Chabot Mathematics
§4.3b AbsVal
InEqualities
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Review § 4.3
MTH 55
 Any QUESTIONS About
• §4.3a → Absolute Value
 Any QUESTIONS About HomeWork
• §4.3a → HW-13
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Solving AbsVal InEqual with <

Solving Inequalities in the
Form |x| < a, where a > 0
1. Rewrite as a compound inequality
involving “and”: x > −a AND x < a.
•
Can also write as: −a < x < a
2. Solve the compound inequality.

Similarly, to solve |x|  a, we would
write x  −a and x  a (or −a  x  a)
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & <
 Given InEquality: |x − 3| < 6
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
• |x – 3| < 6 → x − 3 > −6 and x − 3 < 6
• thus
−6 < x − 3 < 6
– ReWritten as Compound InEquality
• So
Chabot College Mathematics
4
−3 < x < 9
(add +3 to all sides)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & <
 SOLUTION: |x − 3| < 6
• Thus the Solution → −3 < x < 9
• Solution in Graphical Form
-10 -9
-8
-7
-6
-5
-4
(
-3
)
-2
-1
0
1
2
3
4
5
6
7
8
9
10
• Set-builder notation: {x| −3 < x < 9}
• Interval notation: (−3, 9)
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x − 3| + 8 < 5
 Given InEquality: |2x − 3| + 8 < 5
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
 Isolate the absolute value
• |2x − 3| + 8 < 5
• |2x − 3| < −3 ???
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x − 3| + 8 < 5
 The InEquality Simplified to:
|2x − 3| < −3
 Since the absolute value cannot be less
than a negative number, this inequality
has NO solution: Ø
• No Graph
• Set-Builder Notation → {Ø}
• Interval notation: We do not write interval
notation because there are no values
in the solution set.
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Solving AbsVal InEqual with >

Solving Inequalities in the
Form |x| > a, where a > 0
1. Rewrite as a compound inequality
involving “or”: x < −a OR x > a.
2. Solve the compound inequality

Similarly, to solve |x|  a, we would
write x  −a or x  a
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & >
 Given InEquality: |x + 7| > 5
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION:
 convert to a compound inequality and
solve each
• |x + 7| > 5 →
• x + 7 < −5
Chabot College Mathematics
9
or
x+7>5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & >
 SOLUTION : |x + 7| > 5
• x + 7 < −5
or
x+7>5
– The Addition Principle Produces Solutions
•
x < −12
• The
Graph
or
)
x > −2
(
-15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
1
2
3
4
5
• Set-builder notation: {x| x < −2 or x > −2}
• Interval notation: (−, −12) U (−2, ).
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |4x + 7| – 9 > –12
 Given InEquality: |4x + 7| − 9 > −12
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
 Isolate the absolute value
• |4x + 7| – 9 > –12
• |4x + 7| > –3 ???
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |4x + 7| − 9 > – 12
 The InEquality Simplified to:
|4x + 7| > −3
 This inequality indicates that the absolute
value is greater than a negative number.
Since the absolute value of every real
number is either positive or 0, the solution
set is All Real Numbers, 
• The Graph is then the entire Number Line
• Set-builder notation: {x|x is a real number}
• Interval notation: (−, )
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Summary: Solve |ax + b| > k
 Let k be a positive real number, and p
and q be real numbers.
 To solve |ax + b| > k, solve the following
compound inequality
ax + b > k OR ax + b < −k.
 The solution set is of the form
(−, p)U(q, ), which consists of two
Separate intervals.
Chabot College Mathematics
13
p
q
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| > 5
 By the Previous Slide this absolute
value inequality is rewritten as
2x + 3 > 5
or
2x + 3 < −5
 The expression 2x + 3 must represent a
number that is more than 5 units from 0
on either side of the number line.
• Use this analysis to solve the compound
inequality Above
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| > 5
 Solve the Compound InEquality
2x + 3 > 5
or
2x + 3 < −5
2x > 2
or
2x < −8
x>1
or
x < −4
 The solution set is (–, –4)U(1, ). Notice
that the graph consists of two intervals.
–5
–4
Chabot College Mathematics
15
–3
–2
–1
0
1
2
3
4
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Summary: Solve |ax + b| < k
 Let k be a positive real number, and p
and q be real numbers.
 To solve |ax + b| < k, solve the threepart “and” inequality
–k < ax + b < k
 The solution set is of the form (p, q),
a single interval.
p
Chabot College Mathematics
16
q
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| < 5



By the Previous Slide this absolute
value inequality is rewritten as
−5 < 2x + 3
and
2x + 3 < 5
In 3-Part form
−5 < 2x + 3 < 5
Solving for x
−5 < 2x + 3 < 5
−8 < 2x < 2
−4 < x < 1
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| < 5
 Thus the Solution: −4 < x < 1
 We can Check that the solution set is
(−4, 1), so the graph consists of the
single Interval:
–5
–4
Chabot College Mathematics
18
–3
–2
–1
0
1
2
3
4
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Caution for AbsVal vs <>

When solving absolute value
inequalities of the types > & <
remember the following:
1. The methods described apply when the
constant is alone on one side of the
equation or inequality and is positive.
2. Absolute value equations and absolute
value inequalities of the form |ax + b| > k
translate into “or” compound statements.
Chabot College Mathematics
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Caution for AbsVal vs <>

When solving absolute value
inequalities of the types > & <
remember the following:
3. Absolute value inequalities of the form
|ax + b| < k translate into “and” compound
statements, which may be written as
three-part inequalities.
4. An “or” statement cannot be written in
three parts. It would be incorrect to use
−5 > 2x + 3 > 5 in the > Example, because
this would imply that − 5 > 5, which is false
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Absolute Value Special Cases
1. The absolute value of an
expression can never be
negative: |a| ≥ 0 for ALL
real numbers a.
2. The absolute value of an
expression equals 0 ONLY when
the expression is equal to 0.
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |2n + 3| = −7
 Solution:
• See Case 1 in the preceding slide.
• Since the absolute value of an expression
can never be negative, there are NO
solutions for this equation
• The solution set is Ø.
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |6w − 1| = 0
 Solution:
• See Case 2 in the preceding slide.
• The absolute value of the expression
6w − 1 will equal 0 only if
6w − 1 = 0
• The solution of this equation is 1/6. Thus,
the solution set of the original equation is
{1/6}, with just one element.
– Check by substitution.
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x| ≥ −2
 Solution:
• The absolute value of a number is
always greater than or equal to 0.
• Thus, |x| ≥ −2 is true for all real numbers.
• The solution set is then entire number line:
(−, ).
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x + 5| − 1 < −8
 Solution:
• Add 1 to each side to get the absolute
value expression alone on one side.
|x + 5| < −7
• There is no number whose absolute value
is less than −7, so this inequality has no
solution.
• The solution set is Ø
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x − 9| + 2 ≤ 2
 Solution:
• Subtracting 2 from each side gives
|x − 9| ≤ 0
• The value of |x − 9| will never be less than
0. However, |x − 9| will equal 0 when x = 9.
• Therefore, the solution set is {9}.
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
WhiteBoard Work
 Problems From §4.3 Exercise Set
• 88 (ppt), 94 (ppt), 98 (ppt), 56, 62, 78

Albany, NY
Yearly
Temperature
Data
Chabot College Mathematics
27
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-88  |4 − x| < 5
 Solve |4 − x| < 5 by x-y Graph
 GRAPH SOLUTION:
 On graph find the region where the
y = f(x)= |4 − x| function lies
BELOW (i.e., is Less Than) the
y = f(x) = 5 Horizontal Line
Chabot College Mathematics
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-88  |4 − x| < 5
 Ans in SET & INTERVAL Form
• {x| −1 < x < 9}
• (−1, 9)
(
Chabot College Mathematics
29
)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-88  |4 − x| < 5
 Solve |4 − x| < 5
• 4 −x < 5
and
4−x > −5
• −x < 1
and
−x > −9
– Multiply Both expressions by −1, REMBERING to
REVERSE the direction of the InEqual signs
• x>−1
and
x < 9 OR
• −1 < x
and
x<9
• So the Compound Soln: (−1 < x < 9)
• In Set notation: {x| −1 < x < 9}
• Interval notation: (−1, 9)
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-94  Albany, NY Temps
 InEquality for Albany, NY Monthly Avg
Temperature, T, in °F
|T − 50 °F| ≤ 22 °F
 Solve and Interpret
 SOLUTION
• |T−50| ≤ 22 → −22 ≤ T−50 and T−50 ≤ 22
• Or in 3-part form −22 ≤ T−50 ≤ 22
• Add 50 to each Part (addition principle)
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-94  Albany, NY Temps
 Albany, NY Temps → |T − 50 °F| ≤ 22 °F
 SOLUTION
• (50−22) ≤ (T−50+50) ≤ (22+50)
• Or ANS → 28 ≤ T ≤ 72
 INTERPRETATION:
• The monthly avg temperature in Albany, NY
ranges from 28 °F in Winter to 72 °F in
Summer
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-98  Machining Tolerance
 Length, x, of a Machine Part in cm
|x − 9.4cm| ≤ 0.01cm
 Solve and Interpret
 SOLUTION
• |x−9.4| ≤ 0.01 →
−0.01 ≤ x−9.4 and x−9.4 ≤ 0.01
• Or in 3-part form −0.01 ≤ x−9.4 ≤ 0.01
• Add 9.4 to each Part (addition principle)
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-92  Machining Tolerance
 Machine Part Tolerance →
|x − 9.4cm| ≤ 0.01cm
 SOLUTION
• (−0.01+9.4) ≤ (x−9.4+9.4) ≤ (0.01+9.4)
• Or ANS → 9.39 ≤ x ≤ 9.41
 INTERPRETATION:
• The expected dimensions of the finished
machine part are between 93.9mm and
94.1mm (i.e. the dim is 94mm ±0.1mm)
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
All Done for Today
JR Ewing
From
“Dallas”
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
37
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
4
5
6
Weather UnderGnd – Albany, NY
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt