Holt Algebra Solving Absolute-Value Equations and Inequalities Solve compound inequalities. Write and solve absolute-value equations and inequalities. Objectives

Holt Algebra 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 A disjunction is true if and only if at least one of its parts is true.

Holt Algebra Solving 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 Solving 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 < –24y + 5 ≥3 y < –4y ≥ –2 or –6 –5 –4 –3 –2 – (–∞, –4) U [–2, ∞)

Holt Algebra Solving 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 ≥ –4c < 0 –6 –5 –4 –3 –2 – [–4, 0) Example 1B: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. and 2c + 1 < 1

Holt Algebra Solving 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 – x – 5 < –2 OR –2x ≤ –10 x < 3 x ≥ 5 x – 5 < –2 or –2x ≤ –10 (–∞, 3) U [5, ∞)

Holt Algebra Solving Absolute-Value Equations and Inequalities Solve both inequalities for x. 2x ≥ –6 and –x > –4 –4 –3 –2 – x ≥ –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 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 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 Algebra 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 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.

Holt Algebra Solving 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 Solving Absolute-Value Equations and Inequalities Note: The symbol ≤ can replace, and the rules still apply.

Holt Algebra Solving 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 Solving 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 Solving 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 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 Solving 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 Solving Absolute-Value Equations and Inequalities Example 3A Continued {q ≤ –2 or q ≥ 3} –3 –2 – (–∞, –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 Solving 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 Solving Absolute-Value Equations and Inequalities {x 5} –3 –2 – (–∞, –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 Solving 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 – (–∞, ∞) The solution is all real numbers, R.

Holt Algebra Solving 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 Solving Absolute-Value Equations and Inequalities Example 4A Continued The solution set is {–5 ≤ x ≤ 2}. –6 –5 –3 –2 –

Holt Algebra Solving 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 Solving Absolute-Value Equations and Inequalities The solution set is {–3 ≤ x ≤ 13}. Check It Out! Example 4 –10 –

Holt Algebra Solving 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 x 0 Because no real number satisfies both x 0, there is no solution. The solution set is ø. Check It Out! Example 4b |x + 5| < –5