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hendricks_intermediate_algebra_1e_ch1_3

132 Chapter 2 Linear Equations and Inequalities in One Variable Step 3: Select a number (a test point) from each of the resulting intervals and substitute it into the original inequality to determine if it is a solution. Step 4: If the test point is a solution of the inequality, shade the interval of the number line that includes this point. If the test point is not a solution of the inequality, do not shade this interval of the number line. Step 5: Put the appropriate symbol on the endpoints of the intervals. a. If the absolute value inequality contains ≤ or ≥, the endpoints are solutions of the inequality. Brackets are used to represent this. b. If the absolute value inequality contains < or >, the endpoints are not solutions of the inequality. Parentheses are used to represent this. Step 6: Write the solution set in interval notation. Objective 5 Examples Use test points to solve each absolute value inequality. 5a. ux - 2u < 5 5b. u2y - 7u ≥-1 Solutions 5a. Solve the equation ux - 2u = 5. x - 2=-5 or x - 2 = 5 x - 2 + 2=-5 + 2 or x - 2 + 2 = 5 + 2 x=-3 or x = 7 Place the solutions of the equation on a number line to form intervals. –3 7 –10 –8 –6 –4 –2 0 2 4 6 8 10 Test a point from intervals A, B, and C. Interval Test Point ux - 2u < 5 True/False A: (-∞, -3) -6 u-6 - 2u < 5 u-8u < 5 8 < 5 False B: (-3, 7) 0 u0 - 2u < 5 u-2u < 5 2 < 5 True C: (7, ∞) 9 u9 - 2u < 5 u7u < 5 7 < 5 False We shade interval B since its test point is a solution of the inequality. The endpoints are not included since the original inequality is <. –3 7 –10 –8 –6 –4 –2 0 2 4 6 8 10 So, the interval notation for the solution set is (-3, 7). 5b. Solve the equation u2y - 7u =-1. This equation has no solution since the constant on the right side is negative.


hendricks_intermediate_algebra_1e_ch1_3
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