Solving AbsoluteValue 2 8 Equations and Inequalities Warm
Solving Absolute-Value 2 -8 Equations and Inequalities Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2 2
2 -8 Solving Absolute-Value Equations and Inequalities Warm Up Solve. 1. y + 7 < – 11 y < – 18 2. 4 m ≥ – 12 m ≥ – 3 3. 5 – 2 x ≤ 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 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 Example 1 A: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. 6 y < – 24 OR y +5 ≥ 3 Solve both inequalities for y. 6 y < – 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 1 B: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. Solve both inequalities for c. and 2 c + 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 1 C: Solving Compound Inequalities Solve the compound inequality. Then graph the solution set. x – 5 < – 2 OR – 2 x ≤ – 10 Solve both inequalities for x. x – 5 < – 2 or – 2 x ≤ – 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 1 a Solve the compound inequality. Then graph the solution set. x – 2 < 1 OR 5 x ≥ 30 Solve both inequalities for x. x– 2<1 5 x ≥ 30 x≥ 6 or x<3 The solution set is all points that satisfy {x|x < 3 U x ≥ 6}. – 1 Holt Algebra 2 0 1 2 3 4 5 6 7 8 (–∞, 3) U [6, ∞)
2 -8 Solving Absolute-Value Equations and Inequalities Check It Out! Example 1 b Solve the compound inequality. Then graph the solution set. 2 x ≥ – 6 AND –x > – 4 Solve both inequalities for x. 2 x ≥ – 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 1 2 3 4 5 [– 3, 4)
Solving Absolute-Value Equations and Inequalities 2 -8 Check It Out! Example 1 c Solve the compound inequality. Then graph the solution set. x – 5 < 12 OR 6 x ≤ 12 Solve both inequalities for x. x – 5 < 12 or x < 17 6 x ≤ 12 x≤ 2 Because every point that satisfies x < 2 also satisfies x < 2, the solution set is {x|x < 17}. (-∞, 17) 2 Holt Algebra 2 4 6 8 10 12 14 16 18 20
Solving Absolute-Value Equations and Inequalities 2 -8 Check It Out! Example 1 d Solve the compound inequality. Then graph the solution set. – 3 x < – 12 AND x + 4 ≤ 12 Solve both inequalities for x. – 3 x < – 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 Holt Algebra 2 3 4 5 6 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 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 Assignment Page 154 # 1 -7, 14 -19 Due Tuesday Quiz Tomorrow 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 2 A: 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 2 B: 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 Check It Out! Example 2 a 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 Holt Algebra 2 or 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 2 b Solve the equation. |6 x| – 8 = 22 Isolate the absolutevalue expression. |6 x| = 30 6 x = 30 or 6 x = – 30 x=5 Holt Algebra 2 or 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 Algebra 2
2 -8 Solving Absolute-Value Equations and Inequalities Example 3 A: Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |– 4 q + 2| ≥ 10 – 4 q + 2 ≥ 10 or – 4 q + 2 ≤ – 10 – 4 q ≥ 8 q ≤ – 2 Holt Algebra 2 or – 4 q ≤ – 12 or q ≥ 3 Rewrite the absolute value as a disjunction. Subtract 2 from both sides of each inequality. Divide both sides of each inequality by – 4 and reverse the inequality symbols.
2 -8 Solving Absolute-Value Equations and Inequalities Example 3 A 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 3 B: Solving Absolute-Value Inequalities with Disjunctions Solve the inequality. Then graph the solution. |0. 5 r| – 3 ≥ – 3 Isolate the absolute value as a disjunction. |0. 5 r| ≥ 0 Rewrite the absolute value as a disjunction. 0. 5 r ≥ 0 or 0. 5 r ≤ 0 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 3 a Solve the inequality. Then graph the solution. |4 x – 8| > 12 4 x – 8 > 12 or 4 x – 8 < – 12 Rewrite the absolute value as a disjunction. 4 x > 20 or 4 x < – 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 3 a 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 3 b Solve the inequality. Then graph the solution. |3 x| + 36 > 12 Isolate the absolute value as a disjunction. |3 x| > – 24 3 x > – 24 or x > – 8 or 3 x < 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 4 A: Solving Absolute-Value Inequalities with Conjunctions Solve the compound inequality. Then graph the solution set. |2 x +7| ≤ 3 Multiply both sides by 3. 2 x + 7 ≤ 3 and 2 x + 7 ≥ – 3 Rewrite the absolute value as a conjunction. 2 x ≤ – 4 and x ≤ – 2 and Holt Algebra 2 2 x ≥ – 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 4 A 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 4 B: 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 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 4 a 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 4 b 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 Algebra 2
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