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Absolute-Value Equations and Inequalities

Absolute-Value Equations and Inequalities. 9.3. Equations with Absolute Value Inequalities with Absolute Value. 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 < -2 and x + 3 > 6

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Absolute-Value Equations and Inequalities

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  1. Absolute-Value Equations and Inequalities 9.3 Equations with Absolute Value Inequalities with Absolute Value

  2. 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 < -2 and x + 3 > 6 • 5. 5x > 15 or 6x - 4 < 32

  3. 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. Find each absolute value. • |-3| • |0| • |4|

  5. Example a) |x| = 6; b) |x| = 0; c) |x| = –2 Solution 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. 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. Example: a) |2x +1| = 5; b) |3 – 4x| = –10 Solution a) We use the absolute-value principle, knowing that 2x + 1 must be either 5 or –5: |x| = p |2x +1| = 5 Substituting 2x +1 = –5or 2x +1 = 5 2x = –6 or 2x = 4 x = –3 or x = 2 The solution set is {–3, 2}. The check is left for the student.

  8. Solution (continued) b) |3 – 4x| = –10 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.

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

  10. When more than one absolute value expression appears… 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 numberor they are opposites. So solve x+1 = 2x and x+1 = -2x separately to get your answers. As always check by substitution.

  11. This equation sets both sides the same. This equation sets them opposite. Notice parenthesis. 3x – 5 = –(8 + 4x) 3x – 5 = 8 + 4x Solve |3x – 5| = |8 + 4x|. or You are not done until you solve each equation.

  12. The Absolute-Value Inequalities Principle • For any positive number p and any expression x: • |x| < p is equivalent to –p < x < p. (conjunction) • |x| > p is equivalent to x < -p or x > p. (disjunction) • Similar rules apply for less than or equal, greater than or equal.

  13. ) ( • 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 –p p

  14. ( -3 0 3 Solve |x| < 3. Then graph. Solution 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: )

  15. ] [ -3 3 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:

  16. ( ) –5 1/3 Solve |3x + 7| < 8. Then graph. Solution The number 3x + 7 must be less than 8 units from 0. |X| < p Replacing X with 3x + 7 and p with 8 |3x + 7| < 8 –8 < 3x + 7 < 8 –15 < 3x < 1 −5 < x < 1/3 The solution set is {x|–5 < x < 1/3}. The graph is as follows:

  17. ( ) –5 1/3 Solve |3x + 7| > 8. Then graph. Solution The number 3x + 7 must be greater than 8 units from 0. |X| < p Replacing X with 3x + 7 and p with 8 |3x + 7| < 8 –8 < 3x + 7 < 8 –15 < 3x < 1 −5 < x < 1/3 The solution set is {x|–5 < x < 1/3}. The graph is as follows:

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