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miller_intermediate_algebra_4e_ch1_3

80 Chapter 1 Linear Equations and Inequalities in One Variable Solving a Linear Inequality Solve the inequality Solution: Multiply by 3 to clear fractions (reverse the inequality sign). Add 3x to both sides. Subtract 2 from both sides. Divide by 2 (the inequality sign is reversed again). Simplify. 5x 2 5x 2 3 7 x 2. 3 7 x 2 5x 2 6 3x 6 2x 2 6 6 2x 6 8 2x 2 7 8 2 Set-builder notation: Interval notation: Skill Practice Solve the inequality. 4. In Example 4, the inequality sign was reversed twice: once for multiplying the inequality by 3 and once for dividing by 2. If you are in doubt about whether you have the inequality sign in the correct direction, you can check your final answer by using the test point method.That is, pick a point in the proposed solution set, and verify that it makes the original inequality true. Furthermore, any test point picked outside the solution set should make the original inequality false. Pick x 0 as a test point Pick x 5 as a test point Because a test point to the right of x 4 makes the inequality true, we have shaded the correct part of the number line. 723 7 ? 7 True 23 3 7 ? 7 2 3 7 ? 2 False 5152 2 3 7 ? 152 2 5102 2 3 7 ? 102 2 5x 2 3 7 x 2 5x 2 3 7 x 2 ( 3 2 1 0 1 2 3 4 5 6 x 1 3 x 1 14, 2 ( 4 5x 0 x 7 46 x 7 4 3a5x 2 3 b 6 31x 22 Example 4 Answer 4. 32, 2 5x 0 x 26 2


miller_intermediate_algebra_4e_ch1_3
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