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1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Absolute-Value Equations and Inequalities Equations with Absolute Value Inequalities with Absolute.

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Presentation on theme: "1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Absolute-Value Equations and Inequalities Equations with Absolute Value Inequalities with Absolute."— Presentation transcript:

1 1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Absolute-Value Equations and Inequalities Equations with Absolute Value Inequalities with Absolute Value 9.3

2 2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Warm Up – Solve and graph each compound inequality. – 1. -3x – 4 < -16 and 2x - 5 < 15 – 2. 4x + 2 > 10 or -2x + 6 > 16 – 3. 5 < -3x+2 < 8 – 4. 2x 6 – 5. 5x > 15 or 6x - 4 < 32

3 3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Absolute Value The absolute value of a number is its distance from 0. Since absolute value is distance it is never negative. Definition of absolute value.

4 4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Find each absolute value. |-3| |0| |4|

5 5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution Example a) |x| = 6; b) |x| = 0; c) |x| = –2 a) |x| = 6 means that the distance from x to 0 is 6. Only two numbers meet that requirement. Thus the solution set is {–6, 6}. b) |x| = 0 means that the distance from x to 0 is 0. The only number that satisfies this is zero itself. Thus the solution set is {0}. c) Since distance is always nonnegative, |x| = –2 has no solution. Thus the solution set is

6 6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. This brings us to… The Absolute-Value Principle for Equations If |x| = p, then x = p or x = -p. Note: The equation |x| = 0 is equivalent to the equation x = 0 and the equation |x| = –p has no solution.

7 7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example: a) |2x +1| = 5; b) |3 – 4x| = –10 The solution set is {–3, 2}. The check is left for the student. x = –3 or x = 2 2x = –6 or 2x = 4 Substituting Solution a) We use the absolute-value principle, knowing that 2x + 1 must be either 5 or –5: |2x +1| = 5 |x| = p 2x +1 = –5 or 2x +1 = 5

8 8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution (continued) The absolute-value principle reminds us that absolute value is always nonnegative. The equation |3 – 4x| = –10 has no solution. The solution set is To apply the absolute value principle we must make sure the absolute value expression is ISOLATED. b) |3 – 4x| = –10

9 9 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution Since we Given that f (x) = 3|x + 5| – 4, find all x for which f (x) = 11 are looking for f(x) = 11, we substitute: Replacing f (x) with 3|x + 5| − 4 x + 5 = –5 or x + 5 = 5 x = –10 or x = 0 The solution set is {–10, 0}. The check is left for the student. 3|x+5| – 4 = 11 f(x) = 11 3|x + 5| = 15 |x + 5| = 5

10 10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Sometimes an equation has two absolute-value expressions like |x+1| = |2x|. This means that x+1 and 2x are the same distance from zero. Since they are the same distance from zero, they are the same number or they are opposites. So solve x+1 = 2x and x+1 = -2x separately to get your answers. As always check by substitution. When more than one absolute value expression appears…

11 11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solve |3x – 5| = |8 + 4x|. You are not done until you solve each equation. 3x – 5 = 8 + 4x This equation sets both sides the same. This equation sets them opposite. Notice parenthesis. 3x – 5 = –(8 + 4x) or

12 12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc.

13 13 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. The Absolute-Value Inequalities Principle For any positive number p and any expression x: a) |x| < p is equivalent to –p < x < p. (conjunction) b) |x| > p is equivalent to x p. (disjunction) c) Similar rules apply for less than or equal, greater than or equal.

14 14 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. So the solutions of |X| < p are those numbers that satisfy –p < X < p. And the solutions of |X| > p are those numbers that satisfy X < –p or p < X. –p p ( ) ( )

15 15 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution Solve |x| < 3. Then graph. The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions. The solution set is {x| –3 < x < 3}. In interval notation, the solution set is (–3, 3). The graph is as follows: ( )

16 16 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution The solutions of are all numbers whose distance from zero is at least 3 units. The solution set is In interval notation, the solution set is The graph is as follows: [ -3 3 ]

17 17 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution The number 3x + 7 must be less than 8 units from 0. |3x + 7| < 8 |X| < p Replacing X with 3x + 7 and p with 8 –8 < 3x + 7 < 8 The solution set is {x|–5 < x < 1/3}. The graph is as follows: Solve |3x + 7| < 8. Then graph. −5 < x < 1/3 –15 < 3x < 1 –5 1/3 ()

18 18 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solution The number 3x + 7 must be greater than 8 units from 0. |3x + 7| < 8 |X| < p Replacing X with 3x + 7 and p with 8 –8 < 3x + 7 < 8 The solution set is {x|–5 < x < 1/3}. The graph is as follows: Solve |3x + 7| > 8. Then graph. −5 < x < 1/3 –15 < 3x < 1 –5 1/3 ()


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