176 CHAPTER 1 Equations and Inequalities Set the numerator and denominator of -13x - 21 3x + 4 equal to 0 and solve the resulting equations to find the values of x where sign changes may occur. -13x - 21 = 0 or 3x + 4 = 0 x = - 21 13 or x = - 4 3 Use these values to form intervals on the number line. Use an open circle at - 21 13 because of the strict inequality, and use an open circle at - 4 3 because it causes the denominator to equal 0. See Figure 18. –1.5 Interval A (–∞, – ) Interval B (– , – ) Interval C (– , ∞) 4 3 21 13 21 13 21 13 4 3 4 3 – – –1 –2 Test Value ? , 5 2(–2) – 1 3(–2) + 4 , 5 5 2 True Test Value ? , 5 2(–1.5) – 1 3(–1.5) + 4 8 , 5 False Test Value ? , 5 2(–1) – 1 3(–1) + 4 –3 , 5 True Figure 18 Choosing a test value from each interval shows that the values in Intervals A and C satisfy the original inequality, 2x - 1 3x + 4 65. So the solution set is the union of these intervals. a-∞, - 21 13b ´a- 4 3 , ∞b S Now Try Exercise 79. 1.7 Exercises CONCEPT PREVIEW Match the inequality in each exercise in Column I with its equivalent interval notation in Column II. I 1. x 6 -6 2. x … 6 3. -2 6x … 6 4. x2 Ú 0 5. x Ú -6 6. 6 … x 7. –2 0 6 8. 0 8 9. –3 3 0 10. –6 0 II A. 1-2, 64 B. 3-2, 62 C. 1-∞, -64 D. 36, ∞2 E. 1-∞, -3) ´(3, ∞2 F. 1-∞, -62 G. 10, 82 H. 1-∞, ∞2 I. 3-6, ∞2 J. 1-∞, 64
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