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3.7 Absolute Value Inequalities:

Inverse Operations: Operations that undo another operation. . 3.7 Absolute Value Inequalities:. Isolate: The use of inverse operations used to leave a variable by itself. . Absolute Value: The distance from zero to that place. It can be on the left (-) or on the right (+) of zero. GOAL:.

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3.7 Absolute Value Inequalities:

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  1. Inverse Operations: Operations that undo another operation. 3.7 Absolute Value Inequalities: Isolate: The use of inverse operations used to leave a variable by itself. Absolute Value: The distance from zero to that place. It can be on the left (-) or on the right (+) of zero.

  2. GOAL:

  3. Absolute Value Inequalities: The absolute value is treated as any other inequality: 1. First isolate the absolute value by using inverse operations | | <, >, ≤, ≥ something. 2. Set the inside of the absolute value as –( )  left side of zero +( )  Right side of zero and isolate the variable. Remember to switch the sign if you divide OR multiply by a negative number…!

  4. Solving Absolute Value Inequalities: EX: Solve :| x+ 3| < 1

  5. SOLUTION: Here the absolute value is already isolated. | x+3 |< 1 - (x + 3) < 1 x+ 3 < 1 -x - 3 < 1 - 3 -3 +3 +3 x< - 2 -x < 4 Divide by -1 an switch the sign x> - 4 Set Notation: {x| -4 < x < -2} Interval: (-4, -2)

  6. YOU TRY IT: Solve: | x – 5 | > – 4

  7. SOLUTION: Here the absolute value is already isolated. | x-5 |< -4 - (x-5) < -4 x- 5 < -4 -x +5 < -4 + 5 +5 -5 -5 x< 1 -x < -9 Divide by -1 an switch the sign x> 9 Set Notation: {x| x < 1 or x > 9} Interval: (-∞, 1) U (9, ∞)

  8. Solving Absolute Value Inequalities: Remember to use inverse operations to isolate the absolute value EX: Solve : 2| x+ 3| - 7 < 1

  9. SOLUTION: Here we must isolate the absolute value first. 2| x+3 |-7 < 1 + 7 +7 2| x+3 | < 8 2 2 | x + 3| < 4

  10. SOLUTION: Here the absolute value is now isolated and we continue with the process. | x+3 |< 4 - (x + 3) < 4 x+ 3 < 4 -x - 3 < 4 - 3 -3 +3 +3 x< 1 -x < 7 Divide by -1 an switch the sign x> - 7 Set Notation: {x| -7 < x < 1} Interval: (-7, 1)

  11. SOLUTION: - 2 ≤ 2m – 4 < -1 Given (and) 4 - 2 ≤ 2m <-1 + 4 Inverse of subt. 2 ≤ 2m < 3 Like terms Inverse of mult. ≤ m < 1 ≤ m < 1.5 Interval: [1, 1.5)

  12. Real-World: The official weight of a nickel is 5 g, but the actual weight can vary from this amount up to 0.194 g. Suppose a bank weights a roll of 40 nickels. The wrapper weights 1.5 g. What is the range of the possible weights for the roll of nickels.

  13. Real-World: (SOLUTION) Nickel= 5 g |N – 5 | ≤ 0.194 Weight varies up to  0.194 | N -5|< 0.194 N -5 ≤ 0.194 - (N -5) ≤ 0.194 + 5 + 5 -N + 5 ≤ 0.194 - 5 - 5 N≤5.194 -N≤ - 4.806 Divide by -1 an switch the sign N ≥4.806

  14. Real-World: (SOLUTION CONTINUE) Roll  40 nickels Paper  1.5 g Total = 40 (nickel) + 1.5 Low Weight  4.806 High Weight  5.196 Low Total = 40 (4.806) + 1.5  193.74  209.26 High Total = 40 (5.196) + 1.5 Set Notation: {N| 193.74 < N< 209.26} Interval: (193.74, 209.26)

  15. VIDEOS: Absolute Value Inequalities http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities-example-1

  16. VIDEOS: Compoundinequalities http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities-example-2

  17. CLASSWORK:Page 211-213 Problems: As many as needed to master the concept.

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