Section 1-6 Absolute Value
In this lesson you will write and solve equations and inequalities involving
absolute value.
Absolute Value is the distance a number is from zero on a
number line.
5 5
5  5
This is asking how far
away from 0 is 5 & -5 on
the number line
5 432 112 34 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Absolute value equation can have two answers, because
opposites have the same absolute value.
Solving absolute value
The value inside the
absolute value can
equal 15 or -15
What
does x
equal??
x  5  15
x  5  15
x  5  15
x  10
x  20
Solving an Absolute Value Equation
Ex)
3 x  2 1  8
1  1
3x2 9
3
3
STEPS to solving
Abs. Val. Equation
1.
2.
3.
4.
Get the absolute value alone
Rewrite as two equations
Solve
check
x2 3
x23
x  2  3
x 1
x  5
Solving absolute value
equation 6:00 video 1
ex) 1 try)
Solving an Absolute Value Equation
3x  2  4 x  5
Check each solution…
3x  2  4x  5
3x  2  4x  5
 3x
 4x
 3x
2  x5
5
5
3  x
x  3
 4x
7 x  2  5
2
2
7 x  7
7
7
x  1
An extraneous solution
is a solution that does
not work out when
checked in the original
problem.
Since -3 does not work
it is an extraneous
solution. The only
solution is -1
Practice solving Absolute value
equations:

Worksheet Practice C – #1,2,5,6
1) 6 ANSWERS:
& -6
2) ANSWERS:
7 & -7
5) 3 ANSWERS:
& -2
6) ANSWERS:
0&4

Pg 46 #11-23 odd
Solving an Absolute Value Inequality
STEPS to solving
Abs. Val. inequality
2x 1  5
2x 1  5 AND 2x 1  5
1
1
1
2 x  4
2x  6
2
2
1
\
2
2
x  2
x 3
2 x 3
AND
1. Get the absolute value alone
2. Rewrite as two inequalities(flip
inequality sign)
3. Solve
4. graph
Solving absolute value inequality
“OR” 6:30 video 1ex) 1try)
AND b/c inequality sign was <
-2
0
3
0
0
Most of the time the original inequality sign will tell you what the graph should look like
Practice

Worksheet Practice D- #1,2,5,6

1) y < 0ANSWERS:
AND y > -6
2) t ≤ 4ANSWERS:
AND t ≥ -1

5) x ≤ ½ANSWERS:
and x ≥ -1
6) z > 2 ANSWERS:
OR z < -2/3
Practice
Standard Pg. 46 #25-35odd 43-81
 Honors Pg. 46 #25-35odd 43-89


Standardized test prep: 90-93
LITERAL EQUATIONS

Define: literal equation – a literal equation is an
equation that uses at least two different variables.
• Example solve for the indicated variable
Circumference of a circle:
C  2r ; for r
C
r answer

2
P  2W
P  2W  2 L ; for L L  answer
2
What is the length of a rectangle with
a perimeter of 30ft and the width is 7ft.
Distance = rate * time
D  rt
Perimeter of rectangle:
solve for r
D
T
ranswer

If you need to travel 500 miles in 8 hours, what rate should you
maintain?
Practice B- #1-4

1)  S Answers
1  r
L

3)
9g
3  y
Answers
4
2)
3V
Answers
h
2
r
4) t 3 Answers
3 y
5
C  ( f  32)
 Example:
9
Fahrenheit = F and Celsius = C.

Describe what the formula does. (what’s the
input?)
Finds Temperature in
Celsius given Degrees
Answer
Fahrenheit. (input is F & output is C)

What is F in terms of C? (in other words
solve the function for F)
What dos this new formula do? (what’s the
input?)
Practice F #1-3

Answers
1) x=7

2) max: 16.02
Answers
Min: 15.98

3) 513 milesAnswers
(513.66….)
Word problems

Ex: A manufacturer has a 0.6 oz
tolerance for a bottle of salad dressing
advertised as 16oz. Write and solve an
absolute value inequality that describes
the acceptable volumes for “16oz”
bottles.
actual amt. - ideal amt.  tolerance
x 16  0.6
What does this mean?
x  16  0.6 AND x  16  0.6
x  16.6 AND x  15.4
15.4  x  16.6
If you buy a bottle of 16oz dressing, you might get one
that actually has 15.4 to 16.6oz instead.

Ex: A manufacturer has a tolerance of
0.36 lb for a bag of potting soil
advertised as 9.6 lb. Write and solve an
absolute value inequality that describes
acceptable weights for “9.6 lb” bags.
w  9.6  0.36
W - 9.6 ≤ 0.36
W ≤ 9.96
What does this mean?
or
or
w - 9.6 ≥ -0.36
w ≥ 9.24
Practice F #4 & 6

4) t Answers
350  5

.5  0.05
6) A  0Answers