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hendricks_intermediate_algebra_1e_ch1_3

Section 2.7 Absolute Value Inequalities 129 The last property deals with absolute value inequalities in which the number opposite the absolute value expression is zero. Property: Property 4 for Absolute Value Inequalities a. uXu < 0 This inequality has no solution since the absolute value of any real number is nonnegative. b. uXu ≤ 0 The only solution that satisfies this inequality is X = 0. c. uXu > 0 This inequality is solved using property 2. X > 0 or X < 0. The only point not included is X = 0. d. uXu ≥ 0 This inequality has all real numbers as its solution since the absolute value of any real number is always positive or zero. Objective 3 Examples Solve each absolute value inequality. 3a. uxu <-5 3b. uy + 1u ≤-7 3c. uau >-5 3d. u3b + 5u + 4 ≥ 1 3e. ` 4x + 1 3 ` > 0 Solutions 3a. The solution set is the empty set, or , since the absolute value of a number is never less than -5. 3b. Since the absolute value of a number is never less than or equal to -7, the solution set is the empty set, or . 3c. Since the absolute value of a number is always greater than -5, the solution set is all real numbers, or (-∞, ∞). The graph of the solution set is –5 –4 –3 –2 –1 0 1 2 3 4 5 3d. We must first isolate the absolute value by subtracting 4 from each side. u3b + 5u + 4 ≥ 1 u3b + 5u + 4 - 4 ≥ 1 - 4 u3b + 5u ≥-3 Since the absolute value of a number is always greater than or equal to -3, the solution set is all real numbers, or (-∞, ∞). The graph is –5 –4 –3 –2 –1 0 1 2 3 4 5 3e. ` 4x + 1 3 ` > 0 4x + 1 3 > 0 or 4x + 1 3 < 0 3a4x + 1 3 b > 3(0) 3a4x + 1 3 b < 3(0) 4x + 1 >0 4x + 1 < 0 4x + 1 - 1 > 0 - 1 4x + 1 - 1 < 0 - 1 4x>-1 4x<-1 4x 4 > -1 4 4x 4 < -1 4 x>- 1 4 x<- 1 4 Subtract 4 from each side. Simplify. Apply property 4. Multiply each side by 3. Simplify. Subtract 1 from each side. Simplify. Divide each side by 4. Simplify.


hendricks_intermediate_algebra_1e_ch1_3
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