175 1.7 Inequalities Step 2 The quotient 1 - x x + 4 possibly changes sign only where x-values make the numerator or denominator 0. This occurs at 1 - x = 0 or x + 4 = 0 x = 1 or x = -4. These values form the intervals 1-∞, -42, 1-4, 12, and 11, ∞2 on the number line, as seen in Figure 17. Interval A (–`, –4) Interval B (–4, 1) Interval C (1, `) 0 1 –4 Figure 17 Use a solid circle on 1 because the symbol is Ú. The value -4 cannot be in the solution set because it causes the denominator to equal 0. Use an open circle on -4. Step 3 Choose test values. Interval Test Value Is 5 x +4 # 1 True or False? A: 1-∞, -42 -5 5 -5 + 4 Ú ? 1 -5 Ú 1 False B: 1-4, 12 0 5 0 + 4 Ú ? 1 5 4 Ú 1 True C: 11, ∞2 2 5 2 + 4 Ú ? 1 5 6 Ú 1 False The values in Interval B, 1-4, 12, satisfy the original inequality. The value 1 makes the nonstrict inequality true, so it must be included in the solution set. Because -4 makes the denominator 0, it must be excluded. The solution set is the interval 1-4, 14. SNow Try Exercise 67. CAUTION Be careful with the endpoints of the intervals when solving rational inequalities. EXAMPLE 9 Solving a Rational Inequality Solve 2x - 1 3x + 4 65. SOLUTION 2x - 1 3x + 4 - 5 60 Subtract 5. 2x - 1 3x + 4 - 513x + 42 3x + 4 60 The common denominator is 3x + 4. 2x - 1 - 513x + 42 3x + 4 60 Write as a single fraction. 2x - 1 - 15x - 20 3x + 4 60 Distributive property -13x - 21 3x + 4 60 Combine like terms in the numerator. Be careful with signs.
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