students have
demonstrated mastery and achieved competency
more quickly than students taught through the
traditional two-case approach. Collaborating with
colleagues and getting student feedback has allowed
for valuable refinements and resulted in the method
described here.
Let’s first examine the approach that is commonly used to introduce students to solving absolute value equations and inequalities. Consider the
following problem: |2x – 1| = 7. Typically,
students are taught to make this into two equations, 2x – 1 = 7 and 2x – 1 = –7, and solve each
for x. The justification given is that since absolute
value means “distance from zero,” this distance
could be in either a positive or a negative direction.
Although mathematically valid, this method does
not make sense conceptually beyond the simple
case of |x| = c where c is any positive real number.
For instance, what does it mean for an expression
such as 2x – 1 to have a distance of 7 from zero?
When absolute value problems become more complex, students often do not have sufficient conceptual understanding to make any sense of what is
happening mathematically.
We suggest that the absolute value concept can
be more powerfully leveraged by teaching students
to use a transformational approach to make sense
592 MatheMatics teacher | Vol. 104, No. 8 • april 2011
Copyright © 2011 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
hioB/istockPhoto.coM
Absolute Value
and
–6
–4
–2
0
2
4
Equations
Inequalities
of the meaning of the solution set. This method
has been used with middle school and high school
students with remarkable success. When students
understand the meaning of absolute value and
can transform an equation or inequality so that
the coefficient of the variable is 1, they are able to
find the solution set easily by (a) determining the
location of the critical points that are equidistant
from the “anchor point” (which represents an
offset from zero) and (b) identifying whether the
solutions fall on, between, or beyond these points.
What follows is a primer for using this approach.
INTRODUCING THE CONCEPT
OF ABSOLUTE VALUE
As mentioned, students’ first introduction to
absolute value is as the distance from zero. Thus,
|–5| = 5 and |3| = 3 might be interpreted, respectively, as “the distance from –5 to zero is 5” and
“the distance from 3 to zero is 3.” This interpretation provides a conceptual basis for understanding
absolute value. However, many textbooks (and
teachers) will move away from this conceptual basis
when students begin learning to solve absolute value
equations and inequalities. The transformational
approach extends the initial conceptual understanding to realizing that the distance between two values
is the absolute value of their difference—that is, the
distance from x to b is |x – b|.
6
Mark W. ellis and
Janet L. Bryson
distance apart. For instance, “My friend Khiem
lives at 15 Sycamore Lane, and I live 8 houses away
[assuming that house numbers change by units of
1]. Where could my house be located?” Students
readily figure out that the solution is a matter of
moving 8 units from 15 in either the positive or the
negative direction. To help students connect this
idea to absolute value, ask them to make a visual
representation of this problem on a number line
(see fig. 1).
Next, ask students how we might write a mathematical equation for this visual representation.
With some help, students are usually able to come
up with |x – 15| = 8 as a translation of the sentence
“The distance from my house x to house 15 is 8.” In
this way, students have an opportunity to connect
a meaningful context to a visual model, a verbal
description, and a symbolic representation, thus
leading to deeper conceptual understanding (NCTM
2000). The appendix (p. 596) provides practice
with this idea. (These additional problems are also
posted online at www.nctm.org/mt.)
When creating the visual model on a number
line, it is essential to locate the “anchor point,”
b. Conceptually, this anchor point represents the
“offset” from zero and is the point from which the
two critical values are found by moving c units in
the positive and negative directions (see
WORKING WITH ABSOLUTE VALUE
EQUATIONS
Students must first learn to connect the symbolic
expression |x – b| = c to the verbal phrase “x is c
units from b in either direction.” One analogy that
teachers have used to help students with this concept is to talk about friends’ homes being a certain
Fig. 1 a graph helps students visualize the solution.
Vol. 104, No. 8 • april 2011 | MatheMatics teacher 593
Fig. 2 Points can be found c units from b.
Fig. 4 The distance from 3 must be greater than 4.
|x – b| = |– 1(b – x )|
= | –1| • |b – x|
= 1 |b – x|
= |b – x|
Fig. 3 Here we move two units from –5.
fig. 2). Students can then write the solution from
the critical values plotted: x = {b – c, b + c}. This
conceptual approach first solves the equation visually as distances on a number line and then records
the solution in algebraic notation if needed.
Following this introduction of the case |x – b|
= c when b > 0, students are asked to consider a
problem in the format |x + b| = c. For instance, how
might |x + 5| = 2 be interpreted verbally as a problem of distance? Given time to discuss their thinking in pairs or small groups, students will typically
arrive at the idea that the expression x + 5 can be
thought of as x – (–5), thus establishing the anchor
point of –5 from which some distance is being
measured. If students do not readily arrive at this
idea, it may be helpful to ask a simpler question:
“When x is located at the anchor point, the distance
between x and the anchor point is 0. Determining
the anchor point is equivalent to determining the
value of x such that |x + 5| = 0. What value of x will
make the absolute value equation equal 0?” From
here, students are quickly able to determine that
the anchor point is –5.
Transforming the original equation, we get
|x – (–5)| = 2, which can be expressed verbally as
“x is 2 units from –5 in either direction.” Visually,
the anchor point is placed at –5, and the solutions
are found by moving 2 units in either direction (see
fig. 3). Algebraically, x = {(–5 – 2), (–5 + 2)} or,
simplified, x = {–7, –3}.
The connection between absolute value and
distance also provides an unexpected benefit: It
eliminates the difficulty students have in equating
|x – b| and |b – x|. Students whose understanding
is restricted to the procedural algorithm must mentally think through the distributive property and
the property of opposites and apply the definition
of absolute value before they can grasp or retain the
equivalence:
594 Mathematics Teacher | Vol. 104, No. 8 • April 2011
In contrast, when set in the context of distance, students naturally “know” that these expressions are
equal, just as they “know” without question that
the distance from home to school is the same as the
distance from school back to home.
WORKING WITH ABSOLUTE VALUE
INEQUALITIES
This work extends naturally to absolute value
inequalities. Again, there is no need to create
two inequalities when using the transformational
approach. Take a simple example: |x – 3| > 4. This
is read as, “The distance from x to 3 is greater than
4.” Visually, it is modeled as shown in figure 4.
From the representation, students can determine
that the solution is x > 7 or x < –1. When the inequality symbol is the is-less-than symbol, the problem
becomes finding points whose distance is less than
a given value from the anchor point. Students readily catch on to this approach as an extension to their
understanding of absolute value equations. The
appendix (p. 597) provides practice with this idea.
Encourage students to check their solutions by substituting values within each of the graph’s intervals,
a technique they will use later when they learn to
examine critical points of a function in calculus.
HANDLING NONZERO COEFFICIENTS
OTHER THAN 1
When the coefficient (let’s call this A) of the variable is other than 1, it is simplest for students to
treat A as if it were 1 to begin with. This approach
can also be thought of as substituting a single variable with coefficient 1 for something more complex,
a strategy students will use later in trigonometry
and calculus. The problem is then worked out in
the manner described above. Then, because these
solutions represent not x but Ax, the original coefficient is brought back. The last step is to divide each
solution by the coefficient, A.
Let’s examine this approach with the example
|3x – 7| = 5. First, take 3x to be a single unit and
Fig. 5 Taking 3x as a single unit will help students.
CONNECTING THE TRANSFORMATIONAL
APPROACH TO LATER TOPICS
think, “The distance from 3x to 7 is 5.” This step
leads to the representation shown in figure 5.
This represents the solution for 3x, so the solutions 2 and 12 are then divided by the coefficient 3
to obtain x = {2/3, 4} (with each of these representing a distance of 5/3 from 7/3). We have found that
students have more success doing the division after
setting up the visual representation with the coefficient A intact. Errors from incorrectly adding or
subtracting fractional values are virtually eliminated.
The variable substitution approach also eliminates errors that commonly result when the coefficient of x is negative. Students read problems in
the form |b – Ax| > c as they would those in the
form |Ax – b | > c—that is, as “Ax is farther than c
units away from b.” They also realize that because
Ax + b = b + Ax by the commutative property of
addition, |b + Ax| < c and |Ax + b| < c may both be
understood as the values of Ax that are closer than
c units from –b. The appendix (p. 598) provides
practice in solving absolute value inequalities when
the coefficient of x is not equal to 1.
The transformational approach, which builds
understanding by starting from a parent function,
is often advocated for students’ learning about
functions (e.g., Heid and Blume 2008; Ward 2001).
The approach we describe here precedes students’
formal study of functions and can potentially set
the stage for later applications of such an approach.
Teachers who have used this approach have
reported tremendous success with students, who
have commented that this approach is often perceived to be “too easy to be math.”
REFERENCES
Heid, M. Kathleen, and Glendon W. Blume, eds.
Research on Technology and the Teaching and Learning of Mathematics: Syntheses, Cases, and Perspectives. Charlotte, NC: Information Age Publishing/
National Council of Teachers of Mathematics,
2008.
National Council of Teachers of Mathematics
(NCTM). Principles and Standards for School Mathematics. Reston, VA: NCTM, 2000.
Ward, Cherry D. “Under Construction: On Becoming
a Constructivist in View of the Standards.” Mathematics Teacher 94, no. 2 (February 2001): 94–96.
EXTENDING STUDENTS’ UNDERSTANDING
As a way to allow students to develop a sense of
ownership of their newly formed knowledge, ask
them to build their own absolute value equality or
inequality and show the solution set. The following
steps can be used for this process. Afterward, have
students challenge one another to solve the equations or inequalities created by their classmates.
After students explore the impact of coefficients
between 0 and 1 and those greater than 1, ask them
to generate a conjecture about what happens and
a justification as to why. A further extension asks
students to use the visual representation on the
number line and the verbal descriptions that they
generated to explain conceptually why the solution
set for |b – Ax| > c is equal to that of |Ax – b | > c.
Meiko Shimura
Step 1: Choose a “parent” inequality centered at zero.
Step 2: Use a coefficient to expand (stretch) or compress (shrink) the solution set.
Step 3: Use a constant to translate (slide) the solution set.
Step 4: Verify that the final solution set matches the
original equality or inequality created.
For PDFs of the additional problems given in the
appendix, go to the Mathematics Teacher Web site:
www.nctm.org/mt.
MARK W. ELLIS, [email protected],
a National Board Certified Teacher in
early adolescence mathematics, has
taught middle school and high school
mathematics in California public
schools. He is an associate professor
and chair of the secondary education
department at California State University, Fullerton. JANET L. BRYSON, mathcoach
[email protected], a former high school mathematics teacher and mathematics coach, provides
professional development training by sharing her
expertise and passion for mathematics content
and pedagogy done differently.
Vol. 104, No. 8 • April 2011 | Mathematics Teacher 595
Appendix
Introduction to Absolute Value Equations
I know that …
• The absolute value of a number means its distance from zero.
• The distance from x to b is equivalent to the distance from b to x because |x – b| = |b – x|.
• If |x – b| = c, then x is c units from b in either direction.
For each equation, write a sentence in words, draw a graph, and identify the solution(s).
1. | x | = 3
–6 –4 –2
2. | x | = 2
0
2
4
6
–6 –4 –2
Words:_The distance from x
to 0 is
x = _____ or x = ______
.
4. |x – 4| = 2
–6 –4 –2
0
2
4
3. | x – 0 | = 5
0
4
6
–6 –4 –2
0
2
4
6
Words: ______________________ _____________________________ x = _____ or x = ______
Words: ______________________
_____________________________
x = _____ or x = _____
5. |4 – x| = 2
6. |x + 1| = 5
6
–6 –4 –2
Words:_______________________
_____________________________
–6 –4
–4 –2
–2 00 22 4 56 8
2x –6
x = _____ or x = ______
2
0
2
4
6
–6 –4 –2
0
2
4
6
Words: ______________________
_____________________________
–6 –4 –2 0 2 4 6
x = _____ or x = ______
Words: ______________________
_____________________________
–6 –4 –2 0 2 4 6
x = _____ or x = _____
7. |x + 2| = 3
8. |x – 5| = 0
9. |x + 3| = –4
Words:_______________________
Words: ______________________
Words: ______________________
2x
4x
0
x
1
3x
2x
–6 –4 –2
0
x
x
2x
–6 –4 –2
00 21
2
5/2
4
5
8
5/2
5
84
x = _____ or x = ______
x = _____ or x = ______
x
–6 –4 –2
x = _____ or x = _____
0
2
4x 6
–6 –4 –2
0
10. | x – b | = c
Words: _______________________________________________________________________________________
0 1 first?
5/2 _____
4
xWhich value is plotted
Which value tells the distance? ____ Graph the solution.
2x
3x
What
can x equal? x =_____ or x = ______
3x
3x
x
2x
x
4x
x
x
3x
x
2x
x
4x
x
6x
x
5x
2x
x
3x
x
2x
–6 –4 –2
0
2 x 5
2x
0
1
596 Mathematics Teacher | Vol. 104, No. 8 • April 2011
x
3x
x5/2
8
4
2
4
6
Absolute Value Inequalities
I know that …
• The distance from x to b is equivalent to the distance from b to x because |x – b| = |b – x|.
• If |x
– b|–2< c,0then
from
in either
–6 –4
2 x4is less
6 than c units –6
–4b–2
0 2 direction.
4 6
–6 –4 –2 0 2 4 6
• If |x – b| > c, then x is greater than c units from b in either direction.
For each inequality, write a sentence in words, draw a graph, and identify the solution(s).
1. |x| > 2
2. |x – 0| < 5
3. |x – 3| > 4
Words:_The
–6 –4 –2distance
0 2 from
4 6x to 0 is
Solution: x < _____ or x > _____
Words:
_____________________
–6 –4
–2 0 2 4 6
____________________________
Solution: _______ < x < _______
Words:
____________________
–6 –4
–2 0 2 4 6
___________________________
Solution: x < ____ or x > _____
4. |x–6+ 1|
3 0
–4<–2
2x
2
5
8
5. |x – 4| > 2
6. |x + 1| < 4
0
1
5/2
4
Words:_____________________
–6 –4 –2 0 2 4 6
____________________________
Solution: ___________________
Words:
_____________________
–6 –4
–2 0 2 4 6
____________________________
Solution: ___________________
Words:
____________________
–6 –4
–2 0 2 4 6
___________________________
Solution: ___________________
7. |x–6– 2|
3 0
–4>–2
2x
3x
2
8. |2 – x| > 3
9. |x + 3| > –1
0
1
x
x
x
5
8
2x
5/2
4
Words:_____________________
____________________________
Solution: ___________________
4x
x
–6 –4 –2 0 x2Words:
4 6 ____________________
–6 –4 –2 0 2 4
Words: _____________________
____________________________
___________________________
Solution: ___________________
Solution: ___________________
10. –6
Graph
the solutions
–4 –2
0 2
5for x8if |x – b| < c.
2x
2x
3x
Which value (b or c) is plotted first? ____
3x
2x
–6 –4 –2
0
23x
4x 4
6
–6 –4 –2
0
Which value tells the distance? ___
x
1 5/2x-values
4
xxExplain why the0 possible
are xin one continuous region or two separate
regions. __________________
x
x
x
_____________________________________________________________________________________________
11. Graph the solutions for x if |x – b| > c.
–6 –4
–2 separate
0 2x
2
5
8 ________________
2xregion
Explain why the possible x-values are5x
in one continuous
or two
regions.
6x
2x
3x
3x
_____________________________________________________________________________________________
3x
2x
4x
_____________________________________________________________________________________________
x
x
x
x
x
x
x
2x
2x
6x
2xx:
5x
3x
x
3x
x
x
x
x
x
–6 –4 –2
0
1x
x
x
5/2
4
0
2
5
8
0
12x 5/2
3x
4
2x
Vol. 104, No. 8 • April 2011 | Mathematics Teacher 597
x
x
x
2
4
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–4 –2 0 2 Inequalities
4 6
More–6Complicated
Example: |5 – 2x | < 3
2x
–6 –4 –2
x
0
2
5
8
0
1
5/2
4
2x
–6 –4 –2
0
2
5
8
2x
–6 –4 –2
0
2
5
8
x
2x
–6 –4 –2 00 12
5/2
5
48
Graph 2x : All points 3 or fewer units away from 5; 5 – 3 = 2 and
5 + 3 = 8 mark the endpoints
Solution for 2x : All points between and including 2 and 8.
Graph x. Divide the solution for 2x by 2: 2/2 = 1 and 8/2 = 4.
Solution for x : All points between 1 and 4, inclusive
|5 – 2x | < 3 → 5 – 3 < 2x < 5 + 3
2 < 2x < 8
1< x <4
Solve
the following
0 absolute
1 5/2 value
4 inequalities first for Ax and then for x. Draw the graph and write the
x
3x
4x
algebraic
solution statement for each of2xthe following problems:
x
0
1
5/2
4
x |3x| > 2
1.
2.x |2x – 0| < 5
3.x |4x – 2| ≥ 6
3x
2x
4x
3x
x
3x
2x
x
Solution:
___________________
2x
x
2x
3x
x
Solution:
___________________
4x
x
4x
3x
x
Solution:
___________________
x
x
x
x
x
x
4.
2x |2x – 1| > 3
5.
3x |3x – 4| < 4
6. |4 – 3x| < 4
3x
2x
x
2x
6x
x
3x
x
3x
5x
x
3x
x
3x
2x
x
Solution:
___________________
x
x
Solution:
___________________
x
x
Solution:
___________________
x
6x
5x
2x
7.
6x |6x + 2| > 4
8.
5x |7 – 5x| < 3
x
5x
x
9. |2x + 3| > 5
2x
x
2x
x
x
x
x
2x
6x
x
x:
x
2x
Solution:
___________________
Solution: ___________________
2x
x:
2x
x:
x:
598 MatheMatics teacher | Vol. 104, No. 8 • april 2011
x
Solution: ___________________
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© 2011 The College Board.
A Conceptual Approach to Absolute Value
Equations and Inequalities Mark W. Ellis and Janet L. Bryson
Appendix
Introduction to Absolute Value Equations
I know that …
• The absolute value of a number means its distance from zero.
• The distance from x to b is equivalent to the distance from b to x because |x – b| = |b – x|.
• If |x – b| = c, then x is c units from b in either direction.
For each equation, write a sentence in words, draw a graph, and identify the solution(s).
1. | x | = 3
–6 –4 –2
2. | x | = 2
0
2
4
6
–6 –4 –2
Words:_The distance from x
to 0 is x = _____ or x = ______
.
4. |x – 4| = 2
–6 –4 –2
0
2
4
0
2
4
6
–6 –4 –2
0
2
4
6
Words: ______________________ _____________________________ x = _____ or x = ______
Words: ______________________
_____________________________
x = _____ or x = _____
5. |4 – x| = 2
6. |x + 1| = 5
6
–6 –4 –2
Words:_______________________
_____________________________
–6 –4
–4 –2
–2 00 22 4 56 8
2x –6
x = _____ or x = ______
3. | x – 0 | = 5
0
2
4
6
–6 –4 –2
0
2
4
6
Words: ______________________
_____________________________
–6 –4 –2 0 2 4 6
x = _____ or x = ______
Words: ______________________
_____________________________
–6 –4 –2 0 2 4 6
x = _____ or x = _____
7. |x + 2| = 3
8. |x – 5| = 0
9. |x + 3| = –4
Words:_______________________
Words: ______________________
Words: ______________________
2x
4x
0
x
1
3x
2x
–6 –4 –2
0
x
x
2x
–6 –4 –2
00 21
2
5/2
4
5
8
5/2
5
84
x = _____ or x = ______
x = _____ or x = ______
x
–6 –4 –2
x = _____ or x = _____
0
2
4x 6
–6 –4 –2
0
10. | x – b | = c
Words: _______________________________________________________________________________________
0 1 first?
5/2 _____
4
xWhich value is plotted
Which value tells the distance? ____ Graph the solution.
2x
3x
What
can x equal? x =_____ or x = ______ 3x
3x
x
4x
x
2x
x
Mathematics
Teacher | Vol. 104, No. 8 • April 2011x
x
3x
2x
x
4x
2
4
6
Absolute Value Inequalities
I know that …
• The distance from x to b is equivalent to the distance from b to x because |x – b| = |b – x|.
• If |x
– b|–2< c,0then
from
in either
–6 –4
2 x4is less
6 than c units –6
–4b–2
0 2 direction.
4 6
–6 –4 –2 0 2 4 6
• If |x – b| > c, then x is greater than c units from b in either direction.
For each inequality, write a sentence in words, draw a graph, and identify the solution(s).
1. |x| > 2
2. |x – 0| < 5
3. |x – 3| > 4
Words:_The
–6 –4 –2distance
0 2 from x
4 6
to 0 is Solution: x < _____ or x > _____
Words:
_____________________
–6 –4
–2 0 2 4 6
____________________________
Solution: _______ < x < _______
Words:
____________________
–6 –4
–2 0 2 4 6
___________________________
Solution: x < ____ or x > _____
4. |x–6+ 1|
3 0
–4<–2
2x
2
5
8
5. |x – 4| > 2
6. |x + 1| < 4
0
1
5/2
4
Words:_____________________
–6 –4 –2 0 2 4 6
____________________________
Solution: ___________________
Words:
_____________________
–6 –4
–2 0 2 4 6
____________________________
Solution: ___________________
Words:
____________________
–6 –4
–2 0 2 4 6
___________________________
Solution: ___________________
7. |x–6– 2|
3 0
–4>–2
2x
3x
2
8. |2 – x| > 3
9. |x + 3| > –1
0
1
x
x
x
5
8
2x
5/2
4
Words:_____________________
____________________________
Solution: ___________________
4x
x
–6 –4 –2 0 x2Words:
4 6 ____________________
–6 –4 –2 0 2 4
Words: _____________________
____________________________
___________________________
Solution: ___________________
Solution: ___________________
10. Graph
the solutions
–6 –4 –2
0 2
5for x8if |x – b| < c. 2x
2x
3x
Which value (b or c) is plotted first? ____
3x
2x
–6 –4 –2
0
23x
4x 4
6
–6 –4 –2
0
Which value tells the distance? ___ x
1 5/2x-values
4
xxExplain why the0 possible
are xin one continuous region or two separate
regions. __________________
x
x
x
_____________________________________________________________________________________________
11. Graph the solutions for x if |x – b| > c. –6 –4
–2 separate
0 2x
2
5
8
2xregion
Explain why the possible x-values are5x
in one continuous
or two
regions. ________________
6x
2x
3x
3x
_____________________________________________________________________________________________
3x
2x
4x
_____________________________________________________________________________________________
x
x
x
x
x
x
x
2x
2x
6x
2xx:
5x
3x
x
3x
x
x
x
x
x
–6 –4 –2
0
1x
x
x
5/2
4
0
2
5
8
0
12x 5/2
3x
4
2x
Vol. 104, No. 8 • April 2011 | Mathematics Teacher 3
x
x
x
2
4
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–6 –4 –2
0
2
4
6
–4 –2 0 2 inequalities
4 6
More–6Complicated
Example:|5–2x |<3
2x
–6 –4 –2
x
0
2
5
8
0
1
5/2
4
2x
–6 –4 –2
0
2
5
8
2x
–6 –4 –2
0
2
5
8
x
2x
–6 –4 –2 00 12
5/2
5
48
Graph2x :Allpoints3orfewerunitsawayfrom5;5–3=2and
5+3=8marktheendpoints
Solutionfor2x :Allpointsbetweenandincluding2and8.
Graphx. Dividethesolutionfor2xby2:2/2=1and8/2=4.
Solutionforx:Allpointsbetween1and4,inclusive
|5–2x |<3→5–3<2x <5+3
2<2x <8
1< x <4
SolvethefollowingabsolutevalueinequalitiesfirstforAxandthenforx.Drawthegraphandwritethe
0 1 5/2 4
x
3x
2x
4x
algebraicsolutionstatementforeachofthefollowingproblems:
x
0
1
5/2
4
x
1.|3x|>2
x
2.|2x–0|<5
x
3.|4x–2|≥6
3x
2x
4x
3x
x
3x
2x
x
Solution:___________________
2x
x
2x
3x
x
Solution:___________________
4x
x
4x
3x
x
Solution:___________________
x
x
x
x
x
x
4.|2x–1|>3
2x
5.|3x–4|<4
3x
6.|4–3x|< 4
3x
2x
x
2x
6x
x
3x
x
3x
5x
x
3x
x
3x
2x
x
Solution:___________________
x
x
Solution:___________________
x
x
Solution:___________________
x
6x
5x
2x
7.|6x+2|>4
6x
8.|7–5x|<3
5x
x
5x
x
9.|2x+3|>5
2x
x
2x
x
x
x
x
2x
6x
x
x:
x
2x
Solution:___________________
Solution:___________________
2x
x:
2x
x:
x:
MatheMatics teacher | Vol. 104, No. 8 • april 2011
x
Solution:___________________