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Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve compound inequalities.
Write and solve absolute-value equations and inequalities.
Objectives
Holt Algebra 2
2-8Solving 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
A disjunction is true if and only if at least one of its parts is true.
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
A conjunction is a compound statement that uses the word and.
Conjunction: x ≥ –3 AND 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
2-8Solving Absolute-Value Equations and Inequalities
Example 1A: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for y.
The solution set is all points that satisfy {y < –4 or y ≥ –2}.
6y < –24 OR y +5 ≥ 3
6y < –24 y + 5 ≥3
y < –4 y ≥ –2or
–6 –5 –4 –3 –2 –1 0 1 2 3 (–∞, –4) U [–2, ∞)
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve both inequalities for c.
The solution set is the set of points that satisfy both c ≥ –4 and c < 0.
c ≥ –4 c < 0
–6 –5 –4 –3 –2 –1 0 1 2 3[–4, 0)
Example 1B: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
and 2c + 1 < 1
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Example 1C: Solving Compound Inequalities
Solve the compound inequality. Then graph the solution set.
Solve both inequalities for x.
The solution set is the set of all points that satisfy {x < 3 or x ≥ 5}.
–3 –2 –1 0 1 2 3 4 5 6
x – 5 < –2 OR –2x ≤ –10
x < 3 x ≥ 5
x – 5 < –2 or –2x ≤ –10
(–∞, 3) U [5, ∞)
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve both inequalities for x.
2x ≥ –6 and –x > –4
–4 –3 –2 –1 0 1 2 3 4 5
2x ≥ –6 AND –x > –4
x ≥ –3 x < 4
The solution set is the set of points that satisfy both {x ≥ –3 x < 4}.
[–3, 4)
Solve the compound inequality. Then graph the solution set.
Check It Out! Example 1b
Holt Algebra 2
2-8Solving 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
2-8Solving 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 Algebra 2
2-8Solving 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
2-8Solving 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
2-8Solving Absolute-Value Equations and Inequalities
Think: Greator inequalities involving > or ≥ symbols are disjunctions.
Think: Less thand inequalities involving < or ≤ symbols are conjunctions.
Helpful Hint
Holt Algebra 2
2-8Solving 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
2-8Solving Absolute-Value Equations and Inequalities
Solve the equation.
Example 2A: Solving Absolute-Value Equations
Rewrite the absolute value as a disjunction.
This can be read as “the distance from k to –3 is 10.”
Add 3 to both sides of each equation.
|–3 + k| = 10
–3 + k = 10 or –3 + k = –10
k = 13 or k = –7
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Check It Out! Example 2a
|x + 9| = 13
Solve the equation.
Rewrite the absolute value as a disjunction.
This can be read as “the distance from x to +9 is 4.”
Subtract 9 from both sides of each equation.
x + 9 = 13 or x + 9 = –13
x = 4 or x = –22
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Check It Out! Example 2b
|6x| – 8 = 22
Solve the equation.
Isolate the absolute-value expression.
Rewrite the absolute value as a disjunction. Divide both sides of each equation by 6.
|6x| = 30
6x = 30 or 6x = –30
x = 5 or x = –5
Holt Algebra 2
2-8Solving 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
2-8Solving Absolute-Value Equations and Inequalities
Example 3A: Solving Absolute-Value Inequalities with Disjunctions
Solve the inequality. Then graph the solution.
Rewrite the absolute value as a disjunction.
|–4q + 2| ≥ 10
–4q + 2 ≥ 10 or –4q + 2 ≤ –10
–4q ≥ 8 or –4q ≤ –12
Divide both sides of each inequality by –4 and reverse the inequality symbols.
Subtract 2 from both sides of each inequality.
q ≤ –2 or q ≥ 3
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Example 3A Continued
{q ≤ –2 or q ≥ 3}
–3 –2 –1 0 1 2 3 4 5 6
(–∞, –2] U [3, ∞)
To check, you can test a point in each of the three region.
|–4(–3) + 2| ≥ 10
|14| ≥ 10
|–4(0) + 2| ≥ 10
|2| ≥ 10 x
|–4(4) + 2| ≥ 10
|–14| ≥ 10
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve the inequality. Then graph the solution.
Rewrite the absolute value as a disjunction.
|4x – 8| > 12
4x – 8 > 12 or 4x – 8 < –12
4x > 20 or 4x < –4
Divide both sides of each inequality by 4.
Add 8 to both sides of each inequality.
x > 5 or x < –1
Check It Out! Example 3a
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
{x < –1 or x > 5}
–3 –2 –1 0 1 2 3 4 5 6
(–∞, –1) U (5, ∞)
To check, you can test a point in each of the three region.
|4(–2) + 8| > 12
|–16| > 12
|4(0) + 8| > 12
|8| > 12 x
|4(6) + 8| > 12
|32| > 12
Check It Out! Example 3a Continued
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve the inequality. Then graph the solution.
|3x| + 36 > 12
Divide both sides of each inequality by 3.
Isolate the absolute value as a disjunction.
Rewrite the absolute value as a disjunction.
Check It Out! Example 3b
3x > –24 or 3x < 24
x > –8 or x < 8
|3x| > –24
–3 –2 –1 0 1 2 3 4 5 6
(–∞, ∞)
The solution is all real numbers, R.
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve the compound inequality. Then graph the solution set.
Example 4A: Solving Absolute-Value Inequalities with Conjunctions
|2x +7| ≤ 3 Multiply both sides by 3.
Subtract 7 from both sides of each inequality.
Divide both sides of each inequality by 2.
Rewrite the absolute value as a conjunction.
2x + 7 ≤ 3 and 2x + 7 ≥ –3
2x ≤ –4 and 2x ≥ –10
x ≤ –2 and x ≥ –5
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Example 4A Continued
The solution set is {–5 ≤ x ≤ 2}.
–6 –5 –3 –2 –1 0 1 2 3 4
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve the compound inequality. Then graph the solution set.
|x – 5| ≤ 8 Multiply both sides by 2.
Add 5 to both sides of each inequality.
Rewrite the absolute value as a conjunction.x – 5 ≤ 8 and x – 5 ≥ –8
x ≤ 13 and x ≥ –3
Check It Out! Example 4a
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
The solution set is {–3 ≤ x ≤ 13}.
Check It Out! Example 4
–10 –5 0 5 10 15 20 25
Holt Algebra 2
2-8Solving Absolute-Value Equations and Inequalities
Solve the compound inequality. Then graph the solution set.
–2|x +5| > 10
Divide both sides by –2, and reverse the inequality symbol.
Subtract 5 from both sides of each inequality.
Rewrite the absolute value as a conjunction.
x + 5 < –5 and x + 5 > 5
x < –10 and x > 0
Because no real number satisfies both x < –10 and x > 0, there is no solution. The solution set is ø.
Check It Out! Example 4b
|x + 5| < –5