Dmth3018 03

Section 1.1, Slide 1Copyright © 2014, 2010, 2006 Pearson Education, Inc. Section 8.3, Slide 1
Equations, Inequalities, and Systems
Revisited
8

R.1 Fractions
1. Use the distance definition of absolute value.
2. Solve equations of the form |ax + b| = k, for k > 0.
3. Solve inequalities of the form |ax + b| < k and of
the form |ax + b| > k, for k > 0.
4. Solve absolute value equations that involve
rewriting.
5. Solve equations of the form |ax + b| = |cx + d|.
6. Solve special cases of absolute value equations
and inequalities.
Objectives
8.3 Absolute Value Equations and Inequalities

Absolute Value
Recall that the absolute value of a number x, written |x|,
represents the distance from x to 0 on the number line.
0 5–5
For example, the solutions of |x| = 5 are 5 and –5.
5 units from zero. 5 units from zero.
x = –5 or x = 5

Absolute Value
0 5–5
More than
5 units from zero
More than
5 units from zero
Because the absolute value represents the distance from 0,
it is reasonable to interpret the solutions of |x| > 5 to be all
numbers that are more than 5 units from 0.
The set (-∞, –5) U (5, ∞) fits this description. Because the
graph consists of two separate intervals, the solution set is
described using or as x < –5 or x > 5.

Absolute Value
0 5–5
Less than 5 units from zero
The solution set of |x| < 5 consists of all numbers that are
less than 5 units from 0 on the number line. Another way of
thinking of this is to think of all numbers between –5 and 5.
This set of numbers is given by (–5, 5), as shown in the figure
below. Here, the graph shows that –5 < x < 5, which means
x > –5 and x < 5.

Absolute Value
|ax + b| = k, |ax + b| > k, |ax + b| < k,or
ere k is a positive number.
|x| = 5 has the same solution set as x = –5 or x = 5,
|x| > 5 has the same solution set as x < –5 or x > 5,
|x| < 5 has the same solution set as x > –5 and x < 5.
e previous examples, we see that
The equation and inequalities just described are examples of
absolute value equations and inequalities. They involve the
absolute value of a variable expression and generally take the
form

Summary:
Solving Absolute Value Equations and Inequalities
1. To solve |ax + b| = k, solve the following compound
equation.
et k be a positive real number, and p and q be real numbers.
ax + b = k or ax + b = –k.
The solution set is usually of the form {p, q}.
p q

Summary:
Let k be a positive real number, and p and q be real
numbers.
p q
The solution set is of the form (-∞, p) U (q, ∞), which
consists of two separate intervals.
ax + b > k or ax + b < –k.
2. To solve |ax + b| > k, solve the following compound
inequality

Summary:
p q
The solution set is of the form (p, q), a single interval.
–k < ax + b < k.
3. To solve |ax + b| < k, solve the three-part inequality
Let k be a positive real number, and p and q be real
numbers.

Solving an Absolute Value Equation
= 5. Graph the solution set.
For |2x + 3| to equal 5, 2x + 3 must be 5 units from 0 on the number line.
his can happen only when 2x + 3 = 5 or 2x + 3 = –5. Solve this compound
quation as follows.
2x + 3 = 5 or 2x + 3 = –5
2x = 2
x = 1
2x = –8
x = –4
or
or
eck by substituting 1 and then –4 in the original absolute value equation
verify that the solution set is {–4, 1}.
–5 –4 –3 –2 –1 0 1 2 3 4 5
Example 1

Writing Compound Statements
OTE
ome people prefer to write the compound statements in Cases 1 and 2 of
e summary on the previous slides as the equivalent forms
ax + b = k or –(ax + b) = k
and ax + b > k or –(ax + b) > k.
These forms produce the same results.

Solving an Absolute Value Inequality with >
> 5. Graph the solution set.
By part 2 of the summary, this absolute value inequality is rewritten as
2x + 3 > 5 or 2x + 3 < –5
2x > 2
x > 1
2x < –8
x < –4
or
or
cause 2x + 3 must represent a number that is more than 5 units from 0 on
her side of the number line. Now, solve the compound inequality.
–5 –4 –3 –2 –1 0 1 2 3 4 5
2x + 3 > 5 or 2x + 3 < –5,
he solution set is (–∞, –4) U (1, ∞). The graph consists of disjoint intervals.
Example 2

EXAMPLE
3
Solving an Absolute Value Inequality with <
lve |2x + 3| < 5.
–5 < 2x + 3 < 5.
–8 < 2x < 2
–4 < x < 1
The expression 2x + 3 must represent a number that is less than 5 units
om 0 on either side of the number line. 2x + 3 must be between –5 and 5.
s part 3 of the summary shows, this is written as the three-part inequality
–5 –4 –3 –2 –1 0 1 2 3 4 5
Check that the solution set is (–4, 1), so the graph consists of the single
nterval shown below.
Subtract 3 from each part.
Divide each part by 2.
Example 3

CAUTION
When solving absolute value equations and inequalities of the types in
Examples 1, 2, and 3, remember the following.
The methods described apply when the constant is alone on one side
of the equation or inequality and is positive.
Absolute value equations and absolute value inequalities of the form
|ax + b| > k translate into “or” compound statements.
Absolute value inequalities of the form |ax + b| < k translate into
“and” compound statements, which may be written as three-part
inequalities.
An “or” statement cannot be written in three parts. It would be incorrect
to use –5 > 2x + 3 > 5 in Example 2, because this would imply that
–5 > 5, which is false.
Caution

EXAMPLE
4
Solving an Absolute Value Equation That
Involves Rewriting
quation |x – 7| + 6 = 9.
|x – 7| + 6 – 6 = 9 – 6
|x – 7| = 3
x – 7 = 3 or x – 7 = –3
First, rewrite so that the absolute value expression is alone on one side
f the equals sign by subtracting 6 from each side.
Now use the method shown in Example 1.
Subtract 6.
x = 10
heck that the solution set is {4, 10} by substituting 4 and then 10 into the
riginal equation.
or x = 4
Example 4

Solving |ax + b| = |cx + d|
To solve an absolute value equation of the form
|ax + b| = |cx + d|,
solve the compound equation
ax + b = cx + d or ax + b = –(cx + d).

EXAMPLE
5
Solving an Equation with Two Absolute
Values
lve the equation |y + 4| = |2y – 7|.
y + 4 = 2y – 7 or y + 4 = –(2y – 7).
y + 4 = 2y – 7 or y + 4 = –(2y – 7)
This equation is satisfied either if y + 4 and 2y – 7 are equal to each other,
f y + 4 and 2y – 7 are negatives of each other. Thus,
Solve each equation.
11 = y
Check that the solution set is {1, 11}.
3y = 3
y = 1
Example 6

Special Cases for Absolute Value
Special Cases for Absolute Value
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.

Solving Special Cases of Absolute Value
Equations
a) |2n + 3| = –7
(b) |6w – 1| = 0
1
6
1
6
Example 7

Equations
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.
a) |2n + 3| = –7
See Case 2 in the preceding slide. The absolute value of the
expression 6w – 1 will equal 0 only if
6w – 1 = 0.
(b) |6w – 1| = 0
The solution of this equation is . Thus, the solution set of the original
equation is { }, with just one element. Check by substitution.
1
6
1
6
Example 7

Inequalities
Solve each inequality.
(a) |x| ≥ –2
) |x + 5| – 1 < –8
Example 8

Inequalities
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 (–∞, ∞).
(a) |x| ≥ –2
Add 1 to each side to get the absolute value expression alone on one
side.
|x + 5| < –7
) |x + 5| – 1 < –8
There is no number whose absolute value is less than –7, so this inequality
has no solution. The solution set is Ø.
Example 8

Inequalities
Subtracting 2 from each side gives
|x – 9| ≤ 0.
(c) |x – 9| + 2 ≤ 2
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}.
Continued

Dmth3018 03

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