Solving Solving Absolute-Value Absolute-Value 2-8 6-5 Equations Equations and and Inequalities Inequalities Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2 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 Algebra 2 2-8 Solving Absolute-Value Equations and Inequalities Learning Targets 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 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 Algebra 2 0 1 2 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 Algebra 2 0 1 2 3 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 Algebra 2 0 1 2 3 4 5 6 (–∞, 3) U [5, ∞) 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 Algebra 2 [–3, 4) 1 2 3 4 5 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 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 2-8 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 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 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 Algebra 2 Multiply both sides of each equation by 4. 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 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 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 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 Algebra 2 0 ø 1 2 3 4 5