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Solving
Solving Absolute-Value
Absolute-Value
2-8
2-8 Equations
Equations and
and Inequalities
Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
2-8
Solving Absolute-Value
Equations and Inequalities
Warm Up
Solve.
1. y + 7 < –11
y < –18
2. 4m ≥ –12
m ≥ –3
3. 5 – 2x ≤ 17
x ≥ –6
Use interval notation to indicate the graphed
numbers.
4.
(-2, 3]
5.
(-, 1]
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Objectives
Solve compound inequalities.
Write and solve absolute-value
equations and inequalities.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Vocabulary
disjunction
conjunction
absolute-value
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
A compound statement is made up of more than
one equation or inequality.
A disjunction is a compound statement that uses
the word or.
Disjunction: x ≤ –3 OR x > 2
Set builder notation: {x|x ≤ –3 U x > 2}
A disjunction is true if and only if at least one of its
parts is true.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
A conjunction is a compound statement that uses
the word and.
U
Conjunction: x ≥ –3 AND x < 2
Set builder notation: {x|x ≥ –3
x < 2}.
A conjunction is true if and only if all of its parts
are true. Conjunctions can be written as a single
statement as shown.
x ≥ –3 and x< 2
–3 ≤ x < 2
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Reading Math
Dis- means “apart.” Disjunctions have two
separate pieces. Con- means “together”
Conjunctions represent one piece.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Example 1A: Solving Compound Inequalities
Solve the compound inequality. Then graph the
solution set.
6y < –24 OR y +5 ≥ 3
Solve both inequalities for y.
6y < –24
y + 5 ≥3
or
y < –4
y ≥ –2
The solution set is all points that satisfy
{y|y < –4 or y ≥ –2}.
–6 –5 –4 –3 –2 –1
Holt McDougal Algebra 2
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3
(–∞, –4) U [–2, ∞)
2-8
Solving Absolute-Value
Equations and Inequalities
Example 1B: Solving Compound Inequalities
Solve the compound inequality. Then graph the
solution set.
Solve both inequalities for c.
and
2c + 1 < 1
c ≥ –4
c<0
The solution set is the set of points that satisfy both
c ≥ –4 and c < 0.
[–4, 0)
–6 –5 –4 –3 –2 –1
Holt McDougal Algebra 2
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2-8
Solving Absolute-Value
Equations and Inequalities
Example 1C: Solving Compound Inequalities
Solve the compound inequality. Then
graph the solution set.
x – 5 < –2 OR –2x ≤ –10
Solve both inequalities for x.
x – 5 < –2
or
–2x ≤ –10
x<3
x≥5
The solution set is the set of all points that satisfy
{x|x < 3 or x ≥ 5}.
–3 –2 –1
Holt McDougal Algebra 2
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6
(–∞, 3) U [5, ∞)
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 1a
Solve the compound inequality. Then graph the
solution set.
x – 2 < 1 OR 5x ≥ 30
Solve both inequalities for x.
x–2<1
5x ≥ 30
x≥6
or
x<3
The solution set is all points that satisfy
{x|x < 3 U x ≥ 6}.
–1
0
1
Holt McDougal Algebra 2
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6
7
8
(–∞, 3) U [6, ∞)
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 1b
Solve the compound inequality. Then graph the
solution set.
2x ≥ –6 AND –x > –4
Solve both inequalities for x.
2x ≥ –6
and
–x > –4
x ≥ –3
x<4
The solution set is the set of points that satisfy both
{x|x ≥ –3 x < 4}.
U
–4 –3 –2 –1 0
Holt McDougal Algebra 2
[–3, 4)
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Solving Absolute-Value
Equations and Inequalities
2-8
Check It Out! Example 1c
Solve the compound inequality. Then graph the
solution set.
x –5 < 12 OR 6x ≤ 12
Solve both inequalities for x.
x –5 < 12
or
x < 17
6x ≤ 12
x≤2
Because every point that satisfies x < 2 also satisfies
x < 2, the solution set is {x|x < 17}.
(-∞, 17)
2
4
6
8
Holt McDougal Algebra 2
10 12 14 16 18 20
Solving Absolute-Value
Equations and Inequalities
2-8
Check It Out! Example 1d
Solve the compound inequality. Then graph the
solution set.
–3x < –12 AND x + 4 ≤ 12
Solve both inequalities for x.
–3x < –12
and
x < –4
x + 4 ≤ 12
x≤8
The solution set is the set of points that satisfy both
{x|4 < x ≤ 8}.
(4, 8]
2
3
4
5
Holt McDougal Algebra 2
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7
8
9 10 11
2-8
Solving Absolute-Value
Equations and Inequalities
Recall that the absolute value of a number x,
written |x|, is the distance from x to zero on the
number line. Because absolute value represents
distance without regard to direction, the absolute
value of any real number is nonnegative.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Absolute-value equations and inequalities can be
represented by compound statements. Consider
the equation |x| = 3.
The solutions of |x| = 3 are the two points that
are 3 units from zero. The solution is a
disjunction: x = –3 or x = 3.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
The solutions of |x| < 3 are the points that are
less than 3 units from zero. The solution is a
conjunction: –3 < x < 3.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
The solutions of |x| > 3 are the points that are more
than 3 units from zero. The solution is a disjunction:
x < –3 or x > 3.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Helpful Hint
Think: Greator inequalities involving > or ≥
symbols are disjunctions.
Think: Less thand inequalities involving < or
≤ symbols are conjunctions.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Note: The symbol ≤ can replace <, and the rules
still apply. The symbol ≥ can replace >, and the
rules still apply.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Example 2A: Solving Absolute-Value Equations
Solve the equation.
|–3 + k| = 10
This can be read as “the
distance from k to –3 is 10.”
–3 + k = 10 or –3 + k = –10
Rewrite the absolute
value as a
disjunction.
k = 13 or k = –7
Add 3 to both sides of
each equation.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Example 2B: Solving Absolute-Value Equations
Solve the equation.
Isolate the absolute-value
expression.
Rewrite the absolute value as a
disjunction.
x = 16 or x = –16
Holt McDougal Algebra 2
Multiply both sides of each equation
by 4.
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 2a
Solve the equation.
This can be read as “the
distance from x to +9 is 4.”
|x + 9| = 13
x + 9 = 13 or x + 9 = –13
x=4
or
Holt McDougal Algebra 2
x = –22
Rewrite the absolute
value as a disjunction.
Subtract 9 from both
sides of each equation.
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 2b
Solve the equation.
|6x| – 8 = 22
Isolate the absolutevalue expression.
|6x| = 30
6x = 30 or 6x = –30
x=5
or
Holt McDougal Algebra 2
x = –5
Rewrite the absolute
value as a disjunction.
Divide both sides of each
equation by 6.
2-8
Solving Absolute-Value
Equations and Inequalities
You can solve absolute-value inequalities using
the same methods that are used to solve an
absolute-value equation.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Example 3A: Solving Absolute-Value Inequalities with
Disjunctions
Solve the inequality. Then graph the solution.
|–4q + 2| ≥ 10
–4q + 2 ≥ 10 or –4q + 2 ≤ –10
–4q ≥ 8
q ≤ –2
Rewrite the absolute
value as a disjunction.
or –4q ≤ –12
Subtract 2 from both
sides of each inequality.
or q ≥ 3
Divide both sides of
each inequality by –4
and reverse the
inequality symbols.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Example 3A Continued
{q|q ≤ –2 or q ≥ 3}
(–∞, –2] U [3, ∞)
–3 –2 –1
0
1
2
3
4
5
6
To check, you can test a point in each of the
three region.
|–4(–3) + 2| ≥ 10
|14| ≥ 10 
Holt McDougal Algebra 2
|–4(0) + 2| ≥ 10
|2| ≥ 10 x
|–4(4) + 2| ≥ 10
|–14| ≥ 10 
2-8
Solving Absolute-Value
Equations and Inequalities
Example 3B: Solving Absolute-Value Inequalities with
Disjunctions
Solve the inequality. Then graph the solution.
|0.5r| – 3 ≥ –3
Isolate the absolute value as
a disjunction.
|0.5r| ≥ 0
0.5r ≥ 0 or 0.5r ≤ 0
Rewrite the absolute
value as a disjunction.
Divide both sides of each
r ≤ 0 or r ≥ 0
inequality by 0.5.
The solution is all real numbers, R.
(–∞, ∞)
–3 –2 –1
Holt McDougal Algebra 2
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2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 3a
Solve the inequality. Then graph the solution.
|4x – 8| > 12
4x – 8 > 12 or 4x – 8 < –12
Rewrite the absolute
value as a disjunction.
4x > 20 or 4x < –4
Add 8 to both sides of
each inequality.
x>5
or
Holt McDougal Algebra 2
x < –1
Divide both sides of
each inequality by 4.
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 3a Continued
{x|x < –1 or x > 5}
(–∞, –1) U (5, ∞)
–3 –2 –1
0
1
2
3
4
5
6
To check, you can test a point in each of the
three region.
|4(–2) + 8| > 12
|–16| > 12 
Holt McDougal Algebra 2
|4(0) + 8| > 12
|8| > 12 x
|4(6) + 8| > 12
|32| > 12 
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 3b
Solve the inequality. Then graph the solution.
|3x| + 36 > 12
Isolate the absolute value
as a disjunction.
|3x| > –24
3x > –24 or
x > –8
or
3x < 24
Rewrite the absolute
value as a disjunction.
x<8
Divide both sides of each
inequality by 3.
The solution is all real numbers, R.
(–∞, ∞)
–3 –2 –1
0
Holt McDougal Algebra 2
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2-8
Solving Absolute-Value
Equations and Inequalities
Example 4A: Solving Absolute-Value Inequalities with
Conjunctions
Solve the compound inequality. Then graph
the solution set.
|2x +7| ≤ 3
Multiply both sides by 3.
2x + 7 ≤ 3 and 2x + 7 ≥ –3
Rewrite the absolute
value as a conjunction.
2x ≤ –4 and
x ≤ –2 and
Holt McDougal Algebra 2
2x ≥ –10
Subtract 7 from both
sides of each inequality.
x ≥ –5
Divide both sides of
each inequality by 2.
2-8
Solving Absolute-Value
Equations and Inequalities
Example 4A Continued
The solution set is {x|–5 ≤ x ≤ 2}.
–6 –5 –3 –2 –1
Holt McDougal Algebra 2
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2-8
Solving Absolute-Value
Equations and Inequalities
Example 4B: Solving Absolute-Value Inequalities with
Conjunctions
Solve the compound inequality. Then graph
the solution set.
Multiply both sides by –2, and
reverse the inequality symbol.
|p – 2| ≤ –6
|p – 2| ≤ –6 and p – 2 ≥ 6
p ≤ –4 and
p≥8
Rewrite the absolute value
as a conjunction.
Add 2 to both sides of
each inequality.
Because no real number satisfies both p ≤ –4 and
p ≥ 8, there is no solution. The solution set is ø.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 4a
Solve the compound inequality. Then graph
the solution set.
|x – 5| ≤ 8
Multiply both sides by 2.
x – 5 ≤ 8 and x – 5 ≥ –8
Rewrite the absolute
value as a conjunction.
x ≤ 13 and
Add 5 to both sides
of each inequality.
Holt McDougal Algebra 2
x ≥ –3
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 4
The solution set is {x|–3 ≤ x ≤ 13}.
–10
Holt McDougal Algebra 2
–5
0
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10
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20
25
2-8
Solving Absolute-Value
Equations and Inequalities
Check It Out! Example 4b
Solve the compound inequality. Then graph
the solution set.
–2|x +5| > 10
|x + 5| < –5
x + 5 < –5 and x + 5 > 5
x < –10 and x > 0
Divide both sides by –2, and
reverse the inequality symbol.
Rewrite the absolute value
as a conjunction.
Subtract 5 from both
sides of each inequality.
Because no real number satisfies both x < –10 and
x > 0, there is no solution. The solution set is ø.
Holt McDougal Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Lesson Quiz: Part I
Solve. Then graph the solution.
1. y – 4 ≤ –6 or 2y >8
–4 –3 –2 –1 0
{y|y ≤ –2 ≤ or y > 4}
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2
3
4
5
2. –7x < 21 and x + 7 ≤ 6 {x|–3 < x ≤ –1}
–4 –3 –2
–1 0
1
2
3
4
5
Solve each equation.
3. |2v + 5| = 9
2 or –7
Holt McDougal Algebra 2
4. |5b| – 7 = 13
+4
2-8
Solving Absolute-Value
Equations and Inequalities
Lesson Quiz: Part II
Solve. Then graph the solution.
5. |1 – 2x| > 7 {x|x < –3 or x > 4}
–4 –3 –2
–1 0
1
2
3
4
5
6. |3k| + 11 > 8 R
–4 –3 –2 –1
7. –2|u + 7| ≥ 16
Holt McDougal Algebra 2
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