Page 135

hendricks_intermediate_algebra_1e_ch1_3

Section 2.7 Absolute Value Inequalities 133 Since there are no solutions of the equation, the number line is not separated into any additional intervals. The entire number line is the interval we need to test. –5 –4 –3 –2 –1 0 1 2 3 4 5 Test a point from interval A. Interval Test Point u2y - 7u ≥-1 True/False A: (-∞, ∞) 0 u2(0) - 7u ≥-1 u-7u ≥-1 7≥-1 True Since the test point is a solution of the inequality, we shade interval A. –5 –4 –3 –2 –1 0 1 2 3 4 5 The solution set is (-∞, ∞) or all real numbers. Student Check 5 Use test points to solve each absolute value inequality. a. ux + 5u ≥ 4 b. u6a - 1u <-2 Troubleshooting Common Errors Some common errors associated with absolute value inequalities are shown. Objective 6 Examples A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 6a. Solve ux + 2u < 5. Incorrect Solution Correct Solution and Explanation ux + 2u < 5 ux + < 5 and ux + 2u >-5 uxu < 3 and uxu >-7 So, the solution set is 3). ux + 2u < 5 x + 2 < 5 and x + 2>-5 x < 3 and x>-7 6b. Solve uy - 5u + 3 < 2. Incorrect Solution Correct Solution and Explanation uy - 5u + 3 < 2 y - 5 + 3 < 2 or y - 5 + 3>-2 y - 2 < 2 or y - 2>-2 y < 4 or y > 0 Solution: (-∞, ∞) Objective 6 ▶ Troubleshoot common errors. While the solution set is correct, the notation used to solve the problem is wrong. When we write the compound inequality that satisfies the absolute value inequality, we no longer include the absolute value bars. We must first isolate the absolute value expression. We should subtract 3 from each side. uy - 5u + 3 < 2 uy - 5u + 3 - 3 < 2 - 3 uy - 5u <-1 The absolute value of a real number is never less than a negative number. So, the solution set is . 2u a > x nd - the (-7, or > r > on: (


hendricks_intermediate_algebra_1e_ch1_3
To see the actual publication please follow the link above