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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
Holt
Algebra
Holt
Algebra
2 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}
1
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 Algebra 2
4. |5b| – 7 = 13
+4
2-8
Solving Absolute-Value
Equations and Inequalities
Objectives
Solve compound inequalities.
Write and solve absolute-value
equations and inequalities.
Holt Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Vocabulary
disjunction
conjunction
absolute-value
Holt 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 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 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 Algebra 2
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 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 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 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 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 Algebra 2
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 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
Holt Algebra 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.
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 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
(always true?? – all reals)
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 Algebra 2
0
1
2
3
4
5
6
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
Holt Algebra 2
or
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 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
Holt Algebra 2
0
1
2
3
4
5
6
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 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 Algebra 2
0
1
2
3 4
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 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 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 Algebra 2
–5
0
5
10
15
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
(Never True??-no sol.)
Divide both sides by –2, and
revrse 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 Algebra 2
2-8
Solving Absolute-Value
Equations and Inequalities
Assignment
Page 154 # 8-13 all,
20-36 extra credit if you want
it
Due Wednesday
Quiz Tomorrow
Holt Algebra 2